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3

The z-transform is the discrete version of the Laplace transform and in both cases z and s are the set of all complex numbers, and as such we map with the transform the time domain function into the domain of complex frequencies; signals that change in rotation only which is the Fourier Transform and in addition to that such signals that can grow and decay ...


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Yes, of course. Any delay will look like this where the size of the delay determines the slope of the phase. If the the delay turns out to be an integer number of samples, than this very easy to implement. You can also do fractional delays but that's more work and can only be done approximately.


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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 ...


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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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Does the number of taps affect a cost of implementation of FIR? Well, insert your definition of "cost", and your question should really answer itself. FIR filters belong to the class of linear filters, the combination of N lower order filters can create the desired FIR filter of the higher order. So I was thinking If it makes sense to implement ...


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