EECE 301 Signals & Systems Prof. Mark Fowler

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1 EECE 30 Signals & Systems Prof. Mark Fowler Note Set #26 D-T Systems: Transfer Function and Frequency Response /

2 Finding the Transfer Function from Difference Eq. Recall: we found a DT system s freq. resp. H() by analying for input e jn or by taking DTFT of the Diff Eq. Here we can either analye the system for input n or take the ZT of the Diff Eq here we do the later: yn [ ] ayn [ ]... a yn [ N] bxn [ ] bxn [ ]... b xn [ M] N 0 M ZT ZT yn [ ] ayn [ ]... a yn [ N] ZT b x[ n] b x[ n ]... b x[ n M ] 0 N M Delay Prop. & Algebra Algebra Y( ) a... an N X( ) b0 b... bm M M b0 b... bm Y( ) X( ) N a... an So can just write H() by inspection of D.E. coefficients! From this H() we know how to compute the output: y = filter(b,a,x); 2/

3 Poles and Zeros of Transfer Function M b0 b... bm N a... an b b... b M M ( NM) 0 M N N a... an Define the polynomials A() and B() so that: ( NM) B ( ) A( ) Assume there are no common roots in the numerator B() and denominator A(). (If not, assume they ve been cancelled and redefine B() and A() accordingly) Poles of H(): The values on the complex -plane where H() Zeros of H(): The values on the complex -plane where H() = 0 The roots of the denominator polynomial A() determine N poles. The roots of the Numerator polynomial B() determine M eros. The term (N M) gives poles/eros at the origin according to: If N > M : N M Origin If N < M : M N Origin 3/

4 Example: Finding Poles and Zeros yn [ ] yn [ ] yn [ 2] 2 xn [ ] xn [ ] 2 4 p = ero at origin eros: 0, 2 poles: j 2 2 Using MATLAB: >> plane([2 ],[ -(/sqrt(2)) /4]) Coeff. Vectors MUST be rows Conjugate Pair Imaginary Part Real Part 4/

5 Impulse Response of System Sometimes looking at how a system responds to the impulse function (i.e., delta sequence) [n] can tell much about a system. Hitting a system with [n] is lot like ringing a bell to hear how it sounds [n] n LTI D-T system ICs = 0 Note: If system is causal, then h[n] = 0 for n < 0 h[n] n The symbol h[n] means the impulse response. Noting that the ZT of [n] = and using the properties of the transfer function and the Z transform: hn [ ] Z hn [ ] Z H( Z ) [ n] hn [ ] IDTFT H( ) From PFE and Poles/Zeros we see that a TF like this: will have an impulse response with terms like this: hn [ ] kpun [ ] k pun [ ] k p un [ ] n n n 2 2 N N ( NM) B ( ) A( ) Some simplifying assumptions made here! Now we almost always want this to decay (like a bell!): all poles p i < 5/

6 Stability of System Definition: A system is said to be stable if its output will never grow without bound when any bounded input signal is applied and that seems like a good thing!!! Without going into all the details a system with an impulse response that decays fast enough is said to be stable. From our exploration of the effect of poles on the impulse response we say that: j Im{} Re{} For a Stable System Poles must be inside unit circle Zeros can be anywhere j unit circle 6/

7 Relationship: Transfer Function and Freq. Resp. Recall: DTFT = ZT evaluated on Unit Circle if UC is inside ROC Fact: For causal systems UC is inside ROC if all poles are inside UC H ) H ( ) ( If all poles are inside the UC e j We saw how to use freq before to plot the Frequency Response this just shows how to plot the Frequency Response from the Transfer Function coefficients: b b b... b a a a a 2 M 0 2 M 2 N N num bb... b 0 den aa... a 0 omega -pi:?:pi M N H freq(num, denom, omega) plot(omega/pi, abs(h)) plot(omega/pi, angle(h)) must put any ero b i into the vector must put any ero a i into the vector Pick appropriate spacing 7/

8 Visualiing Relationship Between TF & FR = 0 j 0.3 j 0.8e 0.8e 0.3 = 0.8e j0.3 Pole-Zero Plot For This H() Im{} Re{} H() And we know that the Frequency Response is just the Transfer Function evaluated on the Unit Circle. H ( ) H ( ) e j 8/

9 Now plot just those values on the unit circle: Now Cut here and unwrap This shows after it has been cut and unwrapped and plotted on the axis: This shows the Frequency Response H() where is the angle around the unit circle this explains why H() is a periodic function of 9/

10 Effect of Poles & Zeros on Frequency Response of DT filters Im Note: Including a pole or ero at the origin Im doesn t change the magnitude but does change the phase Placing a ero at Im makes H() = 0 Im Placing more eros/poles Im gives sharper transitions. Figure from B.P. Lathi, Signal Processing and Linear Systems 0/

11 Cascade of Systems Suppose you have a cascade of two systems like this: x[n] H () H () y[n] H 2 () H 2 () [n] Y( ) H ( ) X( ) Y( ) H ( ) X( ) Z( ) H ( ) Y( ) Z( ) H ( ) Y( ) 2 2 Z( ) H ( ) H ( ) X( ) 2 Z( ) H ( ) H ( ) X( ) 2 Thus, the overall frequency response/transfer function is the product of those of each stage: H ( ) H ( ) H ( ) total 2 H ( ) H ( ) H ( ) total 2 Obviously, this generalies to a cascade of N systems: H ( ) H ( ) H ( ) H ( ) total 2 H ( ) H ( ) H ( ) H ( ) total 2 N N /

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