# Ideal Discrete Time Hilbert Transformer with constant group delay

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!

## 1 Answer

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.

• 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?
– MJ13
Commented May 4, 2019 at 12:52
• Any idea why I have to keep only the real part of it?
– MJ13
Commented May 4, 2019 at 18:58
• 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. Commented May 4, 2019 at 20:20
• :@ 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?
– MJ13
Commented May 5, 2019 at 8:02
• @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. Commented May 5, 2019 at 10:21