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113_1_chapter14 - Chapter 14 ALL-PASS AND MINIMUM-PHASE...

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Unformatted text preview: Chapter 14 ALL-PASS AND MINIMUM-PHASE SYSTEMS Copyright c 1996 by Ali H. Sayed. All rights reserved. These notes are distributed only to the students attending the undergraduate DSP course EE113 in the Electrical Engineering Department at UCLA. The notes cannot be reproduced without written consent from the instructor: Prof. A. H. Sayed, Electrical Engineering Department, UCLA, CA 90095, [email protected] We now use the frequency characterization of LTI systems in order to provide further insight into their behavior, and in order to introduce two important subclasses of stable LTI systems: minimum-phase systems and all-pass systems. Group Delay The group delay (measured in units of time) of a stable LTI system is defined as the negative derivative of its phase response with respect to ω : τ (ω ) = − d dω H (ejω ) Clearly, a system with affine phase, i.e., with a phase of the form aω + b for some a and b, has a constant group delay that is equal to −a. For example, Consider an LTI system whose impulse response h(n) is the rectangular pulse 0≤n≤L−1 1 h(n) = 0 otherwise Its frequency response is the DTFT of h(n), which we already know is given by L ω=0 jω H (e ) = −jω (L−1) sin(ω L ) e 2 otherwise . sin ω2 (2) Figure 14.1 shows the magnitude and phase plots of H (eω ) for L = 5. Observe that the phase of H (ejω ) is −ω (L − 1)/2 up to a correction factor that depends on the sign of the ratio sin(ωL/2)/ sin(ω/2). No correction is needed when the ratio is positive, and a correction of ±π is used when the sign is negative. Whether we use π or −π in the latter case is not relevant except to guarantee that the resulting value for the phase lies between [−π, π ] in accordance with our convention for plotting phase graphs. 146 147 magnitude of DTFT 5 4 3 0 2 −1 1 0 −4 −2 −2 0 2 4 −3 −4 −2 0 sin(ω L/2)/sin(ω/2) 5 4 3 2 1 0 −1 −2 −4 −2 0 2 4 2 4 3 2 1 phase of DTFT Figure 14.1. Plots of the magnitude and phase (in radians) of the DTFT of a rectangular pulse sequence with L = 5 over [−π, π ]. We can therefore write the phase of H (ejω ) more explicitly as H (ejω ) = −ω (L − 1) ± πI (ω ) 2 From the plot of sin(ωL/2)/ sin(ω/2) we can see that I (ω ) is zero everywhere except when the ratio is negative, where it will have the form of two rectangular pulses. Therefore, its derivative with respect to ω is zero. It then follows that the group delay of the system under consideration is L−1 τ (ω ) = = constant. 2 Linear Phase Systems The group delay of a stable LTI system is a function of the frequency ω . Its value at ωo provides a measure of the time delay that a signal component of frequency ω o undergoes as it passes through the system. This interpretation is obvious in the case of systems with linear phase characteristics as we first explain. Consider again the above example of a rectangular pulse h(n). It corresponds to a stable LTI system since h(n) is absolutely summable. The group delay was found to be constant sin(ω L ) and equal to (L − 1)/2. Choose any frequency ωo within an interval where the sign of sin ω2 (2) is positive. Then ejωo n −→ |H (ejωo )|e[jωo n+ H (ejωo )] where I (ω ) is an indicator function defined as sin(ω L ) 0 if sin ω2 ≥ 0 (2) I (ω ) = 1 otherwise = |H (ejωo )|ejωo (n− L−1 2 ). This shows that the sequence at the output is a scaled and delayed version of the input sequence. The amount of the delay is (L − 1)/2. 148 All-Pass and Minimum-Phase Systems Chapter 14 When the group delay of a system is not constant, then exponential sequences of different frequencies will undergo different delays. In particular, if an input sequence is a combination of two exponential sequences, say x(n) = ejωo n + ejω1 n then the response will be a combination of both sequences y (n) = |H (ejωo )|e[jωo n+ H (ejωo )] + |H (ejω1 )|e[jω1 n+ H (ejω1 )] . Consider the case in which it happens that |H (ejωo )| = |H (ejω1 )| = A, say. Then we can rewrite y (n) in the form y (n) = A e[jωo n+ H (ejωo )] + e[jω1 n+ H (ejω1 )] . If the group delay were a constant, then both exponential sequences would be delayed by the same amount and the output sequence will consequently be a delayed version of the input. On the other hand, when the group delay is dependent on ω , then both exponential components will be delayed