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Solutions to Problems from Chapter 4
Problem 4.
By the definition of the Laplace transform we have
! "#$ % '&)(#*,+,-.
Using the change of variables as /
0. 1 2 43 , we obtain
8
( #9+;:
# /
'=<>/ ?
& ( @*
+;:'A=B C
/
ED
&(#*,+ :
/
& (
GF C
'H
ID
&(@*8+ :J <#K ?
Hence, we have the following Laplace transform pair
8 L % ' ! NM D
&)(#*,+ :"J <>K ?
3 O5P
Problem 4.
Using the fact that
#Q RM J K
and
#SQTU' VM K
J K
W X#6 ( , we have
VY
[Z#Q -. (^) Z]\
^ _Y
.
S 8TU K
\
K `aK
J K
b ced,6f(hgiN j S,TU d,6f(#gk K
ZJ K
K
d6l(hg= m ST8U d,6f(g
and
Y
4n; - ) (^) no\
p Y
)
S ;ZU K
\
K q$K
Z J K
K
d 6 l( g m S8TU d 6 f( gr j S;ZU d 6 f( g
K
nJ K
K
Zed6 ( g= K
S'TU d,6 ( g= m SZU d,6 ( g
Continuing the same procedure, we obtain the required result
Y
^ Y
.
S s ( T U (^) \
K q K
s( TJ K
K
s( Z d 6 l( g ^uuu" m S$s^ ( Z'U^ d 6 f( gr m Svs^ ( T,U^ d 6 f( g
K
sJ K
K
s ( T d6l(@g Xuuu j Ss^ ( TU^ d,6f(g
Problem 4.
By the definition of the Laplace transform we have
w ! "'#,
! "#; Q &.(@*8+-.
Using the change of variables as /
x L 3 5y , we obtain
w)# '#, ! zw
( #9+;:
/
/
& (@* B{4|C^ :
(^) /
(^) D
(^) &( CF+:
(^) # /
'& ( FCH^ - /
(^) D
(^) &( CF+:J <
K ?
CHAPTER 4
Hence, we have the following Laplace transform pair
}~,) %'#~,.L%N L.
~ f
8;%@
^
Using the same line of proof, we can also show that
}~. ~.L% [~ N .
~ v.
(^) ,Q2 N
¡
.
~ ¢);
(^) ,;c
¡
This result will be needed for Problem 4.10.
Problem 4.
Using the Euler formula we have
[~ £I N¤ ¥'¦
¥z¦
,§ (^)
(^) N ¨¡^
£ (^) )© ©
Problem 4.
By the definition of the Laplace transform we have for a semiperiodic signal (
} ~Q£]}#~Q (^) «ª
m¬]^ and }~;£®
¯° )
±²³}~Q´w£
~N£¶μ
·
}#~Q #¸
,¹ (^) 1£»º
·
}~; #¸
,¹ (^)
(^) μ
·
º
(^) }~Q @¸
¹ (^) £ ¼ ~
(^)
(^) μ
·
º
(^) }~ #¸
,¹ (^)
Introducing in the last integral the change of variables as ½
£ ¾1 (^) ª , we obtain
μ
·
º
}#~Q )#¸
¹ b£¶μ
·
}~ ½
(^) «ª .@¸'¿À)Á º LÂ
¹ ½
£ )#¸ º
μ
·
} ~ ½
(^) 2ª )#¸À
¹ ½
£ .@¸ º
μ
·
} #~ ½
.@¸À
¹ ½
£ .@¸ º
~
In the next to the last step in the above derivations we have used the fact that the signal is semiperiodic. Hence,
we have the required result
~
1£ ¼~
.@¸ º
~
à ~
£
@¸ º
¼ ~ N£ μ
Ä
Å ÇÆ
.
Å
º
¸
¼ ~
(a) Applying this formula to the signal
È$É ~ L'#~ with
ª £ 4Ê#Ë , we obtain
¼ ~e£ÍÌ,Î
·
È$É ~ .@¸
¹ b£ #¸
© ©
~^ 'È$É ~;Ï[~Ð^
Æ ©Ñ4Ò Æ ¦
£ @¸ º © ©
~ N1 (^) #¸
© ©
m£
© © ¤
#¸^ º
§
