# EE313 Linear Systems and Signals - Homework #4 Hints

• Problem 4.1(a): This problem involves solving a second-order differential equation for t ≥ 0-. Because the output is causal, its form is O(t) = f(t) u(t), as mentioned on slide 5-10. Here are the steps in finding the impulse response:
• Solve the second-order differential equation for x(t) = 0 and leave the two constants undefined.
• Set the input signal x(t) = d(t), where d(t) is the continuous-time impulse (a.k.a. the Dirac delta)
• Substitute the solution for O(t) into the left-hand side.
• Match the d(t) and d'(t) terms on the left-hand and right-hand sides to solve for the two constants.

The Roberts' textbook Signals and Systems find the impulse response of a first-order differential equation using three different methods in Section 3.6 on pages 170-174.

• Problem 4.2: I am asking that you solve for the range of K for which the LTI system is bounded-input bounded-output (BIBO) stable.

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].

```rootLocusPoints = [];
for K = -10:1:10,
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).

Last updated 11/05/10. Send comments to bevans@ece.utexas.edu