You do not recognise it, you BUILD it, in three forced moves. The question hands you only a fragment, the ramp $h(t)=t$ on $0<t<\pi$, and two instructions: make it EVEN, and make it PERIODIC. Each instruction fixes part of the graph with no freedom left, and what comes out at the end happens to be a zigzag.

Step 1: draw only what you are given

Step 2: “even” forces the left half, it is a mirror

Even means $h(-t)=h(t)$: whatever height the graph has at $t$, it must have the SAME height at $-t$. So the blank left side is not a choice; it is the mirror image of the ramp, as if the page were folded along the vertical axis. The ramp rising to the right becomes a ramp falling from the left: at $t=-\pi$ the height is $\pi$, sliding down to $0$ at the origin.

Step 3: “periodic” forces everything else, copy the V forever

The V between $-\pi$ and $\pi$ is one full period. Periodicity stamps identical copies to the left and right, end to end. Watch what happens at the seam $t=\pi$: the V ends at height $\pi$, and the next copy BEGINS at height $\pi$ (its own left edge). The heights match, so the copies join without any jump, and each joint becomes a sharp peak. Valleys sit at $0,\pm 2\pi ,\dots$ and peaks at $\pm \pi ,\pm 3\pi ,\dots$

Why a triangle and not a curve or a staircase

Because every piece of the final graph is a copy of the original ramp, and the original ramp is a STRAIGHT LINE. Mirroring a straight line gives a straight line; repeating straight lines gives straight lines. The only shape you can build out of straight segments that alternately rise and fall is a zigzag, and a zigzag is what “triangle wave” means. If the given half-range had been $t^2$ instead, the same two moves would produce repeating parabola arcs (Q1(b)'s scallops), not triangles: the shape of the output is always inherited from the shape of the input fragment.

The contrast: the ODD extension of the same ramp is a sawtooth

Odd means mirror AND flip: the left half becomes $h=t$ continuing straight through the origin (at $t=-\pi$ the height is $-\pi$). Now check the seam: one period ends at height $\pi$ and the next begins at $-\pi$. The heights do NOT match, so every seam is a cliff, and the result is the jumping sawtooth. Same fragment, different symmetry instruction, completely different waveform. This seam check is also why the triangle's coefficients decay fast, like $1/n^2$ (continuous, no jumps), while the sawtooth's decay slowly, like $1/n$, with Gibbs ringing at every cliff.

The recipe, reusable for any half-range question

1. Draw the given fragment on its half range.

2. Even: fold it across the vertical axis. Odd: fold and flip below the axis.

3. Copy the completed period end to end forever; check the seams for jumps.

Whatever appears is the waveform, named afterwards. Triangle was the output here, never an assumption.

Two definitions, each a picture operation:

$$\text{even: }h(-t)=h(t)\text{odd: }h(-t)=-h(t)$$

Even: reflect the graph in the vertical axis and it must land exactly on itself (mirror). Odd: reflect in the vertical axis AND flip upside down, and it must land on itself (a 180° rotation about the origin). The test is to actually perform these operations on the sketch and look for any mismatch. One mismatch anywhere is a complete disproof.

First, know what the wave does LEFT of the axis

The definition only covers $0$ to $2\pi$, but the wave repeats. Slide the stretch $(-\pi ,0)$ forward by one period and it lands on $(\pi ,2\pi )$, where $h=0$. So immediately left of the axis the wave is in a GAP, while immediately right of it the wave is on a PULSE. Hold onto that asymmetry; it is about to kill both tests.

Test 1: mirror it (even?)

Test 2: rotate it (odd?)

The one-point knockout

You do not need whole graphs; a single sample settles it. Pick $t=0.5$:

$$h(0.5)=1\text{but}h(-0.5)=0$$

Not equal, so not even. And $0\ne -1$, so not odd. Both symmetries demand their equation at EVERY $t$, so one failing point is a full proof of “neither”.

The zero-work shortcut for “not odd”

An odd function's areas above and below the axis cancel, so its mean over a period must be $0$. This wave never goes below zero and its mean is $\frac{1}{2}$, so odd was impossible before any reflection was drawn. (No such shortcut exists for ruling out even; that one needs the mirror check.)

Why the question even asks

Symmetry is a labour-saving report: even would have meant every $b_n=0$ (no sine integrals to do), odd would have meant $a_0=a_n=0$ (no cosine integrals). “Neither” is the verdict that no coefficient dies for free and all three must be computed. One subtlety worth knowing from the mean entry: THIS wave minus its mean of $\frac{1}{2}$ IS odd, which is the hidden reason all its $a_n$ turned out zero anyway even though the wave itself is “neither”.

The cases block is NOT built by testing $n=1,2,3,\dots$ one at a time and stopping somewhere. No list is ever walked and there is no stopping decision. The whole thing is TWO substitutions.

The only varying part

In the derived formula, exactly one thing changes its value with $n$:

$$b_n=-\frac{1}{n\pi }(\cos (n\pi )-1)$$

The letter $n$ is never replaced by a number. Only the cosine gets substituted, and the cosine of a whole number of half-turns can only be $-1$ or $+1$. Two possible values, so the infinite family of $n$ splits into exactly two groups with nothing left over: every whole number is odd or even.

The odd bucket, slowly

For every odd $n$ the cosine is $-1$. Put the VALUE $-1$ where the cosine sits, touching nothing else:

$$b_n=-\frac{1}{n\pi }(-1-1)=-\frac{1}{n\pi }(-2)=+\frac{2}{n\pi }$$

Trace each piece of the result. The $2$ on top is the bracket's $-2$ after its minus cancels the leading minus. The $n\pi$ underneath is the SAME $n\pi$ that was already in the formula, untouched, still a letter. That is why the answer is a formula rather than a number: it is one calculation that covers $1,3,5,7,\dots$ all at once. Reading it AT a particular $n$ gives the individual values: at $n=1$ it reads $2/\pi$, at $n=3$ it reads $2/3\pi$. Those values come OUT of the formula; the formula was never built from them.

The even bucket, slowly

For every even $n$ the cosine is $+1$:

$$b_n=-\frac{1}{n\pi }(1-1)=-\frac{1}{n\pi }\times 0=0$$

The bracket is one minus one, which is zero, and zero times anything is zero. Done for every even $n$ simultaneously.

Why any earlier “n = 1, 2, 3, 4” list appeared at all

Purely as a demonstration that the two bucket-formulas give the right individual numbers. Stopping that list at 4 (or 2, or 100) carried no meaning, because the list is not part of the method. The two cases ARE the calculation; the cases block just staples them together with labels:

$$b_n=\{\begin{array}{ll}0& n\text{ even}\\ \frac{2}{n\pi }& n\text{ odd}\end{array}$$

Related: which n, and when to stop.

None of it is trial and error, and the confusion dissolves once one fact lands: you never pick $n$ at all. The series uses ALL of them at once.

1. n is not a choice, it is a label

Look at what the sheet's template actually says:

$$f(t)=\frac{a_0}{2}+\sum _{n=1}^{\infty }(a_n\cos (n\omega t)+b_n\sin (n\omega t))$$

The $\sum$ runs $n=1$ to $\infty$: the series contains a term for $n=1$ AND $n=2$ AND $n=3$ AND every whole number forever. Each $n$ is one harmonic, one ingredient wave. Nobody selects some of them; they are all in the recipe from the start.

2. The coefficient formula is a machine, computed once for ALL n

When you derived

$$b_n=-\frac{1}{n\pi }(\cos (n\pi )-1)$$

you did the integral with the SYMBOL $n$ left in, not with any particular number. That one calculation therefore covers every harmonic simultaneously: it is a machine where you feed in any $n$ and it returns that harmonic's coefficient. Feed it 1, get $2/\pi$. Feed it 2, get $0$. Feed it 3, get $2/3\pi$. No trial, no error: the answer for $n=1000$ is already inside the formula.

The odd/even split is not you choosing which $n$ to use. It is you NOTICING a pattern in the machine's output, because $\cos (n\pi )$ only ever takes two values. Infinitely many outputs get described in two lines: zero for every even $n$, $2/n\pi$ for every odd $n$. The cases block is a compressed TABLE of all outputs, not a selection.

3. What the series physically contains

4. So when do you stop? You don’t. You stop WRITING.

The series is infinite and stays infinite; that is what the $\cdots$ at the end of an answer means. The complete, full-marks description of all infinitely many terms is the GENERAL formula plus the assembled series:

$$b_n=\{\begin{array}{ll}0& n\text{ even}\\ \frac{2}{n\pi }& n\text{ odd}\end{array},f(t)=\frac{1}{2}+\frac{2}{\pi }(\sin t+\frac{\sin 3t}{3}+\frac{\sin 5t}{5}+\cdots )$$

Listing terms is only illustration, and how many you list is decided by the QUESTION, not by the maths:

“Find the Fourier series” (Tut1 style): give the general coefficient formulas, the cases block, and the series with the first two or three surviving terms plus $\cdots$. Done.

“Write the series for $n=-3$ to $3$” (Tut2 style): the question names the exact range, so you evaluate the machine at precisely those seven values, zeros included, and stop at the named edge.

“First three non-zero terms”: feed the machine $n=1,2,3,\dots$ in order, keep what survives, stop after three survivors. For the square wave that is $n=1,3,5$.

The one thing you never do is stop because the numbers got small. They always get small (that is why the series converges); smallness is expected, not a stopping signal.

5. The order of work, start to finish

1. Compute each coefficient ONCE with the symbol $n$ left in. This covers all harmonics in one integral.

2. Simplify using $\sin (n\pi )=0$ and $\cos (n\pi )={(-1)}^n$; if the result depends on the parity of $n$, record it as a two-line cases block. That block IS the answer for every $n$.

3. Assemble the sheet's template with those coefficients, writing enough terms to show the pattern, or exactly the terms the question names.

Related entries: why f must be substituted before integrating, why the limits split, where the constant term comes from.

Because $f(t)$ in that line is a NAME, not a formula. Integration rules act on expressions: you can integrate $t$, $t^2$, $\sin (nt)$, because each has an antiderivative rule. There is no rule for the letter $f$. So the line ${\int }_0^{2\pi }f(t)dt$ is not yet solvable; it is a placeholder waiting for you to substitute what $f$ actually IS.

And for this square wave, what $f$ actually is depends on where you are:

$$f(t)=\{\begin{array}{ll}1& 0<t<\pi \\ 0& \pi <t<2\pi \end{array}$$

Two different expressions on two different stretches. There is no single formula valid across all of $(0,2\pi )$ to substitute, so a one-shot integration is impossible for this function. The split at $t=\pi$ is precisely the move that lets the substitution happen: after it, each segment holds exactly one formula, and each becomes solvable.

When you CAN do it in one go

If the question's function is a single expression over the whole period, there is no split and you integrate directly. The sawtooth $f(t)=t$ on $(0,2\pi )$ is the model case:

$${\int }_0^{2\pi }tdt={\left[\frac{t^2}{2}\right]}_0^{2\pi }=2{\pi }^2$$

One formula, one antiderivative, one evaluation. Same sheet line, no splitting anywhere.

The decision rule

Read the question's definition of $f$ FIRST, before copying any integral:

One formula for the whole period → integrate directly, done.

A piecewise list (“$f=\dots$ on this bit, $\dots$ on that bit”) → the integral must be split at every breakpoint of the list, one segment per formula, then each segment solved on its own. See the splitting entry for why the split changes nothing about the period.

So the split is not an extra convention someone chose. It is forced by the shape of the function: you split exactly when, and exactly where, $f$ changes formula.

No. The period is still $2\pi$ and the calculation still covers all of $0\to 2\pi$. What happened between those two lines is a SPLIT, not a change. This is the additivity rule of integrals:

$${\int }_0^{2\pi }f(t)dt={\int }_0^{\pi }f(t)dt+{\int }_{\pi }^{2\pi }f(t)dt$$

The two segments tile the full period with no gap and no overlap. Walking the whole distance is the same as walking it in two stages.

Why cut at π specifically

Because that is where $f$ changes formula. One integrand cannot describe the whole range, since the function is not “$1$” everywhere. After the cut, each segment finally has a single formula you can substitute: $1$ on the first piece, $0$ on the second. Rule: split a piecewise integral at the breakpoints of its definition, never anywhere else.

Why the range then seems to shrink to 0 → π

The $\pi \to 2\pi$ piece was not discarded, it was evaluated. Its integrand is $0$, so it contributes exactly $0$:

$${\int }_{\pi }^{2\pi }0dt=0,\text{so}{\int }_0^{\pi }1dt+0={\int }_0^{\pi }1dt$$

Writing “$+0$” is simply not worth the ink, so the next line shows only the surviving segment. That ${\int }_0^{\pi }$ is not a new period; it is the only segment that contributed anything.

