ECE2260F09_HW3p1soln

ECE2260F09_HW3p1soln - 2260 F 09 HOMEWORK #3 prob 1...

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2260 F 09 HOMEWORK #3 prob 1 solution E X : + C= 1 μ F R= 100 k v o v i + a) Determine the transfer function V o /V i . b) Plot |V o /V i | versus ω . c) Find the cutoff frequency, ω c . S OL ' N : a) Filters are voltage-dividers. The output is measured across the C , so the transfer function is the impedance of the C divided by the total impedance of R and C . H ( j ω ) = V o V i = 1/ j ω C R + j ω C or H ( j ω ) = 1 1 + j ω RC b) For the plot, we compute the numerical value of RC : RC = 100k *1 μ s = 0.1s The following Matlab code makes the plot: % ECE2260F09_HW3p1Matlab.m % % Plot of filter's frequency response curve figure(1) omega = 0:1:100; s = j * omega; FilterResp = 1./(1 + j * (1/10)*omega); plot(omega,abs(FilterResp)) axis([0, max(omega), 0, 1]) xlabel('omega') ylabel('|H|') title('HW 3 prob 2 Frequency Response')
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c) The cutoff frequency, ω c , occurs where the magnitude of H ( j ω ) is 1/ times the maximum value attained by H ( j ω ) for any ω > 0. From the plot in (b), the maximum magnitude of H ( j ω ) occurs when ω = 0 and has a value equal to one.
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ECE2260F09_HW3p1soln - 2260 F 09 HOMEWORK #3 prob 1...

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