Real and Imaginary Frequency Responses of a single complex pole

Question

I'm filtering a real signal with a single complex pole with a complex coefficient (a_re [real part] and a_im [imaginary part]), I also have a gain coefficient but I'm gonna leave it out for the sake of simplicity. The filter has two outputs, one for the real and another for the imaginary part. Here's the filter equation in time for both outputs

Real part output: $$y_1[n] = x_1[0] + a_{R} y_1[-1] - a_{I} y_2[-1]$$ Imaginary part output: $$y_2[n] = a_{I} y_1[-1] + a_{R} y_2[-1]$$

Where $n$ is the sample number, $x_1[0]$ is the real input, $a_{R}$ & $a_{I}$ are, respectively, the real and imaginary part of the coefficient, $y_1[-1]$ is the previous output of "real part output" and $y_2[-1]$ is the previous output of "imaginary part output".

So, I need to plot the frequency response based on the Z transform. I've already succeeded in dealing with the real version of all this (a real pole), and I'm struggling with the complex version. Anyway, the equation for the real pole is:

$$y[n] = x[0] + a y[-1]$$

Where $a$ is the coefficient of the real pole.

Now, the Z-transform of the real pole is:

$$\tag{1}H(z) = \dfrac{1}{ 1 - a z^{-1}}$$

Where, again, $a$ is the coefficient of the real pole.

In my code I am solving it and dealing with the z-transform in this way, the input of the function is the $a$ coefficient and the $\omega$ variable (which is the angular frequency in radians per sample). I'm only interested in getting the magnitude output, so I don't care about the phase response of the filter. Anyway, here is how I get it:

$a \rightarrow$ coefficient

$w \rightarrow$ angular frequency

$f_{R} = \cos(\omega)$; $\;\;\;\;f_{I} = \sin(\omega)$

$\mathbf{R} = 1 - a f_{R}$; $\;\;\mathbf{I} = -a f_{I}$

$\mathbf{Mag} = \sqrt{\mathbf{R}^{2} + \mathbf{I}^{2}}$

$H = \frac{1}{\mathbf{Mag}}$

The whole thing would be:

$$\tag{2}|H(z)| = \dfrac{1}{\sqrt{(1 - a \cos \omega)^2 + (-a \sin \omega)^2)}}$$

Now my trouble is adapting it to the complex version of this z transform, which is supposed to be the same formula as the real pole's z transform in Eq. (1), but I don't know how to adapt the formula/code for the complex pole. Bearing in mind that I want to have the frequency response of the two outputs (the real and imaginary part of the complex filter).

Hopefully, I'd like some help on reaching the frequency response from the coefficient as a function of angular frequency as I've written above for the real pole version of this in Eq.(2)!

Any thoughts, considerations, hints, help is highly welcome.

Thanks a lot

Answer 1

If I understood you correctly you want a formula for the magnitude of the frequency response of a filter with a single complex pole. As you've already noticed, the system function is

$$H(z)=\frac{1}{1-az^{-1}}\tag{1}$$

which is valid regardless of whether $a$ is real-valued or complex-valued. Let's assume that $a=a_r+ja_i$ is a complex-valued number. If we also assume that $|a|<1$ (i.e. the system is causal and stable), then we get the systems frequency response by replacing $z$ by $e^{j\omega}$ in (1):

$$H(e^{j\omega})=\frac{1}{1-ae^{-j\omega}}\tag{2}$$

Now you just have to compute the magnitude of (2). Using

$$|1-u|^2=1-2\text{Re}\{u\}+|u|^2$$

for any complex $u$, we get with $u=ae^{-j\omega}$ from (2)

$$|H(e^{j\omega})|=\frac{1}{\sqrt{1-2\text{Re}\{ae^{-j\omega}\}+|a|^2}}= \frac{1}{\sqrt{1-2(a_r\cos\omega+a_i\sin\omega)+|a|^2}}\tag{3}$$

EDIT:

