Problems Section 14-2: Laplace Transform P14.2-1 () ( ) () ( ) () () 1 1 2 2 1 1 2 2 cos Af t AF s As Fs s s f t t Fs s ω ω ω ⎫ = ⎡ ⎤ ⎣ ⎦ ⎪ ⇒ = ⎬ + = ⇒ = ⎪ + ⎭ L P14.2-2 () 1 1 1 2 1 ! 1 F n n n t s t s s s + + ⎡ ⎤ ⎡ ⎤ = = = ⎣ ⎦ ⎣ ⎦ L L 1 ! = P14.2-3 ( ) ( ) ( ) ( ) () () () [] () () 1 1 2 2 1 1 2 2 1 2 3 1 1 2 2 2 2 Linearity: a a Here 1 1 3 1 1 1 so 3 t f t af t F s aF s a a f t e Fs s f t t F s s Fs s s − + = + ⎡ ⎤ ⎣ ⎦ = = ⎡ ⎤ = = = ⎡ ⎤ ⎣ ⎦ ⎣ ⎦ + = = = ⎡ ⎤ ⎣ ⎦ = + + L L L L L P14.2-4 () ( ) ( ) () ( ) () () () () () () () () () () () ( ) 1 1 2 2 3 1 2 3 1 () 1 1 1 1 , 1 1 bt bt bt f t A e ut Af t f t e ut ut e ut f t f t F s F s s sb Ab Fs AF s AF s F s A s sb ssb − − − = − = = − = − = + − = = + ⎡ ⎤ 3 ∴ = = + = − = ⎡ ⎤ ⎣ ⎦ ⎢ ⎥ + + ⎣ ⎦ 1
Resolução da 7ª edição do livro do Dorf e Svoboda, Introdução Aos Circuitos Elétricos Cap. 14
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Problems
Section 14-2: Laplace Transform P14.2-1
( ) ( )
( ) ( ) ( )( )
1 1
2 21 1 2 2cos
A f t A F sAsF ss sf t t F s
sωω
ω
⎫=⎡ ⎤⎣ ⎦ ⎪ ⇒ =⎬ += ⇒ = ⎪+ ⎭
L
P14.2-2
( ) 11 1 2
1! 1Fnnnt s t
s s s+ +⎡ ⎤ ⎡ ⎤= = =⎣ ⎦ ⎣ ⎦L L 1
!=
P14.2-3
( ) ( ) ( ) ( )
( ) ( )
( ) [ ] ( )
( )
1 1 2 2 1 1 2 2
1 2
31 1
2 22
2
Linearity: a a
Here 11
31
1 1so3
t
f t a f t F s a F s
a a
f t e F ss
f t t F ss
F ss s
−
+ = +⎡ ⎤⎣ ⎦= =
⎡ ⎤= = =⎡ ⎤⎣ ⎦ ⎣ ⎦ +
= = =⎡ ⎤⎣ ⎦
= ++
L
L L
L L
P14.2-4
( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( )
( ) ( ) ( ) ( ) ( )
1
1 2
2 3
1 2 3
1 ( )
1 1
1 1 ,
1 1
bt
bt bt
f t A e u t A f t
f t e u t u t e u t f t f t
F s F ss s b
AbF s AF s A F s F s As s b s s b
−
− −
= − =
= − = − = +
−= =
+⎡ ⎤
3
∴ = = + = − =⎡ ⎤⎣ ⎦ ⎢ ⎥+ +⎣ ⎦
1
Section 14-3: Impulse Function and Time Shift Property P14.3-1
( ) ( ) ( )f t A u t u t T= − −⎡ ⎤⎣ ⎦
( ) ( ) ( ) ( )1 sTsT eA AeF s A u t A u t T As s s
−− −= − − = − =⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦L L
P14.3-2
( ) ( ) ( ) ( ) ( ) ( )at atf t u t u t T e F s e u t u t T⎡ ⎤= − − ⇒ = − −⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦⎣ ⎦L
( ) ( )
( ) ( )( )
( )
( )
11
sTs a T
at
eu t u t T es F ss ae g t G s a
−−
⎫−− − =⎡ ⎤ −⎪⎣ ⎦ ⇒ =⎬ −⎪⎡ ⎤ = −⎣ ⎦ ⎭
L
L
P14.3-3 (a) ( )
( )323
F ss
=+
(b) ( ) ( ) ( ) ( )sT sTf t t T F s e t eδ δ− −= − ⇒ = =⎡ ⎤⎣ ⎦L
(c) ( )
( ) ( ) ( )2 2 22
5 558 48 16 254 5
F ss ss ss
= = =1+ ++ + ++ +
P14.3-4
( ) ( ) ( ) ( )( (0.5 0.5)) 0.5 ( 0.5)0.5 0.5 0.5t t tg t e u t e u t e e u t− − + − − − −= − = − = −
P14.6-6 After the switch opens, apply KCL and KVL to get
( ) ( ) ( )1 sdR i t C v t v t Vdt
⎛ ⎞+ +⎜ ⎟⎝ ⎠
=
Apply KVL to get
( ) ( ) ( )2dv t L i t R i tdt
= +
Substituting into the first equation gives ( )v t
( ) ( ) ( ) ( ) ( )1 2d d d
2 sR i t C L i t R i t L i t R i t Vdt dt dt
⎛ ⎞⎛ ⎞+ + + +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠=
then ( ) ( ) ( ) ( ) ( )2
1 1 2 1 22d d
sR C L i t R C R L i t R R i t Vdtdt
+ + + + =
Dividing by : 1R C L
( ) ( ) ( )2
1 2 1 2 s2
1 1
R C R L R R Vd di t i t i t1R C L dt R C L R C Ldt
⎛ ⎞ ⎛ ⎞+ ++ +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
=
With the given values: ( ) ( ) ( )2
2 25 156.25 125d di t i t i tdtdt
+ + =
Taking the Laplace transform:
( ) ( ) ( ) ( ) ( ) ( )2 1250 0 25 0 156.25ds I s i s i s I s i I sdt s
⎡ ⎤⎛ ⎞− + + + + − + + =⎡ ⎤⎜ ⎟⎢ ⎥ ⎣ ⎦⎝ ⎠⎣ ⎦
We need the initial conditions. For t < 0, the switch is closed and the circuit is at steady state. At steady state, the capacitor acts like an open circuit and the inductor acts like a short circuit. Using voltage division