Using the formula, we get the familiar result
±w²È$É ~~z´k£ ~
@¸ º
©
© ¤
#¸^ º
§ £
(^) © ©
(b) For the semiperiodic train of rectangular pulses given in FIGURE 3.20 with
ÓW£
©
£ Ô
£ , we have
¼~ £ º
Æ © ·
@¸
¹ (^) £
¤ @¸
© ¸
§
CHAPTER 4
(b) 5 76981: ;=<?>@BAC5BD;E<FHGI>
: %JLKNM OP ; E<QF R"> STA 5 UDV;$<FWRXFYR%>
: %JLKNMZ[O\K^]_V]`\P ; $<FaR%>S
AC5UDcbE;E<QF R">
: FdG(;$<F R">feCR"g
J LKNM"JLKNMZ[O\KV]/`\P ; $<F R%>"STA
J 1K^M%JLKNh 5 UiL<
: J 1K^M,OP ; E<?>%j
FBG
J 1K^M%JLKNh 5 i9<
J 1K^MO\P ; $<?>%j7e
J LKNM JLKNh 5 i
J LKNM,O\P ; E<?>%jkA
J LKNM J1K^hl G
; /mXednI> o
F
G
;meYn>
: e
R
mpenrq
(c)*
8 o
; E<?>A<
J LKsOP ; /G9Avu
<
J K sO < xwyR1z{n
|
<x}yRIz{n~
5 B6% o
;$<?>@BA
]? M
<
J 1KrO/JLKNh O < f
] M
<
J LKsOJLKNh O <
AY5 i<
J LKsOP ; $<?> j F
o "
:
<
J LKsOJLKNh O < A
R
; meR%>
: e
R Iz{n
mXe R
J 1KV]? MZ[h_V]` F
R
m eR
] M
J LKNZ=h_V]`[O <
A
R
;\mpeCR%>
: e
R z n
meR
J LKf] MZh,_V]/` e
R
; /mXeR">
: x
J LKf] MZ[h_V]` F WRA
R
; meR">
:
J LKf] MZ[h_V]` e
RIz{n
mXe R
J 1KV]? MZ[h_V]`
A more elegant solution to this problem was derived by Professor LeBlanc from Lulea University of Technology,
Sweden, using the fact that
P ; \G <FHt>A
P ;$<FaRz{n1>
and the time delay property of the Laplace transform, leading to 57698 o
;$<?>3@BA 5 u <
J LKsOP7 < QF
t
G
A 5 u
< F
t
G
e
t
G
J K ;
O K p %_7 (^) >P7 < F
t
G
A
J1K 5 u
<F
t
G
J K ;
O \Kp (^) >P < QF
t
G
e
t
G
J 1K 5 u
J K ;
O \K (^) >P < F
t
G
A
J LK Z=]_^h` R
; \mpeCR%>
: e
t
G
J 1K %Z=]_^h,` R
m peCR
(d) 5 76%811;E<?>@AC576%n^ <
P ;$<FaR%>Ve R
|
<
P ;[<FHtI>3@BAnL5B61;E<FaReR">
P ;[<FYR%>@peR
|
5 761;$<FHtxeYtI>
P ;E<QFtI>@
An15761;$<FWR">
P ;[<FaR >@peYn75B
P ;$<F R%>@peCR
|
5 B61;E<FHtI>
P ;E3@peat
|
576
P ;E3@
n
JLKNh 5 B6<
P ;E<?>@pen
J LKNh 5 76
P ;E<?>@pe R
|
J LK o
h 5 76<
P ;E<?>@pet
|
J LK o
h 5 76
P ;$<?>@BAn
J LKNh R
m
: e Yn
J LKNh R
m
e R
|
J 1K o
h R
m
: e Yt
|
J 1K o
h R
m
Problem 4.
(a) 5 B6%8^ ];$<?>@BA^576
P ;$<FWR">?U;+<?>@BA 576
P ;$<FWR >([U;\V;$<FWReCR%>,>@
AY
P ;$<FaR%>[U;/V;=<FaR%>,>4 ¡I^;\^>^e
P ;$<F R">4 ¡f;\V;$<FaR >>(,[U;/^>@UAFx5B
P ;E<QF R">+[B;V;E<QF R">>@
AF
J LKNh 5 B
P ;$(,[U;/+@UAyF
J LKNh
m
: e Y
:
SOLUTION MANUAL for LINEAR DYNAMIC SYSTEMS AND SIGNALS by ZORAN GAJIC
(b) ¢7£9¤1¥ ¦=§?¨3©ªC¢ «I¬¦$§?¨,1®r¯?°(±"²1³V¦/´V¦E§Qμd¶¨¨%·kªY¢ «I¬¦$§?¨,1®r¯?°"¸ ±²I³V¦/´+§?¨+±²I³V¦μB¶I´^¨^¹³,º»U¦\´+§?¨(³º[»U¦\¶I´^¨=¼?·
ª ¢ «¬¦$§?¨,1®r¯° ±²³f¦\´+§?¨ · ª ½
¹ ¾
¦ ½
¹a¾¨
¥ ¹´
¥
(c)
