Fourier Series for GATE: Coefficients, Symmetry Shortcuts, Parseval and Solved Examples
Fourier series for GATE: exponential and trig forms, the square wave solved, line spectrum, symmetry shortcuts, Parseval power, properties, traps and practice questions.
At the end of the convolution post we saw that an LTI system is completely described by its impulse response. Here's the other half of the story. Feed an LTI system a complex exponential $e^{j\omega t}$, and what comes out is the same exponential, only scaled:
No new frequencies appear. That one fact is why Fourier's idea matters so much. If we can write a signal as a sum of exponentials, we can push each one through the system separately and simply add up the results. Convolution becomes multiplication.
For periodic signals, that sum is the Fourier series. It's a reliable GATE topic, and most questions can be answered in a minute or two once you know the shortcuts.
The two forms of the Fourier series
A periodic signal with period $T0$ has fundamental frequency $\omega0 = 2\pi / T0$. It can be written as
This is the exponential form. The integral can run over any one full period. The same signal in trigonometric form is
and the two are linked by
The factor of 2 between $ck$ and $ak$, $bk$ is the single most common source of lost marks in this topic. For a real signal, $c{-k}$ is the complex conjugate of $ck$, so the positive and negative harmonics share the amplitude equally.
Worked example: the square wave
Take a square wave of period $T0$ that is $+1$ for the first half of each period and $-1$ for the second half.
It's an odd function, since $x(-t) = -x(t)$, so every cosine term vanishes and $a0 = 0$. Working out the sine terms:
So
Only odd harmonics appear, and they shrink like $1/k$. Here is what happens as we add terms, with $\omega0 = 1$ rad/s:
Notice the little horns next to each jump. However many terms you add, the peak overshoots by about 9% of the jump. This is the Gibbs phenomenon, and it's a favourite one-mark question.
The spectrum: a picture of the coefficients
Plot $ck$ against $k$ and you get the signal's line spectrum. Because $k$ only takes whole-number values, this is a discrete plot: one stick per harmonic.
For the square wave, $ck = \dfrac{2}{k\pi}$ for odd $k$ and $0$ otherwise:
The spectrum is symmetric about $k = 0$, as it must be for any real signal. The even harmonics are missing, and the heights fall off like $1/k$.
Symmetry shortcuts
Before you integrate anything, look at the waveform. Symmetry often tells you half the answer for free.
If $x(t)$ is then ------ even, $x(-t) = x(t)$ only cosines: $bk = 0$, and $ck$ is real odd, $x(-t) = -x(t)$ only sines: $a0 = ak = 0$, and $ck$ is purely imaginary half-wave symmetric, $x(t + T0/2) = -x(t)$ only odd harmonics, and no DC term shifted up by a constant $A$ only $a0$ changes, by $A$
The square wave above ticks both the odd and the half-wave boxes, which is why only odd sine terms survived.
Parseval's theorem: power from the coefficients
The average power of a periodic signal can be found in the time domain or from the coefficients, whichever is easier:
For the square wave, $x(t)^2 = 1$ everywhere, so $P = 1$. The fundamental alone, counting $k = 1$ and $k = -1$, carries
So about 81% of a square wave's power sits in its fundamental. GATE asks exactly this kind of question.
Properties worth memorising
If $x(t)$ has coefficients $ck$:
Time shift: $x(t - t0)$ has coefficients $ck\, e^{-jk\omega0 t0}$. The magnitudes don't change, only the phases. Time reversal: $x(-t)$ has coefficients $c{-k}$. Differentiation: $\dfrac{dx}{dt}$ has coefficients $jk\omega0\, ck$. Higher harmonics get boosted. Real signal: $c{-k} = ck^$, so $ck$ is even in $k$. Through an LTI system: the output has coefficients $ck\, H(jk\omega0)$. This is the convolution result from the last post, seen from the frequency side.
That last one is worth a quick example. Send the square wave through an ideal low-pass filter that passes the fundamental but blocks the 3rd harmonic and above. What's left is
a sine wave with peak value $4/\pi \approx 1.27$. The output peak is larger than the input's, which was 1. Students often assume filtering can only make a signal smaller. It can't make it more powerful, though: the output power is 81% of the input's.
Common traps in GATE questions
These are the mistakes I see most often.
Mixing up $ck$ with $ak$ and $bk$. They differ by a factor of 2. Check which form the question uses. Getting the fundamental frequency wrong. For a sum of sinusoids, $\omega0$ is the greatest common divisor of the frequencies. If their ratio is irrational, the signal isn't periodic at all. Forgetting the DC term. $a0$ is the average value over one period, and it counts towards the power. Counting only positive $k$ in Parseval. In the exponential form, $k$ and $-k$ both contribute. Expecting Gibbs to go away. The overshoot gets narrower with more terms, but its height stays at about 9% of the jump. Ignoring symmetry. Checking for even, odd and half-wave symmetry first saves you most of the integration.
Practice questions
Give yourself about two minutes each.
$x(t) = 3 + 4\cos 2t + 2\sin 5t$. Find the fundamental frequency and the average power. Find the exponential Fourier coefficients of $x(t) = \cos \omega0 t$. What fraction of the square wave's power lies in the 1st and 3rd harmonics together? A real signal is even. What can you say about its coefficients $ck$? If $x(t)$ has coefficients $ck$, what are the coefficients of $x(t - T0/2)$? Find the coefficients of a periodic impulse train, $\sumn \delta(t - nT0)$.
Answers
The frequencies are 2 and 5 rad/s, whose greatest common divisor is 1, so $\omega0 = 1$ rad/s and $T0 = 2\pi$ s. The power is $3^2 + \tfrac{4^2}{2} + \tfrac{2^2}{2} = 9 + 8 + 2 = 19$. $c1 = c{-1} = \tfrac{1}{2}$, and every other $ck$ is zero. $\tfrac{8}{\pi^2}\left(1 + \tfrac{1}{9}\right) \approx 0.81 + 0.09 = 0.90$, so about 90%. They're real and even: $c{-k} = ck$. $ck\, e^{-jk\pi} = (-1)^k ck$. $ck = \tfrac{1}{T0}$ for every $k$. An impulse train contains every harmonic in equal measure.
Question 6 is the bridge to our next topic. Let the period $T0$ grow without limit, and the lines in the spectrum crowd together into a continuous curve. That's the Fourier transform, and it's where we'll go next. Tell me in the comments which question took you longest.