differently and the output sequence will be a distorted version (not simply a delayed version) of the input sequence. Hence, we see that ideally it is preferable for a system to have a constant group delay. Nonlinear Phase Systems When a stable LTI system has nonlinear phase characteristics, we can approximate its phase variation around a given ωo by a linear relation of the form H (ejω ) ≈ −τ (ωo )ω − θ, for ω close enough to ωo . Hence, we again see that for angular frequencies close enough to ωo , the group delay can be interpreted as the delay to be expected in time when exponential sequences passes through the system. Indeed, we can write ejωo n → ≈ |H (ejωo )|ejωo n ej (−τ (ωo )ωo −θ) ≈ |H (ejωo )|e−jθ ejωo (n−τ (ωo )) . This shows that the original sequence ejωo n will be delayed by τ (ωo ), its magnitude will be scaled by |H (ejωo )|, and the output sequence will be further multiplied by the constant complex factor e−jθ . All-Pass Systems An all-pass system will be defined as a causal and stable LTI system that has unit magnitude response over the entire frequency range [−π, π ]. The condition of causality means that its impulse response sequence is a right-sided sequence so that the ROC of its z -transform is the outside of a circular region. Now, by the assumption of stability, we know that the ROC must include the unit circle. Therefore, the poles of such an all-pass system have to lie inside the unit circle. The simplest examples of an all-pass system are H (z ) = 1 or H (z ) = z −k (i.e., pure delays). More generally, a first-order all-pass rational transfer function takes the form H (z ) = 1 − a∗ z z −1 − a∗ = , −1 1 − az z−a |a| < 1. 149 That is, it has a single pole inside the unit circle (at the point a) and a single zero outside the unit circle (at the point 1/a∗ ). The magnitude response of the above first-order system can be easily seen to be unity. Indeed e−jω − a∗ e−jω − a∗ 1 − a∗ ejω = ejω jω = =1 |H (ejω )| = jω − a e e −a ejω − a since the expressions in the numerator and denominator are conjugates of each other. The phase response, on the other hand, can be verified to be H (ejω ) = −ω − 2 arctan r sin(ω − θ) 1 − r cos(ω − θ) where we introduced the polar representation for a, a = rejθ . The associated group delay is obtained by differentiation, which leads to τ (ω ) = 1 − r2 . 1 + r 2 − 2r cos(ω − θ) Now since r < 1, we find that the group delay of an all-pass section is necessarily nonnegative at all ω . The transfer functions of rational all-pass systems of higher orders have the form H (z ) = where A(z ) is a polynomial in z , A(z ) = z N + α1 z N −1 + α2 z N −2 + . . . + αN with roots inside the unit circle, and where A# (z ) denotes the conjugate reversal polynomial that is defined as A# (z ) = z N A 1 z∗ ∗ ∗ ∗ ∗ = αN z N + . . . + α2 z 2 + α1 z + 1. A# (z ) A(z ) That is, the coefficients of A(z ) are conjugated and reversed in order. It is an easy exercise to verify that such transfer functions have unit magnitude response. Indeed, H (ejω ) = ejωN ∗ ∗ ∗ ∗ e−jωN + α1 e−j (N −1)ω + . . . + αN 1 + α1 ejω + . . . + αN ejN ω = ejωN jω (N −1) + . . . + α jωN + α ejω (N −1) + . . . + α + α1 e e N 1 N which allows us to conclude that |H (ejω )| = 1 for all ω . ∗ Note also the useful property that if A(z ) has a zero at zo then A# (z ) has a zero at 1/zo . In particular, since all the zeros of A(z ) are assumed to be inside the unit circle, then the zeros of A# (z ) will be outside the unit circle. Minimum Phase Systems A rational minimum phase system will be defined as causal LTI system with rational transfer function whose poles and zeros are all inside the unit circle. The condition of causality means that its impulse response sequence is a right-sided sequence so that the ROC of the transfer function is the outside of a circular region. Now 150 All-Pass and Minimum-Phase Systems Chapter 14 since the poles are, by definition, inside the unit circle, it follows that the ROC must include the unit circle so that a rational minimum-phase system is necessarily BIBO stable. Such a system also has a stable inverse since its zeros are also inside the unit circle. For example, 1 z−2 H (z ) = z−1 4 is minimum phase, while H (z ) = is not. z−2 z−1 4 A Fundamental Decomposition Not every stable causal system is all-pass. Also, not every stable causal system is minimum phase. But every stable causal system can be expressed as the product of a minimum phase system and