## ECE 2020 Fall 2022 HW10

1.) Find the Laplace transform of the following functions:
a. ��(�) = sin (4�)�(�)
b. ��(�) = ��−2�
�(�)
c. ��(�) = (� − 1)�(� − 1)
d. ��(�) = (� − 1)�(�)
2.) Use partial fraction expansion and the Laplace Transform tables to find the inverse Laplace
Transform for the following functions.
a. ��(�) = �
�2+4�+3
b. ��(�) = 2�
�2+2�+2
3.) In the circuit below, vi(t) = 5e-4000t
cos(3000t) V R = 12Ω, C = 83.3µF, L = 1mH.
a.) Find s.
b.) Find the s-domain equation for voltage across the resistor, VO(s)
c.) Find vO(t)
4.) For the circuit below, the input Vi is a 10V pulse that lasts for 1 second. R = 106Ω, C = 44.4nF, L =
1mH. Assume all initial conditions are zero.
a.) Find the equation for Vi(t)
b.) Find Vi(s)
c.) Find the s-domain equation for voltage across
the capacitor, Vo(s).Find the voltage across the
capacitor, vo(t)
[ a ) h { sin ω t ) :twr
c ( 3 ( Ʃ tu ( t ) } =
h ⼀
TS
4 cttt sC-
) eFcssTuce .
Ts =
a{ ) FS =
s ~+ 16 < { ( t –
7 ) u ( t –
1 | =②

⾼ i
< b ) L [ e

ats = sta |= c
[ } ) =
e .
s

LEtf ( t 1 } = –
ds E [ 3 ) cd ) L ( 5 dit ) } = L ( tuct ) ) – LLu (t ) ]
( ct a
2t
u < t ) } = 皆
< ( sti 1
FB ( S ) =
stzn Fd ( s >=

b …

}
LS [
+
C 品 tan | = simat
CG> FaEb ) = {
+1 ” ) < st 3 )
[ b ) Fn [ { ) =
( s + 1 ^+ 1
= –
z { st) + 点 ( st33
L ”
( ! +
. 1 = cost ”
c +
nau)
= ertost ‘
” c
itowr ) =
coswtuC ☆ ninlkeetcswe
L ”
( Fa ( s > ] =
L

(
– {
ast.) + 童 ( st 3 )
} ( ” ( ☆ea ) ^
tw. ) = e
-at sinwt
L +
[ 焱n
+ t
( }
= [
^ { ☆
t” ^+ 1 –

t 1 ” ) z
+ 、 )
⼆⼀定 e ⽐七 {
e 张
( ) usy = Ze –
t
etsintTu cost –
2 ( t )
5 424 S + 22816 .
6
《 ) Wo < s ) =

0 .
G 243 –
135 1 .
a ☆
5 + 1000 .
4 st 12* ( 0 b
+ g 2 t8000 s tz5 × 10 c

924 c st 500 .
2 ) –
889 .
76 5 GL 4 { ) + 40003 –
87 G .
4
Ca ) { = juw 1 × 10 H =
( st 5 00 .
27 ^
+ s 427 . 79
⼗ 2 [ ☆+ 400033 + 30002
S = j 3000
Vo < t ] = – O
. 代
24 e –
500 .rt
cos ( 3423 . 7 Gt )
– { 859 .
7427.
79
} e
-500 .
rt
cb>÷L
1 s
=

40 oot
co s C 3000 t ) – (
8700 ) e
-40 ot

83 .
3 × 10
~
6 F sincsoω t 3 )
< s =
e

40 t ( 5 . 4 r osot3 1 }-o .2 a sin ( soo .t )
=
LCs ^+ 1
( × 10
~
3 a

s 0 at ( o .
9 os ( s 4 L 7 . 9 I t 3 + 0 . 26 sin ( 342799e) }
=
1 × ( 0
” × 83 . 1003 × > 5 ← 1
=
12004 .
8
z +
12 ×<oc
v 。
cs ) = ( 贤 + 2 ) U . ( ∞ )
5 cst 4000 )
5 { { ^+ ( 2 x (
6
] { st 4000 )
72 [ S ^← 2 × 10 + 1000 . 4 S 7 ( } ~+ 8000 s ,
+ 35 x 0 )
= 12 t02004
. 8 tnx 1 o
6
s
‘ ct 80 osti5 xi0
1× 10

}

[ C ) 2 us ) =÷<
Ca 3 Ui ( t ) = ( 0 { ut ) –
ω< t –
( ) ) Z ( ( 3 ) =

5 uRcLtSht 级
HS 级 E
106 44 .
4 ×<0
61-
Ui ( s ) ,
☆ ( RSE ]
( b 3 vics ) = L τ ( ( o ( U (t ) –
U ( t –
+ ) } U 0 [ s } =
CR / c ] + L
R lo –
coe

s
= 10 C ÷ –

( 0 E 1 –
e ”
) CO 6 lo –
loe

s
= [ –
O < st 106 之 38 . 5
) (
st [ 06218 . 5
) 2 + 16 –
( × 10 ‘ 0
t
( 062 ⻢ 8 .
5
Cs + 100 L38 .
5 ) ^≈ 1 .
1 α 100 + 号 ) C ( –
e

」 =
47 xl 0

astio ”
s s… =
S

1 . 0 % 285× ( 05 dt 11
Uo { t | = [ Lo –
LO ←^ 1 .0621 ∞ 5 × 105 t
Cos C ( × Lo 5 tj } ust ) tl 0 ←
1 . 062 ] 85