Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Time Domain Component - Linear Dynamic Systems and Signals - Solved Exam, Exams of Electronic Circuits Analysis

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

2012/2013

Uploaded on 04/16/2013

maalolan
maalolan 🇮🇳

4.5

(8)

123 documents

1 / 4

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Sample Exam 4: Solutions
#1a)
465879;:=<?>;@BACD
9FE$<HG7
I
D
9FJ$<HG7
@4LK9;:
<
7'M&NPORQTS
465879;:U<H>;@
ORQTS
4VKH9;:
<
76M
WYX[Z\
5]9>^+@
W
A
O
QS
4
K
9:
<
7'M_
@
W
A
O
Q
5
S`Qbacdeac
>
4
K
9;:
<
7Mf_
@
Oa
c
W
A
O
Q
5
S`Qgac
>
4
K
9;:
<
7'M_
@
O
a
cOQih
a
c
Wj
OQTS
465k9>ml+@
C
nFo
C
O
a
cOQih
a
c
Another way to find the Laplace transform of
Z\
5p9>
is to use the Laplace transform definition integral
WVXqZ
\
5]9>^+@
W
A
OQS
4K9:
<
7'M_=r
W
A
OQTS
4K9s:
<
7Mf_
@ut
v
wyx
OQTS
4K9;:
<
7M OQ'hzS/{
9
@t
v
ac
OQi|ph/d
\/}
S/{
9@~:
C
5
nFo
C
>O Q'|]h/d
\/}
Sq S]
t
S-
@
C
5
ngo
C
>O Qi|ph/d
\/}
a
c
WVXqZ
59>^Y@
WVXq-m
589>y 465k9>;:=4'5p9s:>^+@
WVXq
5`9>4'5p9>^e:
WLXm
5z9>4'5p9s:=>^
@
n
n
o
:
WVXq
5/5p9s:m>
o
'>4'5p9s:>^
@
n
n
o
:
WLX

585p9:>>

5z'>:
/
5`5p9s:m>>

5/'>8465]9;:>^
@
n
n
o
:
WVX

585p9:>>:
I
465]9;:m>y^Y@
n
n
o
:
OQ
h
WVXq
5`9>4'5p9>^
n
n
o
:
OQ
h
n
n
o
@
n
C
:
O
Q
h
n
o
Another way to solve the same problem is to use the modulation property
WVX
465]9>:=4'5p9:>z^L@
C
n
:
C
n
O
Q
h
@
C
:
O
Q
h
n
@f5
n
>
465]9>:465]9;:m>

5z9>
C
X
35
nFo
'>
o
35
n
:
>^L@
C
K
C
:
O
Q
|khdHq¡
}
no
o
C
:
O
Q
|khQHq¡
}
n
:
M
C
:
O
Q
|]hdHq¡
}
5
n
:
'>
o
C
:
O
Q
|khQHq¡
}
5
nFo¢
£>
H5
n
o
>
@
C
:
O
Q
h
5
n
:
'>
o
C
:
O
Q
h
5
nFo¢
£>
H5
n
o
>
@
n
:
n
O
Q
h
n
o
WLXmZ
5p9>^L@
W
j
9
o
C
465]9;:
C
>
l
@
W
j
9
o
[9s:9
o
C
465]9;:
C
>
l
@
W+¤*¥
5]9;:
C
>
o
9/¦m4'5p9s:
C
>-§@
W+¤
5]9:
C
>
4'5k9:
C
>y§
o
WLX
9465]9;:
C
>^
4
Docsity.com
pf3
pf4

Partial preview of the text

Download Time Domain Component - Linear Dynamic Systems and Signals - Solved Exam and more Exams Electronic Circuits Analysis in PDF only on Docsity!

Sample Exam 4: Solutions

#1a) 4 65879;:=<?>;@BACD

FE$<HG

I

D

9 FJ$<HG

4LK9;:

'M&NPORQTS

65879;:U;@

O

RQTS

VKH9;:

6M

WYX[Z\

]9>^+@

W

A

O

Q

S

4 K 9 :

'M_

W

A

OQ

S`Qbac deac

K

M

f_

O ac

W

A

OQ

S`Qgac

K

'M_

O acOQih ac

W j

O

Q

TS

65k9>ml+@ C

n Fo

C

O

a c O Q ih ac

Another way to find the Laplace transform of

Z

\

p9>

is to use the Laplace transform definition integral

WVXqZ \

]9>^+@

W

A

O

Q

S

4 K 9 :

'M_=r

W

A

OQTS

K

s:

