0
$\begingroup$

I am trying to learn about the behavior of the FIR filter however with complex coefficients. The filter I am trying to analyze is the following:

$$H(z)=a+jbz^{-1}\quad\text{where the variable}\quad j = \sqrt{-1}$$

Now by setting $a = 1$ and $b = -1$ we obtain the filter $H(z)=1-jz^{-1}$

If we consider a discrete complex input signal $x[k] = \{x[0], x[1],...,x[N]\}$ where $N$ is the signals length and $k$ is the discrete index.

By applying the filter on the input signal $x[k]$ we get the output $y[i] = x[i]-jx[i-1]$ with $i <= N$. And by setting $z = e^{j\omega}$ and by substitution we get $H(z) = a + jb\cos(\omega) + b\sin(\omega)$.

If we evaluate the frequency response now $|H(z)| = \sqrt{(a+b\sin(\omega))^{2}+(b\cos(\omega)^{2})}$ so in the set example we will get $$ |H(z)| = \sqrt{(1+\sin(\omega))^{2}+(-\cos(\omega)^{2})} $$

Phase will be $$ \text{phase}(H(z)) = \tan^{-1}(b\cos(\omega))/(a+b\sin(\omega)) $$ again for the example we will get $$ \text{phase}(H(z)) = \tan^{-1}(-\cos(\omega))/(1+\sin(\omega)) $$

But how do we proceed in the analysis?

  1. What would be the effect of such filter on the complex input signal $x[k]$? What about if it was a real signal?
  2. How do we analyze and interpret the magnitude, phase, and z transfer function with respect to the coefficients $a$ and $jb$?

By modeling the example in MATLAB and plotting the Bode plot using the following commands:

q = tf('z',1);
myf = 1 - 1i*q^-1;
bode(myf)

We get the following output plot:

enter image description here

Thank you it is much appreciated

$\endgroup$
3
  • $\begingroup$ Just to clarify, we should name your initial expression with coefficients filter $a$ and $b$, instead of $x$ and $y$,... right?. Note that you rename then after that and use as signals. I think that all in your confusion comes from there..... At the end, your filter is just a linear polinomial with constant parameters in $z^{-1}$, right?... $\endgroup$ – Brethlosze Feb 16 at 22:23
  • $\begingroup$ Again... note this detail. When you pass from $z$ to $\omega$, you lost your real behavior... you are just looking at the Frequency Response, not at the whole Filter response, i.e. the transient. Do you want to analyze just the FR? If you do, your figures are right, you have Magnitude and Phase for sinusoidal inputs responses, in steady state, only. Or do you want to analyze other aspects, too? $\endgroup$ – Brethlosze Feb 16 at 22:29
  • $\begingroup$ Thank you for the remark on the notations, I have edited the post to reflect the default naming mentioned. Additionally, for the input signal I meant that by applying the filter on the input signal $x[k]$ we will be filtering two points at a time $x[i]$ and $x[i-1]$, however the second one will be multiplied by $j$ and experience a phase delay. I did not understand what is the benefit of such procedure on the signal. This filter is used in a publication to accelerate the behavior of the LMS algorithm, I did not understand well its role: ieeexplore.ieee.org/document/9099039 $\endgroup$ – chaosmind Feb 17 at 21:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.