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Main points are: Time Domain Component, Laplace Transform, Definition Integral, Modulation Property, Final Value Theorem, Steady State Response, Transient Response, Electrical Circuit Definition, Delta Impulse Signal
Typology: Exams
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Sample Exam 4: Solutions
#1a) 4 65879;:=<?>;@BACD
S`Qbac deac
f_
O ac
W
S`Qgac
O acOQih ac
W j
65k9>ml+@ C
n Fo
a c O Q ih ac
Another way to find the Laplace transform of
p9>
is to use the Laplace transform definition integral
WVXqZ \
'M_=r
W
s:
f_
ut
v
wyx
'hzS/{
@ t
v
a c
OQi|ph/d
/}
n Fo
O Q'|]h/d
/}
q S] S- t
n go
O Qi|ph/d
/}
a c
WVXqZ
VXq-m
89>y 4 65k9>;:=4'5p9s:>^+@
VXq
`9>4'5p9>^e:
LXm
z9>4'5p9s:=>^
n
n
o
WVXq
/5p9s:m>
o
'>4'5p9s:>^
n
n
o
85p9:>>
z'>:
`5p9s:m>>
n
n
o
85p9:>>:
465]9;:m>y^Y@
n
n
o
h
W VXq
`9>4'5p9>^
n
n
o
h n
n
o
n
h
n
o
Another way to solve the same problem is to use the modulation property
65]9>:=4'5p9:>z^L@ C
n
n
h
(^) h
n
@f
n
4 65]9>:465]9;:m>
z9> C
X
n Fo
o
n
(^) |khdH q¡ }
no
o
khQH q¡ }
n
|]hdH q¡ }
n
o
khQH q¡ }
n Fo¢
n
o
h
n
o
h
n Fo¢
n
o
n
n
h
n
o
WLXmZ
p9>^L@
j
o
l @
j
o
[9s:
o
l
o
9/¦m4'5p9s: C
4 '5k9: C
y§
o
4
¨&©ªi«¬*®¯°'±p®²q³e´%¬eμm¶H±p®s·&¸´&¸²/°6±]®;·&¸q²¹
¶© ªi«
º»
%¬Lμq¶¼±]®;·&¸q²/°6±]®;·$¸²¹g´%¬Vμq¶°6±]®;·½¸-²y¹L¨
© ªi«
ºy»
Another way to solve this problem
¬Vμq°'±p®s·&¸²¹L¨
©ª'«m¾ ®¯°'±p®·$¸q²b¿P±/·¸q² ¯À
i«
ºÃÂ
© (^) ªi« Á
¬Vμ[Ä »
±]®²¹L¨&¬ ®¯°6±k®·$¸q² ³ ´¬Lμq°6±]®·Å¸q²¹Y¨© ªi« Á
#1b1) ÆÇ
±
i¯«
º±º ´&¸²± º ´¶²± º ´ÅÈm²y± º ´%ÉH²± º ´Ê²
/© ªi¯«
The time domain component corresponding to the pole at the origin is given by
°6±]®;·¶m² ¾ Ì
¨BÍkÎÏ
«/Ð
μ
y¹L¨
¸q¶mÕ
#1b2)
Æ
q²
º¯ ´&¸;Ø «zÙ
*¸qÞ¶
º ´Ö
*¸qÞ¶
º ·¢Ö
#°'±p®²'´$¶ Ø
£ßàá
]®s´ Ü
ß àá
ß àá
®²²/°6±]®²;¨Ä (^) »±p®² â Ä ¯
±p®²¨Ú±¸´
ß àá
p®s·È²/²/°6±]®;·È²
Another way to solve this problem
Æ
¸q²
äã º
´æå
Åç
º¯ ´&¸
èã
ã
å
Åç º
±º¯ ´¸q²
â ã
øb¨~· å
(^) ¾ ç¨Õ
±º²g¨
é°'±p®²·
ß àá
p®²°'±k®²êâ Ä ¯
±p®²¨±-¸F·
ß àá
]®;·È²°'±p®s·È²/²
#2a)
Ä
Çmë Ì ?ì (^) ¨Èí-î ¯±Ì ´%¶²/°
ë
Ì?ì^ ¨ï3Ñ)Èí Ì °
ë
Ìì^ ´ ¸qð3Ñ)Èím°
ë
Ì?ì^ ¿éï
mñ
±zñV·È² ¯
¸qð
ñ
ñ ò·È
qðñ (^) ¯ ·¶óñ
±8ñV·È² ¯
Another way to find the ô -transform for
ë Ì ì
Çë Ìì (^) ¨Èíî ¯±Ì ´%¶²°
ë
Ìì^ ¨Ä
ë
Ì ´%¶ ì°
ë
Ìì^ ¾ Ä
ë Ìì (^) °
ë Ìì (^) ¨Èí Ì °
ë
Ìì^ ¿
ñ
±/ñV·=Ȳ ¯
zñ²
ô
μÄ
ë
Ìì^ ¹¨ñ ¯
±zñ²·ñ ¯Ä
ë
͓ ᖀ
ë
¸ì ¨
ñ »
±8ñ+·=Ȳ ¯
Õe´ÅÈñ+¨
qðñ (^) ¯ ·=¶óñ
±zñV·Èm² ¯
5
The steady state response is
q ¡ ¢G£¤z¥
7¦}§~¨K©,ª )¡ ¢«£¤z¥
¬¦}§~¨¬K®¦§~¨U©!m¯ °
±²¦M³¨´ ¯
m¯ °
The system can reach its steady state value since both poles are in the left half complex plane.
#3b)
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¶ ;·
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±²¦M¶¨U©dÃ
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7