M

f_

ut

v

wyx

O

Q

TS

K

M

O

Q

'hzS/{

@ t

v

a c

OQi|ph/d
/}

S

@~: C

n Fo

C

O Q'|]h/d
/}

S

q S]€ S€ƒ‚-„† t

C

n go

C

O Qi|ph/d
/}

a c

WVXqZ‡

ˆ9>‰^Y@

W

VXqŠ-‹mŒ

8ƒ9>yŽ 4 65k9>;:=4'5p9s:>ˆ‘^+@

W

VXqŠ‹Œ

`ƒ9>†4'5p9>^e:

W

LXmŠ‹Œ

zƒ9‰>†4'5p9s:=>^

n

n

o

WVXqŠ‹Œ

/’5p9s:m>

o

'>†4'5p9s:>‰^

n

n

o

WLX

8’5p9:“>†>

z'>’:

`’5p9s:m>>

/'>8‘˜465]9;:™>^

n

n

o

WVX

8’5p9:“>†>’:

I

‘˜465]9;:m>y^Y@

n

n

o

O

Q

h

W VXqŠ‹Œ

`ƒ9>†4'5p9>^

n

n

o

OQ

h n

n

o

n š

C

OQ

h›

n

o

Another way to solve the same problem is to use the modulation property

WVX

65]9>’:=4'5p9:“>ˆ‘z^L@ C

n

: C

n

O

Q

h

@ C

OQ

(^) h

n

@œf

n

Ž 4 65]9>’:465]9;:m>ˆ‘

zƒ9‰>ž C 

X

n Fo™Ÿ

o

n

’>^L@ C

K C

OQ

(^) |kh†dH q¡ }

no™Ÿ

o

C

O

Q

khQH q¡ }

n

M

C

OQ

|]h†dH q¡ }

n

o š

C

O

Q

khQH q¡ }

n Fo¢Ÿ

H

n

o

C

OQ

n

o š

C

O

Q

h ›

n Fo¢Ÿ

H

n

o

n

n

O

Q

h

n

o

WLXmZ

p9>^L@

W

j š

o

C

4 65]9;:

C

l @

W

j š

o

[9s:

o

C

4 65]9;:

C

l

W+¤*¥

]9;:

C

o

9/¦m4'5p9s: C

W

]9:

C

4 '5k9: C

o

W

LX

9465]9;:

C

>^

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

À

ºÃÂ

© (^) ªi« Á

Â

¬Vμ[Ä »

±]®²‰¹L¨&¬ ®¯°6±k®·$¸q² ³ ´™¬Lμq°6±]®·Å¸q²‰¹Y¨© ªi« Á

Â

Á

Â

#1b1) ƞÇ

±

i¯«

º±º ´&¸²± º ´¶²± º ´ÅÈm²y± º ´%ÉH²± º ´Ê²

Æ

/© ªi¯«

The time domain component corresponding to the pole at the origin is given by

Ì

Ç

°6±]®;·¶m² ¾ Ì

Ç

¨BÍkΖÏ

«/Ð

Ë

μ

Æ

Ë

y¹L¨

+Ѷ3ÑÒÈ3ÑÓÉÔÑ)Ê

¸q¶mÕ

#1b2)

Æ

Æ

Æ

Ì

Ç

Ì

™Ö

Ì

×

º ·)Ö

Ì

Ç

º¯ ´&¸;Ø «zÙ

Ë

Ú¸ ¾ Ì

¢ÖR² Ø«†Ù’ª?Û^

±†·FÖ?²±/·L¶[ÖR²

Ø

Ì

Ø

Ü

Ì

'ÜÝ

Æ

*¸qÞ¶

º ´“Ö

*¸qÞ¶

º ·¢Ö

#°'±p®²'´$¶ Ø

Ì

Ø

Ë

£ßàá

]®s´ Ü

Ì

²/°6±]®²

¨&°6±]®²£´

ß àá

]®£´

Ý

²/°6±]®²;¨~±¸´

ß àá

®²†²/°6±]®²;¨Ä (^) »±p®² â Ä ¯

±p®²¨Ú±¸´

ß àá

p®s·È²/²/°6±]®;·“Ȳ

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

Æ

±º²g¨

é°'±p®²’·

ß àá

p®²°'±k®²êâ Ä ¯

±p®²¨±-¸F·

ß ‰àá

]®;·“Ȳ†°'±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

È

±zñV·È² ¯

¸qð

ñ

ñ ò·“È

qðñ (^) ¯ ·“¶óñ

±8ñV·È² ¯

Another way to find the ô -transform for

Ä

Ç

ë Ì ì

Ä

Çë Ìì (^) ¨Èíî ¯±Ì ´%¶²†°

ë

Ìì^ ¨Ä

ë

Ì ´%¶ ì°

ë

Ìì^ ¾ Ä

ë Ìì (^) °

ë Ìì (^) ¨Èí Ì °

ë

Ìì^ ¿

È

±/ñV·=Ȳ ¯

Æ

zñ²

ô

μÄ

Ç

ë

Ìì^ ¹¨ñ ¯

Æ

±zñ²’·ñ ¯Ä

ë

Õì ·ñÄ

ë

¸ì ¨

È

ñ »

