Baixe Solução Lathi 2a ed - Sistemas Lineares - chapter 05 e outras Resumos em PDF para Engenharia Elétrica, somente na Docsity! Chapter 5 Solutions
511.
5.1-2.
Given the fact that the time-domain signal is finite in duration, the region of con-
vergence should include the entire 2-plane, except possibly z = 0 or z = 00:
oo -n T no-n mn (1/2
X=DE o tln! = Di = Dol 1/0º = "EEB Thus,
. 1-28
X =——; C 0.
(3) = ro RO tel>
In this form, X(z) appears to have eight finite zeros and one finite pole. The eight
zeros are the eight roots of unity, or z = e27k/8 for k = (0,1,...,7). The apparent
pole is at 2 = —1. However, there is also a zero 2 = —1 (k = 4) that cancels this
pole. Thus, there are actually no finite poles and only seven finite zeros, z = elmk/8
for k = (0,1,2,3,5,6,7). MATLAB is used to plot the zeros in the complex plane; the
unit circle is also plotted for reference.
>> k = [0:3,5:7); zz = exp(j+24pi+k/8);
>> ang = linspace(0,2+4pi,201);
>> plot(real(zz), imag(zz), 'ko”,cos(ang) ,sin(ang),"k');
>> xlabel('Re(z)?); ylabel( Im(z)º);
>> axis([-1.1 1.1 -1.1 1.1]); axis equal; grid;
IN
No
rea
Figure $5.1-1: Pole-zero plot for «fn] = (= 1)"(ufn] — uln — 81)
(a)
xt o Dream pt
213
Nm
(b) 4º sinanuln] = 0 for all.
Hence
x[]=0
(e) 4º cosmnu
Hence
x =
=
is a sequence
0,41,0,-42,0,4,0, —7
Hence
x() = (Err) (Gs Da + )
fg pg
; ,
(e) "cos ZEufn] is a sequence
10,-2,0,48,0, 48,0, 98,
AB
2 A 68
xt = 1-5
- a OI
(1-5 =)
= 5 bh
(6)
s 2 6fn — 2h] = ófn] + 46[n — 2) + 166[n — 4) + ---
k=0
214
Therefore
(a)
Therefore
(e)
Therefore
elnj= [o cos lu fn—1] = (2)-" cos ou [n) — 5 tn]
H2-025) 025201)
xtd=
2-052+025 — 22-0,52+0.25
elnj=nyuln-]=nyuln)-0=n9utn]
YZ
Xtf]= =
(f) Because n(n — 1)(n—2)=0forn=0,1, and 2
an) =
n(n = Dn = 92%u fr = m] = n(n— 1)(n— 22)" [n]
n=0,1,or 2. Therefore
and
(g)
Using the
2 fr) =(9"n(n — Dn — 2)2"u [n))
xtg=( [ES a] 6z
E-pCE-
ain] = (1) nufn]
+
ufo
l
E
(2en+1)
Jus i
iba
Pa
IO
dr
I
nim
:
ie
4
“|
Raio)
a
entries in Sec. B.7-4, we obtain
22? 2 2 z
(2-1)? 2-1 2-1
-23 40222 2
xt]
PANA
(h)
ajn]= So tó -2k41]
k=0
xg=D til" =5) (5 kófn — 2k + n) 2"
n=0 n=0 Mk=0
Interchanging the order of summation and noting that
=2k-1
on-m+=[0 oa
We obtain
xt] = Su (Sean)
n=0
= 9 pt
k=0
- Sel)
o. 1/2?
= “aja?
2 e)
51-5. (a)
xt z-4002 1
= FT E-DG- 202 2-3
XE = 255073
el) = Out
(b)
xt Z-4 Ba 1/3
= = m-Delgo o tz-272-3
E 1 z
XE = cstç02737-3
et) = =5ótl+ [2 ace) vtn
(e)
xt e2-2 a 1
| > o) es
xt] = 2
sn) = [0”-2Jutnl
(a)
(2-2 22-24 1 2,1
Xhl= Z o 2 z ata
Hence
zln)=ôln-1)-25n-2]+ 6-3]
(e)
xt o 2+3 o 92
2 O E-DE-DE- z-3
52 2 94
Xe = 57010 "5=a faro
5 9
zm) = 5 =)" + ser uln
(9
xt 42 03
z (G+DG-22 2+
Multiply both sides by 2 and let z — co. This yields
0=34+k+0 => k=3
1a z E
Xb = 3a tico
Zn) = BON -3()" + 2n(2"Ju fr)
(8)
xt 1424008 1 k 2
> “E-ode- 087 7-02"7-08' &-08)
Multiply both sides by z and let z — oo. This yields
O=14+k=> k=-1
z z
EA 2
00 7-08"“(-082
[o2r - (087 + nos] un
4
xt)
k
ztn)
(h) We use pair 12e with 4= 1,B=2,0= 0.5, |y| = 1. Therefore
r=vVi=2 B=
af] = 20)" cos( +oeln]= 2cos(TE +SJelnl
219
Therefore «[0] = 0, 2] =, el] =, 28]=37º,--:, and
a [n] = nyºu [n)
5.1.8. (a) We can express
xt), xp)
xt) = xt) + HU AB
z Z
Let X,[2) and Xal2] be the numerator and the denominator polynomials of X[z]
with powers M and N, respectively. IFM = N, then the long division of Xn with
Xa in power series of 2! yields 2/0) = a nonzero constant. IF N — M = 1, the
term «[0) = 0, but 2(1), «/2], --- are generally nonzero. In general, EN -M = m,
then the long division show that all z[0), 2(1), --:, fm — 1] are zero. Only the
terms from «[m] on are generally nonzero. The difference N — M indicates that
the first N — M samples of zfn] are zero.
(b) In this case the first four samples of xfn) are zero. Hence N — M =4.
521. (a) Xl]=1414& ++ 555. This sum can be found using the result in Sec.
B.7.
,as
(jm 1-2"
*Xd=q/)D1 CI=a
(b)
an) = uln-uln-m)
Po mz o =2"
XxX = ao Zi
5.29.
alnj=6[n=1)+26/n-2]+36[n 3) +46[n-4]+35n —5]+26[n 6) +6[n =]
Therefore
12 3 4,3,/2,1
XxX = ctatatatatata
2 +22 +92! +40 432242241
Z7
Alternate Method:
alo) = ntufnj-ufno5) +(cn+8)fuln-5]-uln—9])
= muln)-2mufn= 5] +nuln—9]4+8u[n 5] 8ufn— 9]
nun) = Mn — 5Jufn — 5] + Sufn— 5]
tn = 9)ufn 9] +9ufn —9] + 8u[n 5] - 8ufn — 9)
nun) -2(n- 5Jufn—5]+(n—9ufn—9)-2ufn — 5] + ufn 9]
|
Therefore
z 2z z 2z z
x = coros toe Ze! FR-D
SE [e 241 2(2-1)+(2—0))
222
- 22 +41]
Reader may verify that the two answers are identical.
5.23. (a)
rêyu [n]
Repeated application of Eq. (5.21) to yºu [n] > 5, yields
=
E Yz
nyuln) >» —s
In (2-9)?
2.m vaz +49)
ryuln) ++»
bl (2-9?
Lety=1 nun] > E
(b) Consider
un) <> E
z-7
Application of multiplication property to this pair yields
nyufn) + e (+) - Es
One more application of multiplication property to this pair yields
a a YZ o vdzÃo)
rêruies cega) = TS
<) Application of Eq. (5.21) to n2yºu [n) <=> JEE5P (found in part a) yields
G-)
a [e + 2] AP A az +)
da | (2-9)? (2-8
Now setting y = 1 in this result yields
a(22 44241)
rêuln] = Toy
(a)
= auln]-ufn—m))
= a"uln)- aa Du
xH = to
(e)
atnj= ne "ufn-m) = (nom+me20mDufa -m
= Ma - me Paufn -m) 4 me re ufn -m]
il
Xe
==
(8
en) = (n- 20.5)" un — 4)
E sn 442705" tufn — 4]
- sn — ay(o5)"-Sufn — 4] + (0.5)"-*ujn — 4]
Application of shift property yields
1 0.5. z
dn 5 Ie lr=055
o 0.25 1 2— 0.25
Xt=5k-05)! 20-05)" 26-05)
5.2-4. Pair2:
ulnj=ó[n)+6fn-1]+6[n—2]+6[n = 3] +++
1 1 1 1
unjes1+5+5 Z
Repeated application of Eq. (5.21) to pair 2 yields pair 3, 4, and 5. Application of Ea.
(5.20) to pair 2 yields pair 6, and application of time-delay (5.15) to pair 6 yields
pair 7. Repeated application of Eq. (5.21) to pair 7 yields pair 8 and 9.
5.2:5. There is a typographic error in the book problem: cos(zn/A) should be cos(an/4).
