The Wigner Function of $x(t)$ is

$$ W(t,f) = \frac{1}{2\pi}\int x\Big(t+\frac{\tau}{2}\Big)x^*\Big(t-\frac{\tau}{2}\Big) e^{-j2\pi f\tau}\;d\tau $$

I know how to get the $W(t,f)$ of $$x(t)=\cos(2\pi f_i t)$$ which is

$$[\delta(f+f_i)+\delta(f-f_i) + 2\delta(f) \cos(4 \pi f_i t)].$$

However, I found a difficulty in deriving the cosine when the phase $\theta$ is included, $x(t)=\cos(2\pi f_i t + \theta)$.

  • $\begingroup$ The wigner function of $x(t)=\cos(2\pi f_i t+\theta)$ $\endgroup$ May 5, 2022 at 14:33

1 Answer 1


Hint: $$\cos(2\pi f_i t + \theta) = \cos(2\pi f_i (t + \theta/2\pi f_i)= x(\tilde t), \\ \hbox{where}~~\tilde t\equiv t + \theta/2\pi f_i. $$

So, just plug into your expression for x(t), if you trust and understand it.

  • $\begingroup$ So I just need to replace the $t$ with this new one that includes the phase? $\endgroup$ May 5, 2022 at 15:01
  • $\begingroup$ Yes. Isn't it evident? $\endgroup$ May 5, 2022 at 15:14
  • $\begingroup$ Great! I was just thinking that the two impulses will also be affected by the phase but I realized maybe not. $\endgroup$ May 5, 2022 at 15:19

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