Discrete-Time Signals and Systems. Frequency Domain Analysis of LTI Systems. The Frequency Response Function. The Frequency Response Function

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1 Discrete-Time Signals and s Frequency Domain Analysis of LTI s Dr. Deepa Kundur University of Toronto Reference: Sections 5., of John G. Proakis and Dimitris G. Manolakis, Digital Signal Processing: Principles, Algorithms, and Applications, 4th edition, Dr. Deepa Kundur (University of Toronto) Frequency Domain Analysis of LTI s / 39 Dr. Deepa Kundur (University of Toronto) Frequency Domain Analysis of LTI s 2 / 39 The Frequency Response Function 5. Frequency-Domain Characteristics of LTI s Recall for an LTI system: y(n) = h(n) x(n). Suppose we inject a complex exponential into the LTI system: y(n) = k= x(n) = Ae jωn h(k)x(n k) Note: we consider x(n) to be comprised of a pure frequency of ω rad/s The Frequency Response Function y(n) = = k= k= = Ae jωn 5. Frequency-Domain Characteristics of LTI s h(k)ae jω(n k) h(k)ae jωn e jωk [ = Ae jωn H(ω) k= h(k)e jωk ] } {{ } H(ω)=DTFT {h(n)} Thus, y(n) = H(ω)x(n) when x(n) is a pure frequency. Dr. Deepa Kundur (University of Toronto) Frequency Domain Analysis of LTI s 3 / 39 Dr. Deepa Kundur (University of Toronto) Frequency Domain Analysis of LTI s 4 / 39

2 The Frequency Response Function 5. Frequency-Domain Characteristics of LTI s LTI Eigenfunction 5. Frequency-Domain Characteristics of LTI s y(n) = H(ω)x(n) output = scaled input A v = λ v Eigenfunction of a system: an input signal that produces an output that differs from the input by a constant multiplicative factor multiplicative factor is called the eigenvalue Therefore, a signal of the form Ae jωn is an eigenfunction of an LTI system. Dr. Deepa Kundur (University of Toronto) Frequency Domain Analysis of LTI s 5 / 39 Dr. Deepa Kundur (University of Toronto) Frequency Domain Analysis of LTI s 6 / 39 LTI Eigenfunction 5. Frequency-Domain Characteristics of LTI s Magnitude and Phase of H(ω) 5. Frequency-Domain Characteristics of LTI s Implications: An LTI system can only change the amplitude and phase of a sinusoidal signal. H(ω) = H(ω) e jθ(ω) An LTI system with inputs comprised of frequencies from set Ω0 cannot produce an output signal with frequencies in the set Ω c 0 (i.e., the complement set of Ω 0 ) H(ω) system gain for freq ω H(ω) = Θ(ω) phase shift for freq ω Dr. Deepa Kundur (University of Toronto) Frequency Domain Analysis of LTI s 7 / 39 Dr. Deepa Kundur (University of Toronto) Frequency Domain Analysis of LTI s 8 / 39

3 5. Frequency-Domain Characteristics of LTI s 5. Frequency-Domain Characteristics of LTI s Example: Example: Determine the magnitude and phase of H(ω) for the three-point moving average (MA) system y(n) = [x(n + ) + x(n) + x(n )] 3 By inspection, h(n) = δ(n + ) + δ(n) + δ(n ). Therefore, H(ω) = n= h(n)e jωn = n= 3 e jωn What is the phase of H(ω) = ( + 2 cos(ω))? 3 See Figure 5.. of text. H(ω) = + 2 cos(ω) { ω 2π Θ(ω) = 3 π 2π ω < π 3 = 3 [ejω + + e jω ] = ( + 2 cos(ω)) 3 Dr. Deepa Kundur (University of Toronto) Frequency Domain Analysis of LTI s 9 / 39 Dr. Deepa Kundur (University of Toronto) Frequency Domain Analysis of LTI s 0 / Frequency Response of LTI s Frequency Response of LTI s 5.2 Frequency Response of LTI s Frequency Response of LTI s z-domain H(z) system function Y (z) = X (z)h(z) ω-domain z=e jω = H(ω) z=ejω = frequency response z=e jω = Y (ω) = X (ω)h(ω) If H(z) converges on the unit circle, then we can obtain the frequency response by letting z = e jω : H(ω) = H(z) z=e jωn = = for rational system functions. M k=0 b ke jωk n= + N k= a ke jωk h(n)e jωn Dr. Deepa Kundur (University of Toronto) Frequency Domain Analysis of LTI s / 39 Dr. Deepa Kundur (University of Toronto) Frequency Domain Analysis of LTI s 2 / 39