Given (Pair 11b with |y] = 1)
sin Bnu[n] <> cos fiz FI
cos Bnuln] = —sin (8n - 5) un
and
cos nun) = —sin (Br = 5) ul
= —sin Go - 2] ufn
= -sin Gt nun — 2] +sin (5) ótnl + sm 3) ófn=1)
= -sm[Sn- 2) uln-2]+0n] + gi =]
E
From Pair 11b, and shift property
CAR 1
fm
sn [5-9 dn] ET VE VET)
2(z
224
with y[O)=-M,a
is lim; nico eAGaay = 0. Thus,
0.75
Plot 1 corresponds to tos
m the time-domain, $25/22 has the form of a sinusoid. Thus, either plot 3
22 -2/V2
or plot 4 is possible. Using the IVT, the initial value is lim; .oo 521
Thus,
Pajvê
224
Plot 4 corresponds to
The apparent repeated root at z = 1 of TER ai suggests a signal that
grows linearly in the time-domain. Thus, plot 6 or plot 12 are possible. To dis-
tinguish between the two, first determine whether or not a root at z = 1 really
- - . 287) 8 28-52 : 28 bas
exists. Pirst, lim; EE SgEÊ = lim, oa fo = lime apso —
- E
0/0. “This is indeterminant so use L'HospitaPs rule: lim, E E
: ftet sara : 245 i É
lim;— qÊRaas = lim misêmo = 0/0. This too is
. : ; a . . E sas 448
indeterminant so use L'Hospital's rule again: lim; Eus =
A (485244)
lim, qi = tim.ni
OTTO a505) sos = 20/10 = 2. Since
Ehebetpar 24 00, no root existsatz=1 and
lim.
srs sao
Plot 6 corresponds to a
In the time-domain, 5 is a growing exponential due to the root outside the
unit circle. Thus, plot 16 or plot 17 is possible. Using the IVT, the initial value
às lim, oo Tg = 1. Thus,
Plot 16 corresponds to ——.
z-11
In the time-domain, 1a is the convolution of a decaying sinusoid
and a unit step. Thus, plot 7 or plot 11 are possible. Using the IVT, the initial
value is lim, oo tros) = O. Thus,
0.257"
Plot 7 corresponds to 557 755]
vira) -qybnl=2[n+1]
Pufn-1]
x=-5
snes rt o gbe)esW+M:
227
The z-transform of the system equation is
= (]+Mz-4Y [e]
"
Grid =
and
-Mz Pz
ve = ot EoDe-n
Ya Mo P MS
Zz Do zoy GE-)E-D) 2-7 v-llzoy 2-1
z P [as z
Yi = nistA
vln] = [air + POD um r=9-1
The lomn balance is zero for n = N, that is, y [N] = 0. Setting n = N in the above
equation we obtain
Pa" — “|
z
yiN]= [my +
“This yields
rM
. Because part (b) requires us to separate the response into zero-input and zero-state
components, we shall start with the delay operator form of the equations, as
vln] + 2ytn — 1) = afn)
“To determine the initial condition y[-1], we set n = O in this equation and substitute
v(0] = 1 to obtain
i+wl=afj=e =u-il=(e-0/2
The z-transform of the delay form of equation yields
vtg+2 [Era + SD -—&
Rearranging the terms yields
stofu-ae 6
z 242 z-e+
“The term (1 — e) on the right-hand side is due to the initial condition, and hence rep-
resents the zero-input component. the second term on the right-hand side represents
the zero-state component of the response. Thus
228
5.3-3.
1-e 2e2 e
le £€ rr
Z42" Qe+DE+3) Qe+r(z-e))
Hence
+ 22; x 4+-*
z+2 2erlz+2 2e+1z—
The first term on the right-hand side is the zero-input component and the remaining
two terms represent the zero-state component. Thus
Ylg=(1-0)
2e?
2e+1
ul
(2) uln) +
vin) = (1-2 ufn) +
zero-input comp
e e
2e+1
zero-state comp.
“The total response is
1
or pet
util = 3 [t+ MD" +e Jin
(a) 'The system equation in delay form is
2oto) = 3yln = 1) + vln = 2] = 4efn] = o(a 1]
Also
vt) es YE vh-ojes iva ln 2] + YE] +
cn) es xbl= nojo
2
z— 0.25 0.25
The z-transform of the equation is
, 3 Io 43
We-Vi+aridtl=co 72-02)
or ,
1
(2-*+5)rta
and
re a(82 2.75)
2 O (22-32+1)(2-0.25)
a(8z — 2.75)
2 —0.5)(2— 1)(z— 0.25)
5/2 1/3 4/3
z-1/2
vt) = E +it05p - 360257] un]
- [riem - St") ut
-. z
ie ros "dE DE+05)
Hero Mem HO
zero-input zero-state
- Ma MB, 1/90 1/9 1/12
- 5" (+05) 2-1 2+05 (2+05)2
A
zero-input zero-state
Therefore
(1/92 (82 (1/92 (0/92 | 0NDz
yYjlzl= LD Sm pI
d="505 "(+055 "= +05 * (+05)
ol
zero-input zero-state
Ja qem MOS 1 (-0.5)"
vin)=4 (5055) +35 5
zero-input zero-state
(c) The terms which vanish as n — oo correspond to the transient component and
the terms which do not vanish correspond to the steady-state component. Hence
3 op Entcosr] un) + Dufn)
Nam
steady-state
transient
5.3-6. The system in delay form is
yin) =3yfn=-]+2yln=2]=<2[n-1]
Also
sb) es YE) vino nes dyid+? vin-2 += Syt+2+3
cn) + XE aln= 1) + AXE]
xt]=
The z-transform of the system equation is
rti-s brta+3) eafortasõ+s]
3 2 4 1
(1-Sna)rei=-S+s = 5-3)
Ye -32+12 o -32+12 so 3/2
+ É El&ADdE-) E DE-DE-3 = t7-3
232
9 z z 3 z
Ya = 55-10 *g-atoz-
uti = [cer + Ser) ut
5.3-7. The system equation in delay form is
vt]=2yln-]+2y[n-2]=2[n-2]
st) e YE un) + Avfg+a vn» Srb+o
1 z
cln-2) > 5XEl and X=
The z-transform of the difference equation is
1 1 1 1
vtg-2 [Brig +] +2 [erta+5] = 5
(22-22+2) — 22-42+43
2 YtH=
re) 22? 4243 1
2 (-N(-2+2) 2-1
rg =: z(2-1)
2112-242
For the second fraction on the right-hand side, we use pair 12e with A=1, B= 1,
a=-1, l2=2. This yieldsr = 1,8 = 5, and 0 =0. Therefore
vtnj= f +(v7r cos(Gm)] ufn]
5.3-8. The equation in advance form is
yln+2]+2y[n+1]+2y[n)=<2[n+1)+20(n)
vt) > Y[ gh+)eszp yin+7]e>2r
en) é X(] en+)eszX]-z and Xt]=
“The z-transform of the difference equation is
2
2vtj-ssdy prot) = Ec dE Hed
Zoe *tgrel are
(2 +22 +42)Y [7] 2+ de+2) A2+2)
z0e
Therefore
Y [2] o z+2 o 0.318 —0.318z — 0.502
z rm ze” 2+%+2
q
233
z(0.318z + 0.502)
z
Y = GE —
ti +22+2
For the second fraction on the right-hand side, we use pair 12c with 4 = 0.318,
B=0.502,0=1,|]?=2and
r=0.367 f=cos”
vin)= [osistor —o.367(V 3)" cos(SEm - 0.525 un)
5.3-9. In transform domain, H(z) = 2125; and Y(2) = 271
Ha =2: Since Y(2) = H(z)X(2),
=+2"
we know X(2) = Y(2)/H(2) = 857. Thus, X(2) = -3512. Using tables, a[n] =
3 (Pulo) — He 2p= tudo — 11) = 8 (at-ayrmtu
1)) or
E
= —36fn) + 71(-2)" un — 3).
5.3-10. Taking the 2-transform of bfm] = (1.01)b[m — 1) + p[m] and solving for B(z) yields
B(z) = P(z)y-rd=- Thus, P(2) is required to solve for b[m]. One way to represent
Sally's deposit schedule is pfm] = 100 (ufm] — Sx-o ô[m — (12k + 11)]). Defined this
way, Sally deposits one hundred dollars on the first day of every month m except for
Decembers, (m = 12k + 11 for k = (0,1,2,...)).
Taking the z-transform yields P(z) = 100 (= — DW=n oo De-o ólm — (12k + mz”) =
100 (= - Di£o Decos 6lm — (126 + 1D]e”) -
100 (, - Dio 0), Substituting P(2) imto the expres
sion for B(z) yields B(z2) = 100 (== — Dio 2 0) -
E oo ra O 191 109 o arts)
100 (m= =) + Bio Eta) = 100 (= Qis=t + 126 r + DDk-o InTois
The first two terms are easily inverted using a table of z-transform pairs, while the
last sum is inverted using tables and the shifting property.
blm] = 100 (ionconyrum — 100ufm] — > (1.017 02H Dufm — (12% + E) .
k=o
53-11. (a) Note, hifn) = (=1 + (0.5)") uln] = —(1)"u[n] + (1/2)"u[n]. Thus, two real poles
are evident at z = 1 and z = 1/2. Since h[n) is not absolutely summable, the
system is not BIBO stable. Thought of another way, the pole on the unit-circle
makes the system marginally stable, at best. Marginally stable systems are not
BIBO stable.