The check that keeps you honest

Whatever splitting and discarding-of-zeros happens in the middle, the prefactor at the front never changes: you still multiply by the $2/T$ (or divide by the $T$) of the FULL period, $2\pi$. The area only accumulated over a width $\pi$, but it is spread over the whole period $2\pi$. That mismatch of widths is exactly why the mean of this square wave is $\frac{1}{2}$ and not $1$. See the mean value entry and the area entry for the two companion doubts.

Two separate things are tangled here, and both are worth untangling because they come back in every integral you will ever do.

1. The dotted verticals are not part of the function

They are drawing aids marking where the signal jumps. They cannot be part of the graph, and there is a hard reason why: at the jump instant a vertical line would mean the function takes every value between $0$ and $1$ at that single value of $t$. A function is only allowed one output per input, so that line would fail the vertical line test. Being dotted is the sketch’s way of saying “this is not a value, it is a jump”.

What the function actually is: $h=1$ for every $t$ strictly between $0$ and $\pi$, and $h=0$ for every $t$ strictly between $\pi$ and $2\pi$. Just the horizontal pieces. Nothing vertical anywhere.

2. “Area under the graph” never needed a closed shape

The phrase is shorthand for: the region trapped between the graph and the $t$-axis, over the stretch of $t$ you are integrating across. Its top is the graph, its bottom is the axis, and its left and right edges are simply where you start and stop measuring. Those edges are decided by the limits on your integral sign, not by whether anyone drew lines there.

3. The picture that removes shapes entirely

If “shape” is the sticking point, drop shapes altogether. An integral is a running total of height times width. Slice the interval into strips of width $dt$; each strip contributes (height of the graph there) $\times$ $dt$; add them all up. That is literally what the $\int$ sign means, and it never asks whether anything is closed or drawn.

So the “missing” second half of the wave is not a shape you failed to close. It is a shape with zero height, which has zero area by definition.

4. Yes, shade it in, but area below the axis counts negative

Shading the region and finding its area is exactly the right working method. Just note that the region is not a fiction you are pretending into existence: only its top edge is the function, while its bottom is the $t$-axis and its sides are the integration limits. Those three boundaries are always there, so the region is always genuinely closed. The useful mental swap is to stop reading “area of the shape drawn on the page” and start reading “area between the curve and the axis, from here to here”.

The one catch, and it arrives in the very next tutorial part: where the curve sits BELOW the axis, its area counts as negative. Strips there have negative height, so they subtract.

Written out, the right-hand case is

$$\text{mean}=\frac{(-1)(2)+(+1)(1)}{3}=-\frac{1}{3}$$

so the shading picture and the level-times-duration arithmetic are the same calculation, one drawn and one written.

Putting it back together

Area over one period $=1\times \pi =\pi$, period length $=2\pi$, so

$$\text{mean}=\frac{\pi }{2\pi }=\frac{1}{2},a_0=2\times \frac{1}{2}=1$$

and the integral route agrees because ${\int }_0^{2\pi }hdt$ is exactly “add up the strips”, which is $\pi$.

One thing that follows and is worth keeping: a jump discontinuity never affects an area. Changing a function at a single instant, or arguing about whether $h(\pi )$ equals $0$ or $1$ or neither, changes a strip of zero width, so it changes nothing about the integral. That is why the tutorial can write the pieces on open intervals and never say what happens exactly at the jump. See the mean value entry for how this feeds the $a_0$ shortcut.

The data sheet prints two readings of the same quantity on one line: $a_0=\frac{2}{T}{\int }_0^{T}f(t)dt$ and $a_0=2\times$ mean value of $f(t)$. The second is not a different formula, it is the first one read backwards, and for flat-topped waveforms it turns a calculus question into arithmetic.

The definition

The mean (average) value of a periodic function over one period is

$$\text{mean}=\frac{1}{T}{\int }_0^{T}f(t)dt=\frac{\text{area under the graph over one period}}{\text{length of that period}}$$

Compare that with the sheet’s integral line. The sheet has $2/T$ out front where the mean has $1/T$, and that single factor of two is the entire relationship: $a_0=2\times$ mean.

The square wave, done by area

One period runs from $0$ to $2\pi$. Over it, $h$ is a rectangle of height $1$ and width $\pi$ (from $0$ to $\pi$), then flat zero for the rest. So the area is $1\times \pi =\pi$, the period is $2\pi$ wide, and

$$\text{mean}=\frac{1\times \pi +0\times \pi }{2\pi }=\frac{\pi }{2\pi }=\frac{1}{2},a_0=2\times \frac{1}{2}=1$$

Picture it as flattening: the mean is the height you would get if you took all the area and spread it evenly across the whole period.

Two routes, not two steps

The $=$ in the middle of the sheet’s $a_0$ line is easy to misread. It is not telling you to do the integral and then do something else to the answer. It is offering two independent ways to reach the same number, and you take whichever is faster:

Route A, the integral. Substitute first, then integrate. $T=2\pi$ so $2/T=1/\pi$, and $f=1$ only on $(0,\pi )$, giving $a_0=\frac{1}{\pi }(\pi )=1$.

Route B, the mean. No integration at all. Read the mean off your sketch, $\frac{1}{2}$, then $a_0=2\times \frac{1}{2}=1$.

Same answer, because the sheet’s equals sign guarantees it. For flat-topped waveforms route B takes seconds; for a sloped or curved $f(t)$ you generally need route A.

The trap: never do both. The $2/T$ sitting in the integral ALREADY contains that factor of two. If you evaluate $\frac{2}{T}{\int }_0^{T}fdt=1$ and then multiply by two again because the sheet mentions “$2\times$ mean”, you get $2$, which is wrong. The bookkeeping in one line:

$$\text{mean}=\frac{1}{T}{\int }_0^{T}fdt\Rightarrow 2\times \text{mean}=\frac{2}{T}{\int }_0^{T}fdt=a_0$$

The mean carries a $1/T$; doubling it turns that into the $2/T$ the integral form already shows. So the factor of two gets spent exactly once, whichever route you took, and both ends of the sheet’s line are $a_0$ itself, not $a_0$ and half of it.

The general recipe for flat-topped waves

Whenever the waveform is piecewise constant, no integration is needed at all. Multiply each level by how long it is held, add, divide by the period:

$$\text{mean}=\frac{\sum (\text{level}\times \text{duration at that level})}{T}$$

Half the time at $1$ and half at $0$ gives $\frac{1}{2}\cdot 1+\frac{1}{2}\cdot 0=\frac{1}{2}$, the same answer read as a time-weighted average rather than as an area.

The trap: weight by duration, never by counting levels

Tut1 Q2(b) is the one that catches people. The wave takes only the values $+1$ and $-1$, so it looks like it should average to zero. It does not, because it sits at $-1$ for a width of $2$ and at $+1$ for a width of only $1$, out of a period $T=3$:

$$\text{mean}=\frac{(-1)(2)+(+1)(1)}{3}=-\frac{1}{3},a_0=-\frac{2}{3}$$

Compare Tut1 Q2(c), where the $-1$ and $+1$ pulses have EQUAL width: there the contributions genuinely cancel and the mean is $0$, so $a_0=0$. Same two levels, different answer, purely because of how long each is held.

Why this matters beyond saving time

The printed series carries $\frac{1}{2}{a}_0$ as its constant term, and since $a_0=2\times$ mean, that constant term is exactly the mean itself:

$$\frac{1}{2}{a}_0=\text{mean value of f}=\text{the DC level of the signal}$$

So the constant term of any Fourier series is the average height of the waveform, and every sine and cosine after it has zero average and only describes the wiggle about that level. This also gives a free check on a finished answer: if your series constant does not match the average height you can see on your own sketch, something is wrong earlier in the working.

When the shortcut stops being free

It only collapses to arithmetic when the pieces are flat. For a sloped or curved piece you are back to the integral, though geometry often still beats it: the sawtooth $f(t)=t$ on $(0,2\pi )$ encloses a triangle of area $\frac{1}{2}\times 2\pi \times 2\pi =2{\pi }^2$, so its mean is $2{\pi }^2/2\pi =\pi$ and $a_0=2\pi$, matching the by-parts result with none of the work.

Exam habit: after sketching, write one line, “mean value $=$ ..., so $a_0=$ ...”, quoting the sheet. It is a mark or two you can bank in seconds, and it protects the rest of the question.

Every ENG2086 paper ends with a Fourier Analysis data sheet. You are not meant to memorise the coefficient formulas; you are meant to read them off that sheet and substitute correctly. So here is the whole question done as a substitution exercise into the printed lines, with the sheet itself shown at each step (snipped from the December 2024 paper).

The lines you will be given

Notice what the sheet does NOT give: no $L$, no $-L$-to-$L$ limits, no half-range forms. Everything is in $T$ with limits $0$ to $T$. That is convenient here, because the question defines $h(t)$ on exactly $0$ to $2\pi$: the question’s interval and the sheet’s interval already agree, so nothing needs shifting.

Step 0: read T and ω out of the question

One period is given on $0<t<2\pi$, so $T=2\pi$, and the sheet’s own heading then gives

$$\omega =\frac{2\pi }{T}=\frac{2\pi }{2\pi }=1\text{ rad/s}$$

So on this question every $\cos (n\omega t)$ printed on the sheet is simply $\cos (nt)$, and the sheet’s prefactor $2/T$ is $1/\pi$. Write those two substitutions down before integrating anything.

Step 1: sketch and symmetry

Reflect it about the vertical axis: $h(-t)$ is neither $h(t)$ nor $-h(t)$, so the answer to “odd, even or neither” is neither. (It could not be odd: an odd function averages to zero, and this one never goes below zero.)

But look at the dashed line. Drop the wave by its mean and $h-\frac{1}{2}$ swings between $\pm \frac{1}{2}$ and IS odd. That predicts the shape of the answer: a constant plus sines only, with every $a_n$ vanishing. Useful as a check, not as a substitute for the integrals.

Step 2: a₀ via the sheet’s own shortcut

The sheet spells out $a_0=2\times$ mean value. The wave sits at 1 for half the period and 0 for the other half, so the mean is $\frac{1}{2}$ by eye and $a_0=1$. The integral route, splitting at the break and dropping the half where $h=0$:

$$a_0=\frac{2}{2\pi }[{\int }_0^{\pi }1dt+{\int }_{\pi }^{2\pi }0dt]=\frac{1}{\pi }(\pi )=1$$

Both routes agree. The series then carries $\frac{1}{2}{a}_0=\frac{1}{2}$ as its constant term, which is the mean, as it must be.

Step 3: aₙ, and why they all die

Copy the sheet’s $a_n$ line with $2/T=1/\pi$ and $\omega =1$ already substituted; only $0$ to $\pi$ survives:

$$a_n=\frac{1}{\pi }{\int }_0^{\pi }\cos (nt)dt=\frac{1}{\pi }{\left[\frac{\sin (nt)}{n}\right]}_0^{\pi }=\frac{\sin (n\pi )}{n\pi }=0$$

because $\sin (n\pi )=0$ for every whole $n$. Exactly what the shifted-odd observation predicted.

Step 4: bₙ, the only survivors

$$b_n=\frac{1}{\pi }{\int }_0^{\pi }\sin (nt)dt=\frac{1}{\pi }{[-\frac{\cos (nt)}{n}]}_0^{\pi }$$

Top minus bottom gives $-\cos (n\pi )/n+1/n$, so

$$b_n=\frac{1}{n\pi }(1-\cos (n\pi ))=\frac{1}{n\pi }(1-{(-1)}^n)$$

That bracket is the standard on/off switch: for even $n$ it is $1-1=0$; for odd $n$ it is $1-(-1)=2$. So even harmonics vanish and the odd ones give $b_n=2/(n\pi )$.

Assemble

$$h(t)=\frac{1}{2}+\frac{2}{\pi }(\sin t+\frac{\sin 3t}{3}+\frac{\sin 5t}{5}+\cdots )$$

Two free checks. At $t=\pi /2$ the bracket is $1-\frac{1}{3}+\frac{1}{5}-\cdots =\pi /4$, so the series returns $\frac{1}{2}+\frac{2}{\pi }·\frac{\pi }{4}=1$, matching the wave. At the jump $t=\pi$ every sine is zero, so the series gives $\frac{1}{2}$, the midpoint of the jump from 1 down to 0, exactly as the convergence rule requires.

If the question asks for the complex series instead

Same substitution, same split:

$$c_n=\frac{1}{2\pi }{\int }_0^{\pi }{e}^{-jnt}dt=\frac{1-{(-1)}^n}{2\pi jn}$$

so $c_n=-j/(\pi n)$ for odd $n$ and zero for even $n$, with $c_0$ handled separately as the mean, $\frac{1}{2}$. Cross-check using the conversion line the sheet also prints, $c_n=\frac{1}{2}(a_n-j{b}_n)$: with $a_n=0$ and $b_n=2/n\pi$ that is $-j/n\pi$. The two halves of the sheet agree, which is the cheapest confidence check available in the exam.

And when a complex answer has to be turned back into sines and cosines (or vice versa), the sheet gives you Euler at the bottom, so nothing there needs memorising either:

The five sheet-reading traps

1. The constant term is $\frac{1}{2}{a}_0$, not $a_0$. The sheet defines $a_0$ with the same $2/T$ as the others, so the halving in the series is what makes the constant come out as the mean.

2. The printed limits are $0$ to $T$, but any full period is legal. If a question defines its period as $-\pi$ to $\pi$, integrate over that instead of forcing the sheet’s limits.

3. $\omega$ is angular frequency in rad/s, tied to the period by $\omega =2\pi /T$. Never put an ordinary frequency into the sheet’s $n\omega t$.

4. Piecewise functions need the integral split at the break. Write the zero piece out anyway before dropping it: it costs one line and shows you handled the whole definition.

5. Complex work uses $1/T$ and a MINUS sign in the coefficient exponent, while the reconstruction uses PLUS. This is the most common lost mark in the complex parts, and the sheet warns about it in its own words.

The habit worth drilling: before integrating anything, write three lines, $T=\dots$, $\omega =2\pi /T=\dots$, $2/T=\dots$, then copy the sheet’s line with those numbers already in place. Every Fourier question in this course opens the same way, and those lines are free marks.

Trick answer: both are happening, simultaneously, with exactly equal strength. Electric forces always come in pairs (Newton’s third law). $q_3$ pulls $q_2$ toward it with 11.25 N, AND $q_2$ pulls $q_3$ toward it with 11.25 N. Neither is “the real one”; they are two ends of the same interaction, like the two ends of a stretched rubber band connecting the charges.

So why did the video only use one of them?

Because of what the question asked. Part (a) says: find the net force ON the +100 μC charge. Net-force accounting has one rule: you only tally the arrows that LAND ON the body you were asked about. The arrow $q_2$ exerts on $q_3$ lands on $q_3$, so it belongs to a different ledger. It is not wrong, it is simply somebody else’s force. (And indeed part (b) of the same question, net force on the +200 μC charge, is exactly where the reaction partners get their turn.)

The notation encodes this: $F_{2,3}$ reads “force ON charge 2 FROM charge 3”. Its partner is $F_{3,2}$, same size, opposite direction, different victim:

$$F_{2,3}=-F_{3,2}$$

The classic follow-up worry: don’t the pair cancel?

No, and this is the mistake the setup protects you from: the two arrows of a pair act on DIFFERENT bodies, so they can never cancel each other. Cancellation only happens between arrows on the SAME body (as in a null-point question, where two forces on one charge balance). The pair arrows each do real work on their own victim: $q_2$ accelerates right because of its arrow; $q_3$ accelerates left because of its own.

Also worth noticing: the pair is equal even though the charges are unequal (100 μC vs 50 μC). The formula $k{q}_2{q}_3/r^2$ contains BOTH charges symmetrically, so each feels the same 11.25 N; the smaller charge does not feel a bigger pull. (What differs is the resulting acceleration, if their masses differ, via $a=F/m$.)

One-liner to keep: every Coulomb force is a two-ended pair; the question’s wording (“force ON …”) tells you which end goes in your ledger, and pair partners never cancel because they act on different bodies.

It doesn’t ignore it. The minus sign gets used, but in the DIRECTION step, not in the arithmetic. This is the standard professional way to run Coulomb problems, and it exists precisely because plugging signs straight into the formula produces wrong answers more often than right ones.

The two-step convention

Step 1: magnitudes only in the formula. Compute every force size with absolute values:

$$F=\frac{k|q_1||q_2|}{r^2}$$

A force SIZE cannot be negative, so the signs stay out of this step entirely.

Step 2: signs decide the arrow. Like signs repel, opposite signs attract. Draw the arrow each force puts on the charge you care about. Only THEN, when adding the forces along your axis, do arrows pointing in $-x$ pick up a minus.

Run the actual example

Charges on a line: $q_1=+200\text{ μC}$ on the left, $q_2=+100\text{ μC}$ in the middle (the one we want the force on), $q_3=-50\text{ μC}$ on the right.

Where the minus on $q_3$ earns its keep: because it is negative, it ATTRACTS the middle charge, so its force pulls $q_2$ to the RIGHT (toward it). Meanwhile $q_1$ is positive and repels, pushing $q_2$ AWAY, which is also to the right. Both arrows point the same way, so the magnitudes ADD:

$$F_{\mathrm{net}}=11.25+11.25=22.5\text{ N to the right}$$

If $q_3$ had been POSITIVE instead, its force on $q_2$ would flip to the left and you would SUBTRACT: $11.25-11.25=0$. Same numbers, opposite physics, that is exactly the information the sign carries, and it never needed to enter the formula to deliver it.

Why not just plug the signs into the formula?

Because a signed formula output only tells you “attractive or repulsive”, not which way that is along YOUR axis. A negative product $q_2{q}_3$ means “attraction”, but whether attraction points in $+x$ or $-x$ depends on where the charges SIT, which the formula’s sign knows nothing about. People who carry the sign through the algebra end up subtracting forces that actually cooperate (as here) or adding forces that oppose. The magnitudes-plus-arrows method makes the geometry explicit and survives 2D problems, where forces at angles must be added by components and a bare minus sign is meaningless anyway.

One-liner to keep: signs never go into Coulomb’s formula; they choose the arrow (like repels, unlike attracts), and the arrows choose the plus or minus when you add along your axis. The exam habit that follows: always draw the force arrows on the charge in question before touching the calculator.

The table describes the four possible ways to WIRE a feedback network onto an amplifier. Every feedback amplifier has to do two jobs, and each job has exactly two options; the four topologies are just the four combinations. Once you see the two jobs, every word in the table unpacks itself.

Job 1 at the output: “samples” = what the feedback network measures

The feedback network must measure the output to know what to correct. There are only two things it can measure, and each forces a wiring style:

Sample the VOLTAGE: connect the network ACROSS the output, in parallel, exactly like a voltmeter. A parallel connection is called SHUNT. Sample the CURRENT: break the output loop and insert the network IN the current’s path, exactly like an ammeter. That is a SERIES connection.

Job 2 at the input: “mixes” = how the returned signal is subtracted

The returned correction must be combined with (subtracted from) the incoming signal. Two ways again: series mixing puts the feedback voltage into the same loop as the input source, so the VOLTAGES subtract around the loop (KVL does the subtraction). Shunt mixing joins the feedback wire and the input wire at one NODE, so the CURRENTS subtract at the junction (KCL does the subtraction).

The names: input word first, output word second

“SERIES-SHUNT” means: series mixing at the input, shunt (voltage) sampling at the output. “SHUNT-SERIES” is the mirror image: currents subtracted at the input node, output current sampled. The other two are the remaining combinations. (Watch out in other textbooks: some name the output first. This course’s convention is input-output.)

Z_in raised/lowered: feedback pushes impedances toward the ideal

The pattern behind those columns is one sentence: negative feedback makes the amplifier look MORE IDEAL for the job its topology defines, each by the familiar factor $(1+A\beta )$.

At the input: series mixing RAISES $Z_{\mathrm{in}}$ (good, a voltage input wants to draw no current), shunt mixing LOWERS it (good, a current input wants to present no obstacle). At the output: voltage sampling LOWERS $Z_{\mathrm{out}}$ (a stiff voltage source that doesn’t sag under load), current sampling RAISES it (a stiff current source that pushes its current regardless of load).

The “stabilises” column: four gains, two of them with odd names

Each topology holds one gain constant, the ratio of (what it samples) to (what its input naturally accepts):

Voltage gain $A_v=V_{\mathrm{out}}/V_{\mathrm{in}}$ (dimensionless). Current gain $A_i=I_{\mathrm{out}}/I_{\mathrm{in}}$ (dimensionless). Transconductance $G_m=I_{\mathrm{out}}/V_{\mathrm{in}}$: voltage in, current out, units A/V (siemens). Transresistance $R_m=V_{\mathrm{out}}/I_{\mathrm{in}}$: current in, voltage out, units V/A (ohms). The prefix TRANS- just says the ratio crosses from input to output with DIFFERENT quantities on top and bottom, so the “gain” carries units.

Familiar faces: the op-amp non-inverting amplifier is series-shunt (input voltage subtracted in a loop against the divider’s feedback voltage; output voltage sampled by the R₁-R_F divider), which is exactly why it has huge $Z_{\mathrm{in}}$ and tiny $Z_{\mathrm{out}}$. The inverting amplifier behaves as shunt-shunt (currents subtract at the virtual-earth node; output voltage sampled), which is why its input impedance is only $R_1$ rather than huge.

One-liner to keep: first word = how the input subtracts (series = voltages in a loop, shunt = currents at a node); second word = what the output measurement wire looks like (shunt = voltmeter → voltage, series = ammeter → current); and feedback drives every impedance toward the ideal for that combination.

They are the SAME quantity, the amplifier’s upper cutoff frequency, measured in two different situations. The subscripts are doing all the work:

fₖ: the cutoff of the bare (open-loop) amplifier

The subscript H stands for HIGH: it marks the high-frequency corner, the frequency (in Hz) where the amplifier’s gain has fallen 3 dB below its flat low-frequency value and begins rolling off. Up to $f_H$ the bare amplifier gives you its full gain $A_0$; beyond it the gain slides away. So $f_H$ is the bandwidth edge of the amplifier WITHOUT any feedback applied.

fₖₗ: the cutoff of the same amplifier WITH feedback

The extra subscript f stands for “with Feedback”, exactly the same convention as $A_f$ (closed-loop gain) versus $A$ (open-loop gain). Close the negative-feedback loop and the corner MOVES UP:

$$f_{\mathrm{Hf}}=f_H(1+A_0\beta )$$

Same amplifier, same physics; the feedback trades away gain (divides it by $1+A_0\beta$) and buys bandwidth with it (multiplies the corner by the SAME factor).

Why the corner moves at all

Intuition, not just algebra: near $f_H$ the bare amplifier’s gain starts to sag. But the feedback loop’s whole job is to hold the output at $v_s/\beta$ by amplifying the error harder whenever the output droops. So as the raw gain sags, the loop compensates, spending its surplus gain to keep the CLOSED-loop gain flat. It can keep this up until the raw gain has sagged all the way down to the closed-loop level, and that happens precisely at $f_{\mathrm{Hf}}$. The visible flat band stretches; the underlying roll-off never moved.

Both corners live on the same falling line, which is why the trade is exact:

$$A_{f0}·f_{\mathrm{Hf}}=A_0·f_H=\text{GBW (constant)}$$

One-liner to keep: fₖ = where the bare amplifier’s gain corner sits; fₖₗ = where that corner sits after feedback, pushed up by exactly the factor (1+A₀β) that the gain gave away. Subscript H = high-frequency corner; trailing subscript f = with feedback.

No, but the confusion is understandable, because the active high-pass filter in this course is BUILT FROM the inverting op-amp. The relationship is template and specialisation: the inverting op-amp is a general machine, and the filters are what you get when you fill in its two slots with particular components.

The template: gain = −Z₂/Z₁

The inverting amplifier is defined by its wiring: input through an impedance $Z_1$ into the − input, feedback impedance $Z_2$ from output back to the − input, + input grounded. Its gain is always

$$A_v=-\frac{Z_2}{Z_1}$$

The template says nothing about WHAT $Z_1$ and $Z_2$ are. That choice is where the personality comes from.

How each specialisation works

Plain amplifier: both slots are resistors. Resistors don’t care about frequency, so the gain $-R_2/R_1$ is flat: every frequency is amplified equally. An amplifier, not a filter.

Active low-pass: put a capacitor ACROSS the feedback resistor. At low frequency the capacitor is effectively open and invisible, so gain is $-R_2/R_1$; at high frequency its impedance collapses toward zero, it short-circuits $R_2$, so $Z_2\to 0$ and the gain dies. Highs killed = low-pass.

Active high-pass: put a capacitor IN SERIES with the input resistor. At low frequency the capacitor’s impedance is enormous, so $Z_1$ is huge and the gain $-Z_2/Z_1$ collapses: nothing gets in. At high frequency the capacitor is nearly a short, leaving $Z_1\approx R_1$ and full gain $-R_2/R_1$. Lows killed = high-pass.

Why the confusion is natural, and the sentence to keep

Every one of these circuits LOOKS like an inverting op-amp, because every one of them IS one; and a question about any of them starts from the same master formula $-Z_2/Z_1$. Also note the high-pass is not the only filter wearing this costume, and the inverting amp is not the only high-pass: the PASSIVE high-pass (just C then R, no op-amp) filters the same shape with no gain and no inversion.

