# EE313 Linear Systems and Signals - Homework #10 Hints

• Problem 10.3. In order for a continuous-time filter to be BIBO stable, each pole must be in the left-hand side of the complex s plane.

Given a set of pole(s) and zero(s) for a transfer function, one can define the transfer function H(s) as per equation (10.153) on page 716. Assume that A = 1 for simplicity.

Each filter in this problem is BIBO stable because all of the poles are in the left-hand side of the Laplace domain. This also means that the region of convergence, which is the plane to the right of the pole with the least negative real component, contains the imaginary axis. As a consequence, one can convert the transfer function into a frequency response by substituting s = j w. Please see equation (10.154) on page 716.

For a worked example, please see example 10.5 on pages 718-719.

Here is a simple example with one pole located in the left-hand side of the Laplace domain at s = p0:

```         1
H(s) = ------
s - p
0
```
Because p0 is in the left-hand side of the Laplace domain, we can substitute s = j w to convert a transfer function into a frequency response:
```               1
H    (w) = --------
freq      j w - p
0
```
We can plot the magnitude response in Matlab as follows:
```p0 = -1;
w = -10 : 0.01 : 10;
Hmagresp = abs( 1 ./ (j*w - p0) );
plot(w, Hmagresp)
```
This filter has a lowpass frequency response.

• Problem 10.4. This problem is Roberts, Chapter 10, problem 35.

Roberts is asking that you solve for the range of K for which the LTI system is bounded-input bounded-output (BIBO) stable.

You may recall that problem 4.2 also involved a root locus plot. Here is the help information from problem 4.2 repeated here for convenience, with one change given in boldface font.

A root locus plots the characteristic roots for different values of K and hence is a way to visualize what values of K lead to a stable system. The horizontal axis corresponds to the real part of the characteristic root, and the vertical axis corresponds to the imaginary part of the characteristic root.

The following Matlab code provides a simple way to generate a root locus plot. In Matlab, the polynomial s^2 + 3 s + 2 is represented as the row vector [1 3 2]. I've changed the step size for values of K from 1.0 to 0.1.

```rootLocusPoints = [];
for K = -10:0.1:10,
pairs = roots( [1 3 K] );      %%% roots of polynomial s^2 + 3 s + K
rootLocusPoints = [rootLocusPoints pairs(1) pairs(2)];
end
scatter( real(rootLocusPoints), imag(rootLocusPoints) );
```

Matlab also has a root locus plotting called `rlocus`, which oddly enough implements the Evans root locus algorithm (no relation).

Now, back to problem 10.4. One range of values for K will yield a BIBO unstable system, and the other values of K will yield a BIBO stable system. For the values of K that yield a BIBO stable system, you are then asked to identify which values of K would give different frequency responses: lowpass, bandpass, bandstop, highpass, notch or allpass.

The system may only exhibit a subset of these types of frequency selectivity. Determining the specific value of K for which the BIBO stable system switches from one type of frequency selectivity to another is determined in an ad-hoc manner by plotting the magnitude of the frequency response for different values of K.

When plotting magnitude responses, you may find it more useful to plot the magnitude in a deciBel scale. The deciBel scale will essentially make the small amplitudes more prominent in the plot. To convert magnitude A to a dB scale, use 20 log10(A):

```p0 = -1;
w = -10 : 0.01 : 10;
Hmagresp = abs( 1 ./ (j*w - p0) );
plot(w, 20*log10(Hmagresp))
```

Last updated 11/11/10. Send comments to bevans@ece.utexas.edu