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Unformatted text preview: Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.341: DiscreteTime Signal Processing Fall 2005 Solutions for Problem Set 2 Issued: Tuesday, September 20 2005. Problem 2.1 (i) Realvalued impulse response: Poles that aren’t real must be in complex conjugate pairs. Zeros that aren’t real must be in complex conjugate pairs. (ii) Finite impulse response: All poles are at the origin. The ROC is the entire zplane, except possibly z = 0. (iii) h [ n ] = h [2 α − n ] where 2 α is an integer: Causality combined with the given symmetry property implies a finitelength h [ n ] that can only be nonzero between time zero and time 2 α . Thus we must have all poles at the origin and at most 2 α zeros. The z transform of h [2 α − n ] is z 2 α H (1 /z ), so any zero of H ( z ) at c negationslash = 0 must be paired with a zero at 1 /c . (iii) Minimum phase: All poles and zeros are inside the unit circle (so that the inverse can be stable and causal). (iv) Allpass: Each pole is paired with a zero at the conjugate reciprocal location. Problem 2.2 (a) False. As a counterexample, consider a filter described by H ( e jω ) = 1 − 1 2 e jω . We know that H ( z ) can describe a causal, stable filter since its single pole is at z = 0. In the vicinity of ω = 0, θ ( ω ) is increasing, so the group delay τ ( ω ) is negative. The group delay evaluated at ω = 0 for this example can be determined using equation (5.67) in OSB: grd [1 − re jθ e jω ] = r 2 − r cos( ω − θ )  1 − re jθ e jω  2 and substituting r = 1 2 and θ = 0. (b) False. Any zerophase FIR filter may be delayed to become causal, and the resulting filter will have the same phase as the delay block which was applied to it. One counterexample is the causal, stable filter described by H ( z ) = 1 + 2 z 1 + z 2 . (c) True. Note that integraldisplay π τ ( ω ) dω = − θ ( ω )  π ω =0 = θ (0) − θ ( π ) . Setting this to 0, we realize that the assertion is true if having minimum phase and all poles and zeros on the real axis implies θ (0) = θ ( π ) . A minimum phase filter with poles and zeros on the real axis will have all poles and zeros between 1 and 1. Factoring H ( e jω ) to find poles and zeros, we notice that for a given pole term 1 1 ae jω , the restriction − 1 < a < 1 yields a phase contribution of 0 when ω = 0 or ω = π . Likewise, a given zero term 1 − be jω gives a phase contribution of 0 when ω = 0 or ω = π as long as − 1 < b < 1. Since all poles and zeros of such a minimum phase filter meet these criteria, θ (0) = θ ( π ) = 0 Therefore, integraltext π τ ( ω ) dω = 0. (Note that integraltext π τ ( ω ) dω = 0 for all real minimum phase filters, even if there are complex poles and zeros.) Problem 2.3 The polezero diagram for the original system is as follows:21.510.5 0.5 1 1.5 21.510.5 0.5 1 1.5 pole at ∞ Original system Re(z) Im(z) (a) One way to carry out the minimumphase and allpass decomposition is as follows. In the(a) One way to carry out the minimumphase and allpass decomposition is as follows....
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This note was uploaded on 10/09/2009 for the course EE 6.341 taught by Professor Xyz during the Fall '05 term at Massasoit.
 Fall '05
 xyz
 Signal Processing

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