(b) Notice, h>[n] = (9)" (u[n] — uln — 10)) is a finite duration, causal signal. Thus,
holn] has no poles (other than at zero). Since h>[n] is absolutely summable, the
system is BIBO stable.
53-12. (a) Let
unJ=55k
Fed
234
Use of the result in Prob. 5.2-8a yields
— 2 Hz+1)
Dto tor RP EP
Hence
co n(n+1)
2
(ii) Let
aln) = n2uln)
Then fe41)
az
E =
=p
Use of the result in Prob. 5.2-8a yields
4z
(E -1t
- p= 3 Bm(n-1) 2n(n-1(n—2) 2º +83? +m m(n+I(n+)
x =" 2 3 = G 6
5.3-16. Let .
afn) = nºufn)
Then
2
42 +41)
xe Cami
Use of the result in Prob. 5.2-8a yields
n ao ,
+42+1) z 7z 122 6
pes EE E
Leer Cuptep ep le-
Hence
n DDD nam EmA?
358 = nt Into son(a- 1-2) PDT =D itiri tor mini
k=0
4 É 4 - 4
5.9-17. Let
aln) = na"ufn)
Then az
xtj= —s
=
Use of the result in Prob. 5.2-8a yields
r az? a Zz z z
Etr S ota plo mat te Dg |
237
Hence
no a A atart “9-1
Dtt=qónali=e +(a— 1)na?] stato al
53-18. (a)
ctn]) = eu] XE] +
ez?
Ye = HH qoogroDE-os
Therefore
vtd ez o 0.186 1.13
= É E-JE+O(-os) z+02 z-08
z z z
YE) = 13255 0186-1135
yin] = [132(e)" — 0.186(-0.2)7" — 1.13(0.8)"] u [nl]
(b) From the given Hlz], we can write
(22 — 0.62 — 0.16)Y [2] = zX[2)
Hence, the corresponding difference equation of the system is
vln + 2) — 0.6y[n + 1] — 0.16y[n] = z[n +)
or
al) = 0.6yln — 1) = 016y[n — 2] = af 1)
5.3-19. (a)
o o. z(22+3)
YH=XBt= o(s)
Therefore
re 9/2
2 o z-3
9
rel = *+37-3
vt) = [E-rer+Fer)ui
(b) From the given H[2], we can write
(2 —52+6)Y [2] = (224 3)X[2]
Hence, the corresponding difference equation of the system is
vln + 2) — 5yfn + 1) + Gy[n] = 2z[n + 1] + 3x[n]
238
or
vln] — Syfn — 1) + Gy[n — 2] = 2e[n. — 1) + 3zfn — 2]
5.3-20. AM cases use the same transfer function. From the given H[2] (after dividing the
numerator and the denominator by 6), we can write
Hence, the corresponding difference equation of the system is
vin 42] Bolo 1]+ gufnl = Soto +) ato]
o
bd Suln = 1] + E -“9=5efn-)-ofn-2]
=4ul
(3)ºu [n] so that X [7] = =p» ond
6z(5z — 1) =(52-1)
Va=HMA = os -s:+ 07 E-De-DE-D
Therefore
1 1 1
= yr as() + 36(5)”
vt) = [12 agr 2065" | uti
= a2[4" = 4(3)" + (2) un]
(b) Here the input is 4--2ufn — 2) which is identical to the input in part (a)
delayed by 2 units. Therefore the response will be the output in part (a) delayed
by 2 units (time-invariance property). Therefore
uln] = 12 [4-0 — a(gy (=D + 36978) un — 2]
(c) Here the input can be expressed as
=4"0-Dy
cf = 16(4)“u [n]
This input is 16 times the input in part (a). Therefore the response will be 16
times the output in part (a) (linearity property). Therefore
vn] = 192 [4º — 4(3)" + 3(2)") un)
(d) Here the input can be expressed as
afj=4"ufln-2]= mta Dun 2]
This input is 4; times the input in part (b). Therefore the response will be 5;
239
(e)
— 22 — =
“H(2-162+08)
1.25z
— 1.62+0.8
For the second fraction on the right-hand side, A = 1.25, B = 0,0 = —0.8,
ht =0.8, l= 3, and
T=2795 B= cos (2ES
)=0.464 0 =tan(-2) = —1.107
Afn) = 1.256fn] + 2195(59º cos(0.464n — 1.107)u [n]
5.3-25. (a) Noting that H(z) = 27*&, H(2) = no =] =78 72. Thus,
t
(b) Since h=:[n] is absolutely summable, the system inverse is stable. However,
hn) 0 for n < 0 so the system is not causal.
ho
= 6[n +3] — ó[n + 2).
(c) For systems that have time as the independent variable, it is only possible to
realize causal systems. Shifting h-![n] by three makes it causal and therefore
realizable. That is, implement Açao") = h7Hn 3] = ôfn]— ó[n—1], as shown
in $5.3-25c. Within a delay factor, this implementation functions as the system
inverse.
x[n) tr]
-
-1
Figure $5.3-25c: Block realization of h
causal
f
5.41. For convenience, let y = “2 and rewrite hfn]
(a) By inspection, the structure is a parallel implementation of two modes, which
are easily identified in Afn). The transfer function of the structure is H(z)
1
Taking the transform of h[n), H(2) = 51 + =
(1º + (19º) ul).
1
FAST É TRAZ
Thus,
and Ag = =
*
() tola] = hln) + (1) = Do blbeto = = (DES + (19 Jaln 3). Thus,
Loans erre) 149
voln] = 4—— un 43), where y= 2
otrl ( 1-4 = mn+3) v8
qria
Written another way, yoln] = 2R (15 Jufn + 3].
242
5.42. (a)
HH 322 — 1.87 32-06) ( 2.) (25)
2-z+016 (2-02(2-08) lz-02/1z-0.
Parallel form: To realize parallel form, we could expand H (2) or H [2] /z into
partial fractions. In our case:
1.22 — 0.48
Ht=+ 7 qa-08)
4 É
Hl]=3+ — SA + id
(2-02) (2-08)
Alternatively we could expand H [2] /z into partial fractions as:
1
Hs 3(2-0.6)
z (2-0.2)(2-0.8) 2
and
Hs
The realizations are shown in Figure S5.4-2.
(b) Refer to Figure S5.4-2.
dz+22 00 S2+22
(2+0.2)(2+0.8) 22+2+016
1 52+22) +
z+02/1 2+08 /.
AM the realizations are shown in Figure S5.4-3.
Refer to Figure S5.4-3.
HE]
(b
5.4-4. (a)
3.82-1.1
HkJ= —
[= 2008203770 005
For a cascade form, we express H [2] as:
1 3.82- 11
Hl= (— 52) (5 a)
For a parallel form, we express H [2] as:
—2 + 22+3
z-02 22-0.62+0.25
Ht=
AM the realizations are shown in Figure S5.4-4.
(b) Refer to Figure $5.4-4.
5.45. (a) Note: the complex conjugate poles must be realized together as a second order
factor
243
| .
XE% E 3 YE33
canonia
“ vectr
x£81
yo Ê YCsT
= E Series
Ga og -D6
x La]
pa
YU3]
p—
Parate)
Po
br
GR 5s54-3b
XI34 Id
A
Carne.
Aee
Yr33
a Parallel
ê -08
est
BB dy
XI]
Fig S54-4
-44
Pavalle
Le
fg. S S4-4b
X0%3
Canonte.
asireel—
XE3T
Series
XT33
2 Yi31
Cualte)
22.
XU33
Canonie.
rec
II-
XI
r E<Ê
A
Series
ta ->
xy AE YBS
Taralte |
Lt
VA do
A-6A
25
Fig. 55 4-6 B “
XE3] YA
RARA! +
seres
pm
XLg]
Lyra?
and
2u [e] = ct ( O4sin Q )
1-0.4cos Q
322 — 1.8z
Hlj= 22 "002.
= 251016
3e%9 1.88% | (3cos 20 — 1.8cos N) + (3sin 2 — 1.8sin N)
e%P ei? +0.16 (cos22-cos NL + 0.16) + j(sin 20 — sin N)
2 359 — 1.829 3e UP
Em = [Eacsaçosl |
É
2.0256 —
=
.8 cos
32 cos 8) + 0.32 cos 20
o 1/2
Therefore Ir tem=[ 12.24 — 10.8 cos Q |
2.0256 — 2.32 cos 9 + 0.32 cos 20
and
, 3sin20 — 1 8sin 2 sin20 — sin O
ZH (9) = tan (Bi “tai
[em] = tam (qem Tr) tm A cos20— cos + 016
05,2
++
05,1
Hd=1+5+5 a
+ 5
z
H[P] = 140508420 U8 4 20789 4 0,5e-188 4 ori
eTiZsA [e?? SM +0.5e159 + 240. so +0.5e"7l 592 +eT32 58
o 32 so
Ze 258 [pes + Jose +cos |
Q 302 52
Aos + cos = 4 2cos
Therefore |H [69]|= 3 3
and LH[e9] = -250
(b) Using the same procedure as in Prob5.45a, we obtain:
.Q 3 50
sing +sin > + 2sin —
rté= pasmo
and (H[P]=-250 7/2.
