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lab 4 - ECE 101 Linear Systems Winter 2009 Lab 4...

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ECE 101 - Linear Systems, Winter 2009 Lab # 4 Solutions (send comments/questions to [email protected]) Problem 4.4(a), BDS The Fourier Transform of differential equation (4.11), assuming y ( t ) = h ( t ) and x ( t ) = δ ( t ), is z dh ( t ) dt + a 0 h ( t ) = a 0 z { δ ( t ) } ( + a 0 ) H ( ) = a 0 Thus, H ( ) = a 0 + a 0 In the following figure we plot the magnitude and complex phase ( θ ) of H ( ). 0 2 4 6 8 10 0.2 0.4 0.6 0.8 1 ω |Η( j ω)| a = 3 a = 1/3 0 0 0 2 4 6 8 10 -80 -60 -40 -20 0 ω θ 1
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Problem 4.4(b), BDS The following figure plots the magnitude and complex phase of H ( ) 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 0 = 3), and “fb” is system II ( a 0 = 1 / 3). 10 -2 10 0 10 2 10 -3 10 -2 10 -1 10 0 Frequency (rad/s) Magnitude fa fb 10 -2 10 0 10 2 -80 -60 -40 -20 0 Frequency (rad/s) Phase (degrees) fa fb 0 5 10 0 0.2 0.4 0.6 0.8 1 Frequency (rad/s) Magnitude fa fb 0 5 10 -80 -60 -40 -20 0 Frequency (rad/s) Phase (degrees) fa fb 2
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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. 0 1 2 3 4 5 10 -10 10 -5 10 0 10 5 t (sec) ya yb 0 1 2 3 4 5 0 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 a 3
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Problem 4.4(e), BDS
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