(3 intermediate revisions by 2 users not shown) | |||
Line 1: | Line 1: | ||
+ | [[Category:ECE301Summer08asan]] | ||
[[Category: ECE]] | [[Category: ECE]] | ||
[[Category: ECE 301]] | [[Category: ECE 301]] | ||
Line 5: | Line 6: | ||
[[Category: asan]] | [[Category: asan]] | ||
[[Category: Homework]] | [[Category: Homework]] | ||
+ | =Problem 2.28, [[Homework_3_-_Summer_08_%28ECE301Summer2008asan%29|HW3]], [[ECE301]], Summer 2008= | ||
Determine if each system is causal and stable. | Determine if each system is causal and stable. | ||
Line 36: | Line 38: | ||
This system is stable but not causal. | This system is stable but not causal. | ||
+ | ---- | ||
+ | [[Homework_3_-_Summer_08_%28ECE301Summer2008asan%29|Back to HW3]] |
Latest revision as of 10:21, 30 January 2011
Problem 2.28, HW3, ECE301, Summer 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.