MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6 003 Signals and Systems Spring 2004 Tutorial 7 Monday March 29 and Tuesday March 30 2004 Announcements Problem set 6 is due this Friday Today s Agenda Additional Notes on Partial Fraction Expansion Interpretation of Phase in the Frequency Response Linear phase and constant time shift Nonlinear phase and group delay First and Second Order Systems Bode Plots First order system Bode plots Second order system Bode plots 133 1 Additional Notes on Partial Fraction Expansion Last week we saw that to do partial fraction expansion on rational functions with repeated roots in the denominator there was a complicated formula we could apply However this is confusing and non intuitive Fortunately there is another method that is easy to follow Let s say we want to do a partial fraction expansion on the following function with repeated roots in the denominator 2x 1 F x x 3 3 x 2 The partial fraction expansion will be in the form 2x 1 A0 B A1 A2 3 3 2 x 3 x 2 x 3 x 3 x 3 x 2 We can determine that A0 5 and B 3 using last week s method i e the cover up method However Fortunately we can determining A1 and A2 using cover up results in the left side becoming 2x 1 0 easily x this by subtracting the highest order term from both sides 5 3 2x 1 A1 A2 x 3 3 x 2 x 3 3 x 3 2 x 3 x 2 Simplifying the left side we see that the order of the denominator has been reduced 3 A1 3 A2 2 2 x 3 x 2 x 3 x 3 x 2 We can now determine that A1 3 using cover up as before How do we determine A2 You guessed it Just as before we can subtract the highest order term from both sides to reduce the order of the denominator Using cover up we can now determine that A2 3 Therefore the partial fraction expansion of F x is 5 3 3 3 2x 1 x 3 3 x 2 x 3 3 x 3 2 x 3 x 2 So in general to do partial fraction expansion on rational functions with repeated roots in the denominator determine all the highest order residuals A0 B0 C0 etc using the cover up method subtract the highestorder terms from both sides simplify and then recurse 2 Interpretation of Phase in the Frequency Response In past weeks we ve seen that examining the magnitude of a system s frequency response H j allows us to intuit the behavior of the system by matching it to one of several lter types For review those lter types were lowpass highpass bandpass bandstop and all pass We ll now see that we can also interpret the phase of a system s frequency response H j 2 1 Linear phase and constant time shift We know from the Fourier transform representation of signals that certain CT signals can be expressed as superpositions of complex exponentials of the form ej t without loss of generality we ll use CT notation here Suppose such a signal x t has Fourier transform X j We know from the transform tables that the FT of the shifted version of the signal x t t0 is e j t0 X j But if we lost our tables and even forgot the formula for the Fourier transform how can we re derive this And what does it mean 134 Each of the complex exponentials ej t has a cosine in the real component and a sine in the imaginary component If we time delay the entire signal x t by t0 then each exponential would also be delayed by the same amount in time However they would not by shifted by the same amount in phase Let s consider the cosine component xc t cos t A time shift produces xc t t0 cos t t0 cos t where t0 is the phase shift for that signal Note that the phase added is proportional to the frequency of the cosine This is consistent with our idea that a constant time shift corresponds to a larger phase shift for high frequency components than it does for low frequency ones recall the diagrams drawn in recitation and tutorial So the complex exponential ej t gets mapped to ej t t0 ej t e j t0 Since the Fourier transform X j of a signal is simply the scale factor ignoring the 2 factor that sits in front of the complex exponentials when they are integrated together to form x t this means that a time shift multiplies the FT by e j t0 But we know that the phase of a complex number is the in the ej portion of the number in the magnitude phase representation z rej Multiplying complex numbers adds their phases so we have A constant shift in time corresponds to the addition of a linear phase in frequency 135 Problem 7 1 The magnitude and phase of the frequency response of six di erent all pass systems are plotted below Given the input x t cos t cos 2t nd an expression for the output of each system and summarize in words the behavior of each system a H j H j 2 1 2 1 1 1 2 2 1 1 2 1 2 b H j H j 2 1 2 1 1 1 2 2 1 1 1 2 1 2 2 1 2 2 c H j H j 1 3 100 1 2 1 1 2 2 2 1 1 1 2 d H j H j 2 1 2 1 1 2 2 2 e H j H j 2 1 2 1 1 1 2 2 1 1 2 f H j H j 2 1 2 1 1 1 2 2 1 1 2 136 2 1 Work space 137 2 2 Nonlinear phase and group delay As we saw in Problem 7 1 when the phase of the frequency response H j is linear the system simply d H j When the phase of the frequency shifts the input by an amount equal to the slope t0 d response is nonlinear however the behavior of the system is di cult to the interpret because the system really messes up the signal In engineering lingo one would say Systems with nonlinear phase cause distortion So at this point its appears that H j can be interpreted as either a simple shift or a messy distortion Surely there is some middleground that might appear on an exam right Well there is a special case of interest when the input is narrowband In short when the input is narrowband nonlinear phase can be approximated as linear phase Part f of Problem 7 1 hints at how this narrowband approximation works More speci cally if the frequency content of the input is tightly grouped around some frequency 0 i e the input is narrowband with center frequency 0 H …
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