( ) ( )90 20 14.754 V
9 16 || 4v − = =
+
Then, using current division
( ) ( )040 0.328 A16 4 9
vi
−⎛ ⎞− = =⎜ ⎟+⎝ ⎠
The capacitor voltage and inductor current are continuous so ( ) ( )0 0v v+ = − and ( ) ( )0 0i i+ = − . After the switch opens
( ) ( ) ( ) ( ) ( ) ( ) ( )2
0 9 0 9 0.32814.7540 29.5080.4 0.4 0.4 0.4
v id dv t L i t R i t idt dt
+ += + ⇒ + = + = + =
Substituting these initial conditions into the Laplace transformed differential equation gives
( ) ( ) ( ) ( )2 12529.508 0.328 25 0.328 156.25s I s s s I s I ss
⎡ ⎤− + + − + =⎡ ⎤⎣ ⎦⎣ ⎦
( ) ( ) ( )2 12525 156.25 29.508 0.328 25 0.328s s I s ss
Comparing this to the given equation for vo(t), we see that 12 0.25 F2
CC
= ⇒ = .
(Checked using LNAP, 12/29/02)
14-7
P14.7-11 We will determine , the Laplace transform of the output, twice, once from the given equation and once from the circuit. From the given equation for the output, we have
( )oV s
( )o10 5
100V s
s s= +
+
Next, we determine from the circuit. For , we represent the circuit in the frequency domain using the Laplace transform. To do so we need to determine the initial condition for the capacitor.
( )oV s 0t ≥
When and the circuit is at steady state, the capacitor acts like an open circuit. Apply KCL at the noninverting input of the op amp to get
0t <
( ) ( )1
3 00 0
vv
R− −
= ⇒ − = 3 V
The initial condition is
( ) ( )0 0 3v v+ = − = V
Now we can represent the circuit in the frequency domain, using Laplace transforms.
Apply KCL at the noninverting input of the op am to get
( ) ( )6
1
2 3
10
V s V ss s
Rs
− −=
Solving gives
( )
6
16 6
1 1
103 22 1
10 10
sR
V ss
s s sR R
+
= = +⎛ ⎞ ⎛ ⎞
+ +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
Apply KCL at the inverting input of the op amp to get
P14.7-12 For t < 0, The input is constant. At steady state, the capacitor acts like an open circuit and the inductor acts like a short circuit. The circuit is at steady state at time 0t = − so
( )C 0v − = 0 and ( )L 0i B− = The capacitor voltage and inductor current are continuous so ( ) ( )C C0 0v v+ = −
)and
. ( ) (L L0 0i i+ = − For t < 0, represent the circuit in the frequency domain using the Laplace transform as shown. is the node voltage at the top node of the circuit. Writing a node equation gives
( )CV s
( ) ( ) ( )C C
C
V s V sA B B C sV ss R s L s+
= + + +
so
( )2
CA L s R R LC s V ss R L s
+ +=
Then
( )C 22 1 1
AA R L CV s
R LC s L s R s sRC LC
= =+ + + +
and
( ) ( )CL
2 1 1
AV s B BLCI s
L s s ss s s
RC LC
= + = +⎛ ⎞
+ +⎜ ⎟⎝ ⎠
a.) When 12 , 4.5 H, F, 5 mA and 2 mA9
R L C A B= Ω = = = = − , then
( ) ( )L 2
5 4010 2 3 7 7
144.5 22
I ss s ss s s s
−= + = +
++ +−
+
Taking the inverse Laplace transform gives
( ) 4 0.5L
5 53 mA for 07 7
t ti t e e t− −= + − ≥
b.) When , then 1 , 0.4 H, 0.1 F, 1 mA and 2 mAR L C A B= Ω = = = = −
( ) ( ) ( ) ( )L 2 22
25 2 25 2 1 5 1510 25 5 5
I ss s ss s s s s s
⎛ ⎞− −= + = + = − +⎜ ⎟
⎜ ⎟++ + + +⎝ ⎠s+
Taking the inverse Laplace transform gives
( ) ( )5 5L 1 5 mA for 0t ti t t e e t− −= − + − ≥
c.) When , then 1 , 0.08 H, 0.1 F, 0.2 mA and 2 mAR L C A B= Ω = = = = −
( ) ( ) ( )( )5L 1.8 0.2cos 10 0.1sin 10 mA for 0ti t e t t t−= − − + ≥
P14.7-13 For t < 0, the switch is open and the circuit is at steady state. and the circuit is at steady state. At steady state, the capacitor acts like an open circuit.
( )2Ai tR
= and ( )C 2Av t =
Consequently,
( )02AiR
− = and ( )C 02Av − =
Also
( )C 0 0i − = The capacitor voltage and inductor current are continuous so ( ) ( )C C0 0v v+ = −
)and
. ( ) (L L0 0i i+ = − For t > 0, the voltage source voltage is 12 V. Represent the circuit in the frequency domain using the Laplace transform as shown.
( ) ( )L C and I s I s are mesh currents. Writing a mesh equations gives
P14.7-14 For t < 0, The input is 12 V. At steady state, the capacitor acts like an open circuit. Notice that v(t) is a node voltage. Express the controlling voltage of the dependent source as a function of the node voltage:
va = −v(t) Writing a node equation:
( ) ( ) ( )12 3 08 4 4v t v t
v t−⎛ ⎞ ⎛ ⎞− + + −⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠=
( ) ( ) ( ) ( )12 2 6 0 4 Vv t v t v t v t− + + − = ⇒ = −
( ) ( )0 0 4v v+ = − = − V
For t < 0, represent the circuit in the frequency domain using the Laplace transform as shown. ( )V s is a node voltage. Express the controlling
voltage of the dependent source in terms of the node voltages
( ) ( )aV s V s= − Writing a node equation gives
( ) ( ) ( ) ( )6
3 4 0.758 4 40
V s V s ss V s V ss
− ⎛ ⎞+ + + =⎜ ⎟⎝ ⎠
Solving gives
( ) ( ) ( ) ( )10 10 4 2 2 4 1 15 4 2
5 5 5 5s V s V s
s s s s s s s s− ⎛ ⎞− = − ⇒ = − = + − = − +⎜ ⎟5s− − − − ⎝ ⎠−
Taking the inverse Laplace transform gives
( ) ( )52 1 V for 0tv t e t= − + ≥ This voltage becomes very large as time goes on.
P14.7-15 For t < 0, the voltage source voltage is 2 V and the circuit is at steady state. At steady state, the capacitor acts like an open circuit.
( ) 3 3
2 00 0.010 10 40 10
i −− = =
× + ×4 mA
and
( ) ( )( )3 3C 0 40 10 0.04 10 1.6 Vv −− = × × =
The capacitor voltage is continuous so ( ) ( )C C0 0v v+ = −
o
. For t > 0, the voltage source voltage is 12 V. Represent the circuit in the frequency domain using the Laplace transform as shown.
( ) ( )C and V s V s are node voltages. Writing a node equation gives
The 40 kΩ resistor, 50 kΩ resistor and op amp comprise an inverting amplifier so
( ) ( ) ( )62.5 62.5o C
50 50 9.6 8 12 10 V for 040 40
t tv t v t e e t− −= − = − − = − + ≥
so
( )o 62.5
2 V for 012 10 V for 0t
tv t
e t−
− ≤⎧= ⎨− + ≥⎩
(checked using LNAP 10/11/04)
P14.7-16 For t < 0, the voltage source voltage is 5 V and the circuit is at steady state. At steady state, the capacitor acts like an open circuit. Using voltage division twice
( ) 32 300 5 5 032 96 120 30
v − = − =+ +
.25 V
V
and ( ) ( )0 0 0.25v v+ = − =
For t > 0, the voltage source voltage is 20 V. Represent the circuit in the frequency domain using the Laplace transform as shown. We could write mesh or node equations, but finding a Thevenin equivalent of the part of the circuit to the left of terminals a-b seems promising.
