Electrical – Phase shift of shifted impulse response

frequency responsephase shiftsignal processing

I have the following frequency response:
$$H(e^{jkw_0}) = \begin{cases}0, & -4\leq k \leq-3\\ 1, &-2 \leq k \leq 2\\ 0, & 3\leq k \leq 4\end{cases}$$

And I have graphed \$h[n]4\$ as the following:
enter image description here

and \$h[n-4]\$ as the following:
enter image description here
I would like to argue that \$H_c (e^{jw}) = e^{-4jw}H(e^{jw}) \$ have the same magnitude about a strong phase shift. Would it be sufficient to say that the have the same magnitude based on "inspection?" And, for the phase shift, I can once again clearly see that it has time shifted by \$ 4 \$. However I am not sure how would I prove that the phase shift is \$ 4w\$.

Best Answer

A time delay T is represented in frequency domain as a multiplicator. That multiplicator is exp(-2*Pi * f * T *j). You have 4 unit interval delay, so your formula is right. But its interpretation falters.

The complex exponential multiplicator is only phase shift, no affect to amplitude. The phase lag (=negative phase shift) caused by that delay is not constant, but increases linearly as the freguency increases. At frequency 1/T the phase lag is full 2*Pi. You see that easily: A sine voltage with period=T is delayed exactly one period. It means phase lag = 2*Pi (or phase shift -2*Pi). Higher frequencies lag more and lower frequencies less.

Actually your expression 4W is ok for the phase lag, if W=2* Pi *f. Frequency f must be given in proportion to the sampling frequency. In the beginning you used frequency indexes k= -10....10, so to be notationally consistent, we must write in this formula f=k/10.