5.5-3. The advance operator form of this equation is
Etyn]= : [E!4+ ES 4 E2+ E +41]
252
and
utgj=) drir2+z+]
5 A
1 [ei + ei30 + 620 pe 41
sy
H[oº] = Potente
= gerson [8 pe 414 e 4 ei]
geri [1+2c0sN + 2c0529]
5.5-4. (a) The z-transform of the two equations yield:
(1). ( + e) Ye =X]
(ii. (1 - E) vYtg=X()
Hence the transfer functions of these filters are
(1).H[2] = = (ii).H[2] = .
Consider the first system.
()
sa :
mJ==E ==
2709 140958 1+09cos = 70.9sm A
1 -09sinQ
H[9]| = >=, H [&º] = — tan! |
ta [e] TBI + LêcosO” te Dm |7409c0sn
Eu 1 1
Hliy= EL 2 Als A
[= 50-05" 109058 0.9c0s + 0.9sin A
1 0.9sin QL
SM] — PM] = tan
la [e] VÍBI= IAcon” 2H [28] = —ton [rega
Filter(i) has a zero at the origin and a pole at —0.9. Because the only pole is
near Q = m(z = —1), this is a highpass filter, as verified from the frequency
response shown in Figure $5.5-4a.
Filter(ii) has a zero at the origin and a pole at 0.9. Because the only pole is near
Q = 0(z = 1), this is a lowpass filter, as verified from the frequency response
shown in Figure $5.5-4b.
(b) (i) For 2 =0.017
H (90) = = 0.5966
v1.8
0.017
253
For 9 = 0.997
H (e5989*]
(ii) For 2 = 0.017
H [901=] =
For 9 = 0.997
H [099%] =
Filter(i) gain at No is
tl =
Filter(ii) gain at 7 — Ng is
1
1
=== = 9.58
v1.81 +1.8cos 0.997
1 = 9.58
1.81 — 1.8cos0.01m
1 = 0.5966
v1.81 — 1.8cos 0.997
+
1.81 — 1.8cos No
1
|H| = ===E----===--+= = >
Vi81=18cos(r- No) 1.81 + L8cos(do)
5.55
z+0.8
H =
El--05
(a)
Efe] 2 +08 (cosN+0.8) +jsinQ
EP] = ST = CET St Ismi
05 (csN-05) +jsmQ
Les fem) = 1 [89] Hi (e-59]
LH [9]
(b) 2=05
IH [605] |?
IH [e] )
(2º +0.8)(e12 + 0.8) 1.64+1,6cosN
(0º -0.5)(e-12 0,5) 1.25-cosQ
sin Q sinQ
= tan) — tan MS
(naõ) aos)
1.64 + 1.6cos(0.
64 +1.6c05(0.5) 4,
1.25 — cos(0.5)
2.86
254
(b)
fs > 2fn = 100 kHz, T<7j=10us.
5.5-10. Taking the ztransform of yjn] = S2(0.5)tzfn — k) yiclds Y(2)
Dolo. SEX (e)a=* = X(z) Dyco(0.5/z)*. For |z] > 1/2, this becomes Y(z)
X(2) 55" Thus, the system function is H(z) = 8) = br.
(a) Using H(z) and letting z = e!º, the magnitude response is |H(e)] =
—- em À = À .
1-0.5e=> - (1-0.5 cos(N))2+(-0.5 sin(N))? 1-cos(N)+0.25(cos?(N)+sinZ(9))
Thus,
(b)
Ee] =
MATLAB is used to plot |H (e).
>> Omega = linspace(-pi,pi,501);
>> Hmag = 1./sqrt(5/4-cos(Omega));
>> plot(Omega,Hmag,ºk?); axis([-pi pi O 2.5]); grid;
>> xlabel("'NOmega”); ylabel(” IH(eLjNOmega)) |);
>> set(gca, 'xtick”, [-pi:pi/4:pil,'xticklabel?, [?-p E
POOL OG OD O GD Gr aço apr po,
*fontname”,º symbol);
*
N
Figure 85.5-10a: |H(eº)] = FE:
Using H(z) and letting z = e2º, the phase response is LH(eº) = Lila =
=L(1- 0.5cos(9) — 0. Em Thus,
-0.5sin(9)
IM = (me
LH(e) = acta (a)
MATLAB is used to plot ZH(e?º).
>> Omega = linspace(-pi,pi,501);
>> Hang = -atan2(-0.5+sin(Omega),1-0.5+cos (Omega) );
>> plot(Omega,Hang,'k');axis([-pi pi -0.6 0.6]); grid;
>> xlabel(ºNOmega'); ylabel('angle H(e"(jNOmega)) [rad]');
>> set(gca,xtick”, [-pi:pi/4:pi],'xticklabel?, [º-p *;...
257
Doo Op OP OP op op Sp po.
*fontname”, symbol);
(c) Since H(2) = HE) = s=9551» an equivalent difference equation description is
vln] — 0.5y[n — 1] = a[n). From this equation, an efhicient block representationis
found, as shown in Figure $5.5-10c
xr] Ra
Z1
12
Figure 85.5-10c: Block representation of y[n] — 0.5y[n — 1] = zfn).
To O M—s
vlLHLe'2a (Q>
NL.
Figure 55.6-1
5.6-1. Figure $5.6-1 shows a rough sketch of the amplitude and phase response of this filter.
For the case (a), the poles are in the vicinity of 2 = Z. Therefore, the gain |H [ei]
is high in the vicinity of Q = 7/4. In the case (b), the poles are in the vicinity of
Q=7. therefore, the gain [H[ci2) is high in the vicinity of Q = 7. For case (a), the
phases of the two poles are equal and opposite at 92 = 0. Hence ZH[e?) starts at O
258
5.6-2.
5.6-4,
5.6-5
5.6-6.
(for Q = 0). As 9 increases, the angle due to both poles increase. Hence, ZH[e'º)
increases in negative direction until it reaches the value —27 at Q = 7. For case (b),
similar behavior is observed. Note that angle —27 is the same as 0.
The two systems are very similar and have identical steady-state characteristics. There
is an important difference, however, between the two systems. The system yfn] —
ylr — 1] = z[n] — afn — 1) is first-order and can support an initial condition; the
system y[n] = z[n) is zero-order and cannot support an initial condition. If the initial
condition of the first system is non-zero, the output of the two systems can be quite
different.
(a) From the magnitude response plot, it is clear that this is a lowpass filter. Low
frequencies near Q = O are attenuated, and high frequencies near 9 = 7 are
passed with unity gain.
(b) From the magnitude and phase response plots, H(e7"/2) = L.e/87/2, Thus, the
output to ai[n] = 2sin(n + 3) is
x
2"
tn]
vZsin(çn +71) = -v2sin(
(c) Notice, H(e777/4) = H(e-3*/4). From the magnitude and phase response plots,
H(e717/4) x: 0.071e72:43, Thus, the output to zo[n] = cos(ZEn) is
x
7:
ya[n] = 0.071 costtn +2.43)
Refer to Figure $5.6-4 and the solution to 5.M-1.
Refer to Figure 85.6-5 and the solution to 5.M-4.
Hty= Kit
z
-—a
Figure 85.6-6a shows the realization, Figure $5.6-6b shows the pole-zero configuration,
and Figure $5.6-6c shows the amplitude response of the filter. Observe that the pole
at a is close to = 0. Hence, there is the highest gain at de. There is a zero at —1,
which represents 9 = 7. Hence, the gain is zero at 9.= 7. This is a lowpass filter.
58 o
nfesj=n (é (a)
a
en cosfi-a+jsin
2
WH [é] |=K (L+cosf)
For a = 0.2
2 cos)
39) = Ky) TOS
va [e 1.04-0.4cosQ
The de gain is
LH] = 2.5
For 3 dB bandwidth |H[e9)p? = 1 H(e79]]? = 3.125K?. Hence
2(1+cos9)
1.04 —0.4cos A
3.125K2 = K? [ ] => N = 1.176
259
5.6-9.
5.6-10.
(a)
(b
(a
(b
)
han] = (Diuln]
= eim hn]
Use of the property in Eq. (5.20) with y = e$i”, we obtain Hal Hifetra.
Hence
Ho(9] = Hi[e/(P+r]
Figure $5.6-9 shows the frequency response of an ideal lowpass filter with cutoff
frequency Nc. Figure $5.6-9 also shows the same frequency response shifted by
a (with 2m-periodicity). It is clear that the shifted response corresponds to an
ideal high-pass filter with cutoff frequency 7 — Pc.
