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=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.
  
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'''B'''
 
'''B'''
  
h[n] = (0.8)<math>^n</math> u[n+2]
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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>^n</math> < <math>\infty</math> since lim<math>_{n->\infty} (0.8)<math>^n</math> = 0 the system is stable.
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<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.
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This system is stable but not causal.
 
This system is stable but not causal.
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[[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.


Back to HW3

Alumni Liaison

has a message for current ECE438 students.

Sean Hu, ECE PhD 2009