| Line 19: | Line 19: | ||
'''B''' | '''B''' | ||
| − | h[n] = (0.8) | + | h[n] = <math>(0.8)^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. | 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>\Sigma_{n = -2}^\infty</math> <math>(0.8)^n</math> < <math>\infty</math> since <math>lim_{n->\infty} (0.8)^n = 0</math>, the system is stable. |
The system is not causal and stable. | The system is not causal and stable. | ||
Revision as of 09:55, 21 November 2008
Determine if each system is causal and stable.
A
h[n] = (1/5)$ ^n $ u[n]
For n < 0 h[n] = 0 therefore h[n] is causal.
$ \Sigma_{n=0}^\infty $ (1/5)$ ^n $ < $ \infty $ since lim$ _{n->\infty} $ = 0
The system is both causal and stable.
B
h[n] = $ (0.8)^n $ u[n+2]
Since u[n+2] = 1 for n >= -2 and 0 for n < -2 the system is not causal because h[n] $ \neq $ 0 for t < 0.
$ \Sigma_{n = -2}^\infty $ $ (0.8)^n $ < $ \infty $ since $ lim_{n->\infty} (0.8)^n = 0 $, the system is stable.
The system is not causal and stable.
D
h[n] = 5$ ^n $u[3-n]
Since u[3-n] = 1 for n <= 3 and 0 for n > 3, h[n] $ \neq $ 0 for t < 0.
$ \Sigma_{-\infty}^\infty 5^n u[3-n] = \Sigma_{-\infty}^3 5^n < \infty $, therefore the system is stable.
This system is stable but not causal.