an all-pass system. Indeed, let H (z ) = N (z ) D(z ) denote an arbitrary proper transfer function of a causal stable system. We can factor N (z ) as the product of two polynomials, N (z ) = N1 (z )N2 (z ), with N1 (z ) having all its zeros inside the unit circle and N2 (z ) having all its zeros outside the unit circle. Then we can write # N1 (z )N2 (z ) N2 (z ) H (z ) = · . # D(z ) N2 (z ) minimum phase More compactly, we write H (z ) = Hmin (z ) · Hap (z ) This also shows that H (z ) and Hmin (z ) have the same magnitude response. Moreover, τH (ω ) = τmin (ω ) + τap (ω ) . But since, for any all-pass system τap (ω ) ≥ 0, we conclude that among all systems with the same magnitude response, the minimum phase system is the one with the smallest group delay! all-pass 151 PROBLEMS 152 All-Pass and Minimum-Phase Systems Chapter 14 Problem 14.1: The DTFT of a sequence x(n) has the triangular form shown in the figure below over the interval [−π, π ]. The sequence x(n) is transmitted over a channel (or system) that consists of a series cascade of three relaxed sub-channels (or sub-systems): an FIR system with transfer 1 function z −2 − 2 z −1 , an LTI system with frequency response H (ejω ), and a single pole IIR system. The output of the last subsystem is further modulated by the sequence (−1)n in order to generate y (n). H (ejω ) 1 1/2 x(n) z −2 y (n) − 1 −1 z 2 −π −π 2 π 2 π z −1 (−1)n 1/2 X (ejω ) ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¡ ¡ ¡ ¡ 2 ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ Figure 14.2. A sequence x(n) with DTFT X (ejω ) is transmitted over the channel depicted in the figure for Prob. 14.1. (a) Plot |Y (ejω )| over [−π, π ]. Is y (n) a real sequence? [Hint. Recall the form of a first-order all-pass filter.] (b) How would your answer to part (a) change if the roles of H (ejω ) and X (ejω ) are interchanged? (c) How would your answer to part (a) change if the modulator is moved from the end to the beginning of the channel? That is, x(n) is first modulated by (−1)n before being fed into the cascade of three sub-channels. [The output of the last sub-channel now becomes y (n).] (d) Determine the energy of the impulse response sequence of the overall system shown in the figure. (e) Is the system time-invariant? linear? causal? (f) Give an example of a DTFT plot, X (ejω ), that would result in a zero output sequence {y (n)}. Problem 14.2: Consider the block diagram shown in the figure below, where the circles denote adders, and where the transfer functions H (z ) and G(z ) are given by H (z ) = 1 z− 1 2 These transfer functions denote stable and causal LTI systems. Let {Y (ejω ), X (ejω ), E (ejω )} denote the DTFTs of the signals indicated in the figure. Let also H (ejω ) and G(ejω ) denote the frequency responses of the above systems. ¡ ¡ ¡ −π ¡ ¡ ¡ ¡ π , G(z ) = 1 − ¢ £ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ £ ¢ ω 1 −1 z. 2 152 154 Figure 14.3. A block diagram for Prob. 14.2. All-Pass and Minimum-Phase Systems Chapter 14 All-Pass and Minimum-Phase Systems Chapter 14 e(n) x(n) H (z ) G(z ) y (n) (a) The DTFTs of the signals {x(n), e(n)} are shown in the figure below. Compute the energies of these sequences. Compute also the signal-to-noise energy ratio at the input of the system, (b) Show that which is defined as H (ejω )G(ejω ) 1 ∆ Y (ejω ) = X (SNR+ energy of xejω) . = X (ejω ) + E (ejω ) ejω ) = E ( (n ) 1 − G(ejω ) 1 − G(ejω ) energy of e(n) Figure 14.4. Signals x(n) and e(n) for Prob. ??. X (ejω ) E (ejω ) Figure 14.3. A block diagram for Prob. 14.2. (a) The DTFTs of the signals {x(n), e(n)} are shown in the figure below. Compute the energies of these sequences. Compute also the signal-to-noise energy ratio at the input of the system, which is defined as energy of x(n) . SNR = energy of e(n) where X (ejω ) refers to the contribution of the input signal x(n) at the output, while E (ejω ) refers to the contribution of the interfering signal e(n) at the output. jω X (ejω ) (c) Compute the signal-to-noise energy ratio at the output of theE (e ) which is defined as system, SNR = energy of x (n) . energy of e (n) 2 (d) Assume instead that 1 2 π π . e(n) = sin n+ 4 6 π n, x(n) = cos 3 ω π π π response y (n). π Compute 4 − 8 − the steady-state 4 ss 8 −π 4 π 4 ω Figure 14.4. Signals x(n) and e(n) for Prob. ??. Problem 14.3: Consider the block diagram shown in the figure below where x(n) and y (n) are the input and output sequences, respectively, and