Challenge02_Linear_Systems

# Challenge02_Linear_Systems - -0.5kT s-0.5 e-2.5 kT s = 0.5...

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Digital Signal Processing Dr. Fred J. Taylor, Professor Introduction to Discrete-Time Systems Lesson: 02 Challenge A continuous-time system with an impulse response h ( t )=0.5 e -0.5t -0.5 e -2.5t is to be converted into a discrete time system sampled at a rate of 10 Sa/s. Q: What is the approximate impulse response to the induced discrete-time system? Response The discrete-time impulse response is: h [ k ] = h [ kT s ]= 0.5 e
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Unformatted text preview: -0.5kT s-0.5 e-2.5 kT s = 0.5 (e-0.05 ) k-0.5(e-0.25 ) k = 0.5 (0.951) k 0.5 (0.779) k ~ 0.5 (0.95) k 0.5 (0.8) k The difference equation that corresponds to (0.95) k is y[k]=0.95y[k-1]+x[k]. The difference equation that corresponds to (0.8) k is y[k]=0.8y[k-1]+x[k]. Therefore, from linearity, one obtains: T T x[k] & & & & & & & & i && i i i LINEAR_SYSTEMS - 1...
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## This note was uploaded on 08/21/2010 for the course EEL 5525 taught by Professor Yang during the Summer '09 term at University of Florida.

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