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How would one go about calculating the fourier transform of the signal? $$ y(t) = x(t-t_0)cos(\omega_0t) $$

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2 Answers 2

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In summary, you would:

  1. Apply the Fourier Transform convolution theorem $$\mathcal{F}\left\{x(t-t_0)\cos(\omega_0 t)\right\} = \dfrac{1}{2\pi}\left[\mathcal{F}\left\{x(t-t_0)\right\}*\mathcal{F}\left\{\cos(\omega_0 t)\right\}\right]$$
  2. Apply the Fourier Transform shift/modulation theorem for $x(t-t_0)$ $$\mathcal{F}\left\{x(t-t_0)\right\} = e^{-it_0 \omega}\mathcal{F}\left\{x(t)\right\}= e^{-it_0 \omega}X(\omega)$$
  3. Look up the Fourier Transform of $\cos(\omega_0 t)$ in a table $$\mathcal{F}\left\{\cos(\omega_0 t)\right\} = \pi\left[\delta(\omega-\omega_0) + \delta(\omega+\omega_0)\right] $$
  4. Apply the convolution properties of the Dirac Delta function $$ \dfrac{1}{2\pi} e^{-it_0 \omega}X(\omega) * \pi\left[\delta(\omega-\omega_0) + \delta(\omega+\omega_0)\right] = \dfrac{1}{2}e^{-it_0 (\omega-\omega_0)}X(\omega-\omega_0)+\dfrac{1}{2}e^{-it_0 (\omega+\omega_0)}X(\omega+\omega_0) $$

You end up with a Fourier transform that is the sum of two shifted, scaled, and modulated copies of $\mathcal{F}\{x(t)\}$

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I'm not sure if this is the solution, but I'll try to help with that:

$$\int x(t+t_0)cos(w_0t)e^{-i\omega t}dt$$

re express it as follows

$$\tau=t+t_0$$

$$\int x(\tau)cos(w_0(\tau-t_0))e^{-i\omega (\tau+t_0)}d\tau$$

$$e^{-i\omega t_0}\int x(\tau)cos(w_0(\tau-t_0))e^{-i\omega (\tau)}d\tau$$

operating inside the $cos$ you get

$$e^{-i\omega t_0}\int x(\tau)cos(w_0\tau- w_0t_0)e^{-i\omega (\tau)}d\tau$$

where $w0t0$ is a number representing a phase.

Once there you should figure out hot to integrate it in order to obtain the final result.

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  • $\begingroup$ shouldn't $$ t $$ be equal to $$ \tau - t_0 $$ instead of what you put in the earlier steps (inside the cos)? $\endgroup$ Oct 27, 2018 at 10:29
  • $\begingroup$ You are right! I'm editing it! $\endgroup$
    – Chaos
    Oct 27, 2018 at 11:24
  • $\begingroup$ thanks for the help, the mistake doesn't effect the final result anyway I believe $\endgroup$ Oct 27, 2018 at 13:24

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