P14.8-5 The transfer function can also be calculated form the circuit itself. The circuit can be represented in the frequency domain as
We can save ourselves some work be noticing that the 10000 ohm resistor, the resistor labeled R and the op amp comprise a non-inverting amplifier. Thus
( ) ( )a c110000
RV s V s⎛ ⎞⎟⎜= + ⎟⎜ ⎟⎟⎜⎝ ⎠
Now, writing node equations,
2
( ) ( ) ( ) ( ) ( ) ( )c i o a o
c 0 and 01000 5000
V s V s V s V s V sCsV s
Ls− −
+ = + =
Solving these node equations gives
( )
1 500011000 10000
1 50001000
RC LH s
s sC L
⎛ ⎞⎟⎜ + ⎟⎜ ⎟⎜⎝ ⎠= ⎛ ⎞⎛⎟ ⎟⎜ ⎜+ +⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎝ ⎠⎝⎞⎠
Comparing these two equations for the transfer function gives
( ) ( )1 12000 or 50001000 1000
s s s sC C
⎛ ⎞ ⎛ ⎞⎟ ⎟⎜ ⎜+ = + + = +⎟ ⎟⎜ ⎜⎟ ⎟⎟ ⎟⎜ ⎜⎝ ⎠ ⎝ ⎠
( ) ( )5000 50002000 or 5000s s s sL L
⎛ ⎞ ⎛ ⎞⎟ ⎟⎜ ⎜+ = + + = +⎟ ⎟⎜ ⎜⎟ ⎟⎟ ⎟⎜ ⎜⎝ ⎠ ⎝ ⎠
1
10001
100005000
15 106
CR
L+
⎛⎝⎜
⎞⎠⎟ = ×
The solution isn’t unique, but there are only two possibilities. One of these possibilities is
( )1 2000 0.5 F1000
s s CC
μ⎛ ⎞⎟⎜ + = + ⇒ =⎟⎜ ⎟⎟⎜⎝ ⎠
( )sL
s L+⎛⎝⎜
⎞⎠⎟ = + ⇒ =
50005000 1 H
( )6
61 50001 15 10
10000 11000 0.5 10R R
⎛ ⎞⎟⎜ + = × ⇒ = Ω⎟⎜ ⎟⎟⎜⎝ ⎠×5 k
(Checked using LNAP, 12/29/02)
3
P14.8-6 The transfer function of the circuit is
( )
2
2 1
1
2
11
1
RR C s R C
H sR s
R C
+= − = −
+
The give step response is ( ) ( ) ( )2504 1 Vtov t e u t−= − − . The correspond transfer function is
Comparing the two representations of the transfer functions let 1 13 F6 1
CC= ⇒ =
8,
4 2 2LL= ⇒ = H and . 2 3 12 2 V/Vk k× × = ⇒ =
(Checked using LNAP, 12/29/02)
P 14.9.9 From the circuit:
5
( ) ( )( )
o
i1212
RsV s R L s LH s RV s R L s sL
++= = =
++ + +
From the given step response:
( ) ( ) ( ) ( ) ( )4 0.5 0.5 2 20.5 14 4
tH s s se u t H ss s s s s
−
4s+ +⎡ ⎤= + = + = ⇒ =⎣ ⎦ + + +
L
Comparing these two forms of the transfer function gives:
2 12 2 4 6 H, 12 12 4
RLL L R
R LL
⎫= ⎪ +⎪ ⇒ = ⇒ = =⎬+ ⎪=⎪⎭
Ω
(Checked using LNAP, 12/29/02) P14.8-10
Mesh equations:
( ) ( ) ( )
( ) ( )
1 1
2 1
1 1 1
1 10
V s R I s I sCs Cs Cs
R R I s I sCs Cs
⎛ ⎞= + + −⎜ ⎟⎝ ⎠
⎛ ⎞= + + −⎜ ⎟⎝ ⎠
2
Solving for I2(s): ( )( )
2
1 2
1
2 12( )
V sCsI s
R RCs Cs Cs
⎛ ⎞⎜ ⎟⎝ ⎠=
⎛ ⎞ ⎛ ⎞+ + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
1
Then gives ( ) ( )o 2V s R I s=
( ) ( )( ) [ ] [ ]
( )
0
12 1
1 22 21 1
2 2 1 1 4 122 2
V s RCs sH sV s R Cs RCs RC R CR C s s
RR C RR C
= = =+ + − ⎡ ⎤+⎢ ⎥+ +
⎢ ⎥⎣ ⎦
6
P14.8-11 Let
2
1
1
1 1
x x
RRCsZ
RCsRCs
Z R L s
⎛ ⎞⎜ ⎟⎝ ⎠= =
++
= +
Then
( )
( )
2 22
1 1 2 x x x xx x
2 x
x x21 x
x x
1
11
RV Z RRCs
RV Z Z L RCs L R RC s R RR L sRCs
V L CL R RCV R Rs s
L RC L RC
+= = =+ + ++ +
+
=+ +
+ +
+ +
P14.8-12
Node equations:
( ) ( )1 out1 in 1 11 1
1
out11 2 2 out
2
2
0 1
0 1
V VV V sC R C s V R C sV VR
VV V R C sVR
sC
−− + = ⇒ + = +
− − = ⇒ = −
1 1 in out
Solving gives:
( ) 1 1 2 2out2
2in 1 2 1 2 2 2
1 1 1 2 1 2
1
1 1 1
sR C s R CVH s
V R R C C s R C s s sR C R R C C
−−
= = =+ + + +
7
P14.8-13
Node equations in the frequency domain:
1 11
1 2 3
0iV V V VVR R R
0− −+ + =
01
1 2 3 3
1 1 1 iV VV1R R R R R
⎛ ⎞⇒ + + − =⎜ ⎟
⎝ ⎠
1
2 0 1 2 2 02
0V sC V V sC R VR−
− = ⇒ = −
After a little algebra:
( ) 0 3
2 2 3 2 1 3 2 1 2 1i
V RH sV sC R R sC R R sC R R R
−= =
+ + +
P14.8-14
( )( ) 2
1 1
( ) 1 1o
i
V s Cs LCH s RV s Ls R s sCs L LC
= = =+ + + +
L, H C, F R, Ω H(s)
2
0.025
18 ( )( )2
20 209 20 4 5s s s s
=+ + + +
2
0.025
8 ( )22 2
20 204 20 2 4s s s
=+ + + +
1
0.391
4 ( )( )2
2.56 2.564 2.56 0.8 3.2s s s s
=+ + + +
2
0.125
8 ( )22
4 44 4 2s s s
