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Practical - Chapter 14 practice problem solution
Electric Circuits 2 (McGill University)
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Section 14.2 Laplace Transforms
P14.2‐1 Determine the Laplace Transform of v ( t ) = (17 e − 4 t − 147 e− 5 t ) u ( t ) V
Answer: V ( s ) =
3 s + 29
s + 9 s + 20
2
Solution:
L ⎡⎣17 e− 4 t − 14 e − 5 t ⎤⎦ =
17
14 17 ( s + 5 ) − 14 ( s + 4 )
3 s + 29
−
=
= 2
s+4 s+5
s + 9 s + 20
( s + 4 )( s + 5)
P14.2‐2 Determine the Laplace Transform of v ( t ) = 13cos ( 6 t − 22.62° ) V.
Answer: V ( s ) =
12 s + 30
s 2 + 36
Solution:
v ( t ) = 13cos ( 6 t − 22.62° ) = 13cos ( −22.62° ) cos ( 6 t ) − 13sin ( −22.62° ) sin ( 6 t )
= 12 cos ( 6 t ) + 5sin ( 6 t ) V
V ( s ) = L ⎣⎡12 cos ( 6 t ) + 5sin ( 6 t ) ⎦⎤ = L ⎣⎡12 cos ( 6 t ) ⎦⎤ + L ⎣⎡ +5sin ( 6 t ) ⎦⎤
s
6
12 s + 30
= (12 ) 2
+ ( 5) 2
= 2
s + 36
s + 36 s + 36
P14.2‐3 Determine the Laplace Transform of v ( t ) = 10 e −5 t cos ( 4 t + 36.86° ) u ( t ) V.
Answer: V ( s ) =
8 s + 16
s + 25 s + 41
2
Solution:
10 cos ( 4 t + 36.86° ) = 10 cos ( 36.86° ) cos ( 4 t ) − 10sin ( 36.86° ) sin ( 4 t )
= 8cos ( 4 t ) − 6sin ( 4 t ) V
4
8s − 4
s
− ( 6) 2
= 2
L ⎡⎣10 cos ( 4 t + 36.86° ) ⎤⎦ = ( 8 ) 2
s + 16
s + 16 s + 16
V ( s ) = L ⎡⎣10 e −5 t cos ( 4 t + 36.86° ) ⎤⎦ = L ⎡⎣10 cos ( 4 t + 36.86° ) ⎤⎦
=
s ← s +5
8 ( s + 5) − 4
8s − 4
8 s + 16
=
= 2
2
2
s + 16 s ← s +5 ( s + 5 ) + 16 s + 25 s + 41
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P14.2‐4 Determine the Laplace Transform of v ( t ) = 3 t e − 2 t u ( t ) V
Answer: V ( s ) =
3
s + 4s + 4
2
Solution:
L ⎡⎣3 t e − 2 t ⎤⎦ = L [3 t ] s ←s + 2 =
3
3
3
=
= 2
2
2
s s←s + 2 ( s + 2 )
s + 4s + 4
P14.2‐5 Determine the Laplace Transform of v ( t ) = 16 (1 − 2 t ) e − 4 t u ( t ) V.
Answer: V ( s ) =
16 ( s + 2 )
s 2 + 8 s + 16
Solution:
16
32
⎛ 16 32 ⎞
L ⎡⎣16 (1 − 2 t ) e − 4 t ⎤⎦ = L [16 − 32 t ] s ←s + 4 = ⎜ − 2 ⎟
=
−
2
⎝ s s ⎠ s ←s + 4 s + 4 ( s + 4 )
=
16 ( s + 4 ) − 32
16 ( s + 2 )
= 2
2
s + 8 s + 16
s + 8 s + 16
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Section 14-3: Pulse Inputs
P 14.3-1 Determine the Laplace transform of f(t) shown
in Figure P 14.3-1.