±8ñ+·=Ȳ ¯

“Õe´ÅÈñ+¨

qðñ (^) ¯ ·=¶óñ

±zñV·™Èm² ¯

5

The steady state response is

›œœŸžq ¡ ¢G£¤z¥

› 7¦}§~¨K©,ªž )¡ ¢«£¤z¥

› ¬œ¦}§~¨‰›¬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)

± μ¦M¶P¨

¶ ;·

¶¸z¶,º¹ »

¼%½¿¾cÀ (^) Á( ¹ ¥

±²¦M¶¨U©dÃ

·

±μ¦S¶¨

°

¶ ÅÄ

°

(^) ¶;ƹ ¸

Ä

·

Ƕ;h¹ ¸È

¸

¼€½ ¾dÀ (^) •ÁÉ ¹Ê ° Ä

° ¶

¶ Nº¹ ¸

Ä

Ƕ;º¹ ¸PÈ

¸ WË

 Ì ½¿¾PÀ^ ÎÍ Ä

°„Ï^ Ä

·

Ð ‰Ñ,Ò

Ð ¾ Ï (^) Ä

·

ЃÑNÓÕÔ

½ ¾ PÀ

ÁÅÖ›¬K® ½¿¾PÀ^ y›¬K® ½¾ Ä

·À

·

°

›¬K® ½¾ Ä

ÐÀ ³a×

ØÚÙ (^) ¬K® ¦S¶¨‰

·

Ù ¬K® ¦S¶¨€y› ½ Ä

·À

·

°

Ï

·

¶ ¸

Ù ¬ ~® ¦M¶P¨%

·

›¬K® ½ Ä

·Ày› ¬K® ½ Ä

ÐÀ Ñ

³

Ù ¬K® ¦S¶¨ Ï ·

·

·

° ¶ Ñ

 Ä

›¬K® ½ Ä

·À Ä

·

›¬K® ½ Ä

·À Ä

›¬K® ½ Ä

ÐÀ  Ä ÜÛÐ Ä

·

°¶

Ù ¬~® ¦M¶P¨ Ä

·

Ð

¦

Û

¶;•·¨

¶ (^) Ç ¶ ¸ z¶, ¹ » È

·

Ù¬K®¦S¶¨ Ä

³0Ý ¯ Û

¶ ;·

Ç ¶¸ ¶; (^) »¹ È



¾ ¹ ¶, ¹ ¸

¾ ¸ Ƕ; ¹ ¸ È

¸

 Ä ÛPÞ^

Ð

¶; ¹ ¸

·

Þ

°

Ƕ; ¹ ¸ È

¸

›¬~® ½¿¾PÀ^ Á  ¹Ê Ä Û Ð

¶ (^) Ç ¶ ; ¹ ¸ È

·

°

Ƕ; ¹ ¸ È

¸

´

Ç

Ä

¹ ¸ È Ç

Ä

¹ ¸ È

Ë

 Ä Û

Ð

Ï

Ä

·

Ð ”Ñ ÒÔ

½ ¾ PÀ Ä

·

Ð

¾ Ï Ä

·

Ð ƒÑ ÒÔ

½ ¿¾PÀ

ØÚÙ (^) ¬Uœß¦M¶¨™±μ¦M¶P¨Kà ¦G¶¨

¶ ;· Ç ¶¸ z¶,º¹» È

¶ Ç ¶;h¹» È

·

Ù ¬œ ¦M¶P¨

¾ ¹ ¶ ;º¹»

¾ ¸ ¹ ¶;á¹ ¸

¾ ¸ K¸ Ç ¶; ¹ ¸ cÈ

¸



· Ð

¶ ;h¹» Ä

· Ð

¶;h¹ ¸

Ä

Ð

Ƕ, ¹ ¸ PÈ

¸

›¬œ ½¾PÀ^ Á  ¹Ê

· Ð

¶ ;ƹ» Ä

· Ð ¶

¶;h¹ ¸

Ä

Ð ¶

Ƕ; ¹ ¸ cÈ

¸

´

Ç

Ä

¹ ¸ È Ç

Ä

¹ ¸È

Ë

(^) 8· Ð Ï Ä

·

° ‰Ñ ÒÔ

½ ¿¾PÀ Ä

(^) ·Ð Ï Ä

·

Ð ‰Ñ ÒÔ

½ ¾ dÀ ° ¾ Ï Ä

·

Ð ƒÑ ÒÔ

½ ¿¾PÀ

7