So: “inverting op-amp” names a circuit TEMPLATE; “high-pass filter” names a BEHAVIOUR. The active high-pass is the template wearing a series input capacitor; with the capacitor moved to the feedback it becomes a low-pass; with plain resistors it’s no filter at all.

This formula answers one question: when a sine wave passes through a high-pass filter, by how much is the output shifted in time relative to the input? Piece by piece:

φₖₚ (phi): the phase shift of the output

${\phi }_{\mathrm{HP}}$ is an ANGLE, measured in degrees (or radians), that describes how far the output waveform is slid along the time axis compared to the input. One full cycle of the sine is 360°, so a shift of 90° means the output is displaced by a quarter of a cycle. The subscript HP just says this is the high-pass filter’s version of the formula (the low-pass has its own, with a minus sign and the fraction flipped).

The + sign is information, not decoration: positive phase means the output LEADS the input (its peaks happen earlier in time). The high-pass filter’s capacitor-first structure makes the output current, and hence the output voltage across R, peak before the input voltage does. A low-pass does the opposite: its output LAGS (negative φ), because the capacitor needs time to charge.

f: the frequency of the signal you are feeding in

$f$ is the frequency, in Hz, of the sine wave currently going through the filter. It is the variable you plug in: ask “what does the filter do at 200 Hz?” and 200 Hz is $f$. Different frequencies get different phase shifts, which is the whole point of the formula.

fₔ: the filter’s cutoff frequency

$f_c$ is the corner/cutoff frequency of the filter itself, fixed by the components: $f_c=1/(2\pi RC)$. It is a property of the circuit, not of the signal. In the formula it acts as the yardstick the signal frequency is compared against: only the RATIO $f_c/f$ matters.

arctan: the angle-recovery function

$arctan$ (inverse tangent, the tan⁻¹ button on your calculator) converts a ratio into the angle whose tangent equals that ratio. It appears because $R$ and the capacitor’s reactance $X_C$ sit at right angles in the complex plane; the phase angle of the divider is the angle of that right triangle, and “angle from a ratio of two perpendicular sides” is exactly what arctan computes. Make sure the calculator is in DEGREE mode if you want degrees out.

Reading the formula’s behaviour

Sanity anchors to memorise with it, all straight from the formula:

Far below cutoff ($f\ll f_c$): the ratio $f_c/f$ is huge, and arctan of a huge number approaches 90°: the output leads by nearly a quarter cycle (and is also tiny, since this is the stop band).

Exactly at cutoff ($f=f_c$): the ratio is 1 and $arctan(1)=45°$. The corner is always the 45° point (and the 0.707 amplitude point, both at once).

Far above cutoff ($f\gg f_c$): the ratio shrinks to zero, arctan(0) = 0°: in its pass band the high-pass hands the signal through unshifted and at full size.

Worked micro-example: high-pass with $f_c=1\text{ kHz}$, signal at $f=500\text{ Hz}$. Ratio $=1000/500=2$; $arctan(2)\approx 63°$. Output leads by about 63° (and is attenuated, since 500 Hz is below the corner).

An inductor is a coil of wire, and it does one thing: it gives current inertia. The current flowing through the coil builds a magnetic field around it; that field stores energy, and the stored energy resists any attempt to change the current, in either direction. A capacitor is a bucket for CHARGE whose voltage resists change; an inductor is a flywheel for CURRENT whose current resists change. Almost everything inductors are used for is this one property wearing different clothes.

The defining law, read slowly

$$v=L\frac{di}{dt}$$

The voltage across an inductor depends not on the current but on how fast the current is CHANGING. $L$ (henries, H) is how much inertia the coil has. Three consequences fall straight out, and every one of them is exam material:

1. Inductor current cannot jump. A step change in current would need $di/dt\to \infty$, hence infinite voltage. So current through an inductor is always a smooth, continuous ramp. (The exact dual of the capacitor, whose VOLTAGE cannot jump.)

2. To steady current it is a wire; to fast wiggles it is a wall. If $i$ is constant, $di/dt=0$ so $v=0$: at DC an ideal inductor is a short circuit. For AC its impedance is $Z_L=j\omega L$, growing with frequency: the higher the frequency, the harder it pushes back. Again the mirror image of the capacitor (which blocks DC and passes high frequency). That is why a series inductor used as a “choke” lets the DC through and chokes the ripple.

3. Interrupt the current and it bites. Open a switch feeding an inductor and you force a huge negative $di/dt$; the coil answers with a huge voltage spike of whatever polarity keeps its current flowing, arcing across the switch if necessary. This is why every relay coil and every converter has a freewheeling (flyback) diode: a path for the inductor current to continue into while it winds down gently.

The energy view (why PE2 is built on them)

The magnetic field holds energy

$$E=\frac{1}{2}L{I}^2$$

and unlike a resistor, an ideal inductor never wastes any of it; it only borrows and returns. That makes it the working component of every switching converter: close the switch and the inductor charges up (current ramps, energy in); open the switch and it insists on keeping its current flowing, pushing that energy on into the load through the diode. Do this tens of thousands of times a second and you move power from one voltage to another with almost no loss. Buck, boost and buck-boost converters differ only in WHERE the inductor sits and hence what the release path does; the duty-cycle formulas all come from the inductor ramp equation $\Delta i=(V/L)\Delta t$ balancing over one cycle.

Where you meet them

PE2: the converter inductor (energy shuttle), freewheeling diodes protecting switches, chokes smoothing rectifier output current, and transformers/motors, which are inductors with company (two coupled coils; coils on a rotating machine). AE2: LC tuned circuits and higher-order filters (though AE2 mostly builds filters from RC + op-amps instead). EM2: where $L$ itself comes from, computed from geometry via flux linkage, $L=N\Phi /I$.

One-liner to keep: an inductor stores energy in its magnetic field and uses it to keep its current smooth and continuous: a short to DC, a wall to fast change, a spike generator when interrupted, and an energy shuttle when switched on purpose.

Mains hum is the 50 Hz signal from the power grid leaking into circuits where it doesn’t belong. The UK grid alternates at 50 Hz (60 Hz in the US), and every mains cable, transformer and appliance around you is effectively a weak 50 Hz transmitter. If any of that couples into an audio or sensor circuit, you hear or measure a continuous low buzz at 50 Hz and its harmonics (100 Hz, 150 Hz…). That buzz is the “hum”: the deep drone you hear from cheap speakers, guitar amps with poor cables, or any circuit with its input lead dangling.

How it gets in (three doors)

1. Capacitive (electric-field) pickup. A mains wire and your signal wire form a tiny capacitor through the air. Tiny capacitance, but a high-impedance input needs almost no current to develop millivolts across it. This is why touching a scope probe with your finger displays a 50 Hz wave: your body is the antenna.

2. Magnetic (inductive) pickup. Mains current creates an alternating magnetic field; any LOOP of wire in your circuit is a one-turn transformer secondary in that field, and an EMF at 50 Hz is induced around it. Bigger loop area = more hum, which is why signal and return wires are kept together (twisted pairs make the loop area almost zero and alternate its sign every twist).

3. Power-supply ripple. Your own DC supply is made FROM the mains by rectifying and smoothing it. Imperfect smoothing leaves a residue riding on the DC rail; after a full-wave bridge rectifier the residue is at 100 Hz (both half-cycles get folded up, so the ripple frequency is twice the mains). If the rails wobble, the wobble leaks into the output. This is exactly the ripple calculation in the rectifier exam question: a bigger smoothing capacitor means smaller ripple, less hum.

Why AE2 cares: small signals lose

A sensor signal can be a few millivolts. If the wiring picks up even 10 mV of hum, the interference is bigger than the information, and an amplifier is loyal to neither: it amplifies both by the same gain. Filtering is awkward here because 50 Hz is often INSIDE the band you want (audio, ECG). The real fix exploits geometry instead:

Run the signal on TWO wires and subtract them at a differential amplifier. The wanted signal is applied BETWEEN the wires (differential), but hum couples onto both wires almost identically (common mode) because they sit side by side in the same field. Subtraction keeps the difference and cancels what is common: signal survives, hum vanishes. How well a real amplifier pulls this off is its CMRR (common-mode rejection ratio), and that is precisely why the instrumentation-amplifier exam question quotes a CMRR and asks for the smallest differential input you can still discern against interference.

The defence checklist

Keep loops small (twisted pair), shield cables (grounded screen intercepts capacitive pickup), avoid ground loops (two paths to ground = one big loop = one-turn secondary), smooth and regulate supplies (kills the 100 Hz ripple door), and measure differentially with high CMRR when the signal is small. Musicians’ folklore, “it stopped buzzing when I touched the metal chassis”, is this physics: your touch grounded the shield.

One-liner to keep: mains hum = 50 Hz leakage from the grid, entering by capacitance, induction, or supply ripple; it is common-mode on a wire pair, which is why differential amplifiers with high CMRR are the cure.

The slide is using one direction of a two-way pair. For VOLTAGE gain, the definitions are:

Ratio → dB (the definition):

$$\text{gain in dB}=20{\log }_{10}(A_v)$$

dB → ratio (the same equation solved for $A_v$, which is what the slide used):

$$A_v={10}^{\text{dB}/20}$$

To undo “20 log”, divide by 20 and raise 10 to it; the two operations are inverses, nothing more. With 40 dB: ${10}^{40/20}={10}^2=100$.

The power version (the 20 vs 10 trap)

For POWER ratios the factor is 10, not 20:

$$\text{dB}=10{\log }_{10}(P_2/P_1)\leftrightarrows \frac{P_2}{P_1}={10}^{\text{dB}/10}$$

The decibel is DEFINED on power with the 10. Voltage gets 20 only because power goes as voltage squared ($P=V^2/R$), and the square jumps out of the log as a factor of 2: $10{\log }_{10}{(A_v)}^2=20{\log }_{10}(A_v)$. Exam rule of thumb: voltages and currents use 20, powers use 10.

Landmarks that make the calculator optional

Because logs turn multiplication into addition, a handful of anchors covers most numbers you meet:

×10 = 20 dB, so ×100 = 40 dB, ×1000 = 60 dB (add 20 per factor of 10). ×2 ≈ 6 dB, ×√2 ≈ 3 dB, and ×1 = 0 dB (unity). Negative dB is attenuation: −20 dB = ÷10, −3 dB = ÷√2 ≈ 0.707, which is exactly why the bandwidth edge is called the −3 dB point: the voltage has fallen to 70.7%, i.e. HALF the power.

Combine by adding: 46 dB = 40 + 6 → 100 × 2 = 200. And 37 dB (the slide’s corner) = 40 − 3 → 100 ÷ √2 ≈ 70.7.

So the general machinery is just: dB = 20 log₁₀(ratio) for voltage (10 for power), and ratio = 10^(dB/20) to come back (10^(dB/10) for power). Everything else, cascades adding in dB, Bode plots being straight lines, the −3 dB bandwidth, is that one pair of formulas doing its multiplication-into-addition trick.

The two little arrows ($V^{+}$ pointing up, $V^{-}$ pointing down) ARE connections; they are just drawn in shorthand. An arrow like that means: “this pin goes off to the supply rail of that name, wherever it is drawn (or not drawn) on the page.” Typically $V^{+}=+15\text{ V}$ and $V^{-}=-15\text{ V}$ from a dual (split) supply. Nothing is floating; the wire to the batteries is simply not drawn, because if every diagram drew both supply wires to every chip, the signal path you are supposed to study would be buried in plumbing.

Why the op-amp must have these pins at all

An op-amp is an ACTIVE device. The energy in its output does not come from the input signal; a microvolt-level input could never light up a big output swing by itself. The output energy comes from the supply rails, and the input merely steers how much of it is delivered. Think of the input as the driver’s foot and the supply as the fuel tank: no tank, no motion, however hard you press. Remove the supply and the op-amp is an inert lump of silicon; every gain formula silently assumes the rails are there.

Two places the “invisible” rails still bite you in the exam

Saturation: the output physically cannot go above $V^{+}$ or below $V^{-}$; in practice it stops a volt or two inside them, at $\pm V_{\mathrm{sat}}$. Ask a gain-100 amplifier on ±15 V rails to amplify 0.5 V and it will NOT give 50 V; it slams into ≈±13–14 V and the waveform's tops get clipped flat. Comparator, Schmitt-trigger and oscillator questions run on exactly this: their output is only ever at one rail or the other.

Single vs split supply: with a split supply ($+15/-15$) the output can swing both sides of 0 V, which is why AC signals centred on ground work directly. Power a circuit from a single rail ($0$ to $+15$) and the output can never go negative; the signal has to be biased up to mid-rail instead.