=+ + +
8
a) ( ) ( )( )20
4 5H s
s s=
+ + ( ) ( )
( )
4 5
5 4
20 20 ( ) ( ) = 20 20 ( )4 5
( ) 20 1 5 4 step response ( 4) ( 5) 4 5
step response = 1 4 5 ( )
t t
t t
h t H s h t e e u ts s
H ss s s s s s s
e e u t
− −
− −
= − ⇒ = −+ +
−= = = + +
+ + + +
+ −
L
L ⇒
b) ( )( )2 2
202 4
H ss
=+ +
( ) ( ) 22 2
5(4) (s) 5 sin 4 ( )( 2) 4
th t H h t e t u ts
−= = ⇒ =+ +
L
( ) ( ) ( )
( )( )
( )( )
1 22 2
2 21 2 1 2
1 2
2 22
2
( ) 20 1 step response( 4 20) 4 20
20 4 20 1 4 201, 4
1 421 2 step response2 42 4
1step response 1 cos 4 sin 4 ( )2
t
H s K s Ks s s s s s s
s s s K s K s K s KK K
ss ss
e t t u t−
+= = = +
+ + +
= + + + + = + + + +
⇒ = − = −
−− += + +
+ ++ +
⎛ ⎞⎛ ⎞= − +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
L
L
+
)
c)
( ) ( )(2.56
0.8 3.2H s
s s=
+ +
( ) ( ) ( )
.8t 3.2t
3.2 .8
1.07 1.07( ) 1.07 e e u(t).8 3.2
4 1( ) 2.56 1 3 3step response
( .8) ( 3.2) .8 3.21 4step response 1 ( )3 3
t t
h t H s h ts s
H ss s s s s s s
e e u t
− −
− −
= = − ⇒ = −+ +
−
= = = + ++ + + +
⎛ ⎞= + −⎜ ⎟⎝ ⎠
L
L
d) ( )( )2
42
H ss
=+
( )( )( )
2
2
4 ( )
step response 1 1 2 ( )
t
t
h t te u t
t e u t
−
−
=
= − +
9
P14.8-15 For an impulse response, take ( )1 1 V s = . Then
( ) ( )( ) ( )
*
0
3 23 2 3 2 3 2 3 2
s A B BV ss s j s j s s j s j
+= = +
+ − + + + − + ++
Where
( ) ( ) *0 0 3 20
0.462, ( 3 2) 0.47 119.7 and 0.47 119.7s jsA sV s B s j V s B=− +== = = + − = ∠− ° = ∠ °
Then
( )00.462 0.47 119.7 0.47 119.7
3 2 3 2V s
s s j s j∠− ° ∠
= + ++ − + +
°
The impulse response is
( ) ( )30 ( ) 0.462 2(0.47) cos 2 119.7 Vt ov t e t u t−⎡ ⎤= + −⎣ ⎦
P14.8-16 a. A capacitor in a circuit that is at steady state and has only constant inputs acts like an open circuit. Then
( ) ( )o10 1.5 3.75 V4
v t = − = −
b. Here’s the circuit represented in the frequency domain, using phasors and impedances. Writing a node equation at the inverting input node of the op amp gives
( ) ( )o o3 3
4 30 04 10 10 10 10 10j
ω ω∠ °+ +
× − × ×V V
3 =
or ( ) ( )o10 30 1 0j ω∠ °+ + =V
( )o10 30 7.07 165
1 jω ∠ °
= − = ∠ °+
V
Finally,
10
vo(t) = 7.07 cos(100t +165°) V.
c. Here’s the circuit represented in the frequency domain, using The Laplace transform (assuming zero initial conditions). Writing a node equation at the inverting input node of the op amp gives
( ) ( )o o3 3
6
1
014 10 10 1010
V s V ss
s
+ +× ××
=
( ) ( )3
o10 100 04
s V ss+ + =
( ) ( )o250 2.5 2.5
100 100V s
s s s s−
= = ++ +
Finally, ( ) ( ) ( )1002.5 1 Vt
ov t e u t−= − P14.8-17 Represent the circuit in the frequency domain using the Laplace transform as shown. (Set the initial conditions to zero to calculate the step response.) First,
( )( )
( )
22
2 22
2
11 || 1 1
R L s R L sC sR L sC s C L s C R sR L s
C s
× + ++ = =
+ ++ +
Next, using voltage division,
11
( ) ( )( ) ( )
22
2 2o2
2i 2 1 212
2
2
1 12
1 2 1 22
1 1
11
1
2 44 29
R L sC L s C R s R L sV s
H s R L sV s R L s R C L s C R sRC L s C R s
RsR C R LC s
L R R C R R s ss sR LC R LC
++ + +
= = =+ + + + ++
+ +
++
= =+ + + ++ +
Using ( )i1V ss
= gives
( ) ( )( )
( )
( ) ( )
o 22
2 2
2 22 2
2 4 0.1379 0.1379 1.44834 294 29
0.1379 0.1379 1.44832 5
0.1379 2 50.1379 0.34492 5 2 5
H s s sV ss s s ss s s
ss s
ss s s
+ − += = = +
+ ++ +
− += +
+ +
+= − +
+ + + +
Taking the inverse Laplace transform
( ) ( ) ( )( )( )
2o
2
0.1379 0.1379cos 5 0.3448sin 5
0.1379 0.3713 cos 5 111.8 V
t
t
v t e t t
e t
−
−
= + − +
= + − °
(checked using LNAP 10/15/04)
P14.8-18 First, we determine the transfer function corresponding to the step response. Taking the Laplace transform of the given step response
