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In Boaz Porat's book about signal proccessing, at part 8 he mentions the example: $$ H\left(z\right)=1-z^{-1}\Rightarrow H^{f}\left(\theta\right)=1-e^{-j\theta}=\left(e^{\frac{j\theta}{2}}-e^{-\frac{j\theta}{2}}\right)e^{-\frac{j\theta}{2}}= \\ 2j\sin\left(\frac{\theta}{2}\right)e^{-\frac{j\theta}{2}}=2\sin\left(\frac{\theta}{2}\right)e^{j\left(\frac{\pi}{2}-\frac{\theta}{2}\right)} $$ Therefore the amplitude and phase are given by: $$ \left|H^{f}\left(\theta\right)\right|=\left|2\sin\left(\frac{\theta}{2}\right)\right|$$ $$\psi\left(\theta\right)=\begin{cases} \frac{\pi}{2}-\frac{\theta}{2} & 0<\theta<\pi\\ -\frac{\pi}{2}-\frac{\theta}{2} & -\pi<\theta<0 \end{cases} $$ How did he find the phase and why would it not be the same for all $-\pi <\theta <\pi $?

Also, how would the phase look if instead of $\sin $ he would represent it with a $\cos $? would the regions be $$ \begin{cases} -\frac{\pi}{2}-\frac{\theta}{2} & \frac{\pi}{2}<\theta<\pi\\ \frac{\pi}{2}-\frac{\theta}{2} & -\frac{\pi}{2}<\theta<\frac{\pi}{2}\\ -\frac{\pi}{2}-\frac{\theta}{2} & -\pi<\theta<-\frac{\pi}{2} \end{cases} $$

In general, how do I find the frequency response given a transfer function?

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Say you want to express $x$ as modulus and phase. This is: $x = |x|$ if $x \ge 0$, and $x = -|x|$ when $x < 0$. That minus sign is equivalent to a phase of $\pi$; remember that $-1 = e^{j\pi}$.

For your example above, when $\sin(\cdots)$ is negative, then you have to add (or subtract) a phase of $\pi$, which is why you have 2 expressions for the phase depending on the frequency $\theta$. And there is an extra $\pi/2$ phase that comes from the $j$ multiplying the sine.

For your other question, the difference of exponentials gives a sine, not a cosine. So it is not straightforward how to represent the frequency response if terms of cosines.

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  • $\begingroup$ For your last comment, couldn't we use $sin(\theta) = \cos(\theta - \pi/2)$ ? $\endgroup$ Commented Jan 25, 2023 at 3:11

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