# Proving Fourier transform of cosine multiplied with another function

Cheers, I have a impulse response that looks like this: $$h(t) = h_1(t) \cdot \cos(8 \pi t)$$ and I have to find its frequency response. In order to achieve this I am trying to use the fact that $$x(t)y(t) \to \frac{1}{2\pi} [X(\omega) \ast Y(\omega)]$$.

Using that I end up with $$\mathbb{F}\{ h_1(t) \cdot \cos(\omega_0 t) \} = \frac{1}{2 \pi } \big\{H(\omega) \ast \pi [\delta(\omega - \omega_0) + \delta(\omega + \omega_0) ]\big\}$$

Now I am trying to use the fact that the delta function doesn't change the signal with which it is convolved with,so I end up with:

$$\frac{1}{2}[H(\omega - \omega_0) + H(\omega + \omega_0) ]$$

My questions regarding this are:

1. Is my answer anywhere near the correct one?

2. If it is correct, is there a way to prove that $$H(\omega) \ast \delta(\omega - \omega_0) = H(\omega - \omega_0)$$, because I can visualize it but I don't know if that's right, and how to prove it exactly. I personally would use the fact that the system is time invariant.

3. If there were no delta functions, then this convolution is as difficult to do as it's in the time domain, right?

Thanks =)

What you did is correct, and it's absolutely essential that you know and understand that modulation property.

The fact that convolving a function with a shifted delta impulse results in the shifted function is part of the definition of the Dirac impulse:

$$\int_{-\infty}^{\infty}f(x)\delta(x-x_0)dx=f(x_0)\tag{1}$$

(assuming that $$f(x)$$ is continuous at $$x_0$$). Eq. $$(1)$$ is usually called the sifting property of the Dirac delta impulse.

I can't make sense of your last question, but in the time domain you could just use

$$h(t)\cos(\omega_0t)=\frac12h(t)e^{j\omega_0t}+\frac12h(t)e^{-j\omega_0t}$$

and make use of the modulation property of the Fourier transform.

This is taken from Lapidoth's book on digital communications. It expands on Matt's comment about solving this problem in the time domain, and it may answer your last question.

Let $$y(t) = x(t)\cos(2\pi f_c t)$$ and let $$\hat{y}(f)$$ be the FT of $$y(t)$$. Then, \begin{align*} \hat{y}(f) &= \int_{-\infty}^{\infty} y(t) e^{-j2\pi ft} dt \\ &= \int_{-\infty}^{\infty} x(t)\cos(2\pi f_c t) e^{-j2\pi ft} dt \\ &= \frac12 \int_{-\infty}^{\infty} x(t) \left( e^{j2\pi f_c\,t} + e^{-j2\pi f_c\,t} \right) e^{-j2\pi ft} dt \\ &= \frac12 \int_{-\infty}^{\infty} x(t) e^{-j2\pi (f-f_c)t} dt + \frac12 \int_{-\infty}^{\infty} x(t) e^{-j2\pi (f+f_c)t} dt \\ &= \frac12 \big( \hat{x}(f-f_c) + \hat{x}(f+f_c) \big). \end{align*}

• +1. I like the total avoidance of Dirac deltas and modulation theorems and the like. No muss, no fuss, and no need for any specific knowledge of the properties of Fourier transforms: just the definition suffices! Oh, and I corrected a minor typo, changing $\hat{y}(t)$ to $\hat{y}(f)$. Please change back if you disagree. Feb 11, 2022 at 21:30
• @DilipSarwate Fully agree. Lapidoth's book is remarkable in that it (deliberately) avoids using Dirac deltas. And, thanks for fixing the typo!
– MBaz
Feb 11, 2022 at 23:20