Same convention elsewhere: ground symbols are the identical trick in the other direction. Every little ground pointing down means “connected to the common 0 V node”, and all of them are one node even though no wire is drawn between them. Circuit diagrams omit the boring, always-the-same wiring (power in, common return) so the interesting wiring stands out; the arrows and ground stubs are the breadcrumbs it leaves behind.

Short answer: within the region your exam questions live in, yes, and that is deliberate; but it is not a law of nature. −20 dB/decade is the fingerprint of exactly ONE pole (one RC time constant) being in charge. Wherever one pole dominates, you get exactly that slope; wherever more poles join in, the slope steepens in −20 steps.

Why one pole gives exactly −20 and nothing else

A single pole means the gain behaves like

$$|A|=\frac{A_0}{\sqrt{1+{(f/f_c)}^2}}$$

Well above the corner ($f\gg f_c$) the 1 under the root stops mattering and this collapses to $|A|\approx A_0{f}_c/f$: gain is proportional to $1/f$. Multiply frequency by 10 and the gain divides by exactly 10, which IS −20 dB. The slope cannot be anything else, and it cannot drift: it stays exactly −20 dB/decade for as long as that one pole is the only thing acting. That perfect consistency is precisely why GBW = constant works.

Where it stops being true

Three boundaries, all visible on the sketch:

Below the corner $f_c$: the gain is flat, slope 0. No roll-off yet.

Between the corner and (nearly) $f_T$: the −20 dB/decade region. In a general-purpose op-amp this stretch is enormous, maybe 10 Hz to 1 MHz, five decades, and it is CONSISTENT the whole way because manufacturers build in a compensation capacitor that creates one deliberately dominant pole far below all the others. The op-amp is engineered to look like a textbook single-pole system over its whole useful range. This is the only region your GBW exam questions use.

Near and beyond $f_T$: the transistors’ own parasitic poles catch up. Each extra pole adds another −20, so the plot bends to −40, then −60 dB/decade. (This steepening region is also where feedback can turn unstable, which is exactly WHY the makers force the single dominant pole below it; that story belongs to control theory, not the AE2 exam.)

The rule as a slope bookkeeping system

The generalisation worth keeping: every pole contributes −20 dB/decade, every zero contributes +20, and slopes simply add. You already use this in the filters unit: a first-order RC filter rolls off at −20 dB/decade; a second-order filter at −40; an n-th order at $n\times (-20)$. When an exam asks you to sketch a Bode plot, you are just placing corners and adding ±20 slopes between them.

So: consistent, yes, wherever a single pole dominates, and op-amps are purpose-built so that this covers everything you will calculate. Just remember the slope is per-pole bookkeeping, not a universal constant: more poles, steeper line.

“Unity” is just the formal word for the number 1. Unity gain = a gain of exactly 1: the output is a copy of the input, same size, not amplified, not attenuated. In decibels that is

$$20{\log }_{10}(1)=0\text{ dB}$$

so “unity gain”, “gain of 1” and “0 dB” all say the same thing. The phrase shows up in three different corners of AE2, and each one is exam-relevant.

1. The unity-gain buffer (voltage follower): the amp that “does nothing” on purpose

Tie the op-amp’s output straight back to its inverting input and feed the signal into the + input:

Golden Rule 2 forces $V_{-}=V_{+}$, and here $V_{-}$ IS the output, so $V_{\mathrm{out}}=V_{\mathrm{in}}$. Same result from the non-inverting formula with all of the output fed back ($\beta =1$): $A_v=1+R_F/R_1$ with $R_F=0$ gives 1.

Why build an amplifier that doesn’t amplify? Because gain was never the point; muscle is. The follower has a nearly infinite input resistance (it barely disturbs the delicate circuit driving it) and a tiny output resistance (it can drive a heavy load without sagging). It is an impedance converter: same voltage, vastly better ability to deliver it. Classic use: a sensor that can only supply microamps needs to drive something hungry; put the follower in between and neither side troubles the other (this is the cure for the loading problem in the previous entry).

2. The unity-gain frequency: where an op-amp runs out of gain

A real op-amp’s open-loop gain is huge at DC but falls steadily with frequency. The frequency where it has dropped all the way to 1 (0 dB) is the unity-gain frequency $f_T$, and for the usual one-pole roll-off it equals the gain-bandwidth product:

$$\text{GBW}=\text{gain}\times \text{bandwidth}=f_T$$

This is the number the paper hands you (“GBW = 1 MHz”) so you can trade gain for bandwidth: build a ×100 amplifier from it and your bandwidth is $1\text{ MHz}/100=10\text{ kHz}$. The follower, using gain 1, gets the entire 1 MHz, another reason buffers are everywhere.

3. Cousins worth recognising

The transistor versions of the buffer, the emitter follower (BJT) and source follower (FET), have gain “approximately unity” for the same job: high input impedance, low output impedance, gain ≈ 1 (slightly under, e.g. 0.95, since they lack the op-amp’s feedback precision). And in the oscillator unit, the Barkhausen condition demands unity LOOP gain, $|A\beta |=1$: once around the loop the signal must come back at exactly its own size, so it neither dies out nor blows up.

So whenever you meet the phrase: unity gain = 1× = 0 dB, and the interesting question is never the number itself but WHY a gain of 1 is useful there: copying a voltage with muscle (follower), running out of amplification ($f_T$), or sustaining itself exactly (oscillator).

The load is whatever you connect to a circuit’s output to USE the signal: the speaker on an audio amp, the next amplifier stage, a motor, an ADC input, or in exam questions simply a resistor called $R_L$. From the circuit’s point of view, everything hanging on its output terminals behaves like one resistance that draws current, so we model the whole consumer as that single $R_L$.

Why it has a dramatic name and not just “R3”

Because connecting it COSTS the circuit something, the way loading cargo costs a lorry speed. The amplifier’s output isn’t a perfect voltage source; it has an internal output resistance $R_{\mathrm{out}}$ in series. The moment a load draws current, that current flows through $R_{\mathrm{out}}$ and drops voltage across it, so LESS of the internal voltage reaches the outside. The two resistances form a voltage divider:

$$v_L=K_v{v}_{\mathrm{in}}·\frac{R_L}{R_{\mathrm{out}}+R_L}$$

That reduction is called the loading effect, and “load” also works as a verb: “stage B loads stage A” means B’s input resistance drags A’s output down exactly this way. This is the core mechanic of every cascade question in the paper.

Heavy load vs light load (the counterintuitive bit)

A heavy load is a SMALL $R_L$: small resistance draws lots of current and drags the output far down (divider fraction small). A light load is a LARGE $R_L$: barely any current, barely any drop. The extreme light load is nothing connected at all, called no load / unloaded / open circuit, where the divider fraction is 1 and you get the full $K_v{v}_{\mathrm{in}}$ (that is exactly the condition for measuring $K_v$, two entries up). Battery version of the same physics: a 9 V battery reads 9 V on a voltmeter (light load) but sags to 7 V cranking a motor (heavy load).

Same word elsewhere in AE2

Load line (diode/BJT graph questions): the straight line $I=(V_s-V)/R_L$ expressing what the source-plus-load-resistor part of the circuit permits; the device’s curve crosses it at the operating point. Same $R_L$, same idea: the load constrains the device.

And the design takeaway the course keeps repeating: a good voltage amplifier wants $R_{\mathrm{out}}\ll R_L$ (and $R_{\mathrm{in}}$ huge so IT is a light load on whatever feeds it). Then the divider fractions are ≈1 and the gain barely notices what you connect.

Feedback means: take a sample of the output and send it BACK to the input, so the circuit can react to its own result. That is the whole word. “Feed” a bit of the output “back”. Everything else in the unit is about what happens once you do that, and it splits into two opposite worlds depending on the SIGN with which the returned sample is combined.

Everyday version first

You already use feedback constantly. Steering a car: you look at where the car actually IS (the output), compare it with where you want it (the input), and turn the wheel by the DIFFERENCE. You do not steer blind with a pre-planned sequence of wheel angles; you continuously correct using the result. A thermostat is the same: measure the room, compare with the set temperature, heat by the error. Both are negative feedback: the correction opposes the error, so the error shrinks.

The circuit version: three parts

Three parts, and each has one job:

The amplifier $A$ is powerful but crude: huge gain, but an inaccurate, temperature-dependent, device-to-device-varying number. The feedback network $\beta$ is the opposite: it does nothing clever, it is usually just a resistor divider returning a fixed fraction of the output, but a resistor ratio is precise and stable. The summing node compares: it subtracts the returned sample from the input, and what is left, $x_i=x_s-x_f$, is called the error. Only the error gets amplified.

Why subtracting is magic (negative feedback)

Suppose the output drifts a little too HIGH. Then the returned fraction $x_f$ is too high, so the error $x_s-x_f$ DROPS, so the amplifier drives the output back DOWN. Drift low and the same chain pushes it back up. The loop punishes every deviation of the output from $x_s/\beta$; it is self-correcting, like your steering. With a huge $A$ the loop only settles when the error is almost exactly zero, which forces

$$x_f\approx x_s$$

and since $x_f=\beta x_o$, the output must sit at $x_o\approx x_s/\beta$. The exact formula from the lectures says the same thing without the approximation:

$$A_f=\frac{A}{1+A\beta }\to \frac{1}{\beta }\text{ when }A\beta \gg 1$$

Read what that trade is: you GIVE UP most of the raw gain (from $A$ down to $1/\beta$) and in exchange the gain you keep is set by the trustworthy resistor ratio instead of the untrustworthy transistor. That one trade buys all four advertised benefits: stable gain, less distortion, wider bandwidth, and tamed input/output impedances.

This is also the secret behind the op-amp golden rule “$V_{-}=V_{+}$”: it is not a property of the op-amp alone. The op-amp is just an enormous $A$; it is the NEGATIVE FEEDBACK loop around it that drives the error between the two inputs to essentially zero. No feedback (or feedback to the wrong pin) and the rule is dead.

The other sign: positive feedback

Flip the summing node to ADD the sample instead of subtracting it and every deviation is now rewarded: output rises → returned sample rises → input to the amplifier rises → output rises harder. A runaway. That is useless for amplifying but exactly what you want for two other things the unit covers: circuits that SNAP decisively between two states (Schmitt trigger, astable), and oscillators, where the loop feeds the signal back to sustain itself. An oscillator holds itself going when one trip around the loop returns the signal at exactly its original size and phase, the Barkhausen condition:

$$A\beta =1\text{ (and }0°\text{ phase shift round the loop)}$$

So one sentence to keep: feedback = the circuit listening to its own output; subtract the sample and it self-corrects into a precise amplifier with gain $1/\beta$; add the sample and it self-reinforces into a switch or an oscillator. The quantity deciding everything is the loop gain $A\beta$: negative feedback wants it large, oscillation wants it exactly 1.

Because the diagram is using + and − for two completely different jobs, and only one of them is the “inverting / non-inverting” labelling you were thinking of.

Job 1: the +/− INSIDE the triangle = which input is which

On the op-amp symbol, the − printed at one input and the + printed at the other are names, not voltages. They say: this pin is the inverting input ($V_1$ in the lecture’s model), that pin is the non-inverting input ($V_2$). One symbol per pin, fixed forever. They tell you which input gets the minus sign in

$$V_{\mathrm{out}}=A(V_2-V_1)$$

Job 2: the little +/− PAIRS in the model = polarity marks of a measurement

A “voltage” is never a property of one point. It is always a difference between TWO points, so every voltage you draw needs two reference marks: a + on the point you measure, a − on the point you measure from. The pair just defines which direction counts as positive.

In the internal model, $V_1$ means “the voltage of the inverting terminal measured relative to ground”. So the diagram puts a + at the terminal and a − at the ground symbol below it: red voltmeter lead on the pin, black lead on ground. Same for $V_2$. Each input therefore carries a full +/− PAIR, because each is its own little two-point measurement. It is not saying the input “is both positive and negative”.

Reading the numbers

With the pair placed + at the pin and − at ground, “$V_1=3\text{ V}$” means the inverting pin sits 3 V ABOVE ground. If the pin happened to sit below ground you would not move the marks; the number itself just comes out negative, $V_1=-2\text{ V}$. The marks fix the convention; the sign of the value does the rest.

The tall arrow between the two pins in your snip is a THIRD measurement, taken pin-to-pin without involving ground at all:

$$V_{\mathrm{DIFF}}=V_2-V_1$$

and that difference is the only thing the ideal op-amp ever looks at: it appears across $Z_{\mathrm{in}}$ and gets multiplied by the huge gain $A$.

So the rule to carry forward: ± printed inside the triangle = which pin is which (inverting vs non-inverting); a ± pair drawn beside a labelled voltage = the two reference points of that measurement. Every voltage in every circuit diagram in this course carries such a pair, op-amp or not; it is only here that the two conventions sit close enough to collide.

