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My try:

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First of all I tried observing the symmetry but I did'nt find any.So I tried to calculate the fourier series coefficient of the signal like this

First I differentiated the signal $x(t)$ so that $y(t)=\frac{dx(t)}{dt}$ Now suppose $d_k$ is the Fourier series coefficient of the differentiated signal and $a_k$ is the fourier series coefficient of Orignal signal $x(t)$ Now $d_k=(jk \omega_0 )c_k$ so $c_k=\frac{d_k}{jk \omega_0}$.Now calculating $d_k$

That is

\begin{align} d_k &=\frac{1}{3} \int_{-1}^2 \big(\delta(t+1)+\delta(t)-2\delta(t-1)\big) e^{\frac{-jk2\pi t}{T}}dt\\ &=\frac{1}{3}\big(1+e^{\frac{jk2\pi}{3}}-2e^{\frac{-jk2\pi}{3}}\big) \end{align}

Now $$a_k=\frac{\frac{1}{3}\big(1+e^{\frac{jk2\pi}{3}}-2e^{\frac{-jk2\pi}{3}}\big)}{jk \omega_0}$$ $$a_k=\frac{\big(1+e^{\frac{jk2\pi}{3}}-2e^{\frac{-jk2\pi}{3}} \big)}{j2\pi k}$$

Now checking at

$k=1,a_k\ne 0$ and at $k=2,a_k\ne 0$

so for one odd and one even value of n its not zero so no options matching.What is the mistake?


As stated by @Matt L in the commment to check for period T=6 also, so I did like this

I differentiated the signal from -3 to +3 so the fourier series coefficient I got like this

\begin{align} d_k &=\frac{1}{6} \int_{-3}^3 \big(\delta(t+3)-2\delta(t+2)+\delta(t+1)+\delta(t)-2\delta(t-1)+\delta(t-2)\big)e^{\frac{-jk2\pi t}{6}}dt\\ &=\frac{1}{6}\big(1+e^{jn\pi}-2e^{\frac{jk2\pi}{3}}+e^{\frac{jn\pi}{3}}-2e^{\frac{-jk\pi}{3}}+e^{\frac{jk2\pi}{3}}\big) \end{align}

Now at $k=1,d_k=0$ at $k=2,d_k\ne 0$,at $k=3,d_k=0$ but at $k=6$ also $d_k=0$

Now whats the mistake???

  • 2
    $\begingroup$ I haven't checked your calculations, but in any case you've only excluded options A and B. You say nothing about options C and D. $\endgroup$
    – Matt L.
    Nov 19, 2017 at 21:05
  • 1
    $\begingroup$ "haven't found any symmetry": Looks pretty odd-symmetric to t=-0.5 to me. $\endgroup$ Nov 19, 2017 at 21:12
  • $\begingroup$ @MattL. I really did't cared about Option C and D,because I thought as I have to take fundamental period that is T=3,then why should I look at C and D,but really I should be careful.Thank you sir :) $\endgroup$
    – Rohit
    Nov 20, 2017 at 1:38
  • $\begingroup$ @MarcusMüller you are right its showing odd symmetry like but Sir I only check signal symmetry without shifting it right or left,yes I can check it by shifting up or down that I know,but can I really check it by shifting like that? $\endgroup$
    – Rohit
    Nov 20, 2017 at 1:44
  • 1
    $\begingroup$ @Rohit: No, your answer is correct. And it is also the way you should solve such a problem. $\endgroup$
    – Matt L.
    Nov 20, 2017 at 8:23

2 Answers 2


As Matt stated, by your calculation you have excluded options a and b but you should also check for options c and d, from which you would see that the answer is the option d.

You could practically check the result of coefficients $d_k$ from the interpretation that CTFS coefficients of a periodic waveform $\tilde{x}(t)$ can be obtained from an inverse-DTFT (or an inverse-DFT more practically), by treating $\tilde{x}(t)$ as a DTFT waveform. Note that the computation is exact if the CTFS coefficients $d_k$ were of finite length in advance, but otherwise approximate which would only get better by choosing long enough samples to represent the waveform $\tilde{x}(t)$.

