View source for 2.28 (a,b,c) Old Kiwi - Rhea

Determine if each system is causal and stable.

'''A'''

h[n] = (1/5)<math>^n</math> u[n]

For n < 0 h[n] = 0 therefore h[n] is causal.

<math>\Sigma_{n=0}^\infty</math> (1/5)<math>^n</math> < <math>\infty</math> since lim<math>_{n->\infty}</math> = 0

The system is both causal and stable.

'''B'''

h[n] = (0.8)<math>^n</math> u[n+2]

Since u[n+2] = 1 for n >= -2 and 0 for n < -2 the system is not causal because h[n] <math>\neq</math> 0 for t < 0.

<math>\Sigma_{n = -2}^\infty</math> (0.8)<math>^n</math> < <math>\infty</math> since lim<math>_{n->\infty} (0.8)<math>^n</math> = 0 the system is stable.

The system is not causal and stable.

'''D'''

h[n] = 5<math>^n</math>u[3-n]

Since u[3-n] = 1 for n <= 3 and 0 for n > 3, h[n] <math>\neq</math> 0 for t < 0.

<math>\Sigma_{-\infty}^\infty 5^n u[3-n] = \Sigma_{-\infty}^3 5^n < \infty</math>, therefore the system is stable.

This system is stable but not causal.