This is how you can derive the transfer functions relating the real-valued input to the real part and the imaginary part of the output, respectively. Let $y(n)=y_r(n)+jy_i(n)$. Then you have the following two difference equations for the two real-valued output signals:

$$y_r(n)=x(n)+a_ry_r(n-1)-a_iy_i(n-1)\\ y_i(n)=a_ry_i(n-1)+a_iy_r(n-1)$$

Transforming these equations to the $\mathcal{Z}$-domain gives

$$Y_r(z)=X(z)+a_rz^{-1}Y_r(z)-a_iz^{-1}Y_i(z)\\ Y_i(z)=a_rz^{-1}Y_i(z)+a_iz^{-1}Y_r(z)$$

Now you have two equations for the two unknowns $Y_r(z)$ and $Y_i(z)$. After some algebra you arrive at

$$Y_r(z)=H_r(z)X(z)=\frac{1-a_rz^{-1}}{1-2a_rz^{-1}+(a_r^2+a_i^2)z^{-2}}X(z)\\ Y_i(z)=H_i(z)X(z)=\frac{a_iz^{-1}}{1-2a_rz^{-1}+(a_r^2+a_i^2)z^{-2}}X(z)\tag{4}$$

Note that even though

$$H(e^{j\omega})=H_r(e^{j\omega})+jH_i(e^{j\omega})$$

$H_r(e^{j\omega})$ and $H_i(e^{j\omega})$ are not the real and imaginary parts of the total frequency response $H(e^{j\omega})$ of the complex-valued system because they are both themselves complex-valued. You can obtain the magnitudes of the frequency responses by computing the magnitudes of $H_r(e^{j\omega})$ and $H_i(e^{j\omega})$ as given in (4), i.e. by replacing $z$ by $e^{j\omega}$ and computing the magnitudes of the two complex functions.

Another option is to express the frequency responses $H_r(e^{j\omega})$ and $H_i(e^{j\omega})$ using the total frequency response $H(e^{j\omega})$ (2):

$$H_r(e^{j\omega})=\frac12[H(e^{j\omega})+H^*(e^{-j\omega})]\\ H_i(e^{j\omega})=\frac{1}{2j}[H(e^{j\omega})-H^*(e^{-j\omega})]\tag{5}$$

From (5), the magnitudes can be written as

$$|H_r(e^{j\omega})|=\frac{1}{2}\sqrt{|H(e^{j\omega})|^2+2\text{Re}\{H(e^{j\omega})H(e^{-j\omega})\}+|H(e^{-j\omega})|^2\}}\\ |H_i(e^{j\omega})|=\frac{1}{2}\sqrt{|H(e^{j\omega})|^2-2\text{Re}\{H(e^{j\omega})H(e^{-j\omega})\}+|H(e^{-j\omega})|^2\}}$$

where $|H(e^{j\omega})|$ is given by (3).

EDIT 2:

Here's some Matlab/Octave code using equation (4) to plot the different transfer functions. $|H_r(e^{j\omega})|$ and $|H_i(e^{j\omega})|$ are of course symmetrical because they correspond to real-valued systems, whereas $|H(e^{j\omega})|$ is not due to the complex pole.

w = linspace(0,2*pi,300);          % frequency vector
ar = .4; ai = -.3; a = ar+j*ai;    % complex pole
u = exp(-j*w);                     % u = z^(-1)
Hr = (1-ar*u)./(1-2*ar*u+(ar^2+ai^2)*u.^2);
Hi = (ai*u)./(1-2*ar*u+(ar^2+ai^2)*u.^2);
H = 1./(1-a*u);
plot(w/pi,abs(Hr),w/pi,abs(Hi),w/pi,abs(H))
legend('Hr', 'Hi', 'H');

EDIT 3:

If you need to split the numerators and denominators of (4) into real and imaginary parts, here's what you get:

$$H_r(e^{j\omega})=\frac{1-a_r\cos\omega+ja_r\sin\omega}{\left[1-2a_r\cos\omega+(a_r^2+a_i^2)\cos(2\omega)\right]+j\left[2a_r\sin\omega-(a_r^2+a_i^2)\sin(2\omega)\right]}\\ H_i(e^{j\omega})=\frac{a_i\cos\omega-ja_i\sin\omega}{\left[1-2a_r\cos\omega+(a_r^2+a_i^2)\cos(2\omega)\right]+j\left[2a_r\sin\omega-(a_r^2+a_i^2)\sin(2\omega)\right]}$$