Therefore, the following computes (approximately here) $3N$ CTFS coefficients of $x(t)$:

>     N = 128;
>     x = [ones(1,N) , -1*ones(1,N) , zeros(1,N)];
>     d = ifft(x);

On the orther hand the following code would do the same by treating $\tilde{x}(t)$ as periodic in $2 T_0$ rather than $T_0$

>     N = 128;
>     x = [ones(1,N) , -1*ones(1,N) , zeros(1,N)];
>     d = ifft([x x]);

Based on your edit, the following provides you the answer: Given that the CTFS coefficients of the derivative signal is:

\begin{align} d_n &=\frac{1}{6} \int_{-3}^3 [\delta(t+3)-2\delta(t+2)+\delta(t+1)+\delta(t)-2\delta(t-1)+\delta(t-2)]e^{\frac{-jn2\pi t}{6}}dt\\ &=\frac{1}{6}[ e^{j\frac{2\pi}{6}3n} - 2 e^{j\frac{2\pi}{6}2n} + e^{j\frac{2\pi}{6}1n} + 1 -2e^{-j\frac{2\pi}{6}1n} + e^{-j\frac{2\pi}{6}2n}] \\ &=\frac{1}{6} [ e^{j\pi n} - 2 e^{j\frac{2\pi}{3}n} + e^{j\frac{\pi}{3}n} + 1 -2e^{-j\frac{\pi}{3}n} + e^{-j\frac{2\pi}{3}n}] \\ \end{align}

Above three lines were straightforward. Now we shall group those terms to yield something that can be simplified:

$$d_n =\frac{1}{6} [ (1 + e^{j\pi n}) - 2 ( e^{j\frac{2\pi}{3}n} + e^{-j\frac{\pi}{3}n} ) + (e^{j\frac{\pi}{3}n} + e^{-j\frac{2\pi}{3}n})] $$

Now add $2\pi n$ to the negative angles so that they become: $$e^{-j\frac{\pi}{3}n} = e^{j( 2\pi n -\frac{\pi}{3}n) } = e^{j\frac{5\pi}{3}n } = e^{j\pi n }e^{j\frac{2\pi}{3}n } $$ and $$e^{-j\frac{2\pi}{3}n} = e^{j( 2\pi n -\frac{2\pi}{3}n) } = e^{j\frac{4\pi}{3}n } = e^{j\pi n}e^{j\frac{\pi}{3}n } $$ respectively. And plugging them into $d_n$ line, yields:

\begin{align} d_n &=\frac{1}{6} [ (1 + e^{j\pi n}) - 2 ( e^{j\frac{2\pi}{3}n} + e^{j\pi n }e^{j\frac{2\pi}{3}n } ) + (e^{j\frac{\pi}{3}n} + e^{j\pi n}e^{j\frac{\pi}{3}n})]\\ &=\frac{1}{6} [ (1 + e^{j\pi n}) - 2 (1 + e^{j\pi n}) e^{j\frac{2\pi}{3}n} + (1 + e^{j\pi n}) e^{j\frac{\pi}{3}n}]\\ &=\frac{1}{6} [1 + e^{j\pi n}]\cdot[1 - 2 e^{j\frac{2\pi}{3}n} + e^{j\frac{\pi}{3}n}]\\ &=\frac{1}{6} [e_n]\cdot[f_n]\\ \end{align}

Now it can be shown that the product term $e_n$ will be zero for all odd indice $n$, hence $d_n$ will also be zero for all odd $n$. Note that the other product term $f_k$ will be zero for $n = 6m$ that $d_n$ will also be zero for $n = 6,12,18...$ However for your example it sufficies to show that $e_n$ will be zero for odd $n$.

FURTHERMORE below is the theoretical proof that it's not a coincidence to have all the odd indexed terms in CTFS to be zer0, independent of the signal itself, when the period $T_y$ of the periodic signal $x(t)$ is assumed to be twice that of its fundamental period $T_x$. Assume $x(t)$ is a periodic signal with fundamental period of $T_x$ and let $y(t)$ be the same signal $x(t)$ but interpreted to have a period of $T_y = 2 T_x$. Let $a_k$ and $b_k$ denoted the CTFS coefficients of $x(t)$ and $y(t)$ respectively, then we have:

\begin{align} b_k &= \frac{1}{T_y} \int_{0}^{T_y} y(t) e^{-j\frac{2\pi}{T_y} k t} dt = \frac{0.5}{T_x} \int_{0}^{2T_x} x(t) e^{-j\frac{2\pi}{T_x} (k/2) t} dt\\ &= 0.5 \left( \frac{1}{T_x} \int_{0}^{T_x} x(t) e^{-j\frac{2\pi}{T_x} (k/2) t} dt + \frac{1}{T_x} \int_{T_x}^{2T_x} x(t) e^{-j\frac{2\pi}{T_x} (k/2) t} dt \right)\\ \end{align}

