New answers tagged


However, {f~n} has a time shift comparing to the origin signal, and their amplitudes also have a deviation. Any causal system has a non-zero group delay. That's to be expected from systems theory! So, yes, you've got a shift. And, unless your ${h_n}$ is a linear-phase filter, that delay isn't even constant for different frequencies. This has nothing ...


Did you take care of the FFT shift operations while doing IFFT/FFT? Did you take care of the fact that multiplication in the frequency domain is circular convolution in time domain (not linear convolution)? What tool are you using for this experiment? Matlab?


There is nothing preventing you from writing down an expression that has more zeros than poles. In fact, a typical FIR filter may be designed as something like $$\sum_{n = -\frac{N-1}{2}}^{\frac{N-1}{2}} f(n) z^n$$ giving you a filter that's nice and symmetrical around zero delay. The problem is, that in the physical world, $z$ is a lead operator -- it ...


Because for a polynomial of degree $N$, the number of zeros (roots) is no more than $N$.

Top 50 recent answers are included