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Unformatted text preview: Response From Pole-Zero Location 723 lines connecting Z1> Z2, . .. , Zn t o t he p oint eiwT . Similarly, let d 1, d2, . .. , d n b e t he
l engths a nd 01, O2 , . .. , On be t he angles, respectively, o f t he lines connecting '11> '12,
. .. , 'In t o eiwT . T hen 1m z - plane H [e i wT _
(qei<l>. )(T2ei4>2) . .. (rnei<l>n) - H[zllz=e;wT - bn ( dleiO. )(d2ei02) . .. ( dne iOn )
= b Tl T2 . .. Tn ei[(<I>. +<1>2+"·+<I>n)-(O. +02+"'+On)] n d l d 2··· dn (12.17)
(12.18) T herefore
IH[ejwTlI = b T1T2'" Tn
d 1d2'" d n n
=b (a) p roduct of t he d istances of zeros t o eiwT
p roduct of distances o f poles t o eiwT (12.19a) a nd
L H[e iwT ] = (<1>1
F ig. 1 2.3 ( a) vector r epresentation o f c omplex n umbers (b) vector representation o f
f actors o f H[z]. 1 2.2 + <1>2 + ... <l>n) - (0 1 + O2 + ... + On) = s um of zero angles t o e jwT - sum o f pole angles t o eiwT (b) Frequency Response From Pole-Zero l ocation T he frequency response (amplitude and phase response) of a system are determined by pole-zero locations o f t he t ransfer function H [z]. J ust as in continuoustime systems, it is possible t o d etermine quickly t he a mplitude a nd t he phase response a nd t o obtain physical insight into the filter characteristics of a discrete-time
system by using a graphical technique. Consider t he t ransfer function
We c an c ompute H[z] graphically using t he concepts discussed in Sec. 7.3. T he
d irected line segment from Zi t o Z in t he complex plane (Fig. 12.3a) represents the
complex number z - Zi. T he l ength of this segment is Iz - zil a nd i ts angle with the
horizontal axis is L (z - Zi).
I n filtering applications, t he i nputs are often t he s ampled continuous-time sinusoids. Earlier, we showed t hat a s ampled continuous-time sinusoid cos w t a ppears
as a discrete-time sinusoid cos nk (0, = w T). T he a ppropriate function for computing t he frequency response in such a situation, therefore, is H [e jwT ] (0, = w T).
To compute t he frequency response H [e iwT ] we e valuate H[z] a t z = e jwT . B ut for
, Izi = 1 a nd L z = w T so t hat z = eiwT r epresents a point on the u nit circle
a t a n angle w T w ith the horizontal. We now connect all zeros ( Zl, Z2, . .. , z n) a nd
all poles ("n, '12, . .. , 'In) t o t he p oint eiwT , as indicated in Fig. 12.3b. Let T l, T2,
. .. , Tn b e t he l engths and 'Pl, <1>2, . .. , <l>n be t he angles, respectively, of t he s traight (12.19b) In this manner, we c an compute t he frequency response H [e jwT ] for any value of w
by selecting t he p oint on t he u nit circle a t a n angle w T corresponding t o t hat value
of w. T his point is eiwT . I n s ummary, t o c ompute t he frequency response H [e jwT ],
we c onnect all poles a nd zeros t o t his point a nd d etermine IH[ejwTlI a nd L H[e jwT ]
using t he above equations. We r epeat t his procedure for all values of w T from 0 t o
7r t o o btain t he frequency response.
Controlling Gain by P lacement o f Poles and Z...
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