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question: If $x(t)$ and $y(t)$ are two periodic signals(both with period T) with Fourier coefficients $c_{n}$ and $d_{n}$ respectively then, Fourier coefficient of $z(t)=x(t)\cdot y(t)$ is:

(a) $\dfrac{c_{n}d_{n}}{T}$
(b) $T c_{n}d_{n}$
(c) $c_{n}d_{n}$
(d) $c_{n}*d_{n}$

my attempt :

let, $x(t)=\sum_{-\infty}^{\infty} c_{n}\cdot e^{jn\omega_{0}t}$ and $y(t)=\sum_{-\infty}^{\infty} d_{n}\cdot e^{jn\omega_{0}t}$.

Then, fourier cofficient of $z(t)$ will be equal to

$$\begin{align} \dfrac{1}{T}\displaystyle\int_{0}^{T} z(t) e^{-jn\omega_{0} t}dt &=\dfrac{1}{T}\displaystyle\int_{t=0}^{T} x(t)\cdot y(t) e^{-jn\omega_{0} t}dt \\ &=\dfrac{1}{T}\displaystyle\int_{t=0}^{T} \left(\sum_{l=-\infty}^{\infty} c_{l}\cdot e^{jl\omega_{0}t}\sum_{m=-\infty}^{\infty} d_{m} \cdot e^{jm\omega_{0}t} \right)e^{-jn\omega_{0} t}dt\\ &= \dfrac{1}{T}\displaystyle\int_{t=0}^{T}\left(\sum_{l=-\infty}^{\infty}\sum_{m=-\infty}^{\infty}c_{l}d_{m}\cdot e^{j(l+m)\omega_{0}t}\right)e^{-jn\omega_{0}t}dt \end{align}$$

[in above $l$, $m$ are dummy variables]

after changing order of integration and summation i got this:

$$\begin{align} \displaystyle\sum_{l=-\infty}^{\infty}\sum_{m=-\infty}^{\infty}c_{l}d_{m}\left[\dfrac{1}{T}\int_{t=0}^{T}e^{-j(n-l-m)\omega_{0} t}dt\right]=\displaystyle\sum_{l=-\infty}^{\infty}\sum_{m=-\infty}^{\infty}c_{l}d_{m}\cdot\delta[n-l-m]\end{align}$$

where

$$\delta[n] \triangleq \begin{cases} 1 \qquad & \text{for } n = 0 \\ 0 \qquad & \text{for } n \in \mathbb{Z}, n \ne 0 \\ \text{undefined} \quad & \text{otherwise} \\ \end{cases}$$ is the Kronecker delta.

But from here on I'm unable to proceed,

i know answer is option (d)i.e, $c_{n}*d_{n}$ but i am interested in proving the result formally.

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    $\begingroup$ hi! I tried to make your question readable. You were heavily abusing the "implication" symbol $\implies$ (\implies) whenever (I guess) you meant "equality" $=$ (=), as well as the decimal dot $.$ when you meant "multiplication" $\cdot$ (\cdot). Whilst the second really is just a matter of diligence, the first mixup is a bad one: implication works on statements, whereas equality works on values or expressions; I hence can only hope I did catch the meaning of what you wanted to ask. Please check whether your question now reads correctly! $\endgroup$ – Marcus Müller Feb 10 '18 at 22:25
  • $\begingroup$ @ Marcus Miller ...thank you for editing.....yes you're right..i mixed up $\implies$ with $equality$ and multiplication with decimal dot... in future during formatting i will keep track of these mistakes ...meaning of the question is still the same as intended...... $\endgroup$ – Faraday Pathak Feb 10 '18 at 22:33
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You did everything right, and you're almost there. Your last equation becomes

$$\sum_lc_l\sum_md_m\delta[n-l-m]\tag{1}$$

Due to the Kronecker delta the sum over $m$ reduces to a single element:

$$\sum_md_m\delta[n-l-m]=d_{n-l}\tag{2}$$

because $\delta[n-l-m]$ is non-zero for $m=n-l$ and zero otherwise. Plugging $(2)$ into $(1)$ gives the desired result:

$$\sum_lc_ld_{n-l}$$

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