Let H(2) represent the original filter, either highpass or lowpass. The transformed
filter has system function Hp(2) = H(2)| = H(-z). The basic character
of the transformed filter can be assessed by its magnitude response, |Hp(e'?)| =
tH (= = |H(e7 e) = |H(e!º+5)]. That is, the magnitude response of a
transformed filter is just the magnitude response of the original filter shifted in
frequency by 7. If the magnitude response of a digital LPF is shifted in frequency
by 7, it becomes a highpass filter. Similarly, if the magnitude response of a digital
HPPF is shifted in frequency by 7, it becomes a lowpass filter. Put another way,
an original passband centered at 92 = O (LPF) is shifted by the transformation
to a passband centered at 9 = 7 (HPF), and vice-versa.
E H(7) = Dºcohlnzo” is the original filter, Hy(z) = H(-2) =
DE hfn](=2)" = 52 o(=1)"h[n]z””. Thus, the transformed filter has
4 m=00
impulse response
hrln] = (DAL
Put another way, the same transformation is accomplished by simply negating
the values of hfn) for every odd integer n.
5.6-11. The bilincar transformation states s = 24). Rearranging yields z = 1ÉTs/2,
T+) sing Y) I-Ts/2
Thus, s = qu maps to z = XTR,
(a) The magnitude of this transformation is [2]
(b) Using the solution to 5.6-1a, we know that z =
ItmwT/2) UtawT/2] 1
a] = mta = 1. Since
only the unit circle has |2| = 1, the bilinear transform maps s = jw onto the unit
ircle in the z-plane.
ci
LtawT/2
FETiS describes the unit circle
in the complex plane. This allows us to write z = e? = JEMTB. Thus, the
262
bilinear transformation maps (—c0 < w < 00) to (-7/2 < Stx/2) in a monotonic,
although non-linear, manner according to N = £z = «a = arctan(wT/2) —
arctan(—wT'/2) = Zarctan(wT/2).
5.7-1. For the system in Eq. (3.15a), the transfer function H(s) is given by
1
H()=
and
h(t) = eu(t)
Using impulse invariance criterion, we obtain the equivalent digital filter transfer func-
tion from Table 5.3 corresponding to H(s) = «x; as
Tz Tz
(1-7)
HI] assuming T>0
Dea
The approximation used in Chapter 3, Eq. (3.15c), yields
5 z
Mtl= É
z+a
Substituting 4 = Ez and a = jris, we obtain à
Tz
z Ter
Rd=""—
Tea
AsST>0 1-cT. Hence
o
2 I+eT
a TA-cT)z Tz
fgs "0H wu tool
=D 0-0
which is same as that found by impulse-invariance method.
5.72. (a)
15+20 75+20 1 (5/2
AS +7s+10) As+2)(s+5) s+2 s+5
Ho(s) =
Using Table 12.1, we get
HE)
z 5 z
T [=> tazresr cen]
First we select T:
H0)=1 andior s>5=> Hi(s)= =
5
7
and Holjol=o >
(b) We shall choose the filter bandwidth to be that frequency where
7
|Haliwo)] isl%of |Ha(0)]. Hence mo” 0.0! and w)=350, T= E
263
Substituting this value of T in H(z] yields
He] = 0.008976 [A + 2
5 z- 097474
=-—— | = 0.031416:
z-0.9822 "27— Sm] 00814162 [ |
z? — 1.9383z + 0.9391
1) Canonical realization:
0.031416z2 — 0.03062z
Hd = 2 1.95852 + 0.9591
2) parallel realization:
Hj= 0.008976z 0.02244z
2 2709822 *2-0.9561
The canonical and the parallel realizations are shown in Figure $5.7-2.
——o: 3116
canon ica]
321
P
L
(O Y [o-52289] Parallel
0.456)
Figure $5.7-2
5.7-3. Ha(s) = aa Using Table 12.1, we get
Hf]= |>——— ABr
22 — 2ze-Tivicos (E) pe-vZ
We now select T:
Ha(0)=1 and for highs, Hu(s) = 5
5
1
and |Ho(jw)| =
for high w
264
Sing
Figure $5.7-5
The unit step response ga(t) is given by:
gal) = 0 [BM] ga [oe o gafio
Therefore ga()=(1-e“u(t)
and gAnT) = (1- een]
Also, g[n), the response of H[z] to ufn) is given by:
gn) = 27 (ut)
Since gfn] = ga(nT),
ut = AA eua)
z— erwT
1
= erur
Therefore Hl]=120
z
Teu
Using the above argument, we can generalize:
dm
HE)
(b)
and
267
(b)
Hence,
Wit] = z nda Wi<ã
[cosa = 17 + sn? um] V20 = cosuT)] 2sme7]
The ideal integrator amplitude response is
Italo) = À
Observe that this amplitude response is identical to that found by the impulse
invariance method in Prob. 5.7-4. Hence, this amplitude response and the ideal
integrator amplitude response are the same as those in Figure $5.7-4. The only
difference between the answers obtained by these methods is that the phase re-
sponse of the step invariance response difers from that of the impulse invariance
method by a constant Z.
For a differentiator
Ha(s)
The unit ramp response r(t) is given by
MO= LA P(Ha(o) = 045
(8) = u(t) +
Now we must design H[z] such that its response to input nTu[n] is ufn), that is
Zlufn]] = H(2]ZnTufn))
Tp o HE-
z-1)
For an integrator
1
H(s)=—
(=.
The unit ramp response r(t) is given by
= E = Pulo
s
Now we design H[z) such that its response to nTufn] is n2T2u[n), that is
z ( AnêTtul ) = HEZ(nTun))
or ,
Têdz+1) Ta
Deo 2H
22-13 1? te
Hence
T/2+1
He) (5)
268
5.7-9. (a) The w-axis is given by s = jw. Rewriting the transformation s = E
HA;
ki
Por H()=5 0 HE=TD —
= a; + jk, then eNT = emTeikiT, When À; is in LHP, a; < O and [eMT| =
esT < 1. Hence if À; is in LHP, the corresponding pole of He] is within the unit
circle. Clearly if Ha(s) is stable, the corresponding H[] is also stable.
(b)
Z— as
or Thus, we need to show that
=oy describes a circle centered at (1/2,0) with a radius of 1/2.
For a complex variable z, the equation |z — 1/2]? = (1/2)? describes a circle
centered at (1/2,0) with a radius of 1/2. The transformation rule z =
substituted into this expression.
(2-1/2(2 1/2)
1 1
“1 1
2/1 rmT
7 and substituting s = ju yields z =
1
Ir 8
lz=1/2]º
A :
Tre + ago tam td
= 1 + SUSWT)-AnwT) (1
= Trust 2 pa7) 0 =qoT) 4
= trsm + agroro ta
= 1/4=(1/2º
Since the equation is satisfied, the transformation rule z =
to a circle centered at (1/2,0) with a radius of 1/2.
Notice that different values of w can map to the same value of z (aliasing), which
makes an inverse transformation very problematic.
1=20+
T
7 maps the w-axis
asz = +. Next, notice
First, rewrite the transformation s = iz
Po =
1=sT— ss TE
É RCE
> ET rot
For o < 0, the denominator 1 — 20T + (0? + w2)T? > 1 and |z|? < 1. That is,
the left-half plane of s (o < 0) is guaranteed to map to the interior of the unit
circle in the z-plane.
= [n] = (0.8*u[n] + 2ºu[-(n + 1))
mn) zaln)
an) «> as izl> 0.8
z
go[n) > lzl<2
269
>> plot(real(zz),imag(zz),"ko?,real(zp), imag(zp),'kx');
>> xlabel('Re(z)?); ylabelC' Im(z)');
>> axis([-1.5 1.5 -1.5 1.5]); axis equal; grid;
-os]
“35 = “os o , 15 2
[e
Figure $5.9-4a: Pole-zero plot for H(2) =
There are two possible regions of convergence, both of which exclude the thrge
system poles: || < 3/2 or |z| > 3/2.
(b) The poles and zeros of H-(z) are just the zeros and poles, respectively, of H(2).
Thus, the three zeros of H(z) satisfy 2º = 2, or z = 3/2e227/3 for k = (0,1,2).
There are two finite poles at z = 0 and z = 1/2 as well as a pole at infinity.
MATLAB is used to create the corresponding pole-zero plot.
>> k = [0:2]; zz = 3/2xexp(j*2*pi*k/3); zp = [0,1/2];
>> plot(real(zz) imag(zz),'ko”,real(zp), imag(zp),'kx');
>> xlabel('Re(z)'); ylabel(ºIm(z)');
>> axis([-1.5 1.5 -1.5 1.5]); axis equal; grid;
Za oo
“SE AA
5
Figure $5.9-4b: Pole-zero plot for H1(2) = =)
There are two possible regions of convergence, both of which exclude the three
system poles: 0 < |z|<1/20r 1/2< |z] < oo.
272
5.95.
5.9-6.
5.9-8.
5.9-9.
It is known that afn] has a mode (1/2)" and that z1[n] = (1/3)"z[n] is absolutely
summable. For this to be true, the mode at (1/2)" must be right-sided. That is,
(1/2)"(1/3)"u[n] is absolutely summable but (1/2)"(1/3)"u[-n) is not. It is also
known that z2[n] = (1/4)"z[n] is not absolutely summable. For this to be true, there
must be a pole somewhere in the annulus 3 < [2] < 4 that corresponds to a left-sided
signal; such a mode when multiplied by (1/3)" is still absolutely summable but when
multiplied by (1/4) is not absolutely summable. Thus,
ef) is a two-sided signal.
Im polar form, the known pole is at 2 = /18/16€7*/4, To be absolutely summable,
the signal's region of convergence must include the unit circle, |2| = 1.
(a) Yes, the signal can be left-sided. Since the known pole is outside the unit circle,
a region of convergence that includes the unit circle (needed for absolute summa-
bility) implies that the known pole corresponds to a left-sided component of the
signal. .
(d)
No, the signal cannot be right-sided. If the known pole outside the unit circle
is right-sided, the region of convergence cannot include the unit circle and the
signal cannot be absolutely summable as required.
S
Yes, the signal can be two-sided. Let the known pole correspond to a left-sided
component and let there be another pole within the unit circle that corresponds
to a right-sided component. Such a signal is two-sided and has a region bf
convergence that includes the unit circle, which ensures the signal is absolutely
summable as required.
(d) No, the signal cannot be finite duration. Pinite duration signals cannot have
poles in the region 0 < [z] < oo. Such poles, such as the pole known to exist,
correspond to time-domain components with infinite duration.
(a) = a: Next, N(2) = H(2)Xi(2) = a Thus, Yi(2)/z =
5a — ia + a or Yi(2) = 6, + 385. Inverting yields
n 1
nb (1/2)Pufn] + 5 (8/4) und
(b) The idea of frequency response is used to determine the ontput in response to the
everlasting exponential input z2[n] = (3/4). That is, H(z = 3/4) = 34=U3 =
é = É Thus,
1
vali E (3/4)".
The given signal is z[n) = (—1)"u[n — no] + a"u[-n]. 1f Jal = 2, the z-transform
X(2) = (—1)tzno 27 4 =E has region of convergence 1 < |z| < Ja] = 2, as desired.
Thus, the necessary constraint is
lal=
There is no constraint on the integer no, other than it be finite.
(a) Xi(z) = 3 til” = Des (ND TPul-n)+óf-n)z" =
Dino (5) + DX able] = DO (2) 41 Por fe] < 1, this be
comes 0-3/z
Kl)=1+7557=1+ ROC |2]<1.
273
6) Xo(z) = Dezinimo = ol)" cos(n + Duln) =
Do3"0.5 (emu + eimoth(n+1)) a" = 4 0.5€7 (12) +
Dito0.5e7a (22)", por |z] > 1, this becomes
0.5eº =
X(2) = Ta 1; ROC |H> 1.
(0) Xa(z) = tan” = Dito (10.5(€" -euf-n+ 12" =
te-00 10.5 ((£)" — (E)"3. For [2] < e and |2] < €7!, this becomes Xa(2) —
PURE = Ef). Ts,
E -
tr rocpI<e .
(4) Xa(2) =
Dh-oo Dn. oo (2)" ófm — 24] =
4, this becomes
Dic (Etc ool29)"An 26) 2 =
- For |22| <
2) = Disco (É
42
“dz
Xa(z) = ; ROC |z]<2. 4
Note: |2%| =
2, so the region of convergence is |2]? < 4 or |z| < 2.
5.9-10. (a) The signal Xy (2) = ==
one must be real. Using the region of convergence, we know that a real root
must be either 0.5, —0.5, 2, or —2. Trying these values, we find that z = —0.5
and 2 = —2 are both roots of the denominator. The remaining root can be
found by noting that the denominator 1 + d8z-1.4 12-2.. 12-3 must be equal to
(+0.52"D( +220D(14 4273) Equatin the Der of z73 on each side yields
-1/3=A. Thus, Xi(2) = tros nm nnc=3 Expanding yields Xi(2) =
Tr + Er + 1288, Using the region of convergence (0.5 < [2] < 2) and
tables, the inverse is
3 has three finite poles, of which at least
.a n Beomilno 2 n
calo] == (1/2ufo) — (-2"ufon = 1) + So (1/3 un
(b) Xa (2) = = em = “Pam =
ep(s q + Tê) Using the region of convergence (0.5 < |2] < 2)
and tables, die i inverse is
zo[n] = BU ulm +3]- Han -4.
s .
SS (Me) = mms = medos - (=p + =p)
Tr =D
1 (a + xi). Since hn is stable, the region of convergence for H(2)
must be [2] > 1/2. Úsing this ROC and z-transform tables, the inverse transform
is
dalr] = (1/2""ufn =) = (=1/29)"ufn — 1).
274
X(2) (2? + 1). Thus, the system function is H(2) = SE) = ot) = 0.25 (125,
x) 2 dz
(a) MATLAB is used to create the pole-zero diagram.
(b) H(eº) = qa
>> zz = roots([1 O 1]); zp = roots([4 O -1]);
>> theta = linspace(0,2+*pi,201);
>> plot(real(zz), imag(zz), 'ko',real(zp), imag(zp), 'kx”,
cos(theta),sin(theta),ºk”);
>> xlabel(”Re(z)'); ylabel('Im(z)'); grid;
>> axis([-1.1 1.1 -1.1 1.1]); axis equal;
08
04
oa
£o
-08
2H
Figure $5.M-la: Pole-zero diagram for H(2) = Fil.
rea
MATLAB is used to plot the magnitude response
z=e8
(e.
>> Omega = linspace(-pi,pi,501);
>> z = exp(j*Omega); H = (2.72+1)./(4+2.72-1);
>> plot (Omega, abs(H),ºk'); axis([-pi pi O 1]); grid;
>> xlabel("NOmega?); ylabel(” |H(e-(jN0mega)) |”);
>> set(gca,'xtick”, [-pi:pi/4:pil,'xticklabel”, ['-p
POR OP OP O jr apr o po,
*fontname”,' symbol”);
The pole-zero plot of Figure $5.M-la and the magnitude response plot of Figure
S5.M-1b confirm that this is a band-stop system.
Yes, the system is asymptotically stable. Referring to Figure S5.M-la, all the
system poles are within the unit circle.
Yes, the system is real. Since the system is expressed as a constant-coefficient
linear difference equation with real coefficients, the impulse response h[n] and
system are both real.
For an input of the form zf[n) = cos(9m), the greatest possible amplitude of the
output corresponds to the greatest gain shown in the magnitude response plot of
Figure $5.M-Jb. Thus, 2/3 is the greatest output amplitude given an input of
<[n) = cos(Qm). This output amplitude occurs when Q = kr, for any integer k.
Ay
an + 2] + 2!
Figure $5.M-1b: Magnitude response plot for 4yfn + 2] — y[nl)
x[n] Xin
1/4
2
+
A
z
1/4 À -14
E
Figure $5.M-1g: TDFII implementation of y[n] — 0.25yfn — 2) = 0.25a[n] + 0.25cfn — 2].
(g) Inverting H(z) = SE) = 0.257Ht2>, provides y[n] — 0.25yfn — 2] = 0.25x[n) +
0.25x[n — 2], which is a convenient form for implementation. Figure 85.M-1g
illustrates a TDFII implementation of the system.
M.2. (a) H(2) = ofsta = COLE! Thys, there are zeros at z = esr/4 and
—33n/4
z=e , and there are poles at z = 0.9e737/4 and z = 00.
>> zz = roots([1 O -5]); zp = 0.9+exp(j+3+pi/4);
>> theta = linspace(0,2+pi,201);
>> plot(real(zz), imag(zz),'ko”,real(zp) imag(zp),'kxº,...
cos(theta),sin(theta),ºk');
>> xlabel("Re(z)?); ylabel('Im(z)º);
>> axis([-1.1 1.1 -1.1 1.1]); axis equal; grid;
Since this is a complex system, poles and zeros need not occur in complex con-
jugate pairs.
Substituting z = 2 into H(z) = — bon, MATLAB is used to create the
(b 967
magnitude response plot.
>> Omega = linspace(-2+pi,24pi,1001); z = exp(j+Omega);
>> H = (2.02-5)./(2-0.9+exp(j+3+*pi/4));
>> plot(Omega,abs(H),º'k?); axis([-2+*pi 2+pi O 21));
>> xlabel(*WOnega”); ylabel(? |H(e"(jNOmega)) |); grid;
278
ta)
2
Figure $5.M-2a: Pole-zero diagram for H(z) = —Est=a-
1/4:2+pil,?xticklabel”, ['-2p”;
Po 0 09;
5 *5? 2p'], "fontname”, 'symbol?)
>> set(gca, 'xtick”, [-2+pi
Door o; po
Figure $5.M-2b: Magnitude response plot for H(z) = —F=ta-
From Figure $5.M-2b, the system appears to be a type of bandpass filter. This
particular filter tends to pass only positive frequency inputs near Q = 3x/4; the
corresponding negative frequencies near Q = —37/4 are significantly attenuated.
Since the magnitude response is not an even function of O, the output of the
filter will be complex-valued.
5.M-3. (a) H(z) = 555/55) has four finite zeros and two finite poles. The zeros are the four
2(27+0.819)
roots of unity, and the poles are at z = 0.9e77/4 and 2 = 0.9e-337/4, MATLAB
is used to compute the pole-zero plot.
>> zz = roots([1 0 0 0 -1]); zp = roots([1 O j+*0.81));
>> theta = linspace(0,2+pi,201);
>> plot(real(zz), imag(zz),'ko',real(zp), imag(zp),'kx,...
cos(theta),sin(theta),ºk');
279
>> plot(Omega,Hmag,'k'); axis([-pi pi O 2.5]); grid;
>> xlabel(ºNOmega'); ylabel(” |H(e-LjNOmega)) |º);
>> set(gca,'xtick”, [-pi:pi/4:pil,'xticklabel”, [?-p ?;..
POOR OD OD Gr o op po,
*fontname”, symbol”);
Figure 85.M-4b: |[H(eº)| = VE E
Standard filter types do not provide a good description of this filter. The system
appears most like a bandstop filter, but its stopband attenuation is quite poor.
The system boosts the gain of low and high frequencies more than it attenuates
the middle frequencies.
ey
E
, 2 2
(c) Inverting H(z) = bs = ma + a yields
fm) = 0.5 ((1/V2)" + (-1/V3) u
5.M-5. Label the first summation block output as v[n]. Thus, v[n] = zfn — 1] + iytn] or
V(2) = 27! X(2) + 1Y (2). The output of the second summation block is yin +1) =
alnJtuln=1) or zY(z) = X(2)+2"!V(2). Combining the two equations yields zY (2) =
X(2) +20 (2 X(2) + 4Y(2)) or Y (2) (2- 121) = X(2) (1+272). Multiplying
both sides by 2") yields Y(2) (1 — 4z-2) = X(2) (2-2 4.253). Thus, H(2) = HD =
2tpzo 22 - 5/2 52 -1.52/2 -1 52/2 Es
is =mGas + ia+ ce =Prrtda+ ia: Taking
the inverse transform yields
4
bn] = 40 = + 501/2"upn o 1)+ Hatha)
Since hfn] = 0 for n < 0, the system is causal.
The system has three poles located at 2 = 0, 2 = 1/2, and z = —1/2. Since all three
poles are inside the unit circle, the system is stable.
Rr
2
5.M-6. Since h[n] = 6[n—1]+ó[n+1], H(2) = 2!+z and |H (e) = = 2] cos(9).
>> Omega = linspace(-pi,pi,501);
282
>> Hmag = 2*abs(cos(Omega));
>> plot(Omega,Hmag,'k'); axis([-pi pi O 2.5]); grid;
>> xlabel(?NOmega'); ylabel(” IH(e(jNOmega)) |');
>> set(gca, 'xtick*, [-pi:pi/4:pil,?xticklabel”, ['-p *;
Po op o OP jr Op DP po.
*fontname”,*symbol?);
Figure 85.M-6: |H(e'?)| = 2] cos(9)].
Using Figure $5.M-6, this system is Dest described by a bandstop filter with gain.
That is, low and high frequencies have à boosted gain of two and middle frequencies
near Q = 1/2 are attenuated.
5.M-7. (a) >> Omega = linspace(-pi,pi,501); z = exp(j+Omega);
>> H = cos(1./z);
>> plot(Omega,abs(H),'k'); axis([-pi pi O 2]); grid;
>> xlabel (*NOmega'); ylabel(” |H(e-(jNOmega)) |');
>> set(gca,'xtick?, [-pi:pi/4:pil,ºxticklabel”, [?-p *;
Pop op op OP dr op dp po.
“fontname”,º symbol);
Figure $5.M-Ta: Magnitude response for H(z) = cos(z-!)
283
d
From Figure S5.M-Ta, it appears that this system behaves somewhat like a band-
pass filter. Frequencies near Q = 7/2 are boosted while frequencies near Q = km
are somewhat attenuated.
(eat
A Maclaurin series expansion of cos(z) is Jyco “aor- Substituting n = 2"
vields H(z) = cos(2-3) = D$£o CE. Inverting vields
e (1) 1 1 1
hnJ=5> Tr ln — 2] = ófn] — sôfn — 2] + pôln — 4] — gole +...
& Em 1 i i
>> n = [0:10]; h = zeros(size(n));
>> k = [0:5]; h(2*k+1) = (-1).-(k)./(gamma(2+k+1));
>> stem(n,h,ºk?); xlabel('n'); ylabel(?h[n]');
>> axis([-.5 10.5 -.6 1.1]);
Pigure $5.M-7b: Impulse response hfn] for H(z) = cos(z71).
(c) Using Figure S5.M-7b, it is clear that the impulse response quickly decays to zero.
Thus, only the first few terms from h[n) are needed for a good approximation.
Note that h[n] = ó[n] — ófn — 2]/2.+ ó[n — 4]/24. An FIR difference equation is
found by letting h[n] = y[n) and ófn] = zfn). That is,
yin] = efn] — zfn — 2]/2 + afn — aJ/24.
This is a fourth-order FIR filter with only three non-zero coefficients. The mag-
nitude response is casily computed using MATLAB.
>> Omega = linspace(-pi,pi,501); z = exp(j*Omega);
> H=1-2.0(-2)/2 + 2.0(-4)/24;
>> plot(Omega,abs(H),'k); axis([-pi pi O 2]); grid;
>> xlabel(ºNOmega'); ylabel(” |H(e(jNOmega)) |");
>> set(gca, 'xtick”, [-pi:pi/4:pil,*xticklabel?, ['-p *;.
POP O O OP op apr as poT.
*fontname”, 'symbol?);
Visually, Figure S5.M-Tc is indistinguishable from Figure $5.M-7b. Thus, the
FIR filter appears to closely approximate the original system.
Roe)
Figure S5.M-8c: Order-8 Butterworth BPF with passband between 57/24 and 117/24.
(d) Order-8 Butterworth BSF with stopband between 57/24 and 117/24. Notice
that the command butter requires the parameter N = 4 to be used to obtain a
(2N = 8)-order bandstop filter.
>> Omega c = [5*pi/24,11+pi/24]; Omega = linspace(-pi,pi,1001);
>> [z,p,k] = butter(4,0mega c/pi,'stop'); *
>> subplot(121),plot(real(p),imag(p),'kx”,...
real(z) imag(z), ko” ,cos(Omega),sin(Omega),'kº);
>> axis([-1.1 1.1 -1.1 1.1)); axis equal;
>> xlabel(*Re(z)?); ylabel('Im(z));
>> [B,A) = butter(4,0mega c/pi,'stop”);
>> HBS = polyval(B,exp(j+Omega))./polyval(A,exp(j+Omega));
>> subplot(122),plot (Omega, 20+10g10 (abs (HBS)),k');
>> axis([-pi pi -40 2]); grid;
>> xlabel(ºNOmega'); ylabel(” |H (BSH(e"(jNOmega)) |);
>> set(gca,?xtick”, [-pi:pi/3:pil,'xticklabel?,[?-p ?;...
»oDpo op O pr pr po,
*fontname”, symbol”);
Figure S5.M-8d: Order-8 Butterworth BSF with stopband between 51/24 and 117/24.
287
5.M-9. Factored form is used to plot roots, and standard transfer function form is used to
compute magnitude response plots.
(a) Order-8 Chebyshev Type I LPF with Q = 7/3.
>> Omega c = pi/3; Omega = linspace(-pi,pi,1001);
>> [z,p,k] = chebyi(8,3,0mega c/pi);
>> subplot(121),plot(real(p) ,imag(p),'kx',...
real(z),imag(z),'ko”,cos(Omega),sin(Dmega),'k?);
>> axis([-1.1 1.1 -1.1 1.1]); axis equal;
>> xlabel('Re(z)'); ylabelC' Im(z)º);
>> [B,A] = cheby1(8,3,0mega c/pi);
>> HLP = polyval(B,exp(j*Omega))./polyval(A,exp(j*Omega));
>> subplot (122) ,plot (Omega, 20+10g10(abs(HLP)),?k?);
>> axis([-pi pi -40 2]); grid;
>> xlabel("NOmega”); ylabel( IH (LPJ(e-(jNOmega)) |?);
>> set(gca,'xtick”, [-pi:pi/3:pil,'xticklabel”, [?-p ?;...
» Op op Op dp dp pa.
*fontname”, 'symbol”);
reta
Figure S5.M-9a: Order-8 Chebyshev Type I LPF with M = 7/3.
(b) Order-8 Chebyshev Type I HPF with Q. = 1/3.
>> Omega c = pi/3; Omega = linspace(-pi,pi,1001);
>> [z,p,k] = cheby1(8,3,0mega c/pi,"high?);
>> subplot(121),plot(real(p), imag(p),ºkx”,...
real(z),imag(z),'ko”,cos(Omega),sin(Omega),ºk?);
>> axis([-1.1 1.1 -1.1 1.1]); axis equal;
>> xlabel(Re(z)'); ylabel('Im(z)º);
>> [B,A] = cheby1(8,3,0mega c/pi,"high'); .
>> HHP = polyval(B,exp(j+Omega))./polyval(A,exp(j+*Omega));
>> subplot(122) plot (Omega, 20+10g10(abs (HHP)),k);
>> axis([-pi pi -40 2]); grid;
>> xlabel('NOmega”); ylabel(? |H (HPJ(e"(jNOmega)) |");
>> set(gca, 'xtick”, [-pi:pi/3:pil,'xticklabel”, ['-p ?;...
Pop op Op pr pr po),
*fontname”,'symbol?);
288
as dE “ o .
q
Bete, 2
Figure $5.M-9b: Order-8 Chebyshev Type I HPF with Mc = 1/3.
(c) Order-8 Chebyshev Type I BPF with passband between 57/24 and 117/24. No-
(d
tice that the command cheby1 requires the parameter N = 4 to be used to obtain
a (2N = 8)-order bandpass filter.
>> Omega c = [5+pi/24,11+pi/24]; Omega = linspace(-pi,pi,1001);
>> [z,p,k] = cheby1(4,3,0mega c/pi); x
>> subplot(121),plot(real(p), imag(p),'kx?,...
real(z),imag(z),ºko”,cos(Omega) ,sin(Omega),ºk”);
>> axis([-1.1 1.1 -1.1 1.1]); axis equal;
>> xlabel('Re(z)?); ylabel(" Im(z)');
>> [B,A) = cheby1(4,3,0mega c/pi);
>> HBP = polyval(B,exp(j+*0mega))./polyval(A,exp(jtOmega));
>> subplot(122) plot (Omega, 20*+10g10 (abs (HBP)),'k”);
>> axis([-pi pi -40 2]); grid;
>> xlabel(?NOmega”); ylabel(” IH (BPJ(e"(jNOmega)) |);
>> set(gca,'xtick”, [-pi:pi/3:pil,'xticklabel?, ['-p º;.
Doo op Op à; Pp...
*fontname”, *symbol?);
Order-8 Chebyshev Type I BSF with stopband between 57/24 and 117/24. No-
tice that the command cheby1 requires the parameter N = 4 to be used to obtain
a (2N = 8)-order bandstop filter.
>> Omega c = [5+pi/24,11+*pi/24]; Omega = linspace(-pi,pi,1001);
>> [z,p,k] = cheby1(4,3,0mega c/pi,'stop');
>> subplot(121),plot(real(p), imag(p), 'kx”,...
real(z),imag(z),'ko”,cos(Omega),sin(Omega),ºkº);
>> axis([-1.1 1.1 -1.1 1.1]); axis equal;
>> xlabel('Re(z)'); ylabel(' Im(z)');
>> [B,A] = cheby1(4,3,0mega c/pi,'stop');
>> HBS = polyval(B,exp(j*Omega))./polyval(A,exp(j*0mega));
>> subplot(122) plot (Omega, 20+10g10(abs (HBS)),ºk');
>> axis((-pi pi -40 2]); grid;
>> xlabel(? Omega”); ylabel(? IH (BSJ(e-LjN0mega)) |);
>> set(gca, 'xtick”, [-pi:pi/3:pál,'xticklabel?, ['-p *;...
Por OPG Gr PR pI...
289
(c
may
st
E CER a =
Reta, a
Figure S5.M-10a: Order-8 Chebyshev Type II LPF with Lc = 7/3
>> HHP = polyval(B,exp(j+Omega))./polyval (A,exp(j+Omega));
>> subplot (122) plot (Omega, 20+10g10 (abs (HHP)),k”);
>> axis([-pi pi -40 2]); grid;
>> xlabel(º Omega”); ylabel(”? |H (HP)(e"(jNOmega)) |");
>> set(gca, 'xtick', [-pi:pi/3:pil,ºxticklabel?,['-p ;...
POOR O GP Pr 52 pI. +
*fontnamé*,* symbol”);
Man
Reta a
Figure $5.M-10b: Order-B Chebyshev Type IL HPF with Q, = 7/3.
Order-8 Chebyshev Type II BPF with passband between 57/24 and 117/24.
Notice that the command cheby2 requires the parameter N = 4 to be used to
obtain a (2N = 8)-order bandpass filter.
>> Omega c = [5*pi/24,11+pi/24]; Omega = linspace(-pi,pi,1001);
>> [z,p,k] = cheby2(4,20,0mega c/pi);
>> subplot(121),plot(real(p),imag(p),'kx”,...
real(z),imag(z),'ko”,cos(Omega) ,sin(Omega),*k');
>> axis([-1.1 1.1 -1.1 1.1]); axis equal;
>> xlabel('Re(z)”); ylabel( Im(z)');
>> [B,A] = cheby2(4,20,0mega c/pi);
292
>> HBP = polyval(B,exp(j*Omega))./polyval(A,exp(j*Omega));
>> subplot(122) plot (Omega, 20+10g10 (abs (HBP)),?k?);
>> axis([-pi pi -40 2]); grid;
>> xlabel(?NOmega”); ylabel(? |H (BPJ(e(jNOmega)) |");
>> set(gca,'xtick?, [-pi:pi/3:pil,'xticklabel?, ['-p ?;...
Doo Op O apr Op po),
*fontname”,?symbol');
DT fim
ab
Ss d+ a *
not
Figure $5.M-10c: Order-8 Chebyshev Type II BPF with passband between 57/24 and
17/24.
(d) Order-8 Chebyshev Type II BSF with stopband between 57/24 and 117/24.
Notice that the command cheby2 requires the parameter N = 4 to be used to
obtain a (2N = 8)-order bandstop filter.
>> Omega c = [5+pi/24,11+pi/24]; Omega = linspace(-pi,pi,1001);
>> [z,p,k] = cheby2(4,20,0mega c/pi,'stop');
>> subplot (121) ,plot(real(p),imag(p),'kxº,...
real(z), imag(z), 'ko”,cos(Omega), sin(Omega),ºk');
>> axis([-1.1 1.1 -1.1 1.1]); axis equal;
>> xlabel('Re(z)'); ylabel( Im(z)');
>> [B,A] = cheby2(4,20,0mega c/pi,'stop');
>> HBS = polyval(B,exp(j+Omega))./polyval(A,exp(j*Omega));
>> subplot (122) plot (Omega, 20+10g10(abs (HBS)),ºkº);
>> axis([-pi pi -40 2]); grid;
>> xlabel(*NOmega”); ylabel(” |H (BS)(e"(jNOmega)) |');
>> set(gca,xtick”, [-pi:pi/3:pi],'xticklabel?,['-p *;...
POP OP O Op op po.
*fontname”, symbol);
To demonstrate the effect of increasing Rs, consider magnitude response plots
for Chebyshev Type HI LPPs with Rs = (10, 20,30).
>> Omega c = pi/3; Omega = linspace(-pi,pi,1001);
>> [B,A] = cheby2(8,10,0mega c/pi);
>> HLP1 = polyval(B,exp(j+Omega))./polyval(A,exp(j+Omega));
>> [B,A] = cheby2(8,20,0mega c/pi);
>> HLP2 = polyval(B,exp(j+Omega))./polyval(A,exp(j*Omega));
>> (B,4) = cheby2(8,30,0mega c/pi);
o
293
ima)
at
note, a
Figure S5.M-10d: Order-8 Chebyshev Type IL BSF with stopband between 57/24 and
1x/24.
>> HLP3 = polyval(B,exp(j+Omega))./polyval(A,exp(j*Omega));
>> plot (Omega, 20+10g10 (abs (HLP1)),*k-?,...
Omega, 20+10g10 (abs (HLP2)), *k
Omega, 20+10g10 (abs (HLP3)), *Xk
>> axis([-pi pi -40 2]); gria;
>> xlabel("NOmega'); ylabel(” IH (LPHeLjNOmega)) 1º);
>> legend('R s = 10º,ºR.s = 20º,ºR.s = 30",0);
>> set(gca,'xtick”, [-pi:pi/3:pil,'xticklabel?, ['-p 25...
POP OP OD dr Sp poT,.
*fontname”, symbol);
Figure $5.M-10d: Changing R, for a digital Chebyshev Type II filter.
Thus, increasing Rs tends to broaden the transition bands of the filter.
5.M-11. Factored form is used to plot roots, and standard transfer function form is used to
compute magnitude response plots
(a) Order-8 Elliptic LPF with Q. = 7/3.
>> Omega c = pi/3; Omega = linspace(-pi,pi,1001);
294