5 ⎞
5 ⎛ 21 ⎞ ⎛ 21 ⎞
⎛
Hint: f ( t ) = ⎜ 5 − t ⎟ u ( t ) + ⎜ t − ⎟ u ⎜ t − ⎟
3 ⎠
3⎝
5⎠ ⎝
5⎠
⎝
5
f (t)
0
–2
5e−4.2 s + 15s − 5
Answer: F ( s ) =
3s 2
3
t
Figure P 14.3-1
Solution:
⎛ 5
⎞
⎛ 5
⎞
f ( t ) = ⎜ − t + 5 ⎟ u ( t ) − ⎜ − ( t − 4.2 ) ⎟ u ( t − 4.2 )
⎝ 3
⎠
⎝ 3
⎠
⎛ 5 5⎞
⎛ 5 ⎞ 15 s + 5 ( e
F ( s ) = ⎜ − 2 + ⎟ − e−4.2 s ⎜ − 2 ⎟ =
s⎠
3 s2
⎝ 3s
⎝ 3s ⎠
−4.2 s
P 14.3-2 Use the Laplace transform to obtain the
transform of the signal f(t) shown in Figure P14.3-2.
Answer: F ( s ) =
− 1)
3
f (t)
3 (1 − e −2 s )
s
0
2
t
Figure P14.3-2
Solution:
2
3 e − st
3(1−e −2 s )
=
F ( s ) = ∫ 0 f (t ) e dt = ∫ 0 3 e dt =
s
−s 0
∞
− st
2
− st
1
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P 14.3-3 Determine the Laplace transform of f(t) shown
in Figure P 14.3-3.
5
Answer: F ( s ) = 2 (1 − e −2 s − 2 se −2 s )
2s
5
f (t)
0
1
2
t
Figure P 14.3-3
Solution:
⎧5 2 t 0<t < 2
f (t ) = ⎨
otherwise
⎩0
5
5
5
5
t ⎡⎣u ( t ) −u ( t − 2 ) ⎤⎦ = t u ( t ) − t u ( t − 2 ) = ⎡⎣t u ( t ) −( t − 2 )u ( t − 2 )− 2u (t − 2) ⎤⎦
2
2
2
2
−2 s
−2 s
5⎡ 1 e
2e ⎤ 5 1
⎡1− e −2 s − 2 se −2 s ⎤⎦
∴F ( s ) = L ⎡⎣ f ( t ) ⎤⎦ = ⎢ 2 − 2 −
=
2⎣s
s
s ⎥⎦ 2 s 2 ⎣
f (t ) =
f(t)
P 14.3-4
Consider the pulse shown in Figure P 14.3-4,
where the time function follows eat for 0 < t < T. Find
eat
F(s) for the pulse.
Answer: F ( s ) =
1− e ( )
s−a
− s−a T
1
0
T
t
Figure P 14.3-4
Solution:
f ( t ) = ⎡⎣u ( t ) −u ( t −T ) ⎤⎦ e at
⇒ F ( s ) = L ⎡⎣ eat ⎡⎣u ( t ) −u ( t −T ) ⎤⎦ ⎤⎦
1− e − sT ⎫
1− e( s − a )T
s ⎪⎬ ⇒ F ( s ) =
( s −a )
L ⎡⎣e at g ( t ) ⎤⎦ =G ( s − a ) ⎪⎭
L ⎣⎡u ( t ) −u ( t −T ) ⎦⎤ =
2
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P 14.3-5
Find the Laplace transform for g(t) = e–tu(t–0.5).