4 LTI s as Frequency-Selective Filters Ideal Filters Filter: device that discriminates, according to some attribute of the input, what passes through it For LTI systems, given Y (ω) = X (ω)h(ω) H(ω) acts as a kind of weighting function or spectral shaping function of the different frequency components of the signal LTI system Filter Classification: lowpass highpass bandpass bandstop allpass See Figure 5.4. of text. Dr. Deepa Kundur (University of Toronto) Frequency Domain Analysis of LTI s 3 / 39 Dr. Deepa Kundur (University of Toronto) Frequency Domain Analysis of LTI s 4 / 39 Ideal Filters Ideal Filters Suppose H(ω) = Ce jωn 0 for all ω: Common characteristics: unity (flat) frequency response magnitude in passband and zero frequency response in stopband linear phase; for constants C and n0 δ(n) δ(n n 0 ) C δ(n n 0 ) F F e jωn 0 F C e jωn 0 = Ce jωn 0 H(ω) = { Ce jωn 0 ω < ω < ω 2 0 otherwise Therefore, h(n) = Cδ(n n 0 ) and: y(n) = x(n) h(n) = x(n) Cδ(n n 0 ) = Cx(n n 0 ) Dr. Deepa Kundur (University of Toronto) Frequency Domain Analysis of LTI s 5 / 39 Dr. Deepa Kundur (University of Toronto) Frequency Domain Analysis of LTI s 6 / 39

Deepa Kundur (University of Toronto) Frequency Domain Analysis of LTI s 7 / 39 Dr.

5 Ideal Filters Phase versus Magnitude Therefore for ideal linear phase filters: H(ω) = { Ce jωn 0 ω < ω < ω 2 0 otherwise What s more important? signal components in stopband are annihilated signal components in passband are shifted (and scaled by passband gain which is unity) Dr. Deepa Kundur (University of Toronto) Frequency Domain Analysis of LTI s 7 / 39 Dr. Deepa Kundur (University of Toronto) Frequency Domain Analysis of LTI s 8 / 39 Dr. Deepa Kundur (University of Toronto) Frequency Domain Analysis of LTI s 9 / 39 Dr. Deepa Kundur (University of Toronto) Frequency Domain Analysis of LTI s 20 / 39

Dr. Deepa Kundur (University of Toronto) Frequency Domain Analysis of LTI s 2 / 39 Dr. Deepa Kundur (University of Toronto) Frequency Domain Analysis of LTI s 22 / 39 Why Invert?

reverse intersymbol interference in data symbols in telecommunications applications to improve error rate; called equalization correct blurring effects in biomedical imaging applications for more

6 Dr. Deepa Kundur (University of Toronto) Frequency Domain Analysis of LTI s 2 / 39 Dr. Deepa Kundur (University of Toronto) Frequency Domain Analysis of LTI s 22 / 39 Why Invert? There is a fundamental necessity in engineering applications to undo the unwanted processing of a signal. reverse intersymbol interference in data symbols in telecommunications applications to improve error rate; called equalization correct blurring effects in biomedical imaging applications for more accurate diagnosis; called restoration/enhancement increase signal resolution in reflection seismology for improved geologic interpretation; called deconvolution Dr. Deepa Kundur (University of Toronto) Frequency Domain Analysis of LTI s 23 / 39 Dr. Deepa Kundur (University of Toronto) Frequency Domain Analysis of LTI s 24 / 39

7 Invertibility of s Invertibility of LTI s Invertible system: there is a one-to-one correspondence between its input and output signals the one-to-one nature allows the process of reversing the transformation between input and output; suppose y(n) = T [x(n)] where T is one-to-one w(n) = T [y(n)] = T {T [x(n)]} = x(n) Identity Direct Impluse response = Inverse Identity Direct Inverse Direct Inverse Dr. Deepa Kundur (University of Toronto) Frequency Domain Analysis of LTI s 25 / 39 Dr. Deepa Kundur (University of Toronto) Frequency Domain Analysis of LTI s 26 / 39 Invertibility of LTI s Invertibility of Rational LTI s Therefore, h(n) h I (n) = δ(n) For a given h(n), how do we find h I (n)? Consider the z domain H(z)H I (z) = H I (z) = H(z) Suppose, H(z) is rational: H(z) = A(z) B(z) H I (z) = B(z) A(z) poles of H(z) = zeros of H I (z) zeros of H(z) = poles of H I (z) Dr. Deepa Kundur (University of Toronto) Frequency Domain Analysis of LTI s 27 / 39 Dr. Deepa Kundur (University of Toronto) Frequency Domain Analysis of LTI s 28 / 39

