### Z-plane to Frequency Response

There is a simple graphical relationship between the z-plane and the frequency response of a filter.
For example, if the transfer function of a system is: $$H(z)=\frac1{1 - 2r \cos(\theta)z^{-1} + r^2 z^{-2}}$$ with $$r=0.97$$ and $$\theta=\frac23 \pi$$. Multiply top and bottom by $$z^2$$ to get: $$H(z)=\frac{z^2}{z^2 - 2r \cos(\theta)z + r^2}$$ What are the poles and zeros of this $$H(z)$$? Will $$r$$ and $$\theta$$ be the radius and angle of a pole?

Note: the transfer function above can be written with the symbolic toolbox in MATLAB:

EDU>> Hz = sym('1/( 1 - 2*r*cos(ang)*z^(-1) + r^2*z^(-2))')

Hz =

1/( 1 - 2*r*cos(ang)*z^(-1) + r^2*z^(-2) )

EDU>>pretty(Hz)


Here is how you can get MATLAB to show you the poles and zeros. First, express the numerator as a polynomial, then find its roots to get the zeros:

EDU>>numerator = [1 0 0]
numerator =
1     0     0
EDU>>roots(numerator)
ans =
0
0

So the zeros are both at $$z=0$$.

Now check for the poles by finding the roots of the denominator polynomial:


EDU>>r = 0.97;
EDU>>ang = 2*pi/3;
EDU>>denominator = [1 -2*r*cos(ang) r^2]
denominator =
1.0000     0.9700     0.9409
EDU>>roots(denominator)
ans =
-0.4850 + 0.8400i
-0.4850 - 0.8400i
EDU>>zprint(ans)
Z =     X    +     jY     Magnitude    Phase    Ph/pi   Ph(deg)
-0.485        0.84        0.97    2.094    0.667   120.00
-0.485       -0.84        0.97   -2.094   -0.667  -120.00

EDU>>zplane(numerator, denominator);
EDU>>axis([xmin xmax ymin ymax]);


The two poles have a magnitude of 0.97 and their angles are +120 degrees and -120 degrees. Figure 1 is a plot of the pole locations and zero locations in the complex Z-plane.

Figure 1. Pole-Zero plot of $$H(z)=\frac1{1+0.97z^{-1}+0.9409z^{-2}}$$

Figure 2. 3D surface plot of $$H(z)=\frac1{1+0.97z^{-1}+0.9409z^{-2}}$$

Are the poles and zeros where they should be, based on factoring the polynomials?

Figure 2 is a 3D plot of $$H(z)$$ over the entire complex Z-plane. You can see the two peaks caused by the poles and the valley in between formed by the zeros at $$z=0$$. The frequency response is is found by evaluating $$H(z)$$ along the contour defined by $$z$$ equal $$e^{j\hat\omega}$$. In Figure 1, the unit circle is given for reference; the two poles lie just inside the unit circle. Figure 2 shows a blue line that traces out the unit circle.

The frequency response is the curve you get if you ride a bike along the unit circle and plot your altitute. Figure 3 is the frequency response of $$H(z)$$.

Figure 3. Frequency Response plot of $$H(z)=\frac1{1+0.97z^{-1}+0.9409z^{-2}}$$

Figure 4 is a movie showing how the frequency response is found by tracing around the unit circle.

Figure 4. Movie