Note that
³ º[»U¦/´V¦E§μa¿¹¿"¨¨Xª³º[»U¦\´V¦À§μa¿ ¨¨4±3²I³V¦/´^¨f¹Y±²³^¦/´V¦=§μa¿ ¨¨(³,º»U¦\´^¨pªÁμT³º[»k¦/´V¦$§μ ¿"¨¨
¢B£1¤. Then ¯
¦E§?¨©Bª ¢ «§? ®
¥ ° ³º[»U¦/´+§?¨,¬^¦E§μa¿ ¨· ªC¢B¦E§Qμ ¿¹C¿%¨ ®
¥ à ° ® ^Ä ÅVÄÆ ³º[»U¦\´V¦$§μW¿¹C¿%¨,¨?¬¦[§μa¿ ¨"Ç
ªC¢UÂfμB ®
¥ ¦$§μY¿%¨ ®
¥ à ° ® fÄ/Æ ³,º»U¦\´V¦$§μ ¿%¨¨,¬^¦E§QμY¿%¨ÇÈμ¢UÂ^ ®
¥ ®
¥ à ° \®VÄ/Æ ³º[»U¦\´V¦$§μW¿ ¨¨,¬^¦E§Qμ ¿"¨"Ç
ªÉμL®
¥ L®Ê,¢ «§?L®
¥ °(³º[»7¦´+§?¨¬¦$§?¨ · μËL®
¥ L®NÊ¢ «L®
¥ °³º[»U¦/´+§?¨,¬^¦E§?¨ ·
ªyμBL®
¥ 1®^Ê
¶ I´V¦ ½
¹¶I¨
Ì
¦ ½
¹d¶¨
¥ ¹Y´
¥ Í
¥ μ 1®
¥ L®NÊ
´
¦ ½
¹ a¶1¨
¥ ¹Y´
¥
(d) ¤Î¦$§?¨ÏªY§?L®
¥ °I³,º[»U¦$§μ˶¨¬¦$§μH¶¨ªy¦E§Qμ¶x¹Y¶I¨L®
¥ Ã ° ®
¥ Å
¥ Æ ³º[»U¦E§Qμ¶I¨¬¦$§μd¶I¨
ª ®
Î ¦$§μd¶1¨ ®
¥ Ã ° \®
¥ ƦE§μH¶I¨³º[»U¦$§μd¶I¨¬¦$§μH¶¨^¹Y¶I ®
Î ®
à °\®^
¥ Ƴº[»U¦$§μd¶1¨¬¦$§μH¶¨
Ð
®
Î ¢ «§? (^) ®
¥ (^) °³,º»U¦E§?¨¬¦$§?¨ (^) · ®
¥ (^) Ê ¹Y¶I (^) ®
Î ¢ « ®
¥ (^) °³º[»U¦E§?¨¬¦$§?¨ (^) · ®
¥ (^) Ê
ªL®
Î ¶ ¦ ½
¹d¶I¨
Ñ
¦ ½
¹d¶I¨
¥ ¹¿Ò
¥ L®
¥ ÊV¹a¶1®
Î ¿
¦ ½
¹ Y¶I¨
¥ ¹ ¿
1®
¥ Ê
Problem 4.10*
¤ ^¦=§?¨
Ð ÔÓ
¦ ½
¨ XÕ ¢ «Ö L®
¥ °\¤^¦=§?¨ · ª Ö
Ó
¦ ½
¹d¶I¨×Ø¢Ù ÚÛÈÜÝ
Þß
¶ I¤¦À§?¨,àL§ áâ
ã
ª
¶
½ Ó
¦ ½
¨
¢Ù ÚÛ Ü Ý
Þ ß
¤ ¦=§Qμdä4¨(³,º[»B¦ Öä(¨,à äÏá â
ã
ª
Ó
¦ ½
¨
Ö
½
(^) ¥ ¹ Ö
(^) ¥ åIæÁçè4é ±²»(êI²1ë[ì (^) çº[²»yísî,²Ií (^) éîç\æ
Using the second result established in Problem 4.3, we have ¤¦¾ §μH¶¨¬¦/¾ §μH¶¨4±²³f¦EïQ§?¨
Ð
¿
ð
® ¦ñóòLô/õ
ö
¨¥
Ó ø÷
½
¹ Ëù ï
¾ ú
¹
¿
ð
®
(^) ¦Iñ
ß
ô=õ
û
¨¥
Ó ÷
½
μ øù%ï
¾ ú
Problem 4.
(a) ¤ Ä (ü\ý
Å VþBªÿë[º Ê
Ü
£ ½ Ó
Ä
¦ ½
¨3©ª ë[º Ê
Ü
(^) ½ ü
¶ ½
¥ ¹ ¶ ½
¹d¾ (^) þ
¦ ½
¹¿%¨3¦ ½
¹d¶I¨3¦ ½
¹d¾¨
ª ¶
SOLUTION MANUAL for LINEAR DYNAMIC SYSTEMS AND SIGNALS by ZORAN GAJIC
The above results for the coefficients * can be verified using the MATLAB statements
k=5; z=[]; p=[-1 —2 —3 —5 —10 —20]; [num,den]=zp2tf(z,p,k); residue(num,den).
(b) *
b7. Ob7{D
#&n
b
b7.
]
bL8.
]
b7D
b7.
/g (^)
J
5W
F>
>W
F
F
where
b7. K¦ §>¨
¢«^
b
K®
j©
F
j©
b° ©
(^) ¢!¡x±Q²´³ §μ
F
/g
F
b71*
b° ©
*!¡ ±Q²³
§>μ
/g
F
G©
bL
W
b° ©
!¡x±Q²³ §μ
F
b7{ ¦§>¨
Comment: MATLAB 6.1 failed to find the coefficients in this case (den=[1 14 80 240 400 352 128];
num=1; residue(num,den)).
(c)
©
¸L
/D
(^) b7
(^) b7.
W
¸L8.
b7.
W
W
b
I¡
6 W
F
I¡
F £ ¥
b
¸L8.
q¬
¬
bL8.
D* ¦§>¨
G¹
W
D*
b7. (^) ¦§¨
W
bL
F
b7
O ¶
CHAPTER 4
Comment: MATLAB 6.1 failed to find the coefficients »*¼ in this case also.
Problem 4.15 ½¾ ¿&ÀVÁWÂ6Ã@Ä4½#¾ Å ÀVÁWÂ&ÆÇ.ÀVÁKÆ8È,ÂÉ1Ç.ÀVÁKÆgÊ.ÂFÉÅ ÀVÁÆ{Ë*Â5Ã#Ä
È
Ì
ÆÍ ÎFÏ
È
Ì
Ð
É
ÍOÎ ÑÏ
È
Ì
,Ð
É
DÍ ÎFÒWÏ
È
Ì
Problem 4.