( ) ( ) ( )( ) ( ) (( )( )
)
( )( )
o
50 20 0.667 20 1.667 501 0.667 1.66750 20 50 20
100050 20
H s s s s s s sV s
s s s s s s s
s s s
+ + + + − += = + − =
+ + + +
=+ +
Consequently,
( ) ( )( ) ( )( )
o
i
100050 20
V sH s
V s s s= =
+ +
12
Next, we determine the transfer function of the circuit. Represent the circuit in the frequency domain using the Laplace transform as shown. (Set the initial conditions to zero to calculate the transfer function.) Apply KVL to the left mesh to get
( ) ( ) ( ) ( ) ( )ii 1 a a a
1
V sV s L s I s K I s I s
K L s= + ⇒ =
+
Next, using voltage division,
( ) ( ) ( ) ( )( ) ( )o a o2 2 1
R RV s K I s V s V sL s R L s R K L s
= ⇒ =+ + + i
K
Then, the transfer function of the circuit is
( ) ( )( ) ( )( )
1 2o
i 2 1
2 1
R KL LV s R KH s
V s L s R L s K R Ks sL L
= = =⎛ ⎞⎛+ +
+ +⎜ ⎟⎜⎜ ⎟⎜⎝ ⎠⎝
⎞⎟⎟⎠
Comparing the two transfer functions gives
( )( ) ( ) 1 2
2 1
100050 20
R KL L
H ss s R Ks s
L L
= =+ + ⎛ ⎞⎛
+ +⎜ ⎟⎜⎜ ⎟⎜⎝ ⎠⎝
⎞⎟⎟⎠
We require 1 2
1000 R KL L
= and either
2
50 RL
= and 1
20 KL
= or 2
20 RL
= and 1
50 KL
= . These
equations do not have a unique solution. One solution is
L1 = 0.1 H, L2 = 0.1 H, R = 5 Ω and K = 2 V/A
(checked using LNAP 10/15/04)
13
P14.8-19 Represent the circuit in the frequency domain using the Laplace transform as shown. (Set the initial conditions to zero to calculate the step response.) First,
( )222
2 22
2
111||
11
R L s R C L sC sR L s
C s C L s C R sR L s
C s
⎛ ⎞× +⎜ ⎟ +⎛ ⎞ ⎝ ⎠+ = =⎜ ⎟ + +⎛ ⎞⎝ ⎠ + +⎜ ⎟⎝ ⎠
Next, using voltage division twice,
( ) ( )( )
( )
( ) ( )
( )
( )
22
22 2o
2 2i 2 1 2 1 2 1 2
122
2
1 22
1 22
1 2
1 11
111
81 10
R C L sC L s C R s RV s C sH s
V s R C L s R R C L s R R C s R RL sR C sC L s C R s
RR R LC
R R s ss sLCR R L
+
+ += = × =
+ + + + ++++ +
+= =
+ ++ +
+16
Using ( )i1V ss
= gives
( ) ( )( ) ( )( )o 2
2 118 8 3 62
2 8 210 16H s
V ss s s s s ss s s
−= = = = + +
8s+ + ++ + +
Taking the inverse Laplace transform
( ) ( )2 8o
1 2 1 V2 3 6
t tv t e e u t− −⎛ ⎞= − +⎜ ⎟⎝ ⎠
(checked using LNAP 10/15/04)
14
P14.8-20 First, we determine the transfer function corresponding to the step response. Taking the Laplace transform of the given step response
( ) ( )( )
( ) ( )( ) ( )
2
o 2 2
3.2 5 3.2 5 163.2 3.2 16 805 5 5
H s s s s sI s
s s s s s s
⎛ ⎞ + − + += = − + = =⎜ ⎟
⎜ ⎟+ + +⎝ ⎠25s s +
Consequently,
( ) ( )( ) ( )
o2
i
805
I sH s
V s s= =
+
Next, we determine the transfer function of the circuit. Represent the circuit in the frequency domain using the Laplace transform as shown. (Set the initial conditions to zero to calculate the transfer function.) First
11
11
1
11|| 1 1
R RC sRC s R C sR
C s
×= =
++
Next, using voltage division,
( ) ( ) ( )
1
1 1a i
1 1 2 1 22
1
1
1
RR C s R
V s V s V sR R R R R C sRR C s
+= =
+ +++
i
( ) ( ) ( ) ( )( ) ( ) ( )1 2ao o i
3 3 1 2 1 2 3 1 2
1 2
KK R R C LKV s
iI s I s V sL s R L s R R R R R C s R R R
s sL R R C
= ⇒ = =+ ⎛ ⎞+ + + +⎛ ⎞
+ +⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
V s
Then, the transfer function of the circuit is
15
( ) ( )( )
2o
i 3 1
1 2
KR C LI s
H sV s 2R R R
s sL R R C
= =⎛ ⎞+⎛ ⎞
+ +⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
Comparing the two transfer functions gives
( )( ) 2
23 1
1 2
805
KR C L
H s2R R Rs
s sL R R
= =⎛ ⎞++ ⎛ ⎞
+ +⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠C
We require
( )1 2
1 2
40 105 240 10
R RC
R R C C+ +
= = ⇒ =×
5 mF ,
3 205 4 HR
LL L
= = ⇒ =
and
( )2
80 80 V/V10 0.025 4
K K KR C L
= = ⇒ = .
