ELEC 320
Prof. Rich Kozick

Notes for Homework 10

Problem: Prove that for a continuous-time system that is linear and time-invariant, the zero-state response (ZSR) of the system to a sinusoidal input is a sine wave with the same frequency as the input wave, but a different amplitude and phase shift. Also, find an expression for the frequency response $H ( \omega )$ of the system in terms of the impulse response $h(t)$.

Use the following approach.

1. Please explain why it is true that the ZSR of any linear, time-invariant (LTI) system is completely described by the impulse response $h(t)$ of the system. (Are there any LTI systems for which this is not true?) If the impulse response $h(t)$ is known, then the system output $y(t)$ due to any input $x(t)$ is given by
   
   $$y(t) = x(t) \ast h(t) = \int_{-\infty}^{\infty} h(\lambda) x(t - \lambda) \, d \lambda$$
   
   

2. Now consider a particular input $x(t) = \cos ( \omega_0 t)$ that is applied to a LTI system with impulse response $h(t)$. Put this $x(t)$ into the convolution integral, and look at the resulting $y(t)$. You should be able to recognize that $y(t)$ is a sine wave with the same frequency $\omega_0$, but with a different amplitude and phase shift. The trigonometric identities at the bottom of the page will be helpful.

3. In terms of the frequency response of the system $H ( \omega )$, recall that we expect that the system output has the form
   
   $$y(t) = \vert H(\omega_0) \vert \cos (\omega_0 t + \angle H(\omega_0)).$$
   
   
   Use your result from item 2 to relate the frequency response $H ( \omega )$ of the system to the impulse response $h(t)$. This provides a mathematical connection between the frequency domain and time domain descriptions of a system.

4. You now understand the very important result that a sine wave input to a LTI system produces a sine wave output with the same frequency but different amplitude and phase shift!

Here are some useful identities:

$$\cos [\omega_0 (t - \lambda)] = \cos (\omega_0 t) \cos (\omega_0 \lambda ) + \sin (\omega_0 t) \sin (\omega_0 \lambda )$$

$$A \cos (\omega_0 t) - B \sin (\omega_0 t) = H \cos ( \omega_0 t + \theta)$$

where $H = \sqrt{A^2 + B^2}$ and $\theta = \arctan (B/A)$.