Lecture 13: Pole/Zero Diagrams and All Pass Systems

Transcription

1 EE518 Digital Signal Processing University of Washington Autumn 2001 Dept. of Electrical Engineering Lecture 13: Pole/Zero Diagrams and All Pass Systems No4, 2001 Prof: J. Bilmes TA: Mingzhou Song Ex: Group Delay (continuous time) Instantaneous frequency is f x (t) d dt arg{x(t)} = d dt (t) A chirp signal has instantaneous frequency f x (t) = ct j2 ct2 x(t) = e 2 = A(t)e j(t) units of Hz (or radians, etc.) so instantaneous frequency for a linear FM chirp is a linear function of time t. The frequency is increasing linearly with time, with slope c. Note: if there are more than one simultaneous frequency components, at a particular time, then it is hard to analyze this way. But OK for FM signals. x(t) = Acos((t)) so d(t) dt = c + kψ(t) t (t) = c t + kψ(t)dt If this was impulse response of an LTI, we can think of this as, at each time, there is a lag in frequency. δ(t) has all frequency at a given time. Group delay is dual of instantaneous frequency. t x ( f ) d arg{x( f )} d f For the chirp signal, Therefore X( f ) = A( f )e j( f ) X( f ) = 1 c exp[ j( 1 4 f 2 /c)] arg{x( f )} = 4 f 2 /c and d arg{x( f )} = f /k d f units of time (seconds, ), and k is some constant. Note, if this was impulse response of an LTI, this can be seen as, at each frequency, what was the time delay of the resulting system. 13-1

2 Frequency Response of Single Zero and Pole Systems When there is only single zero and pole, we have either or where is a general complex number. H(z) = 1 1 az 1 H(z) = 1 az 1 a = re j Rational z-transforms and FT s For we might want the magnitude squared or plot in db scale H(e j ) = b 0 M (1 c ke j ) N (1 d ke j ) H(e j )H (e j ) = H(e j ) 2 20log 10 H(e j b 0 ) = 20log 10 + Also, system response is additive in log domain, i.e., M 20log 10 1 c k e j N 20log 10 Y (e j ) = 20log 10 H(e j ) + 20log 10 X(e j ) 20log 10 1 d k e j which is important and used for Homomorphic processing (useful in automatic speech recognition.) Phases add in this case H(e j ) = b 0 + zeros add, poles subtract with the phase. A note on arg and ARG. M which is only the important part that can have an effect. arg is the continuous phase. when For some applications, r() takes on integer values. Note Now we look at the individual factors: 1 re j e j. [1 c k e j ] < ARG[H(e j )] arg[h(e j )] = () H(e j ) = H(e j ) e j() N arg[h(e j )] = ARG[H(e j )] + 2r() ARG[H(e j )] = tan 1 H I(e j ) H R (e j ) [1 d k e j ] 1 re j e j 2 = (1 re j e j )(1 re j e j ) = 1 + r 2 2r cos( )

3 13-3 Also, and ARG[1 re j e j ] = tan 1 r sin( ) 1 r cos( ) grd[1 re j e j ] = r2 r cos( ) 1 + r 2 2r cos( ) For a system of single zero/pole and r 0.9, its magnitude and phase responses and group delay is shown in Fig log 10 1 re j e j 0 phase in radian group delay (in samples) Figure 13.1: Frequency domain responses of a single zero/pole system. Geometrical view of frequency response as viewed from the poles, zeros on the z-plane Consider a single zero system H(z) = 1 re j z 1 z re j = z The geometric view of H(z) is shown in Fig < r < 1 So Then =e j =re j = = e j re j 1 re j e j = e j re j e j = = 1 re j e j = = since = 1. The length of tells us the absolute value of the magnitude of the response. Note:

4 13-4 Figure 13.2: Geometric view. Think about as rubber band at the zero, and at z = 0. Streched to current position on unit circle. 1. This is not even at = 0, when time signal is complex. If = 0, magnitude would be even and time signal would be real. 2. is called a zero vector. If we had a pole, we would have a pole vector. We can generalize. To get magnitude response at a given : multiply all zero vector lengths multiply all pole vector lengths divide the result (i.e., divide by all pole vectors) in db domain, add log zero vector lengths and subtract log pole vector lengths. What about phase? As shown in Fig 13.3, Figure 13.3: Geometric view of phase.

5 13-5 (1 re j e j ) = (e j re j ) e j = = Consider an example as shown in Fig 13.4, where =, r < 1 So, as increases, will get smaller until =, where will be the minimum and after >, gets larger Figure 13.4: An example. again. In general, zero at re j =. From that zero will be the smallest at = (that is the closest zero to the unit circle at that point.) For phase, for <, we have >, so that phase is negative. at =, we have =, so zero phase there (at minimum magnitude.) at >, we have >, so positive phase. This works for other s as well, i.e., phase sign shift and minimum amplitude magnitude at =. When r 1, what happens? When r 1, it is possible for = = 0. Hence magnitude is zero at =, as expected. (zero on the unit circle should give zero magnitude). What about the phase? At =, the phase is 3. When <, the phase is shown in Fig When >, the phase is shown in Fig So is going to flip sign instantaneously when = and = + δ. This gives what is shown in Fig 13.7.

6 13-6 Figure 13.5: <. Figure 13.6: >. radian 1 1 Figure 13.7: Phase sign flip when r > 1. When r 1, the phase is smooth.

7 13-7 Relationship between magnitude and phase For rational system function, there are constraints on magnitude and phase (unlike in general system) How to tell? H(e j ) 2 = H(e j )H (e j ) = H(z)H (1/z ) z=e j If Note: zero at z = c k = c k e j c k. H(z) = b 0 M (1 c kz 1 ) N (1 d kz 1 ) H (1/z ) = b 0 M (1 c k z) N (1 d k z) Note: zero at z = 1/c k = 1 c k e j c k. Poles and zeros reflect across unit circle. Define C(z) = ( b0 ) 2 M (1 c k z 1 )(1 c k z) N (1 d kz 1 )(1 d k z) For C(z), pole/zero at z = a implies pole/zero at z = 1/a. But C(z) could come from other H(z) with same magnitude response but different phase response. The key is: given a magnitude response of a linear constant coefficient difference equation system, there exist some finite number of possible phase responses, and correspondingly to different systems (some might be stable/causal, while others do not) All Pass System Consider H(z) = z 1 a 1 az 1 with pole at z = a and zero at z = 1/a, where a = re j. Then H(e j ) = e j a 1 ae j = e j 1 a e j 1 ae j so H(e j ) = 1