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

---

Here is another movie to help you visualize what is happening

---