(a) In the first example of this problem we have first to perform the long division since Ó
Ä
ÕÔÖÄÕ×
, hence the
partial fraction expansion procedure can not be applied directly
ÌÐØ Ù^ ÌÐ
Æ
Ì
Æ
{× Ú#ÄÈÉ
Ì
É
×
Ì Ð
Æ
Ì
Æ
×ÜÛ
Î
FÝOÞ
Ì
Ð
À
Ì
É
BÈ Â6À
Ì
Æ~×.Âß
Ä
4½ Î&ÝnÞ ÈÉ
Ì
É
1×
Ì,Ð
Æ
Ì
Æ
×ß
Ä4½#Î&Ý Þ ÈÉ
Æ
È à.Ê
Ì
ÉGÈ
É
qá
à .Ê
Ì
Æ~×7ß
Ä
Bâ.À¤ÁWÂ&Æäã
È
Ê
ÍOÎå&Æ;á Ê
Í
Ð
åæ#ÅFÀçÁWÂÄ4½#Î&Ý ¾ è Ý
À
Ì
Â!Ã#Ä;¿
Ý
ÀVÁWÂ
(b) ¿
Ð
À¤ÁWÂKÄB½ ÎÝ ¾è
Ð
À
Ì
Â!ÃYÄ4½ ÎÝ Þ
È
Ì
,Ð
À
Ì
É
GÈ Â5À
Ì
É1×*Â(ß
Ä
B½ ÎÝ Þ
Ý5Ý
Ì
É
Ý
Ð
Ì
Ð
É
Ñ
Ì
É
GÈ
É
é Ì
ÉD×ß
Ä4½ ÎÝ Þ
Æ
Yêìë í.Ë
Ì
É
êìëîË
Ì,Ð
É
È
Ì
É
GÈ
É
Æ
#êìë ×.Ë
Ì
ÉD×ïß
Ä
ã Æ
Ê
á
É
È
×
Á
É1Í Î(å Æ
È
á
ÍÎ
Ð
å¤æ Å ÀVÁWÂ!ð
é
Ä
È
Ì
Ð
À
Ì
É
GÈ Â(ñ Ïò&Î^
Ð
Ä
jÆ
È
á
Ñ
Ä
È
Ì,Ð
À
Ì
É
1×*Â(ñ Ï>ò&ÎÝ^
Ä
È*ð
Ý
Ð
Ä
È
À
Ì
É
4È Â5À
Ì
É1×*Â(ñ Ï>òFó^
Ä
È
×
ð
Ý5Ý
Ä
qô
ô
Ì
Þ
È
À
Ì
É
GÈ ÂÀ
Ì
ÉD×*Â(ß#ñ ÏòFó^
Ä
ªÆ
Ê
á
(c) ¿ Ñ
ÀçÁWÂÄD½^ ÎÝ^ ¾ è^ Ñ
À
Ì
Â5Ã^ÄB½ ÎÝnÞ
Ì
É
1×
À
Ì
ÉBÈ,Â5À
Ì
,Ð
É
×*Â(ß
Ä
D½ ÎÝ*Þ
Ý
Ì
É
gõ*ö ×
É
I÷
Ý
Ì
ÆõOö ×
É
Ñ
Ì
É4Èlß
Ñ
Ä
À
Ì
ÉD×*Â
À
Ì Ð
É
×. ñÏò&Î&Ý^
Ä
È
Ê
ð
Ý
Ä
Ì
É
D×
À
Ì
ÉBÈ,Â
Ù
Ì
Æøõ ö ×Ú ñÏ>ò&Îù5ú^
Ð
Ä
Æ
7õ ö ×ûÉ×
Ù
Æ7õ ö ×ûÉBÈ Ú
Ù
Æ#×,õ ö ×Ú
Ä
×Æõ ö ×
Æ
á
Æ~×nõ ö ×ü
Æ
á
É
×nõ ö ×
Æ
á
É×nõ ö ×
Ä
Æ
á
É
gõ*ý ö ×
× á
Ä
ªÆ
È
þ
Égõ
ö ×
Ê
Ä;ÿÉõ
Û
ñ
Ý
ñ
Ä
È
×
ð
Ý
ÄÈ ê
ë á
í
½ ÎÝ Þ
Ý
Ì
É
~õ ö ×
É
ì÷
Ý
Ì
Æõ ö ×
É
Ñ
Ì
É;È^ ß
Ä
×
ñ
Ýñ
Í
å
À
Á
lÉ
Ý
ÂFÉ
Ñ
ÍÎ(å ÅFÀçÁWÂ
Ä
Æ
ö × ,ÁÉGÈ ê
ë á
í
É
È
Ê
Í
Î
(å Å ÀVÁWÂ!ð
Ý
Ä É{õ ÄjÆ7õ ö ×
Û
ÄGêIð ÄªÆ ö ×
(d) ½ ÎÝ^ ¾è
é
À
Ì
Â!ÃYÄ4½ ÎÝ Þ
Ì
É
1Ê
À
Ì
É8×.Â5À
Ì
Ð
ÉD×
Ì
ÉD× Â ß
Ä
4½ Î&Ý Þ
Ý
Ì
É
BÈÉgõ
É
I÷
Ý
Ì
ÉBÈ7Æõ
É
Ñ
Ì
É× ß
Ä
Ù
×
ñ
Ýñ
Í
å
À
Ý
Á
lÉ
Ý
ÂFÉ
Ñ
ÍÎ
Ð
åÚÅFÀçÁW ð
Ý
ÄjÆÈ*ð Ý
ĪÆÈ*ð
Ý
Ä
Ý
Égõ Ý
ÄjÆÈ7Æøõ
CHAPTER 4
O# < 8¡9¢
<£1¡( &¡(8¤ ¥B¦C§C¨
7¡l¢
&¡=¬Q
8¡¬8®¯ A¤ ¥B¦C§C¨M§ °
C±
C¨#¶ ®
©_·^
_±
¸¹aº
¬c»
¼* «y½^ P¾
_ÀaÁcÂ<Ã
¡ ¬¸¹ º^
¬Q» e¡
§£ÀÄeÅ
Z½
(d) Æ
e¨*ÇQÈCÉ
QËÊ^
Æ
§C¨kÌ
*£7¡¸oÍ
Æ
e¨<Ì
UÎ
®k¯
ÏÍ
Ð
O¤
<ÑaÒ
À
UÁ ÂÃ
/Ó
e¡
Ô
Å
«y½ ª Õ
Ö ª Ó
×
where
*£&¡}¸ ¤¥B¦_Ø
8®9*¯ ¤¥B¦C§ £°
Ù
Ù
*Ú
³M´Oμ
e¨
®#^
Ù
¬QÚ
¬¬Q¹ º~¢ Ú»
Hence, the required result is given by
É
«y½ Û
¬QÚ
Á
Â
Ã
¡¬ ¬Q¹ º~¢ Ú»_¡
_Ü
Å
Z½
Problem 4.
(a)
ÞÝ àß|á|â
¥/ãØ1ä
l 8¡¬
7¡¬Q
7¡l £å^
=ÖUª ¼
ÐÖæ Ô çß|á|â
¥Bãè ä
l 8¡¬
7¡¬Q
8¡9 £å^
Using the expression for
y½
obtained in Problem 4.14, that is
Z½ PÐ/¿
À
}é
§ £À ¡(
§AÀ
Ô
Å
Z½
, we see that
«Öæ
¬v¡
and
ÞÝ êÖ
so that the results obtained via the initial and final value theorems are
consistent with the actual function values at zero and infinity.
