ELEC 320
Prof. Rich Kozick
Fall, 1998

Notes for Homework 14

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 $\theta = \arctan (B/A)$.