(checked using LNAP 10/15/04)
P14.8-21 First,
( ) ( ) ( ) ( ) ( ) ( ) ( )6.5cos 2 22.6 6.5 cos 22.6 cos 2 6.5 sin 22.6 sin 2 6cos 2 2.5sin 2t t t+ ° = ° − ° = −t t Consequently, the impulse response can be written as
( ) ( ) ( )( ) ( )2o 6cos 2 2.5sin 2 Vtv t e t t u t−= −
The transfer function is
( )( ) ( ) ( )2 2 2 22 2 2
3 2 6 13 66 2.56 133 2 3 2 3 2
s sH ss ss s s
+ += − = =
13s ++ ++ + + + + +
The Laplace transform of the step response is
( )
( ) ( ) ( ) ( )2 222 2 2
6 13 1 1 1 3 3 26 13 26 13 3 2 3 2 3 2
H s s s s ss s s s s ss s s s s s
+ += = − = − = − + ×
+ ++ + 2 2+ + + + + +
16
Taking the inverse Laplace transform gives the step response:
( ) ( ) ( )( )( ) ( ) ( )( )2 2o 1 1.5sin 2 cos 2 1 1.803 cos 2 123.7 Vt tv t e t t u t e t− −= + − = + − °
P14.8-22 Taking the Laplace transform of the step response,
( )( ) ( ) ( )2 2
1 3 1 1 6 933 3
H s ss s s ss s
⎡ ⎤ += − + = − =⎢ ⎥
++ +⎢ ⎥⎣ ⎦23s s +
The transfer function is
( )( )2
93
H ss
=+
Taking the inverse Laplace transform gives the impulse response:
( ) ( )3
o 9 Vtv t t e u t−=
(checked using LNAP 10/15/04) P14.8-23 Represent the circuit in the frequency domain using the Laplace transform as shown. (Set the initial conditions to zero to calculate the transfer function.) First,
( ) ( )a
1
iV sI s
L s R=
+
The equivalent impedance of the parallel capacitor and inductor is
22
22
2
11|| 1 1
R RC sRC s R C sR
C s
×= =
++
Next, using voltage division,
( ) ( ) ( ) ( )( )( ) ( )3 2 33 3 2 3
o a a2 2 3 2 3 1 2 3 2 3
321
K R R R C sR R R R C sV s K I s K I s V sR R R R R C s L s R R R R R C sR
R C s
++= = =
+ + + + +++
i
17
Then, the transfer function of the circuit is
( ) ( )( )
( )( )( )
2o
i 1 2 3
2 3
15 0.5
5 2.
K sL R CV s s
H sV s s sR R R
s sL R R C
⎛ ⎞+⎜ ⎟⎜ ⎟ +⎝ ⎠= = =
+ +⎛ ⎞+⎛ ⎞+ +⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
5
Using ( )i1V ss
= gives
( ) ( ) ( )( )( )o
5 0.5 0.2 1.8 1.65 2.5 5 2.5
H s sV s
s s s s s s s+ −
= = = + ++ + + +
Taking the inverse Laplace transform
( ) ( ) ( )5 2.5o 0.2 1.8 1.6 Vt tv t e e u t− −= − +
(checked using LNAP 10/15/04)
18
Section 14-9: Convolution Theorem P14.9-1
( ) ( ) ( ) ( ) ( ) ( ) 1 11 1s se ef t u t u t F s u t u t
The poles of the transfer function are 1 8 rad/ss = − and 2 320 rad/ss = − , so circuit is stable. Consequently,
( ) ( ) ( ) ( )102400 408 320 1 1
8 320s j
H sj j j j
ωω
ω ωω ω== = =
+ + ⎛ ⎞ ⎛+ +⎜ ⎟ ⎜⎝ ⎠ ⎝
H⎞⎟⎠
14-2
The network function has poles at 8 and 320 rad/s and has a low frequency gain equal to 32 dB = 40. Consequently, the asymptotic magnitude Bode plot is
P14.10-5
( ) ( ) ( ) ( )( )2 6 60 60 24060
2 6 2t tH s
e e u ts s s
− −⎡ ⎤= − = − =⎣ ⎦ 6s s+ + + +L
so
( ) ( )( )2406 2
sH ss s
=+ +
The poles of the transfer function are 1 2 rad/ss = − and 2 6 rad/ss = − , so circuit is stable. Consequently,
( ) ( ) ( ) ( )240 20
2 6 1 12 6
s j
j jH sj j j j
ω
ω ωωω ωω ω=
= = =+ + ⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
H
The network function has poles at 2 and 6 rad/s. The asymptotic magnitude Bode plot has a gain equal to 40 = 32 dB between 2 and 6 rad/s. Consequently, the asymptotic magnitude Bode plot is
14-3
P14.10-6
( ) ( ) ( ) ( )( )( )