Problem 4.
(a) Æ
C¨cÇkÈ
QÊ^
Æ
§C¨
ä
§ë ¥
8¡ ¬c £ å
Æ
C¨
ä
&¡=¬Q £ å
Æ
e¨
ä
_ëM¥
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SOLUTION MANUAL for LINEAR DYNAMIC SYSTEMS AND SIGNALS by ZORAN GAJIC
Note that the term ñ<ò_óBô only indicates a time delay of five units.
(b) õ
ò
eöc÷<ø1ù<ú/ûQüËýÿþ
õ
òeö
ñ
ò
ó ô
û
ù
ú/û ü
þ
õ
ò
Cö
û
ù
ú
Þû ü
õ
ò
eö
ñ
ò
ó ô
û
ù
úBû<ü
û
ù
ú/û<ü
þ
öö
û
ö
ù
û
ù
û ö
ù
8þ
û
ô ^
þ
öMö
þ
"!
!
û
û
# ô$ % ^
þ
úÞû <ü
ù
ô& % ^
þ
(^)
'
þ
û
ù
ô$ Cò^
þ
' (^)
õ
ò
eö
û
ù
ú
/û ü
þ
õ
ò
eö
)( '
û
)(
û
ù
)( '
û
þ
+*
'
-,
'
ñ
ò
/.1032 ú
,
ü4þ^54 _ú^
,
ü
õ
òCö
û
ù
ú
Bû<ü
ñ
ò
ó ô
û
ù
ú
Þû ü-
þ
4
ú
,
ü
4
_ú
,
ü
þ *
' 7 ,
8' ñò9. 0 2 ú
,
ü
'
ú
,
ü
' ñò ;:<. ò_ó&= 0 2 ú
,
ü
1þ?4<ùOú
,
ü
(c) This function is similar to the function in part (d) whose Laplace inverse is derived in detail below. here, we
use MATLAB to find the Laplace inverse of the required function, which produces
4
ú
,
ü&þ *
@A
ú
,
ü
@ ñò
ù
:<. ò = 0 2 ú
,
ü
(d)
4
CB<ú
,
ü&þ
õ
òeö^ ÷køDB<úÞû*üý#þ
õ
òeö^
û
úÞû
ù
@
üû
ñ
ò
_ô
FE
õ
òeö
û
úÞû
ù
@
üBûG
þ
õ
ò
Cö ö
û
FH
JIö
û
H
û
þ
K4 ú
,
üFLCM-N
õ
òeö
û
úû
ù
@
ü
Bû
ñ
ò
ô þK4 ú
,
ü1þK4 Bú
,
ü
þ
û
û
ù
@
ô
$ %
þ
@
ö
þ
û
úÞû
H
üBûG ô& CòPO^
ù
þ
H
ú
Q
H
ü
ú
H
ü
þ
R
H
@ þ5SQ>HTþ
ö
õ
òCö ö
û
>H
UI ö
û
H
û
þ
WV
- ö
ñ
)X .UY9Z[ ú]\
,
^
ö
ü
`_
2
ú
,
ü
a ö
þ
b R 6
^
ö
þ
6 dce6fCg
4 _ú
,
ü&þih b@ 6 Y9Z[ ú
Q ,
6 Jc 6 f g
ü
%
@%j 2 ú
,
ü
lk ö
þ
H
þ
KmnoHC\
Note that ñò_ô indicates a time delay of one unit. This term must not be included in the procedure for finding the
coefficients
ö
and
. Hence
4 Bú
,
ü1þ
õ
òCö ÷kø Bú/ûQüý#þph b@ 6 Y9Z[ ú
Q
ú
(^) ,
ü
%
6 Jce6f g
ü
D
@ j 2 ú
,
ü
(e) Applying the MATLAB function residue to ø
ó
úÞû*ü}þ^ ú/û^
ü
( rq
û
ù q
û
ù
s )s
, we obtain the coefficients
þ ÷
QtUce6 u
H tJc
6
Cv /tJc 6 J
ý
at the poles k
þxw
u b
H /tJ/tUy
. The corresponding result is given by
4 ú
,
ü&þ{z
t Jc 6
(^) ,
}|~tUc
@ t
(^) Y9Z[ b ,
6Uc
g$%
2
ú
(^) ,
ü
E
4
ó
ú
,
ü4þz
t Uce
(ú
,
ü
% }|~tUc
@ t
(^) YZ[ b
ú
,
ü
%
6Jc
g9%
2
ú
,
ü
SOLUTION MANUAL for LINEAR DYNAMIC SYSTEMS AND SIGNALS by ZORAN GAJIC
É Ê Ë Ì
Í$Î&Ï Ð
Ê Ë Ì
Í9Î ÏCÑ
Í
Ë
Ð )ÒoÓ
Ì
Ô ;Î;ÕÏ Ñ
Ó WÖ× Ì
(^) Í/Î Ï Ø ÒoÓ
Ì
Ô ;Î (^) ÏÕ ÙÚ
Ø Ò oÓ
Ì
Ô ;Î ÕÏ
Ð
Ê Ë Ì
Í9Î ÏCÑ
Í
Ë
Ð )ÒoÓ
Ì
Ô ;Î;ÕÏ Ñ
ÛÜ Ý8ÞÕ§ß ÏCàPáâã#äGå^
Ò
Ó æ
ç Pè Í é
ê&ëíìrîïê/ð
Ó æ
ñè^ é
Ø Ò oÓ
Ì
Ô ;Î (^) ÏÕ
Ü CÝÞÕ ß ÏCàdã&ò<óäGå^
Ò oÓ æ
ç dè Í é
ê&ëôìrîïê/ð`õ
Ò ö æ
ç Pè Í é
where we have used the results
ÜÝ%÷ àã&ò<ó^ î ïøGê/ð9ìrî<ê/ð
Û ø
î Ê Ë ù (^) ð
Í
Ë ø
Í õ
Ü Ý ÷ àá9âã^ î ïøGê/ð&ìrîïê/ð
Û Ê Ë ¿ù
îÊ Ëù^ ð
Í
Ë ø
Í
In the case when
Ò ú æ ç Pè Í é the roots are real and distinct, say û Ì
õ
û Í9ü^ , and the solution has the form ý Ì
Ü $þ Þ
à Ë
ý
(^) ÍÜ&þ Õà .
