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<math> {\color{White}ab} y[n] = h_1[n]\;\ast\;x[n] = \frac{1}{4} y[n-1] + x[n] </math>
 
<math> {\color{White}ab} y[n] = h_1[n]\;\ast\;x[n] = \frac{1}{4} y[n-1] + x[n] </math>
  
Then, find the inverse LTI filter (<math>h_2</math>) of <math>h_1</math>, which satisfies the following relationship for any discrete-time signal <math>x[n]</math>,
+
Then, find the inverse LTI filter <math>h_2</math> of <math>h_1</math>, which satisfies the following relationship for any discrete-time signal <math>x[n]</math>,
 +
 
 +
(assume that the impulse responses are causal and zero when <math>n<0</math>)
  
 
<math> {\color{White}ab} x[n] = h_2[n]\;\ast\;h_1[n]\;\ast\;x[n] </math>
 
<math> {\color{White}ab} x[n] = h_2[n]\;\ast\;h_1[n]\;\ast\;x[n] </math>

Revision as of 11:48, 8 October 2010


Under construction -Jaemin


Quiz Questions Pool for Week 8


Q1. Find the impulse response of the following LTI systems and draw their block diagram.

(assume that the impulse response is causal and zero when $ n<0 $)

$ {\color{White}ab}\text{a)}{\color{White}abc}y[n] = 0.6 y[n-1] + 0.4 x[n] $

$ {\color{White}ab}\text{b)}{\color{White}abc}y[n] = y[n-1] + 0.25(x[n] - x[n-3]) $


Q2. Suppose that the LTI filter $ h_1 $ satifies the following difference equation between input $ x[n] $ and output $ y[n] $.

$ {\color{White}ab} y[n] = h_1[n]\;\ast\;x[n] = \frac{1}{4} y[n-1] + x[n] $

Then, find the inverse LTI filter $ h_2 $ of $ h_1 $, which satisfies the following relationship for any discrete-time signal $ x[n] $,

(assume that the impulse responses are causal and zero when $ n<0 $)

$ {\color{White}ab} x[n] = h_2[n]\;\ast\;h_1[n]\;\ast\;x[n] $


$ \text{Q3.} $


$ \text{Q4.} $


$ \text{Q5.} $


$ \text{Q6.} $


Back to ECE 438 Fall 2010 Lab Wiki Page

Back to ECE 438 Fall 2010

Alumni Liaison

Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010