“Wiggle” is the informal name for the small AC signal part of a voltage or current. In an amplifier, every voltage and current is a big steady DC level with a tiny signal moving up and down on top of it. The steady level is the bias (the Q-point). The moving part is the wiggle, and the wiggle IS your signal; it is the only part the amplifier is trying to amplify.

$$i_C(t)=\underset{\text{DC bias}}{I_C}+\underset{\text{the wiggle}}{i_c}$$

The notation carries the split: capital letter, capital subscript ($I_C$) is the DC value with no signal; lowercase letter, lowercase subscript ($i_c$) is the wiggle alone; lowercase letter with capital subscript ($i_C$) is the total, the sum of both.

Why the word exists: it buys you a linear circuit

A transistor is a NONLINEAR device: its input-output curve is bent (exponential for a BJT). Nonlinear circuits are hard. But zoom in on any smooth curve close enough and it looks like a straight line. If the signal only wiggles a small distance about the Q-point, the transistor only ever uses a tiny, effectively straight piece of its curve, so for the wiggle alone the amplifier behaves as a LINEAR circuit whose gain is the slope of the curve at the Q-point (that slope is where $g_m$ comes from).

How the course uses it

The word appears whenever the analysis splits in two, and the split is the whole method of the BJT and FET amplifier units:

DC analysis (signal off): kill the wiggle, keep the batteries. Find the Q-point ($I_C$, $V_{\mathrm{CE}}$) from the bias resistors. This decides WHERE on the curve the amplifier sits.

Small-signal (AC) analysis (only the wiggle): kill the DC instead. Constant supplies have zero wiggle, so they become AC ground; big coupling capacitors pass the wiggle freely, so they become shorts; the transistor becomes its small-signal model. This linear circuit answers “how does the circuit look TO the wiggle?” and gives the gain.

Because the zoomed-in curve is straight, the two answers superpose: total = bias + amplified wiggle, which is exactly the first equation again.

One catch worth remembering for exam comments: the wiggle must actually BE small. Drive the input hard and the signal reaches the bent parts of the curve (or hits saturation and cut-off at the ends), the straight-line approximation dies, and the output comes out distorted, a flattened, non-sine shape. That is also why a measured gain is taken with a deliberately small input (see the Kᴸ measurement entry above).

One sentence first, then the detail: you feed the amplifier a small sine wave, look at the input and the output on an oscilloscope with nothing connected to the output, and divide the two amplitudes:

$$K_v=\frac{v_{\mathrm{out(open)}}}{v_{\mathrm{in}}}$$

Every word of that sentence is doing a job. Here is why each one is there.

The bench setup

Step by step

1. Drive it with a small mid-band sine. Set the generator to a sine of a few tens of millivolts at a middle frequency like 1 kHz. Small, so the amplifier stays in its linear region; mid-band, so coupling capacitors (which kill gain at low frequency) and stray capacitance (which kills gain at high frequency) are both out of the picture and you are measuring the flat, maximum gain.

2. Put CH1 across the amplifier’s INPUT terminals. Not on the generator’s dial setting. The generator has its own internal resistance $R_s$, and some of its voltage is lost across it before reaching the amplifier (the input divider). $v_{\mathrm{in}}$ in the definition of $K_v$ is the voltage that actually appears at the amplifier’s input pins, so that is where the probe goes.

3. Put CH2 across the output with NO load connected. This is the crucial one. $K_v$ is defined as the open-circuit gain: the gain when no current is drawn, so nothing is dropped across $R_{\mathrm{out}}$. If you attach a load $R_L$, the output divider steals part of the internal voltage and you would measure the smaller loaded gain:

$$\frac{v_{\mathrm{out}}}{v_{\mathrm{in}}}=K_v·\frac{R_L}{R_{\mathrm{out}}+R_L}$$

A scope probe is allowed because its resistance (1 MΩ or 10 MΩ) is thousands of times bigger than a typical $R_{\mathrm{out}}$ of a few kΩ: the divider fraction is essentially 1, so the probe is “open circuit” as far as the amplifier can tell.

4. Check the output is a clean sine, then read both amplitudes. If the tops of the output wave are flattened, the amplifier is clipping against its supply rails and you are measuring the power supply, not the gain: turn the input DOWN until the shape is clean. Then read the same measure off both channels (both peak-to-peak, or both amplitude, or both RMS; it cancels in the ratio) and divide:

$$K_v=\frac{v_{\mathrm{out}}\left(\text{pk-pk}\right)}{v_{\mathrm{in}}\left(\text{pk-pk}\right)}$$

Volts divided by volts, so $K_v$ is a pure number. Quote it in decibels if asked:

$$K_v\left(\text{dB}\right)=20{\log }_{10}{K}_v$$

The same trick measures Rₒᵤᵗ and Rᵢₙ too

The classic lab follow-up, worth knowing because it uses the same setup. First record the open-circuit output $v_{\mathrm{out(open)}}$. Now connect a variable resistor as the load and turn it down until the output reads exactly HALF of that. At that point the divider is $\frac{R_L}{R_{\mathrm{out}}+R_L}=\frac{1}{2}$, which forces $R_L=R_{\mathrm{out}}$: read $R_{\mathrm{out}}$ straight off the dial. Mirror image at the input: put a known resistor in series with the generator and increase it until $v_{\mathrm{in}}$ at the amplifier terminals halves; that resistor now equals $R_{\mathrm{in}}$.

So when a question says an amplifier “has $K_v=100$”, that number was obtained exactly this way: small clean sine in, unloaded sine out, ratio of the two. Everything else in the systems model (the loading formula with the two dividers) exists to predict how much LESS than $K_v$ you get once a real source and a real load are attached.

Two things that look alike but are different kinds of object: $K_v$ is a fixed NUMBER describing the amplifier; $K_v{v}_{\mathrm{in}}$ is that number MULTIPLIED by the input voltage, and it is the value of the internal voltage source in the amplifier’s model. Read the dot that isn’t written: Kᴸvᵢₙ means $K_v\times v_{\mathrm{in}}$.

Kᴸ: one number printed on the box

$K_v$ is the amplifier’s open-circuit (unloaded) voltage gain: how many times bigger the output is than the input when NOTHING is connected to drag it down. If $K_v=100$, the amp is a “multiply by 100” machine in the ideal case. It has no units (volts per volt) and it does not change with the signal; it is a property of the amplifier itself, like a car’s top speed.

The three-part model where Kᴸvᵢₙ lives

Circuit theory replaces the whole complicated amplifier with three parts: an input resistance $R_{\mathrm{in}}$ (what the source sees), a dependent voltage source whose value is $K_v{v}_{\mathrm{in}}$, and an output resistance $R_{\mathrm{out}}$ in series with it:

So $K_v{v}_{\mathrm{in}}$ is not a new symbol to memorise; it is a recipe: measure the voltage at the input terminals, multiply by $K_v$, and that is the voltage the internal source generates. If $v_{\mathrm{in}}=10$ mV and $K_v=100$, the source pumps $1$ V.

Why the distinction earns marks: real gain is LESS than Kᴸ

The exam trick is that $K_v$ only equals the measured gain when nothing loads the amplifier. In a real circuit two voltage dividers eat into it: at the input, the source’s own resistance $R_s$ and $R_{\mathrm{in}}$ split the signal before it is even measured; at the output, $R_{\mathrm{out}}$ and the load $R_L$ split the pumped voltage before it reaches the load:

That is exactly the 2022 Q1(b) pattern: an amplifier with $K_v=100$ (40 dB unloaded) delivers an overall gain of only $96.6$ (39.7 dB) once the dividers take their cut. The unloaded number describes the box; the loaded number describes the circuit you built with it.

The three-line summary

• 1$K_v$: a fixed, unitless NUMBER, the amplifier’s gain with no load (open-circuit gain).
• 2$K_v{v}_{\mathrm{in}}$: a VOLTAGE, the internal dependent source’s value: Kᴸ times whatever vᵢₙ currently is.
• 3Measured gain < $K_v$ in any real circuit, because the input and output dividers each take a slice. Quoting Kᴸ as the circuit gain is the classic lost mark.

The exams: ENGM2 Mon 27 Jul · EM2 Tue 28 Jul · AE2 Thu 30 Jul · PE2 Wed 5 Aug · EP2 Fri 7 Aug · MS2A Mon 10 Aug (all 09:30). Target: a solid 50% in each. The front is brutal (100 of the 155 hours land before 30 Jul, about 8h/day); the back half is comfortable (4–5h/day).

Phase 1, the crunch: 18–29 Jul (~8h/day)

• 1Every morning 18–29 Jul: the 2h AE2 session (list below), fresh brain, before anything else. Its value is the SPACING; missing sessions cannot be made up by cramming.
• 218–21 Jul, ~6h/day: ENGM2 (~26h). Fourier (ride the momentum), then Laplace (most formulaic), then first-order ODEs. Ruthlessly skip the MVC tail (Stokes/divergence proofs) and second-order ODE depth.
• 322–25 Jul, ~6h/day: EM2 (~26h). Vikram chapters + the eight recycled paper patterns ONLY: Gauss sphere/coax, B-fields of wires, induction/emf, capacitor energy, dielectric sphere, Maxwell statements. No derivation completionism.
• 426 Jul: final ENGM2 drill (papers, timed). 27 Jul after the exam: final EM2 drill. 28 Jul afternoon + 29 Jul: AE2 finale (mixed tutorial runs under time).

The 12 AE2 morning sessions (2h each)

• 1Op-amp golden rules + ideal properties (op-amp unit, first sections)
• 2Inverting amplifier: derivation, then gain practice until automatic
• 3Non-inverting amplifier + voltage follower/buffer
• 4Summing and difference amplifiers
• 5Decibels unit: landmarks table + log rules
• 6The no-calculator dB conversions + cascaded stages
• 7Gain-bandwidth product + frequency-response basics
• 8Active filters: first-order low-pass and high-pass
• 9Filter practice + response sketches
• 10Op-amp non-idealities (offset, slew) light pass + tutorial questions
• 11Mixed tutorial run, timed
• 12Mixed past-paper op-amp/dB questions, exam conditions

Deliberately absent: BJT/FET small-signal depth. That is >50% material and it is the sacrifice that makes this schedule possible.

Phase 2, the recovery: 31 Jul–9 Aug (4–5h/day)

• 531 Jul–4 Aug: PE2 (~15h) + start EP2 alongside. PE2 bundle: waveform RMS, rectification with smoothing, transformers, one converter, one thermal chain.
• 65 Aug afternoon + 6 Aug: finish EP2 (~15h total). Bundle: number systems, IEEE-754 encode/decode, ALU traces, memory dumps, reading GPIO C. Pure pattern drilling.
• 77 Aug afternoon + 8–9 Aug: MS2A (~25h). Axial/torsion/bending stress + Mohr’s circle from solved examples until mechanical.

The three rules that decide it

• !The next ten days are FULL-TIME. One lost day before 27 Jul has nowhere to be absorbed.
• !EM2 must not leak backwards into ENGM2’s four days; the maths feeds everything and goes first.
• !The daily AE2 block happens even on bad days. Spacing is the whole strategy for that subject.

The task, from the general recipe (with $\omega =1$ and the integral already shrunk to where the wave is 1):

Step 1: the antiderivative, checked, not assumed

We need a function whose derivative is $\sin (nt)$. The candidate is $-\cos (nt)/n$. Check by differentiating it:

Chain rule pushes out a factor $n$ and a minus; they cancel against the built-in $-1/n$, leaving exactly $\sin (nt)$. So both the minus and the divide-by-$n$ are compulsory.

Step 2: substitute the limits

The bracket notation means “the antiderivative at the TOP limit, minus it at the BOTTOM limit”. Top is $t=\pi$, bottom is $t=0$:

Step 3: subtracting a negative becomes adding

The second bracket was minus a negative, so it flips to plus. This is where most sign errors are born.

Step 4: cos(0) = 1

Zero angle means no turn at all, and cosine of no turn is 1 (the very top of the trig-values table):

Step 5: combine over the shared denominator

Both terms sit over $n$, so they merge into one fraction, written with the positive term first:

Step 6: bring back the prefactor

The $1/\pi$ has been waiting outside the bracket the whole time. Multiplying two fractions multiplies their denominators:

Step 7: split by even and odd n

Now use $\cos (n\pi )={(-1)}^n$ (its own entry below): the numerator is $1-1=0$ for even $n$, and $1-(-1)=2$ for odd $n$:

Concrete values, to make the pattern real:

• 1$b_1=\frac{2}{\pi }$  (odd: numerator 2)
• 2$b_2=0$  (even: numerator 0, the harmonic is simply absent)
• 3$b_3=\frac{2}{3\pi }$  (odd again, and smaller: higher harmonics contribute less)

That is every step between the integral and the answer. The same seven-step shape runs every bₙ and aₙ in this course; only the wave’s constants and limits change.

First, your real question: is there a general formula?

Yes. Every “find the Fourier series” question, this one, Q2(b), the exam’s, is solved by the SAME four lines. Nothing about them changes between questions; you only swap in your wave’s brace and period:

Read the template in words: any repeating wave = a constant + a mix of sines and cosines at frequencies 1×, 2×, 3×… the fundamental. The three integrals (the “Euler formulae”) measure HOW MUCH of each ingredient your particular wave contains. ${\int }_T$ means “integrate across any one full period”. The recipe order is always: find $T$ and $\omega =2\pi /T$ first, then the three integrals, then assemble.

What each symbol is, for a beginner

• ∑$\sum$ means “add up”: the terms for $n=1,2,3,\dots$ one after another, forever.
• n$n$ is the harmonic number: which ingredient we are on. $n=1$ wiggles once per period, $n=3$ three times, etc.
• a,b$a_n,b_n$ are plain NUMBERS (one per $n$): the amount of the $n$-th cosine and sine in the mix. Finding them IS the whole question.
• ∫An integral is an AREA machine: $\int fdt$ totals the area under the curve. Multiplying $f$ by $\cos$ or $\sin$ before integrating makes it measure the OVERLAP between your wave and that ingredient (the deep reason is orthogonality, covered in the Fourier unit; for solving questions you only need to run the machine).

Now run the recipe on Q2(a), every move shown

Ingredients: $h=1$ on $(0,\pi )$, $0$ on $(\pi ,2\pi )$. Window length $T=2\pi$, so $\omega =2\pi /2\pi =1$, which is why every trig argument below is just $nt$.

Coefficient 1: a₀ (four small moves)

• 1Prefactor: $2/T=2/2\pi =1/\pi$. Just arithmetic.
• 2Limits shrink from $(0,2\pi )$ to $(0,\pi )$ because $h=0$ on the second half, and integrating zero adds zero. The brace’s value (1) replaces $h$ on the part that remains.
• 3$\int 1dt$ from 0 to π = area of a rectangle, height 1, width π = $\pi$. No calculus needed once you see the rectangle.
• 4The series uses $a_0/2=1/2$, the wave’s average height (the DC term entry below).

Coefficient 2: aₙ (why every cosine dies)

Moves: same prefactor and limit-shrink as before. The antiderivative of $\cos (nt)$ is $\sin (nt)/n$, divide by $n$ because differentiating $\sin (nt)$ spits out a factor $n$ (chain rule) that must be cancelled. Evaluate top minus bottom: $\sin (n\pi )$ is the sine of a whole number of half-turns, ALWAYS 0 (the trig-values table entry), and $\sin (0)=0$. So every single cosine coefficient is zero, which the hidden-symmetry argument predicted without integrating.

Coefficient 3: bₙ (the one that survives)

Two sign traps: the antiderivative of $\sin$ carries a built-in MINUS, and evaluating the bracket top-minus-bottom turns that into $1-\cos (n\pi )$. Every micro-move of that evaluation is worked one line at a time in the entry above this one. Then the even/odd split (the $\cos (n\pi )={(-1)}^n$ entry): even $n$ gives $1-1=0$; odd $n$ gives $1-(-1)=2$:

Assemble: pour the numbers back into the template

The template wants $a_0/2$ plus all the surviving terms. $a_0/2=1/2$; every $a_n$ is 0 (no cosines); $b_n$ is 0 for even $n$ and $2/n\pi$ for odd $n$, so only $n=1,3,5,\dots$ appear:

Sanity-read the answer: the ½ is the average height (right, the wave spends half its time at 1). Only sines, because wave-minus-mean is odd. Only odd harmonics, coefficients shrinking like 1/n because the wave has jumps. Every feature of the answer was predictable from the graph, which is the real power move in the exam.

The takeaway

• 1ONE general recipe solves every such question: $\omega =2\pi /T$, then the three Euler integrals, then assemble into the template. Memorise those four lines.
• 2What changes per question is only the BOOKKEEPING: your wave’s brace sets the integral limits and the constants inside.
• 3The recurring micro-skills: antiderivatives of $\cos /\sin (nt)$ (divide by $n$, sine’s carries a minus), $\sin (n\pi )=0$, $\cos (n\pi )={(-1)}^n$. Those three facts finish almost every coefficient in this course.

For reading the point ITSELF: correct, the middle of a flat piece always has a clean, unambiguous height, as far from every jump as possible. But there are two ways “pick the middle” can still bite, and one of them happens in your own Q2(b).

Caveat 1: your point is safe, but its MIRROR PARTNER might not be

The symmetry test uses two points: $t$ and $-t$. Choosing $t$ mid-shelf guarantees nothing about where $-t$ lands. Concrete case, the Q2(b) wave: $h=-1$ on $(-2,0)$, $+1$ on $(0,1)$, period 3, jumps at …, −2, 0, 1, 3, …

• ✗Middle of the WIDE shelf: $t=-1$. Perfectly readable ($h=-1$). But its partner is $+1$, which is EXACTLY the seam where the wave jumps from +1 down to −1. Partner unreadable, test dead on arrival.
• ✓Middle of the NARROW shelf: $t=0.5$. Partner $-0.5$ sits comfortably inside the wide shelf. Both readable: $h(0.5)=+1$, $h(-0.5)=-1$. Test works.

So the complete habit is: middle of a shelf for your point, then CHECK where the partner lands. If it hits a seam, slide your test time a little (0.4 instead of 0.5, say) or pick the other shelf, any interior pair works.

Caveat 2: a middle point that PASSES proves nothing

Safety of reading is not the same as strength of conclusion. If your middle-point test FAILS, done, the symmetry is disproved forever. If it PASSES, you have learned almost nothing, symmetry must hold at EVERY point, and one agreeing pair cannot certify that. Q2(b) again: points near the origin pass the odd test (heights flip nicely), yet the wave is not odd, which a second pair (like $t=1.5$) exposes. When a test passes, either test more pairs, or argue from the whole shape (equal widths? never negative? mirror image by construction?).

The complete checklist

• 1Pick $t$ strictly inside a shelf (middle = maximum safety margin). Never $0$, never a jump.
• 2Locate $-t$ (use periodicity if needed) and confirm IT is inside a shelf too. If not, adjust $t$.
• 3A FAIL is final; a PASS demands more evidence (another pair, or a whole-shape argument).

Short answer: the brace IS the formula. There is no single algebraic expression like $h(t)=t^2$ hiding somewhere; the piecewise rule is the complete definition, and “plugging in” works differently for it.

Two kinds of function definitions

• 1Formula-functions, like $f(t)=t^2$: plugging in means COMPUTING. Feed in 3, do the arithmetic, get 9.
• 2Rule-functions (piecewise), like our $h$: plugging in means LOOKING UP. Feed in a time, check which case it falls into, read off that case’s value. No arithmetic on the value at all.

Both are complete, legitimate definitions. Maths does not require a function to be one algebraic expression; it only requires that every input gets exactly one output. The brace does that job by cases.

The FULL definition of h has two parts

The brace covers one window, $0<t<2\pi$. The periodicity statement on the right extends it to every other time: any input outside the window is first slid by whole periods until it lands inside. Together they define $h$ for all $t$.

The plug-in procedure, as a machine

The three plug-ins we actually did, run through the machine

• 1$h(\pi /2)$: already inside the window (1.57 is between 0 and 6.28). Which case? 1.57 < 3.14, first line. Output 1.
• 2$h(-\pi /2)$: outside the window (negative). Slide +2π: lands at $3\pi /2$ (4.71). Which case? Between 3.14 and 6.28, second line. Output 0.
• 3For practice, $h(7\pi /2)$ (≈ 11.0): outside (too big). Slide −2π: $3\pi /2$. Second line. Output 0.

So what did “put in our two numbers” mean in the even test?

The equation $h(-t)=h(t)$ is a CLAIM to be checked, not a formula to substitute into. Checking it at $t=\pi /2$ means: evaluate the left side with the lookup machine ($h(-\pi /2)=0$), evaluate the right side with the lookup machine ($h(\pi /2)=1$), and compare the two finished numbers: $0\ne 1$, claim false. Every “plug in” in the symmetry entries was this lookup, run once per side.

Perspective: plenty of engineering functions are lookup-defined rather than formula-defined, gear ratios per gear, tax bands, a thermostat’s on/off state. The square wave is just the simplest lookup function in your course, and the whole point of the Fourier series is to REBUILD it out of formula-functions (sines and cosines), because formulas are what calculus can differentiate, integrate and manipulate.

The middle is not compulsory, ANY time strictly inside a shelf works ($1$, $2$, $0.4$, all fine, see the “is the 2 arbitrary?” entry). The middle is just the safest spot. But $0$ and $\pi$ specifically are the two WORST choices, each broken for its own reason.

Reason 1: 0 and π are jump instants, there is no height to read there

Look at the brace: $h=1$ for $0<t<\pi$ and $h=0$ for $\pi <t<2\pi$. Both inequalities are STRICT ($<$, not $\le$), so neither line covers $t=0$ or $t=\pi$ at all. The function is literally undefined at the jumps. And you can see why on the graph: approach $\pi$ from the left and the wave sits at 1; approach from the right and it sits at 0. Two different answers collide at the same instant, so “the height at π” has no single value. A symmetry test needs definite heights to compare; a jump cannot supply one.

(Extra wrinkle: the Fourier series itself, evaluated at a jump, converges to the MIDPOINT ½, a third value different from both shelves. Jumps are genuinely ambiguous territory, stay off them.)

Reason 2: t = 0 is useless for a symmetry test even if it were defined

The test compares the height at $t$ with the height at $-t$. Put $t=0$ in and the mirror partner is $-0=0$: the point is its own partner. The even test becomes “is $h(0)=h(0)$?”, which is true for EVERY function ever. It can never expose an asymmetry, so it carries zero information. A useful test point must be strictly away from zero so that $t$ and $-t$ are genuinely different places.

Reason 3: π’s mirror partner is ALSO a jump

Suppose you shrugged off Reason 1 and tried $t=\pi$ anyway. The test then needs the height at $-\pi$, and by periodicity $-\pi$ sits exactly where $+\pi$ does in the repeating pattern: on ANOTHER seam between a 1-shelf and a 0-shelf. Both ends of the comparison are ambiguous. Nothing comparable on either side.

The rule to carry

• 1Pick test times STRICTLY INSIDE a flat stretch, never at a jump, never at $t=0$.
• 2Check where the MIRROR PARTNER lands too: it must also sit inside a shelf, not on a seam.
• 3“Middle of the shelf” is a habit, not a law: it keeps you (and your partner point) as far as possible from every edge, so there is never doubt about which shelf you are reading.

Companion to the graph-building entry below: same wave ($h=1$ on $(0,\pi )$, $0$ on $(\pi ,2\pi )$, repeating every $2\pi$), and now the question “is it odd, even, or neither?” answered with nothing assumed.

Step 0: what the two symmetries even are

“Even” and “odd” are names for two ways a graph can be its own twin:

• EEVEN = mirror twin. Fold the picture along the vertical axis (the line at $t=0$). If the left half lands exactly on the right half, the function is even. In symbols: the height at $-t$ always equals the height at $t$, written $h(-t)=h(t)$.
• OODD = upside-down twin. Spin the whole picture half a turn (180°) about the origin point. If it lands on itself, the function is odd. In symbols: the height at $-t$ is always the NEGATIVE of the height at $t$, written $h(-t)=-h(t)$.

A function might have the first symmetry, the second, or NEITHER. To find out which, you compare heights at mirror-partner times: some time $t$ on the right of zero, and its partner $-t$ the same distance to the left. One pair of heights is enough to kill a symmetry, because a symmetry must hold everywhere; one place where it fails means it is gone.

Step 1: pick a test time and read the right-side height

Pick $t=\frac{\pi }{2}$ (≈ 1.57), a convenient spot in the middle of the ON shelf. From the graph (or the brace: 1.57 is between 0 and 3.14, the “height 1” stretch):

Step 2: read the left-side height, slowly (this is the step people skip)

Now we need the height at the mirror partner, $t=-\frac{\pi }{2}$ (≈ −1.57). Problem: the brace only describes times between $0$ and $2\pi$, and −1.57 is outside that window. The way in is periodicity: the wave repeats every $2\pi$, meaning sliding any time forward by one full period lands on an identical part of the wave. So slide −π/2 forward by $2\pi$:

(The arithmetic: $-\frac{\pi }{2}+2\pi =-\frac{\pi }{2}+\frac{4\pi }{2}=\frac{3\pi }{2}$, about 4.71.) And 4.71 lies between $\pi$ (3.14) and $2\pi$ (6.28), the OFF shelf, height 0:

Step 3: run the EVEN test

Even demands mirror partners have EQUAL heights: $h(-t)=h(t)$. Put in our two numbers:

The left dot is on the floor, the right dot is up at 1. Fold the picture along the vertical axis and those two points do NOT land on each other. Not even.