For ý
É ñ and ý
É ?ÿ , due to double and triple poles, the solution becomes computatioally involved and it is omitted.
Problem 4.
(a) It is known from the table of the Laplace transform common pairs that
ñ
ø Ê
îÊ
Í
Ë ø
Í
ð
Í Û ê/ìrîïê/ð
ã $ò<ó
î
ïøGê/ð
ç
Ê
î Ê
Í
Ë
ç
ð
Í Û ê/ìrîïê/ð
ã $ò<ó
î
ñ
ê /ð
so that
ç
î Ê
Í
Ë
ç
ð
Í É
æ
Ê
ñ
~ñ
Ê
îÊ
Í
Ëñ
Í
ð
Í Û
à
è
ì %î
è
ð
ã &ò<ó
î
ñè
ð
è É
à
è ã &ò<ó
î
ñè
ð
è É
æ
ç î
ã &ò<ó
î
ñ
ê /ð
Ó
ê
á âã
î
ñ
ê /ð&ð
The integral is evaluated by using formula (B10) from Appendix B.
(b) Using the result known from the table of the Laplace transform common pairs given in part (a) of this problem,
we have that
ç
Ê
î Ê
Í
Ë
æ
ð
Í É
ñ
ñ
ç
Ê
î Ê
Í
Ë
ç Í
ð
Í Û
æ
ñ ê /ìrîïê/ð
ã &ò<ó
î
ç
ê /ð
Problem 4.
(a)
Ì
î ê /ð
É Ý
Ì^
Ì
îÊð
É Ý
Ì^
Ê Ë
ÿ
ÊîÊ Ë
æ
ð 9î ÊË^
ñ
ð
É
Ý
Ì
ý
Ì Ê
Ë
ý Í
Ê Ë
æ Ë
ý
ÊË
ñ
É
ý Ì
Ë
ý
(^) ÍÜÝ à Ë
ý
ÜÝ
Í
à ^
ì rîïê/ð
ý Ì
É Ê Ë
ÿ
îÊ Ë
æ
ð9î ÊË^
ñ
ð !
É
ÿ
ñ õ
ý Í É Ê Ë
ÿ
ÊîÊ Ë
ñ
ð !
Ý
Ì
É
Ó ñ
õ
ý
É Ê Ë
ÿ
ÊîÊ Ë
æ
ð" !
Ý
Í É
æ
ñ
Ì
îïê/ð
É
ÿ
ñ
Ó ñÜÝà Ë
æ
ñ
ÜÝ
Í
à %^
ì rîïê/ð
(b)
Í î ê /ð
É
& Ý
Ì
Í
îÊð'
É
& Ý
Ì
(^) Ê
î Ê
(^) Í
Ë
æ
ð 9î Ê Ë
ñ
ð
É
( Ý
Ì
ý Ì Ê
Ë*)
Ë
ý
Ì Ê
Ó^
Ë
ý-
ÊË
ñ
É/. ñ
ý Ì
Ü÷àá9âã^ î
10 -ê Ë^32
ý Ì
(^) ð Ë
ý
Ü Ý
Í à
ì rîïê/ðõ
ý
É Ê
Ê
Í
Ë
(^) æ
Ý
Í É
Ó ñ
CHAPTER 4
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Problem 4.
(a)
s
u
B
I
Q
t
C
<C&Q
I
Q
?C
t
t
C
D
] ª
I
t
<?C
t
t
C
D
] ª
I
d
]
D
u
u
B
(b)
s
vu
B
I
<?C
R
t
C³ª<?C³D
I
CnQ
D
<CD
I
´D }I
t
vu
B
(c)
s
u
B
I
D<
t
C<CD
C³ª-<
t
Cª-<CnQ
I
D
CnQ
<?CQ B
t
C
<?C QMB
e
R
I
u
}I
C
D
u
t
}I
f
vu
B
CHAPTER 4
Problem 4.
The unit ramp responses can be obtained either integrating the system step responses obtained in Problem 4.27 via
the formula (4.48) or by finding "!$#&%'!(').
(a) *,+.-0/
BA
C
!9D@
!E6F;
G
IHJ#K38 $L^
M+J-N/O
"!$#NPQ3SRT4>
BAC
H
U687H (
;'V XW^6 D
V (&WZY[^ ]HJ#
(b) *,+0-0/
^!#3_
^!#3 4
`! ( 6
ba cd> ! @
ace !E6F;
a c
a fce4&g !: g
a fc
a c e4Ng ! @ g
G
+J-N/O
IHJ#3h ^ L
i+J-0/
^!#Pj3lk 4 ; @
$a
V
m(&W (^6) on >
qprs
H
a tfc tu
&v [ ]HJ#
(c) (^) *,+0-0/O
"!$#E
^!#3 4
.w ^!E
t
D
t
D
w
D
t
G
+J-N/O
IHJ#3h ^ L
i+J-0/
^!#Pj3 R @
t
t
H
'a
H
a
H^964
t
Vmx&W^ Y [ eHJ#
(d)
,+J-N/O
w
"!$#3y ! (
w
"!$#E
`!
#0 ^! ( 6 8!