8 Example Determine the inverse system of the system with impulse response h(n) = ( 2 )n u(n). Common Transform Pairs Signal, x(n) z-transform, X (z) ROC δ(n) All z 2 u(n) z z > 3 a n u(n) az 4 na n u(n) az ( az ) 2 5 a n u( n ) az 6 na n az u( n ) 7 (cos(ω 0 n))u(n) 8 (sin(ω 0 n))u(n) 9 (a n cos(ω 0 n)u(n) 0 (a n sin(ω 0 n)u(n) ( az ) 2 z cos ω 0 2z cos ω 0 +z 2 z > z sin ω 0 2z cos ω 0 +z 2 z > az cos ω 0 2az cos ω 0 +a 2 z 2 az sin ω 0 2az cos ω 0 +a 2 z 2 Dr. Deepa Kundur (University of Toronto) Frequency Domain Analysis of LTI s 29 / 39 Dr. Deepa Kundur (University of Toronto) Frequency Domain Analysis of LTI s 30 / 39 Example Determine the inverse system of the system with impulse response h(n) = ( 2 )n u(n). H(z) =, ROC: z > ; note: direct system is causal + 2 z 2 stable. Therefore, By inspection, H I (z) = H(z) = 2 z h I (n) = δ(n) δ(n ) 2 Another Example Determine the inverse system of the system with impulse response h(n) = δ(n) δ(n ). 2 H(z) = H I (z) = n= = 2 z 2 z h(n)z n = n= [δ(n) n δ(n )]z 2 Is the inverse system stable? causal? Dr. Deepa Kundur (University of Toronto) Frequency Domain Analysis of LTI s 3 / 39 Dr. Deepa Kundur (University of Toronto) Frequency Domain Analysis of LTI s 32 / 39

9 Common Transform Pairs Common Transform Pairs Signal, x(n) z-transform, X (z) ROC δ(n) All z 2 u(n) z z > 3 a n u(n) az 4 na n u(n) az ( az ) 2 5 a n u( n ) az 6 na n az u( n ) 7 (cos(ω 0 n))u(n) 8 (sin(ω 0 n))u(n) 9 (a n cos(ω 0 n)u(n) 0 (a n sin(ω 0 n)u(n) ( az ) 2 z cos ω 0 2z cos ω 0 +z 2 z > z sin ω 0 2z cos ω 0 +z 2 z > az cos ω 0 2az cos ω 0 +a 2 z 2 az sin ω 0 2az cos ω 0 +a 2 z 2 Signal, x(n) z-transform, X (z) ROC δ(n) All z 2 u(n) z z > 3 a n u(n) az 4 na n u(n) az ( az ) 2 5 a n u( n ) az 6 na n az u( n ) 7 (cos(ω 0 n))u(n) 8 (sin(ω 0 n))u(n) 9 (a n cos(ω 0 n)u(n) 0 (a n sin(ω 0 n)u(n) ( az ) 2 z cos ω 0 2z cos ω 0 +z 2 z > z sin ω 0 2z cos ω 0 +z 2 z > az cos ω 0 2az cos ω 0 +a 2 z 2 az sin ω 0 2az cos ω 0 +a 2 z 2 Dr. Deepa Kundur (University of Toronto) Frequency Domain Analysis of LTI s 33 / 39 Dr. Deepa Kundur (University of Toronto) Frequency Domain Analysis of LTI s 34 / 39 Another Example There are two possibilities for inverses: Causal + stable inverse: h I (n) = ( ) n u(n) 2 Homomorphic Deconvolution The complex cepstrum of a signal x(n) is given by: c x (n) = Z {ln(z{x(n)}} = Z {ln(x (z))} = Z {C x (z)} Anticausal + unstable inverse: ( ) n h I (n) = u( n ) 2 x(n) cepstrum c x (n) Z Z X (z) C x (z) = ln(x (z)) We say, c x (n) is produced via a homomorphic transform of x(n). Dr. Deepa Kundur (University of Toronto) Frequency Domain Analysis of LTI s 35 / 39 Dr. Deepa Kundur (University of Toronto) Frequency Domain Analysis of LTI s 36 / 39

10 Homomorphic Deconvolution Homomorphic Deconvolution Therefore, Y (z) = X (z)h(z) C y (z) = ln Y (z) = ln X (z) + ln H(z) = C x (z) + C h (z) Z {C y (z)} = Z {C x (z)} + Z {C h (z)} c y (n) = c x (n) + c h (n) In many applications, the characteristics of c x (n) and c h (n) are sufficiently distinct that temporal windows can be used to separate them: where: ĉ h (n) = c y (n)ŵ lp (n) ĉ x (n) = c y (n)ŵ hp (n) ŵ lp (n) = { n N 0 n > N convolution in time-domain addition in cepstral domain ŵ hp (n) = { 0 n N n > N Dr. Deepa Kundur (University of Toronto) Frequency Domain Analysis of LTI s 37 / 39 Dr. Deepa Kundur (University of Toronto) Frequency Domain Analysis of LTI s 38 / 39 Homomorphic Deconvolution Obtaining the inverse homomorphic transforms of c h (n) and c x (n) give estimates of h(n) and x(n), respectively. Dr. Deepa Kundur (University of Toronto) Frequency Domain Analysis of LTI s 39 / 39

Discrete-Time Signals and Systems. The z-transform and Its Application. The Direct z-transform. Region of Convergence. Reference: Sections

Discrete-Time Signals and Systems The z-transform and Its Application Dr. Deepa Kundur University of Toronto Reference: Sections 3. - 3.4 of John G. Proakis and Dimitris G. Manolakis, Digital Signal Processing: