Exercise Lecture: the Frequency Response of LTI Systems

Exercise Lecture: the Frequency Response of LTI Systems

Introduction to Frequency Response Frequency Response of LTI Systems Exercise Assume that the signal x[n]=exp(j2pfnˆ ) is input to a 4-point averager. 1. Give a general expression for the output signal and identify the frequenchy response of the system. 2. Compute the output signals for the specific frequencies fˆ = 0, fˆ = 1/4, and fˆ = 1/2. ©2016, B.-P. Paris ECE 201: Intro to Signal Analysis 233 Introduction to Frequency Response Frequency Response of LTI Systems Lecture: The Frequency Response of LTI Systems ©2016, B.-P. Paris ECE 201: Intro to Signal Analysis 234 Introduction to Frequency Response Frequency Response of LTI Systems Introduction I We have demonstrated that for linear, time-invariant systems I the output signal y[n] I is the convolution of the input signal x[n] and the impulse response h[n]. y[n]=x[n] h[n] ⇤ = ÂM h[k] x[n k] k=0 · − I Question: Find the output signal y[n] when the input signal is x[n]=A exp(j(2pfnˆ + f). ©2016, B.-P. Paris ECE 201: Intro to Signal Analysis 235 Introduction to Frequency Response Frequency Response of LTI Systems Response to a Complex Exponential I Problem: Find the output signal y[n] when the input signal is x[n]=A exp(j(2pfnˆ + f). I Output y[n] is convolution of input and impulse response y[n]=x[n] h[n] ⇤ = ÂM h[k] x[n k] k=0 · − = ÂM h[k] A exp(j(2pfˆ(n k)+f) k=0 · − = A exp(j(2pfnˆ + f) ÂM h[k] exp( j2pfkˆ ) · k=0 · − = A exp(j(2pfnˆ + f) H(fˆ) · I The term M H(fˆ)= Â h[k] exp( j2pfkˆ ) k=0 · − is called the Frequency Response of the system. ©2016, B.-P. Paris ECE 201: Intro to Signal Analysis 236 Introduction to Frequency Response Frequency Response of LTI Systems Interpreting the Frequency Response The Frequency Response of an LTI system with impulse response h[n] is M H(fˆ)= Â h[k] exp( j2pfkˆ ) k=0 · − I Observations: I The response of a LTI system to a complex exponential signal is a complex exponential signal of the same frequency. I Complex exponentials are eigenfunctions of LTI systems. I When x[n]=A exp(j(2pfnˆ + f), then y[n]=x[n] H(fˆ). · I This is true only for exponential input signals, including complex exponentials! ©2016, B.-P. Paris ECE 201: Intro to Signal Analysis 237 Introduction to Frequency Response Frequency Response of LTI Systems Interpreting the Frequency Response I Observations: I H(fˆ) is best interpreted in polar coordinates: ˆ H(fˆ)= H(fˆ) ej\H(f ). | |· I Then, for x[n]=A exp(j(2pfnˆ + f) y[n]=x[n] H(fˆ) · ˆ = A exp(j(2pfnˆ + f) H(fˆ) ej\H(f ) ·| |· =(A H(fˆ) ) exp(j(2pfnˆ + f + H(fˆ)) ·| | · \ I The amplitude of the resulting complex exponential is the product A H(fˆ) . ·| | I Therefore, H(fˆ) is called the gain of the system. | | I The phase of the resulting complex exponential is the sum f + \H(fˆ). I \H(fˆ) is called the phase of the system. ©2016, B.-P. Paris ECE 201: Intro to Signal Analysis 238 Introduction to Frequency Response Frequency Response of LTI Systems Example I Let h[n]= 1, 2, 1 . { − } I Then, H(fˆ)=Â2 h[k] exp( j2pfkˆ ) k=0 · − = 1 2 exp( j2pfˆ)+1 exp( j2pfˆ2) − · − · − = exp( j2pfˆ) (exp(j2pfˆ) 2 + exp( j2pfˆ)) − · − − = exp( j2pfˆ) (2 cos(2pfˆ) 2). − · − I Gain: H(fˆ) = (2 cos(2pfˆ) 2 | | | − | ©2016, B.-P. Paris ECE 201: Intro to Signal Analysis 239 Introduction to Frequency Response Frequency Response of LTI Systems Example 4 3 )| d 2 |H(f 1 0 −0.5 0 0.5 Frequency (f ) d 4 ) d 2 0 Phase of H(f −2 −4 −0.5 0 0.5 Frequency (f ) ©2016, B.-P. Paris ECE 201:d Intro to Signal Analysis 240 Introduction to Frequency Response Frequency Response of LTI Systems Example I The filter with impulse response h[n]= 1, 2, 1 is a high-pass filter. { − } I It rejects sinusoids with frequencies near fˆ = 0, ˆ 1 I and passes sinusoids with frequencies near f = 2 I Note how the function of this system is much easier to describe in terms of the frequency response H(fˆ) than in terms of the impulse response h[n]. I Question: Find the output signal when input equals x[n]=2 exp(j2p1/4n p/2). − I Solution: 1 1 1 jp/2 jp/2 H( )=exp( j2p ) (2 cos(2p ) 2)= 2e− = 2e . 4 − 4 · 4 − − Thus, y[n]=2ejp/2 x[n]=4 exp(j2pn/4). · ©2016, B.-P. Paris ECE 201: Intro to Signal Analysis 241 Introduction to Frequency Response Frequency Response of LTI Systems Exercise 1. Find the Frequency Response H(fˆ) for the LTI system with impulse response h[n]= 1, 1, 1, 1 . { − − } 2. Find the output for the input signal x[n]=2 exp(j(2pn/3 p/4)). − ©2016, B.-P. Paris ECE 201: Intro to Signal Analysis 242 Introduction to Frequency Response Frequency Response of LTI Systems Computing Frequency Response in MATLAB function HH = FreqResp( hh, ff ) % FreqResp - compute frequency response of LTI system % % inputs: % hh - vector of impulse repsonse coefficients % ff - vector of frequencies at which to evaluate frequency response % % output: % HH - frequency response at frequencies in ff. % % Syntax: % HH = FreqResp( hh, ff ) HH = zeros( size(ff) ); for kk = 1:length(hh) HH = HH + hh(kk)*exp(-j*2*pi*(kk-1)*ff); end ©2016, B.-P. Paris ECE 201: Intro to Signal Analysis 243 Complex Numbers Part VII Appendix: Complex Numbers and Complex Algebra ©2016, B.-P. Paris ECE 201: Intro to Signal Analysis 244.

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