C
g z
!E6
g z
g z
{@ g z
G
+0-0/O
w
IHJ#3h L
,+J-0/
w
^!#P3 (^) k @
C68;H 6 F; V |W 6
C
n
V~}
W
p 0r's
H
O
>'a u
v
[
IHJ#
Problem 4.
The unit parabolic responses can be obtained by finding ;' "!$#&%'! 9 ).
(a) *Q1-0+
"!$#K
^!#:
!9 2!K
c A
c d> !(
C
!E
C
!E6F;
G
]HJ#3h L
^!#0P\3D"
c A
c d>
H
m 4 'cd>
H
C
V XW 6
a c
V (`W]
[ IHJ#
(b) *1-0+
^!#
^!#
^! ( 6
#N ^!E6F; #
a c d> !
6 a ce !E6F;
oa c
E
g 'a c
C
!E
g
oa c
(^) @ g'a c
C
@ g
G
1 -N+
]HJ#3 L
^!#0PQ3lk @ a c d>
HU
a c e
V
m(&W 6 <; n >
p 0r's
H
O
t
c
C
u
#"v [ IHJ#
SOLUTION MANUAL for LINEAR DYNAMIC SYSTEMS AND SIGNALS by ZORAN GAJIC
(c) 0
^
b'f$
^
o'
]¡Jh¢£¤.¥
^0¦\S§5
f$
U
f$
&¬®Q¯ I¡J
(d) \N
^°
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N^8
μ±
E
'¸e
K ¤
'¸e
E ¤
I¡JK8¢£¤$¥
\0
"$¦Dº
M»¼f¸d'ª½±¾¿À
¡m
Á
"Â ¯ ]¡J
Problem 4.
According to formula (4.40), the steady state system response to the input signal à ®¡J:ÅÄ^
¯ ]¡J
is obtained via the
application of the final value theorem,
Æ`Æ
`Ä
, which requires that the system transfer function has all poles
strictly in the left half of the complex plane.
(a) The required solution does not exist since ¤
^^ has a pole at the origin.
(b) The required solution does not exist since "$^ has a complex conjugate pole on the imaginary axis.
(c) The required solution does not exist since "$^ has a double pole at the origin.
(d) The required solution exists since "$^ has all poles strictly in the left half of the complex plane. The
solution is given by
Æ&Æ
`Ä·
Ä·
.
Problem 4.
(a)
^
Ç
^
K
$K
È
^
Â
^É":
Ç
Z
È Ê
®¡J
Ê
Ê
®¡J
Ê
¡J
Ê
I¡J
Ê
Ê
à ®¡J
Ê
Ã
®¡J
(b)
"$K
Ç
FE
È
Â
^Å
Ç
È Ê
®¡J
Ê
Ê
Z¡J
Ê
¡J
Ê
]¡J
Ê
®¡J
Ê
Ã
®¡J
Ê
(c)
^
Ç
^
F $ È
8' Â
^
Ç
^
È
Ê
®¡J
Ê
Ê
]¡J
Ê
¡J
Ã
®¡J
SOLUTION MANUAL for LINEAR DYNAMIC SYSTEMS AND SIGNALS by ZORAN GAJIC
Problem 4.34 - .,/
/ 547698:0<;>=?/A@CBDAEF4HG=?/IKJ L&@CBMDNEO4PG=?/IKQ LR@#BDAEF4SB
T
.&U (^8)
T
(^) ;AV
T
Q/ @BD E V
T
/I J L@BD E V/ IKQ L @BD E 1 3
T
U 8
T
;WV 3
/ @BD E 1
U 8
T
;)
G
T U 8
T
;X
G
T
. 1 ! T
1
G OY
T
Q 1
T
1
B 1 3Z
G
T [
4
T
. 1 T
Q 1 ! T
1
G
8
T
. 1! T
1
G +;
T
To find the Laplace inverse (to perform the partial fraction expansion) we have to use MATLAB and its function
residue, which produces/8K0<;X4]^G)V_B`aGRbccdDfe g hRi .h j
1 !35k
B ` cG 3 l
d +e>g Q Q#m n joRpqA8CB` 3 3r
b , 1
bc` s 3 t
; uR6f8v0<;
Problem 4.
Using the MATLAB statements num=1; den=[5 2 3 2 3 1 0]; [k,p]=residue(num,den);
angk=angle(k)*180/pi; absk=abs(k), we obtain the values for the system poles and the coefficients of
the corresponding partial fraction expansion. They are given by
w
4 yxzzzzz
{
zz zz z|
B ` } r
b 1 ~
B `bb l
b
B` } r
b V ~
B`bb l
b
VOB` }b 3 Z1*~
B ` r
GG
VOB` }b 3
V ~
B` r
GG
VOB`cb 3 B
= ?54xzzzzz
{
z z z z z |
V OB`BG&bb
VOB`BG&bbV ~
B`Bb}
VOB`BG 3
s 1 ~
B G+b l VOBBG 3
sV ~
BG+b l VOBsc l l G
= N 4xzzzzz
{
zz zz z|
B Bb r r B Bb r r B`aG&b l
b
B`aG&b l
b
B` sc l l G
= ?X54xzzzzz
{
z z z z z |
G +Bcc t VG+Bcc t s}aG t VOs} G t G l
B t B
It can be seen that there are two real and four complex conjugate poles so that the solution is given by/M8:0<;W4 @
3 5k
B `aG&b l
bdD9e>g hCn i Q jo pqA8CB` r
GG 1
s}`aG t
; 1 !35k
B ` Bb r r
d D9e>g hRm:Q n jo pqf8 B`bb l
b, 1
G&Bc`c t
;E 6 8^0<;
1
@G)VB`sc^ l l
d DfeRg .n iKQ jE 6 f8v0<;
Problem 4.36 - Q/
0 (^) Q 1 r - / - 0 (^1)
s /4c
#
V _=/ @BMD E 4 PG=?/ICJL @BD E 4 7B=?f8K0<;X4dDj 6 f8v0<;
T
QU 8
T
;WV
T
/ @BD E V/ IJ^ L@BD E 1 !r T
U 8
T
; AV r
/ @B D E 1
sU 8
T
;X
c
T
V !G
T
1
G
U 8
T
;X
G
T
Q 1 r T
1
s Y (^) T
1
B 1 rZ
c
(^) T
V SG
T
1
G
[
4
T
(^) Q 1
G+B
T
1
8
T
1
c; Q 8
T
1
GR;
4
J J
T
1
c 1
JQ 8
T
1
c ; QF^
.
T
1
G
/8K0<;X4 DAJ+ U 8
T
;>4 DNJ, (^3)
T
1
c 1
(^) l 8
T
1
c ; Q^
V G
T
1
G N
4 @ 3
dD .j 1 l
0 <d D .j V_d Dj E6f8v0<;
8
T
;X
c
T
V G
T
Q 1 !r T
1
s
= 8^0<;W4ODAJ
8
T
;>O4@CcdD .jVG&B&0<dD .jKE,68^0<;
CHAPTER 4
Problem 4.37 Rv <¡
£¢¤
K <¡
¢
M: <¡¦¥7§^ <¡>¨?A©CªM«N¬O¥P¨
#ª (^) « ¡
¥ ®C¯°&©#ª«A¬O¥ ¤
Applying the Laplace transform, we have
±
R² (^)
±
(^) ¡A³
±
©ª« ¬ ³* ®#¯^ °©ª« ¬ ¢ !¤ ±
²
±
¡ ´³ ¤
©ª« ¬ ¢
²
±
¡X¥¶μ
±
¡X¥
±
The solution for
²
±
¡
is given by²
±
(^) ¡X¥ ²· ¸ (^)
±
(^) ¡ ¢
²·R¹ (^)
±
(^) ¡X¥ ±
(^) ¢º
±
¢
&¡ ¢
±
±
¢
&¡
Applying the inverse Laplace transform we obtain the zero-state and zero-input components of the system response as ·<¸^ ^ <¡¦¥S»¼«N¯R½ ²· ¸^
±
(^) ¡ ¾F¥»¼«N¯,¿ ±
(^) ¢º
±
¢
R¡ ÁÀ^
¥ »O«A¯,¿
±
¢
¢ Â
±
¢
&¡ WÀ^
¥ © #Ã «Ä ¢ Â
Ã«Ä ¬§^ <¡
·Å¹^ ^ <¡X¥»O«A¯R½ ²·R¹^
±
(^) ¡>¾O¥S»¼«N¯ ¿
±
±
¢
+¡ ÁÀ^
¥ »O«A¯ ¿
±
³
±
¢
³
±
¢
+¡ ÁÀ^
¥ © )³ ëÄA³Æ Ã«Ä ¬§^ <¡
The system transfer function and the system impulse response are given by
Ç
±
¡¦¥
²
±
¡
È
±
¡
¥
±
¢
+¡ É Ê
v <¡X¥! Ã«Ä §^ <¡
Problem 4.
(a)
Ç
(^)
±
(^) ¡X¥
²
±
¡
È
±
¡
¥ ±
(^) ¢
±
¢ ¤
¡ Å
±
¢ º
¡
±
¢ !Ë
¡
¥ ±
¢
± RÌ
¢
±
(^) ¢ Â Í (^) ±
¢ º
ª
É
®
Ì
°^ <¡ ¢
& ®°v <¡ ¢ Â Í
® K¯ °^ <¡ ¢ º
ª ^ <¡¦¥7Î ®#¯ °^ <¡ ¢
Î: <¡
The system response due to the forcing function
Î K <¡2¥ ëfÏ<Ä §^ <¡
and the system initial conditions given by
ª« ¡Z¥Ð¨O (^) ®K¯ °Cª (^) « ¡Z¥Ñª¨O (^) ®°#ª (^) « ¡¼¥ ¤
© is obtained by using the Laplace transform
±Ì
¢
±
¢ Â Í (^) ±
¢ º
ª ¬ ²
±
¡ X¥]Ò
±
©ª«N¬ ¢ ±
®¯ °&© ª «f¬ ¢
®°&© ª «f¬&Ó
¢
Ò
±
©ª«f¬ ¢
®¯ °&© ª«f¬ Ó ¢  Í
© ª « N¬ ¢
±
¢
&¡
È
±
¡
Solving for
²
±
¡
(^) ² , we have
±
¡¦¥ ² · ¸^
±
(^) ¡ ¢
² ·Å¹ (^)
±
(^) ¡X¥ ±
(^) ¢
±
¢ *º
ª
±>Ì
¢
±
¢ Â Í (^) ±
¢ *º
ª ¢ ±
¢
±
Ì
¢
±
¢ Â Í (^) ±
¢ º
ª ¡Å
±
¢ º
¡
¥ ±
¢
±
¢ º
ª
±
¢!¤
¡
±
¢ *º
¡ Å
±
¢ Ë
¡ ¢ ±
¢
±
(^) ¢¤
¡
±
¢ º
¡
±
¢Ë
¡
¥Ô
+Õ
Â
±
¢¤
³ Ö
±
¢º
¢
+ªÕ
Â
±
¢ Ë ×
¢ Ø
³
+Õ ¤
±
¢!¤
³ Ë
Õ º
±
¢º
¢ Â
Õ ¤
±
(^) ¢*º
¡ ¢
º
Õ
Â
±
¢!ËÙ
Ú
·<¸^ ^ <¡ ¢
·R¹^ v <¡X¥ Ô
Â
à « Ä A³ Ö
Ã«Ï Ä ¢
&ª
Â
à « NÛ Ä
×
§^ <¡ ¢
Ô ³
¤
ë Äf³ (^) Ë
º
Ã«Ï Ä ¢ Ü ¤
à « Ï Ä ¢
(^) º
Â
ë9Û Ä
×
§^ <¡