90 36 104 32 36 3604 3290 90 90
tH s sse u ts s s s s
− ++⎡ ⎤= + = + = =⎣ ⎦ + + +L
s s
so
( ) ( )( )
1036
90s
H ss+
=+
The pole of the transfer function , so circuit is stable. Consequently, 1 90 rad/ss = −
( ) ( ) ( )( )
110 1036 490 1
90s j
jjH s
j jω
ωω
ωωω=
⎛ ⎞+⎜ ⎟+ ⎝ ⎠= = =+ ⎛ ⎞+⎜ ⎟
⎝ ⎠
H
The network function has a zero at 10 rad/s and a pole at 90 rad/s. The low frequency gain is equal to 4 = 12 dB. Consequently, the asymptotic magnitude Bode plot is
P14.10-7
14-4
( ) ( ) ( ) ( ) ( )5 205 1.67 1.67 25
3 5 20t tH s
e e u ts s s
− −⎡ ⎤= − = − =⎢ ⎥ + + + +⎣ ⎦L
5 20s s
)
so
( ) ( )(25
5 2sH s
s s=
+ + 0
The poles of the transfer function are 1 5 rad/ss = − and 2 20 rad/ss = − , so the circuit is stable. Consequently the network function of the circuit is,
( ) ( ) ( ) ( )25 0.255 20 1 1
5 2s j
j jH sj j j j
ω
0
ω ωωω ωω ω=
= = =+ + ⎛ ⎞⎛+ +⎜ ⎟⎜
⎝ ⎠⎝
H⎞⎟⎠
s
Using ( ) ( ) ( )o ω ω ω=V H V at ω = 30 rad/s gives
( ) ( ) ( ) ( )( )o
0.25 30 9012 0 8.2 47 V30 30 1 6 1 1.51 15 20
j jj jj j
ω = ∠ = =+ +⎛ ⎞⎛ ⎞+ +⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
V ∠− °
Back in the time domain, the steady state response is
( ) ( )o 8.2 cos 30 47 Vv t t= − °
(checked using LNAP 10/12/04)
P14.10-8
( ) ( ) ( )( )
( )( ) ( )
52 2
10 5 5010 50 1010 505 5 5 5
t s sH s e t u ts s s s
− + −⎡ ⎤= − = − = =⎣ ⎦ + + +
L 2+
The poles of the transfer function are 1 5 rad/ss = − and 2 5 rad/ss = − , so the circuit is stable. Consequently the network function of the circuit is,
( ) ( )( )2 2
10 0.45 1
5
s j
j jH sj j
ω
ω ωωω ω=
= = =+ ⎛ ⎞+⎜ ⎟
⎝ ⎠
H
Using ( ) ( ) ( )o sω ω ω=V H V at ω = 10 rad/s gives
14-5
( ) ( ) ( )( )o 2 2
0.4 10 4812 0 9.6 37 V1 2101
5
j jjj
ω = ∠ = = ∠−+⎛ ⎞+⎜ ⎟
⎝ ⎠
V °
Back in the time domain, the steady state response is
( ) ( )o 9.6 cos 10 37 Vv t t= − °
(checked using LNAP 10/12/04)
14-6
Section 14.12 How Can We Check…? P14.12-1
( ) ( ) 2.1 15.93 6 2t tL L
dv t i t e edt
− −= = − −
( ) ( ) 2.1 15.91 0.092 0.57575
t tC C
di t v t e edt
− −= = − −
( ) ( ) 2.1 15.9
1 12 12 6 2t tR Lv t v t e e− −= − = + +
( )( ) ( )( ) 2.1 15.9
2
121 0.456 0.123
6L C t t
R
v t v ti t e e− −
− += = + −
( ) ( ) 2.1 15.93 1 0.548 0.452
6C t t
R
v ti t e e− −= = + +
Thus, ( ) ( ) ( ) ( ) ( )1 212 0 and +L R R C Rv t v t i t i t i t− + + = = 3
as required. The analysis is correct.
P14.12-2
( ) ( )1 218 20 and 3 3
4 4
I s I ss s
= =− −
1
KVL for left mesh: 12 1 18 18 206 03 3 324 4 4
s s s s s
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟
+ + −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟− − −⎝ ⎠ ⎝ ⎠
= (ok)
KVL for right mesh: 18 20 20 186 3 43 3 3 34 4 4 4
s s s s
⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟
− − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟− − − −⎝ ⎠ ⎝ ⎠ ⎝ ⎠
0= (ok)
The analysis is correct.
P14.12-3
Initial value of IL (s): 2
lim 2 15
sss s s
+=
→∞ + + (ok)
Final value of IL (s): 2
lim 2 00 5
sss s s
+=
→ + + (ok)
Initial value of VC (s): ( )( )2
lim 20 20
5s
ss s s s
− +=
→∞ + + (not ok)
Final value of VC (s): ( )( )2
lim 20 28
0 5s
ss s s s
− += −
→ + + (not ok)
Apparently the error occurred as VC (s) was calculated from IL (s). Indeed, it appears that VC (s)