H1 and H2 are two LTI systems with frequency responses H1 (ejω ) and H2 (ejω ). In the upper branch, the sequence x(n) is first multiplied by (−1)n , n (b) filtered that 1 , and then multiplied by (−1) . In the lower branch, the sequence x(n) is simply Show by H filtered by H2 . The output sequence y (n) is the sum of the sequences obtained at the outputs of both branches. H (ejω )G(ejω ) 1 ∆ jω jω jω Y ejω ) = X (ejω ) + E ( i.e., Let H (ejω )(denote the frequency response of the overall system, e ) = X (e ) + E (e ) jω jω 1 − G(e ) X jω jω ( jω ) (ejω ) Y (e ) = H (e )XEe(ejω.) 1 − G(e ) for any input-output pair {x(n), y (n)}. jω where X (e ) refers to the contribution of the input signal x(n) at the output, while E (ejω ) refers to the contribution of the interfering signal e(n) at the output. 1. Prove that H (ejω ) is equal to H (e ) = H1 [e ] + H2 (e ) . (c) Compute the signal-to-noise energy ratio at j (ω−π) the output of the system, which is defined as jω jω energy of x (n) SNR = 2. Plot H (ejω ) (both magnitude and phase in the range [−π,.π ]) when H1 is an ideal low-pass energy of e (n) filter with cutoff frequency π/4 and unit magnitude in the passband (including the frequencies (d) Assumethe passband (including the frequencies π/3 and −π/3). Both filters have linear phases in in instead that their passbands with group delays that are equal to 2. π π π n , e(n) = sin n+ . x(n) = cos 3 4 6 Compute the steady-state response yss (n). π/4 and −π/4), while H2 is a low-pass filter with cutoff frequency π/3 and unit magnitude 153 Problem 14.3: Consider the block diagram shown in the figure below where x(n) and y (n) are the input and output sequences, respectively, and H1 and H2 are two LTI systems with frequency responses H1 (ejω ) and H2 (ejω ). 14.5. Block diagram for Prob. ??. x(n) is first multiplied by (−1)n , Figure In the upper branch, the sequence filtered by H1 , and then multiplied by (−1)n . In the lower branch, the sequence x(n) is simply filtered by H2 . The output sequence y (n) is the sum of the sequences obtained at the outputs of both branches. (−1)n X x(n) H1 (ejω ) (−1)n X y (n) H2 (ejω ) Figure 14.5. Block diagram for Prob. 14.3. 3. Assume H1 is the same as above, while H2 is now a high-pass filter with cutoff frequency at π/2, unit magnitude in the pass band, and group delay that is equal to 4. Determine the π steady-state response of the system to x(n) = cos 34 n + π . 3 jω Let H (e ) denote the frequency response of the overall system, i.e., Problem 14.4: The impulse response ofYa(ejω ) = H (ejω )X (ejω ) . system S is given by causal and stable LTI for any input-output pair {x(n), y (n)}) = h(n . 1. Prove that H (ejω ) is equal to Find the impulse response of another causal and stable LTI system L with a zero at the point z = 3 and such that the magnitude responses of both) S and ej (ω−π)identical. How do the phase responses H (ejω = H1 [ L are ] + H2 (ejω ) . compare? 2. Plot H (ejω ) (both magnitude and phase in the range [−π, π ]) when H1 is an ideal low-pass Problem 14.5: with cutoff frequency π/4 causal and stable all-pass passband (including the frequencies filter Consider an elementary and unit magnitude in the function with a real pole at a, π/4 and −π/4), while H2 is a low-passafilter with cutoff frequency π/3 and unit magnitude z −1 − | |< − B (z the in the passband (including ) = frequencies ,π/3aand 1 .π/3). Both filters have linear phases in 1 − az −1 their passbands with group delays that are equal to 2. 1. ArgueAssumeB (ejω )the a monotonically decreasingnow a high-passstarts with B (ej 0 )frequency at 3. that H1 is is same as above, while H2 is function that filter at cutoff = 0 and attains 2, (ejπ )magnitude in is, the change in phase as delay that is 0 to π to −π . π/ B unit = −π . That the pass band, and group ω goes from equal is 4. Determine the steady-state response of the system to x(n) = cos 3π n 4 1 2 n u(n) . + π 3 . Problem 14.4: The impulse response of a causal and stable LTI system S is given by h(n) = 1 2 n u(n) . Find the impulse response of another causal and stable LTI system L with a zero at the point z = 3 and such that the magnitude responses of both S and L are identical. How do the phase responses compare? Problem 14.5: Consider an elementary causal and stable all-pass function with a real pole at a, B (z ) = z −1 − a , 1 − az −1 |a| < 1 . order (M = 2) all-pass function as above. 