Step 4: run the ODD test

Odd demands mirror partners have OPPOSITE heights: $h(-t)=-h(t)$. Since $h(\pi /2)=1$, oddness would need the partner to be at $-1$:

There is also a zero-effort way to see oddness was never possible: an odd graph must dip BELOW the axis somewhere (every up-height needs a matching down-height after the 180° spin). This wave lives entirely at heights 0 and 1, never negative, so it cannot be odd no matter which points you test. Not odd.

Step 5: the verdict, and what it changes

• 1Even test failed, odd test failed → the wave is NEITHER. That one word is what the exam wants stated (with a reason, e.g. the failing pair of heights).
• 2Consequence for the series: no symmetry shortcut applies directly, so in principle all three coefficient families ($a_0$, $a_n$, $b_n$) must be computed from the brace.
• 3Hidden bonus (worked in the DC-term entry below): subtract the mean ½ and what remains IS odd, which is the deep reason the $a_n$ all come out zero anyway and the final series is a constant plus sines.

One caution worth engraving: a symmetry test that PASSES at one point proves nothing (symmetry must hold at every point), but a test that FAILS at one point is final. That is why one carefully chosen pair of heights settles “not even” and “not odd” here, yet no number of passing points could ever prove a symmetry on its own.

They are the same quantity, not merely related. In mechanics, angular velocity is how fast something turns: radians of angle per second. The Fourier $\omega$ is the angular velocity of an invisible rotating point whose shadow draws the wave. Same symbol, same formula, same units, on purpose.

The mechanics version first

A wheel turning steadily sweeps $2\pi$ radians (one full turn) every revolution. If one revolution takes $T$ seconds, the turning rate is:

That IS the formula from the tutorial. Nothing about it is special to waves yet, it is just “how fast is the turning”.

The bridge: a sine wave is circular motion seen edge-on

Take a point moving round a circle of radius $A$ at a steady $\omega$ rad/s. After $t$ seconds it has turned through the angle $\omega t$. Now watch only its height (its shadow on a vertical wall): that shadow is exactly $A\sin (\omega t)$. Every sinusoid is a rotation viewed from the side:

One full lap is $2\pi$ radians of turning AND one full period $T$ of the wave, which is exactly why $\omega =2\pi /T$ for waves too. The wave inherits its $\omega$ from the circle that generates it.

So what does ω = 1 rad/s mean for the square wave?

• 1The square wave itself does not rotate. But its Fourier series is built from harmonics $\sin (nt)$, $\cos (nt)$, and EACH of those is the shadow of its own spinning point.
• 2$\omega =1$ rad/s says the fundamental’s circle turns 1 radian per second, so a full lap ($2\pi$ rad) takes $2\pi$ seconds, matching the wave’s period $T=2\pi$. Consistent.
• 3The $n$-th harmonic’s circle spins $n$ times faster ($n\omega$ rad/s), completing $n$ laps per wave period. That is what “harmonics at multiples of the fundamental” means physically.

Why engineers keep ω around (and not just ν in Hz)

Frequency $\nu =1/T$ counts full turns per second (Hz); $\omega =2\pi \nu$ counts radians per second. Maths prefers $\omega$ because the argument of sin/cos must be an angle, writing $\sin (\omega t)$ keeps the bookkeeping in radians with no stray $2\pi$ factors. And the rotation picture is not a metaphor you discard later: it is literally the phasor used for AC circuits in PE2/AE2 (UK mains: $\nu =50$ Hz means a phasor spinning at $\omega =2\pi ·50\approx 314$ rad/s), and the same $\omega$ is a motor shaft’s angular velocity in dynamics. One concept, whole degree.

Goal: start from a blank page and end at the app’s graph, with no step assumed. The function is the one from Q2(a):

Step 0: what a graph even is

A graph is a lookup table drawn as a picture. The horizontal axis is the floor: every position on it is a time $t$. The vertical axis is the wall: every position on it is a height. A point drawn at floor-position $t$ and wall-height $y$ is the statement “at time $t$, the function’s value is $y$”. Drawing the whole graph means doing that for every time at once.

Before drawing anything, mark the landmark times the brace mentions. $\pi$ is just the number 3.14…, so the landmarks are $0$, $\pi \approx 3.14$, and $2\pi \approx 6.28$:

Step 1: draw the first line of the brace

Line one says: for every time between $0$ and $\pi$, the height is $1$. Pick any such time and plot its point: at $t=1$, height 1. At $t=2$, height 1. At $t=0.5$, height 1. Every single point lands at the SAME height, so joining them gives a flat shelf at height 1 running from $t=0$ to $t=\pi$:

Step 2: draw the second line, height 0 lies ON the axis

Line two says: for every time between $\pi$ and $2\pi$, the height is $0$. Height zero is not “nothing drawn”, it is a real shelf that happens to sit at the level of the floor itself. So the graph between $\pi$ and $2\pi$ is a flat segment lying exactly ON the t-axis. This is why the app’s picture looks like the wave “disappears” there: it has not disappeared, it is at height 0.

At $t=\pi$ itself, the value jumps: just left of $\pi$ the height is 1, just right of it the height is 0. Nothing is drawn vertically at the jump, a function has exactly one value per time, so there is no wall connecting the shelves. The dashed vertical in the app is only a visual guide marking where the teleport happens. (Fine print: the brace uses strict < signs, so at exactly $t=\pi$ it does not define a value at all; for Fourier purposes the series will later converge to the midpoint ½ there.)

The window we have drawn runs from 0 to $2\pi$, length $2\pi$. That length is the period: $T=2\pi$, hence $\omega =\frac{2\pi }{T}=1$ rad/s.

Step 3: photocopy the block both ways forever

“Periodic” means $h(t+2\pi )=h(t)$: shift the picture by one period and it must look identical. So stamp the Step-2 block at every shift of $2\pi$. Shift right: shelf at 1 on $(2\pi ,3\pi )$, zero on $(3\pi ,4\pi )$. Shift left: shelf at 1 on $(-2\pi ,-\pi )$, zero on $(-\pi ,0)$:

Step 4: verify against the app’s picture

• 1Just right of 0: shelf at 1, width π. Matches (the app’s hump from 0 to π). ✓
• 2Just LEFT of 0: the app shows the wave on the axis from −π to 0. Our copy rule predicts exactly that: $(-\pi ,0)$ is the zero shelf of the left-shifted block. ✓
• 3Spot values: $h(\pi /2)=1$ (inside the first hump), $h(3\pi /2)=0$ (on the zero shelf), matching the earlier reading-a-value entry. ✓

Contrast with Q2(b) below: same recipe exactly (draw the brace once, tile by the period), only the ingredients differ, Q2(b)’s shelves are −1 and +1 with unequal widths and the stamp distance is 3 instead of 2π. Master the recipe once and every “sketch over a few periods” question is the same two moves.

Yes, almost all of them, because the series is computed from the BRACE (the piecewise definition), not from your drawing. But there is exactly one pipe connecting the two, and if you use it carelessly a bad graph can poison the algebra. Know where that pipe is.

What each part is actually marked on

• 1The sketch marks are their own little pot: axes, correct widths and heights, a few periods shown. Lose these and they are gone, but they are only the sketch’s pot.
• 2The series marks come from: correct $T$ and $\omega =2\pi /T$, correctly split integrals with the right limits and the right constant in each piece, correct integration, correct assembly. Every one of those reads straight off the brace: for Q2(b), $b_n=\frac{2}{3}[{\int }_{-2}^0(-1)\sin (n\omega t)dt+{\int }_0^1(+1)\sin (n\omega t)dt]$. No picture required anywhere in that.

The one way a bad graph poisons the series

The symmetry shortcut. If your (wrong) sketch convinces you the wave is odd, you will write “odd, so $a_0=a_n=0$” and skip two integrals. If the wave is actually “neither” (as in Q2(b)), you have now thrown away real, non-zero coefficients, and the final series is wrong for a reason that traces straight back to the graph.

The safe play when unsure of your sketch

• ✓Don’t claim a symmetry you can’t verify from the brace. The algebraic test (compare h(−t) with ±h(t) using the piecewise rule, not the picture) is sketch-proof.
• ✓If in any doubt, compute all three coefficients from the brace. Costs a few minutes, immune to drawing errors, and every correct integral earns its marks even if the sketch above it is nonsense.
• ✓Use the graph only for free CHECKS, never as an input: the DC term should match the eyeballed average, and coefficients should decay like 1/n for a jumpy wave. If they disagree, something is wrong somewhere, which is information, not marks lost.

Bottom line for Q2(b): a mangled sketch costs you the sketch’s own couple of marks and nothing else, PROVIDED your series work starts from the brace and you only state “neither odd nor even” if the algebra (or the widths in the brace: 1 vs 2) backs it.

The question hands you a piecewise rule for ONE period and expects you to turn it into a repeating graph. There are exactly two steps: (1) draw literally what the brace says, once; (2) photocopy that block left and right forever. Here is each step slowly.

Step 1: read the brace, draw one block

The rule is:

Read each line as “flat shelf at this height, over this stretch of floor”: a shelf at height $-1$ from $t=-2$ to $t=0$ (width 2), then a shelf at height $+1$ from $t=0$ to $t=1$ (width 1). Together they cover $-2<t<1$, a window of total length $3$. That length IS the period: $T=1-(-2)=3$, so $\omega =\frac{2\pi }{3}$ rad/s.

Step 2: photocopy the block every 3 units

“Periodic with period 3” means $h(t+3)=h(t)$: slide the whole picture 3 to the right (or left) and it must look identical. So take the block from Step 1 and stamp copies at every shift of 3:

Where each feature of the final graph comes from:

• 1Narrow tops, wide bottoms: straight from the brace, the $+1$ piece is width 1, the $-1$ piece is width 2. (A one-third duty cycle.)
• 2Jump locations: the block’s internal jump at $t=0$ and its seams at $-2$ and $1$, then all of these repeated every 3: …, −2, 0, 1, 3, 4, 6, 7, …
• 3Dashed verticals: not part of the function, just guides marking the instant the value teleports between −1 and +1. A function can only have one value per t, so the jump has no “vertical wall”.

Why the graph shows “neither odd nor even” at a glance

The widths give it away. Mirror the graph in the vertical axis: just right of 0 the wave is $+1$ for a width-1 stretch, but just left of 0 it is $-1$, so it is not even. For oddness, beware: near the origin it LOOKS odd (heights flip across 0). Test one point further out: $h(1.5)=-1$ (in the (1,3) low shelf) but $-h(-1.5)=-(-1)=+1$. Not equal, so not odd. The unequal widths (1 vs 2) are what break both symmetries.

Lesson inside the lesson: a symmetry test that PASSES at one point proves nothing, symmetry must hold everywhere, and this wave passes the odd test near 0 yet fails it at 1.5. One failing point disproves; one passing point does not prove.

Bonus: what the graph already tells you about the series

Mean value by areas: over one period, area $=(2\times (-1))+(1\times (+1))=-1$, divided by the width 3 gives the DC term $-\frac{1}{3}$. The wave spends more time low than high, so its average sits below zero, and with no symmetry, expect both $a_n$ and $b_n$ to survive in the series.

This is from the both-odd-and-even proof below: we reached $f(t)=-f(t)$ and then “added $f(t)$ to both sides” to get $2f(t)=0$. The confusion: it can look like we claimed $-f(t)+f(t)=f(t)$. We did not. The two sides get the same treatment but produce different results.

Watch each side separately

Adding the same thing to both sides of an equation keeps it true. Do it and simplify each side on its own:

• LLeft side: it was $f(t)$, we add $f(t)$, and two copies of the same thing is twice the thing: $2f(t)$. This is where the 2 comes from.
• RRight side: it was $-f(t)$, we add $f(t)$, and anything plus its own negative cancels to $0$. This is where the 0 comes from. (Never $f(t)$: that would be $-1+1=1$, which is false.)

Same move with a plain x

If the $f(t)$ clutter is the problem, rename it: let $x=f(t)$, one unknown number. The proof line is then just ordinary equation-solving:

Sanity check with a number: could $x=5$ satisfy $x=-x$? That would need 5 = −5, false. Could $x=0$? 0 = −0 is true. Zero is the only number equal to its own negative, which is exactly what the algebra concluded, and (applied at every t) why the only both-odd-and-even function is the zero function.