Line 10: Line 10:
  
 
x[n] = <math>2^{n}</math>u[-n]
 
x[n] = <math>2^{n}</math>u[-n]
 +
 +
y[n] = x[n] * h[n]
 +
 +
= <math> \sum^{\infty}{k = -\infty} 2^{k}u[-k]u[-n--k]</math>
 +
 +
since -k > 0 and k < 0 the summation parameters change
 +
 +
= <math> \sum^{0}{k = -\infty} 2^{k}u[-n+k]</math>

Revision as of 09:48, 12 October 2008

Test Correction of # 3

An LTI system has unit impulse response $ h[n] = u[-n] $

Compute the system's response to the input $ x[n] = 2^{n}u[-n] $

(Simplify answer until all summation signs disappear.)

h[n] = u[-n]

x[n] = $ 2^{n} $u[-n]

y[n] = x[n] * h[n]

= $ \sum^{\infty}{k = -\infty} 2^{k}u[-k]u[-n--k] $

since -k > 0 and k < 0 the summation parameters change

= $ \sum^{0}{k = -\infty} 2^{k}u[-n+k] $

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva