Two signals are given to me :

$$x(t)=\cos(4\pi t)$$ $$y(t)=\sin(4\pi t)$$

I have founded their coefficients as follows:

$$a_k = a_1=a_{-1}=\frac{1}{2} $$

$$b_k = b_1=b^*_{-1}=\frac{1}{2j} $$

Now I am stuck at convolution. In the formula of the convolution:

$$\sum_{l=-\infty}^{\infty}a_lb_{k-l} $$

What are these $l$ and $k$ and from the coefficients I found what is their respective value?

  • $\begingroup$ Welcome to DSP.SE! Is the Fourier series coefficients of the signal $z(t)=x(t)y(t)$ what you are looking for? $\endgroup$
    – Tendero
    Commented Nov 9, 2018 at 14:17
  • $\begingroup$ @Tendero yes that's what I am trying to find. $\endgroup$ Commented Nov 9, 2018 at 14:24

1 Answer 1


Following your notation, if we define the coefficients of the Fourier series of $z(t)=x(t)y(t)$ as $c_k$:


Notice that the $k$ in which you evaluate the left side of the equality has to be the same as the $k$ in the summation. Therefore, for example, if you want to find $c_0$, you should calculate:


Then you have to calculate that summation. Given that $a_l$ and $b_{-l}$ are non-zero only if $l$ equals $1$ or $-1$:

$$\begin{align} c_0 &=\sum_{l=-\infty}^{\infty}a_lb_{-l} \\ &=a_{-1}b_1 + a_1b_{-1} \\ &=\frac12 \frac{1}{2j} - \frac12 \frac{1}{2j}=0 \end{align}$$

I think this might have clarified the notation for you. You can go on and calculate the rest of the coefficients $c_k$ for all values of $k$.


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