3

It has to do with the Bilinear Transform, as Hilmar already stated. The theoretical function is defined for the analog domain, so inherently there will be differences when you convert the response to the digital domain. However, you can still generate the "analog" frequency response and yield the expected result. I took your code and added/modified ...


3

There is already a good answer, but here is some extra explanation. This is called "pole/zero cancellation". That means you have a pole, say, $z_p$ but you also have a zero at $z_p$ so the zero cancels the pole. One particular example where this can be useful is a moving average filter of length $N$, The transfer function is $$H(z) = \frac{1}{N} \...


3

You can simplify the z-transform further, $ H(z) = \frac{Y(z)}{X(z)}= \frac{0.5-0.5z^{-2}}{1-z^{-1}}$ $ H(z) = \frac{(1-z^{-1})(1+z^{-1})}{2(1-z^{-1})} $ canceling the common pole and zero $ H(z) = \frac {1+z^{-1}}{2}$ take inverse z-transform $ Y(n) = \frac{x(n)+x(n-1)}{2} $


2

The slope of the lowpass is 20dB/decade or 6dB/octave, that means it's a simple first order lowpass filter. At the corner frequency the gain is -3dB. Same of the phase. It goes from 0 to -90, so it's a first order filter. At the corner frequency, it's 45 degrees. Looking at both phase and level we conclude that it's a first order lowpass with $f_c = 1000 rad/...


2

Considering an LTI filter, one can define its frequency response by evaluating its transfer function H(z) on the unit circle H(ejω). Is this definition correct? This is only correct if the LTI filter discrete in time. But if it is: yes. I also came across the second definition, which says that the filter's frequency response is the Fourier Transform of its ...


1

If you put the transfer function into the $z$ form, you get $$ H(z) = \frac{Y(z)}{X(z)} = \frac{z - 0.5}{z}$$ Then you can immediately see that $Y(z) = z - 0.5$, and $X(z) = z$. Thus, by inspection, the transfer function has a pole at $z = 0$, and a zero at $z = 0.5$.


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