154 All-Pass and Minimum-Phase Systems Chapter 14 1. Argue that B (ejω ) is a monotonically decreasing function that starts at B (ej 0 ) = 0 and attains B (ejπ ) = −π . That is, the change in phase as ω goes from 0 to π is −π . 2. Now consider an all-pass function of order M with real poles, viz., a product of M elementary sections as follows z −1 − a1 z −1 a2 z −1 − aM A(z ) = ... . −1 1 − a z −1 1 − a1 z 1 − aM z −1 2 3. Consider the filter structure A digital notch filter for Prob.(z ) is chosen as a second Figure 14.6. shown in the figure below, where A ??. order (M = 2) all-pass function as above. Prove that the phase response A(ejω ) is also a monotonically decreasing function that starts at A(ej 0 ) = 0 and attains A(ejπ ) = −M π . That is, the change in phase as ω goes from 0 to π is −M π . In particular, what is the value of A(ejπ )? 1 2 x(n) A(z ) y (n) Figure 14.7. A digital notch filter. Figure 14.6. A digital notch filter for Prob. 14.5. The transferThe transfer function from x(n) to y(n) is denoted by G(G(and and is clearly equal to function from x(n) to y (n) is denoted by z ) z ) is clearly equal to 1 G(z ) = [A(z ) + 1] . 12 G(z ) = [A(z ) + 1] component at ω0 from the input signal x(n), by properly choosing a1 and a2 so as to satisfy A(ejω0 ) = −π .] Problem 14.6: Consider the feedback configuration shown in the figure below. The impulse reProblem response sequence, g (n), of the causal system indicated by the rectangular box is also shown. The 1. Find the transfer function of the discrete-time system indicated by the rectangular box, i.e., value of the sample at the Be sure instant n indicate the corresponding and denoted by α, which can assume find G(z ). time to properly = 1 is an unknown region of convergence. What is the an arbitrary real value of the system? Is it stable? Is it minimum phase? What are the poles of G(z )? Do order contrary to what the figure might suggest. they depend on the value of α? What are the zeros of G(z )? Do they depend on α? 2. Find the transfer function, H (z ), of the feedback system that maps x(n) to y (n). How do the zeros of H (z ) compare to the zeros of G(z )? What about the poles? What is the order of H (z )? Find an α for which H (z ) is unstable and verify your answer. Is there any α for which H (z ) is minimum phase? 3. Find a set of conditions that α should satisfy if H (z ) were to be stable and verify whether a solution α exists. sponse sequence, g (n), of the causal system indicated by the rectangular box is also shown. The value of the sample at the time instant n = 1 is an unknown and denoted by α, which can assume 14.6:arbitrary real value feedback what the figure might suggest. the figure below. The impulse Consider the contrary to configuration shown in an Argue that there should exist an angular frequency 0 < ω0 < π such that G(ej 0 ) = G(ejπ ) = 1 and G(ejω0 ) = 0. [In summary, this filter structure can< ω0 < π suchthe single-frequency G(ejπ ) = 1 Argue that there should exist an angular frequency 0 be used to remove that G(ej 0 ) = ω0 and G(ejω0 ) component at π.] from the input signalstructure can be used a1 and a2 so as to single-frequency = (0. 0[In summary, this filter x(n), by properly choosing to remove the satisfy A ejω ) = − 2 . 155 157 Figure 14.8. A feedback configuration for Prob. ??. g (n) α x(n) 1 3 4 1 n−3 3 u(n − 3) y (n) 2 3 4 n Figure 14.9. A feedback system. Figure 14.7. A feedback configuration for Prob. 14.6. 1. Find the transfer function of the discrete-time system indicated by the rectangular box, i.e., find G(z ). Be sure to properly indicate the corresponding region of convergence. What is the order of the system? Is it stable? Is it minimum phase? What are the poles of G(z )? Do they depend on the value of α? What are the zeros of G(z )? Do they depend on α? 2. Find the transfer function, H (z ), of the feedback system that maps x(n) to y (n). How do the zeros of H (z ) compare to the zeros of G(z )? What about the poles? What is the order of H (z )? Find an α for which H (z ) is unstable and verify your answer. Is there any α for which H (z ) is minimum phase? 3. Find a set of conditions that α should satisfy if H (z ) were to be stable and verify whether a solution α exists. ...
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This note was uploaded on 11/07/2009 for the course EE 113 taught by Professor Walker during the Spring '08 term at UCLA.

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