Let's assume we have an ideal discrete time Hilbert Trasnformation system (90-degree phase shifter) with a frequency response over one period:

$$H(e^{jw}) = \left\{ \begin{array}{ll} -j & \mbox{if}\;\;\;\;\;\; 0 < w < \pi \\ \;\;j & \mbox{if}\;\; -\pi < w < 0 \end{array} \right.$$

I want to define an ideal frequency response $H_d(e^{jw})$ of an ideal discrete time Hilbert Trasnformation system that has a non-zero constant group delay.

I consider that $$H_d(e^{jw})= [1-2u(w)]e^{j(\frac{\pi}{2}-τ\cdot w)}$$ since $$\left | H_d(e^{jw}) \right | = 1 \; \; \; \forall w$$ and $$\measuredangle H_d(e^{jw}) = \left\{ \begin{array}{ll} \frac{\pi}{2}-τ\cdot w & \mbox{if} \;-\pi < w < 0 \\ -\frac{\pi}{2}-τ\cdot w & \mbox{if}\;\;\;\;\;\; 0 < w < \pi \end{array} \right.$$

So, it seems an acceptable answer. However, I didn't make any calculations to find it. Just guessed it. So my question is: which is the approach to such a problem is order to define $H_d(e^{jw})$? Thanks in advance!


There's not much need for complicated calculations. If you have the zero-delay frequency response $H(e^{j\omega})$ as defined in the first equation of your question, the frequency response with a linear-phase (constant delay) is given by

$$H_d(e^{j\omega})=H(e^{j\omega})e^{-j\omega\tau}=e^{-j\left(\frac{\pi}{2}\textrm{sgn}(\omega)+\omega\tau\right)},\qquad \omega\in(-\pi,\pi)\tag{1}$$

where $\tau$ is the group delay in samples.

  • $\begingroup$ I also have to define the impulse response $h_d[n]$ assuming that $h[n]=0$ for $n<0$ and $n>M$ ( We assume that we want to use the window method in order to design the linear phase approach for the ideal Hilbert Transform). It is given that the result is : $$h_d[n]=\frac{1-\cos(\pi(n-τ))}{\pi(n-τ)}$$ but the result I get is $$h_d[n]=\frac{1-e^{j\pi(n-τ)}}{\pi(n-τ)}$$. Is there any reason that $sin(\pi(n-τ))$ is considered to be zero? $\endgroup$
    – MJ13
    May 4 '19 at 12:52
  • $\begingroup$ Any idea why I have to keep only the real part of it? $\endgroup$
    – MJ13
    May 4 '19 at 18:58
  • $\begingroup$ The impulse response must be real-valued. I don't know how you came up with your result, so I can't know where you've gone wrong. $\endgroup$
    – Matt L.
    May 4 '19 at 20:20
  • $\begingroup$ :@ Matt L. I just applied the definition of inverse fourier transform: $$ h_d[n]=\frac{1}{2\pi}\int_{-\pi}^{0}e^{j(\frac{\pi}{2}-τw)}e^{jwn}dw-\frac{1}{2\pi}\int_{0}^{\pi}e^{j(\frac{\pi}{2}-τw)}e^{jwn}dw \Rightarrow $$ $$h_d[n]=\frac{e^{\frac{j\pi}{2}}}{2} \cdot \left ( \int_{-\pi}^{0}e^{jw(n-τ)}dw-\int_{0}^{\pi}e^{jw(n-τ)}dw \right ) \Rightarrow $$ $$h_d[n]=\frac{1-e^{j[\pi(n-τ)]}}{\pi(n-τ)}$$ Can you spot the mistake? $\endgroup$
    – MJ13
    May 5 '19 at 8:02
  • 1
    $\begingroup$ @MJ13: I really think you should try to solve such elementary problems yourself, but here is one more (final) hint: don't you see that the integral limits are $-\pi$ in one integral and $+\pi$ in the other? Might it be that that's how the different signs in the exponent come about? If you still don't reach the final result, you could ask a new question, because then the solution would at least be to the benefit of all other users instead of being lost in the comments here. $\endgroup$
    – Matt L.
    May 5 '19 at 10:21

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