Make the substitution $t' = t - T_x$ in the second integral and recognize that $x(t'+T_x) = x(t')$ as $x(t)$ is periodic with $T_x$ , yielding:

$$b_k = 0.5 \left( \frac{1}{T_x} \int_{0}^{T_x} x(t) e^{-j\frac{2\pi}{T_x} (k/2) t} dt + e^{-j\frac{2\pi}{T_x} (k/2) T_x} \frac{1}{T_x} \int_{0}^{T_x} x(t') e^{-j\frac{2\pi}{T_x} (k/2) t'} dt' \right) $$

Now recognize the integrals as $a_{k/2}$; i.e, the CTFS of the signal $x(t)$ evaluated at $k/2$. And simplify the sum: \begin{align} b_k &= 0.5 \left( a_{k/2}+ e^{j\pi k} a_{k/2} \right)\\ &= 0.5 \left( 1+ e^{j\pi k} \right) a_{k/2} \\ &= \begin{cases}{ a_{k/2} ~~, \text{ for k even, k=2m , m=1,2,...}\\ 0 ~~~~~~~, \text{ for k odd , k=2m+1, m=1,2,...}}\end{cases} \\ \end{align}

  • $\begingroup$ Please check the edit.sir $\endgroup$
    – Rohit
    Nov 20, 2017 at 4:50
  • $\begingroup$ Sir I am really not familiar with the coding part. $\endgroup$
    – Rohit
    Nov 20, 2017 at 9:04
  • $\begingroup$ @Rohit I hope you knw about the dsp programs MATLAB or its opensource clone OCTAVE. Just get these lines and see by yourself. $$d_k = \frac{1}{T} \int_{T/2}^{T/2} \tilde{x}(t) e^{-j \frac{2\pi}{T} k t} dt$$ Replace $t$ by $\omega$ and let $T=2\pi$ and then treat $x(t)$ as a DTFT of $d[k]$ via inverse DTFT : $$d[-k] = \frac{1}{2\pi} \int_{-\pi}^{\pi} \tilde{x}(\omega) e^{j k \omega} d\omega$$ This would yield $d[-k]$ whcih would be $d[-k] = d[k]^*$ for real $x(t)$. $\endgroup$
    – Fat32
    Nov 20, 2017 at 9:53
  • $\begingroup$ @Rohit now you can see how to reach the correct conclusion from the method you have initially chosen. $\endgroup$
    – Fat32
    Nov 20, 2017 at 15:09
  • 1
    $\begingroup$ $b_k = \frac{1}{T_y} \int_{0}^{T_y} y(t) e^{-j\frac{2\pi}{T_y} k t} dt = \frac{2}{T_x} \int_{0}^{2T_x} x(t) e^{-j\frac{2\pi}{T_x} (k/2) t} dt$ here you have multiplied with 2 I think it should be divided by 2 because $T_y=2T_x$ $\endgroup$
    – Rohit
    Nov 20, 2017 at 16:50
  • A signal is called Odd half wave symmetry where only odd harmonics are present.

for this conditon is,


  • similarly a signal is called Even half wave symmetry where only even harmonics are present.

and for this condition is


Now the given signal is Even half wave symmetric with period $T=6$

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Now shifting half the period of the signal


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As we can see signal is completely overlapping the other half of the signal so the signal is even half wave symmetry so only even harmonic will present and odd harmonic will be zero,that is option D.

  • $\begingroup$ Eventhough odd harmonics are missing, all even harmonics are not present as well... so some even harmonics are available only. You better work this explicitly by putting $k = 2m + 1$ to see that $d_k$ will be zero then. $\endgroup$
    – Fat32
    Nov 20, 2017 at 9:36

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