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Unformatted text preview: ECE 101  Linear Systems, Winter 2009 Lab # 4 Solutions (send comments/questions to kquest@ucsd.edu) Problem 4.4(a), BDS The Fourier Transform of differential equation (4.11), assuming y ( t ) = h ( t ) and x ( t ) = ( t ), is z dh ( t ) dt + a h ( t ) = a z { ( t ) } ( j + a ) H ( j ) = a Thus, H ( j ) = a j + a In the following figure we plot the magnitude and complex phase ( ) of H ( j ). 2 4 6 8 10 0.2 0.4 0.6 0.8 1 ( j ) a = 3 a = 1/3 2 4 6 8 1080604020 1 Problem 4.4(b), BDS The following figure plots the magnitude and complex phase of H ( j ) in both linear and logarithmic scale, using the MATLAB routine FREQS (see code at end). Note that the linear scale plots agree with the analytic predictions. fa is system I ( a = 3), and fb is system II ( a = 1 / 3). 102 10 10 2 103 102 101 10 Frequency (rad/s) Magnitude fa fb 102 10 10 280604020 Frequency (rad/s) Phase (degrees) fa fb 5 10 0.2 0.4 0.6 0.8 1 Frequency (rad/s) Magnitude fa fb 5 1080604020 Frequency (rad/s) Phase (degrees) fa fb 2 Problem 4.4(c), BDS In the following figure (plotted in both linear and log scale) ya is h ( t ) for system I, and yb is h ( t ) for system II. 1 2 3 4 5 1010 105 10 10 5 t (sec) ya yb 1 2 3 4 5 1 2 3 t (sec) ya yb Problem 4.4(d), BDS The rate of decay of the impulse response has an inverse response with respect to the rate of decay of the frequency response. This can be most easily seen from the property of time and frequency scaling z { x ( at ) } = 1  a  X j a 3 Problem 4.4(e), BDS The following figure compares the frequency response of system I to that...
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This note was uploaded on 02/24/2010 for the course ECE ECE25 taught by Professor Bill lin during the Spring '09 term at UCSD.
 Spring '09
 bill lin

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