The fourier series coefficients is given as:

$$c_k= \begin{cases} 1 \qquad & k \ \text{ even} \\ 2 \qquad & k \ \text{ odd} \\ \end{cases}$$

the period of the signal is $T=4$, what is signal $x(t)$?

Attempt: when i am trying to find the signal by applying the general formula at the end i am getting exponential term now i am finding that hard to convert it into impulse again because i know answer of this signal will be a impulse but i am not getting any idea to convert it into impulse again? Please help with this?

  • 1
    $\begingroup$ what does your textbook say for $x(t)$ given the information you are given? $\endgroup$ Commented Jul 15, 2017 at 4:02
  • 1
    $\begingroup$ i guess i was waiting for you to respond with $$ x(t) = \sum\limits_{k=-\infty}^{\infty} c_k \, e^{j k \frac{2\pi}{T} t } $$ have you seen this equations before? $\endgroup$ Commented Jul 15, 2017 at 4:27

1 Answer 1


You should use the synthesis equation of an impulse train with period $T$ (which is easy to derive):

$$x(t)=\sum_{k=-\infty}^{\infty}\delta(t-kT)=\sum_{k=-\infty}^{\infty}\frac{1}{T}e^{jk\frac{2\pi}{T} t}\tag{1}$$ That is: the Fourier coefficients for all terms is a constant ($\frac{1}{T}$).

Now assume that there are two impulses with different amplitudes $a$ and $b$ per period, or $$x(t)=a\delta(t)+b\delta(t+2),\ -4< t \le0$$ In such case we have

$$\begin{align} x(t)&=\sum_{k=-\infty}^{\infty}a\cdot\delta(t-kT)+b\cdot\delta(t+\frac{T}{2}-kT)\\&=\sum_{k=-\infty}^{\infty}a\cdot\delta(t-4k)+b\cdot\delta(t+2-4k)\\ &=\sum_{k=-\infty}^{\infty}\frac{a}{4}e^{jk\frac{2\pi}{4} t}+\sum_{k=-\infty}^{\infty}\frac{b}{4}e^{jk\frac{2\pi}{4} (t+2)}\\ &=\sum_{k=-\infty}^{\infty}\frac{1}{4}e^{jk\frac{\pi}{2} t}\left(a+be^{jk\pi}\right)\\ &=\sum_{k=-\infty}^{\infty}\frac{1}{4}e^{jk\frac{\pi}{2} t}\left(a+b(-1)^k\right)\\ &=\begin{cases} \displaystyle\sum_{k=-\infty}^{\infty}\frac{a+b}{4}e^{jk\frac{\pi}{2} t},& k\text{ even}\\[10pt] \displaystyle\sum_{k=-\infty}^{\infty}\frac{a-b}{4}e^{jk\frac{\pi}{2} t},& k\text{ odd} \end{cases} \end{align}$$ Now refering again to $(1)$ and comparing it with the question, we should have $$c_k=\begin{cases} \frac{a+b}{4}=1\\[10pt] \frac{a-b}{4}=2 \end{cases}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.