Solution:
g ( t ) = e − t u ( t − 0.5 ) = e − ( t + (0.5−0.5))u ( t − 0.5 ) = e−0.5 e − (t −0.5)u ( t − 0.5 )
L ⎡⎣e −0.5 e − (t −0.5)u ( t − 0.5 ) ⎤⎦ = e−0.5 L ⎡⎣e− (t − 0.5)u ( t − 0.5 ) ⎤⎦ = e−0.5 e−0.5 s L ⎡⎣ e− t u ( t ) ⎤⎦ =
e0.5 e −0.5 s e0.5 − 0.5 s
=
s +1
s +1
Find the Laplace transform for
− (t − T )
f (t ) =
u (t − T )
T
−1e − sT
Answer: F ( s ) =
Ts 2
Solution:
P 14.3-6
− sT
e − sT
⎡ t −T
⎤
⎡ t
⎤ e
L ⎢−
L ⎡⎣ − t u ( t ) ⎤⎦ = − 2
u ( t − T ) ⎥ = e− sT L ⎢ − u ( t ) ⎥ =
T
Ts
⎣ T
⎦
⎣ T
⎦
3
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Section 14-4: Inverse Laplace Transform
P 14.4-1
Find f(t) when
F (s) =
s+3
s + 3s 2 + 6s + 4
3
2
2
1 −t
e × sin 3t , t ≥ 0
Answer: f ( s ) = e − t − e −t cos 3t +
3
3
3
Solution
F (s) =
Where
s +3
s +3
A
Bs + C
=
=
+ 2
2
2
s + 3s + 6s + 4 ( s +1) ⎡( s +1) + 3⎤ s +1 s + 2 s + 4
⎣
⎦
s +3
2
A=
=
2
( s +1) +3 s =−1 3
3
Then
2
Bs +C
2
8
⎛4
⎞
= 3 + 2
⇒ ( s +3) = ( + B) s 2 + ⎜ + B + C ⎟ s + + C
2
3
3
⎝3
⎠
( s +1) ( s + 2s + 4 ) s +1 s + 2s + 4
( s + 3)
Equating coefficient yields
2
2
+ B ⇒ B= −
3
3
4 2
1
s : 1= − + C ⇒ C =
3 3
3
s2 : 0 =
Then
1
2
2 1
2
2
3
− s+
− ( s +1)
3
3
3
3
3
3
F (s) =
+
=
+
+
s +1 ( s +1)2 + 3 s +1 ( s +1)2 + 3 ( s +1)2 + 3
Taking the inverse Laplace transform yields
f (t ) =
2 −t 2 −t
1 −t
e − e cos 3t +
e sin 3 t , t ≥ 0
3
3
3
1
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P 14.4-2
Find f(t) when
s 2 − 2s + 1
s 3 + 3s 2 + 4s + 2
F (s) =
Solution:
s 2 − 2s + 1
s 2 − 2s + 1
a
a*
b
F (s) = 3
=
=
+
+
2
s + 3s + 4s + 2 ( s +1) ( s +1− j )( s +1+ j ) s +1− j s +1+ j s +1
where
s 2 − 2 s +1
b=
( s +1) +1 s =−1
2
a=
=4
3− j 4 3
s 2 − 2s +1
=
=− + j 2
( s +1) ( s +1+ j ) s =−1+ j −2 2
3
a* = − − j 2
2
Then
3
3
− + j2 − − j2
4
+ 2
+
F (s) = 2
s +1− j
s +1+ j s +1
Next
m=
From Equation 14.5-8
( −3 2 ) + ( 2 )
2
2
⎛
⎞
⎜
⎟
5
2
=
= 126.9°
and θ = tan −1 ⎜
3⎟
2
⎜− ⎟
⎝ 2⎠
f ( t ) = ⎣⎡5 e − t cos ( t + 127° ) + 4 e −t ⎦⎤ u ( t )
2
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P 14.4-3
Find f(t) when
F (s) =
5s − 1
s − 3s − 2
3
Answer: f(t) = – e– t + 2te– t + e2t, t ≥ 0
Solution:
F ( s) =
where
B=
Then
A=
Finally
F (s) =
5 s −1
( s +1) ( s − 2 )
2
=
A
B
C
+
+
2
s +1 ( s +1)
s −2
5 s −1
5 s −1
= 2 and C =
=1
2
s − 2 s =−1
( s +1) s =2
d ⎡
−9
2
( s +1) F ( s )⎦⎤ s =−1 =
2
⎣
ds
( s −2)
−1
2
1
+
+
⇒
2
s +1 ( s +1)
s −2
= −1
s =−1
f ( t ) = ⎡⎣ −e − t + 2 t e −t + e2t ⎤⎦ u ( t )
3
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P 14.4-4
Find the inverse transform of
Y(s) =
1
s + 3s + 4s + 2
3
2
Answer: y(t) = e–t(1 – cos t), t ≥ 0
Solution:
Y (s) =
1
1
A
Bs +C
=
=
+
2
( s +1) ( s + 2s + 2 ) ( s +1) ⎡⎣( s +1) + 1⎤⎦ s +1 ( s +1)2 +1
2
A=
where
Next
1
=1
s + 2 s + 2 s =−1
2
1
1
Bs +C
=
+ 2
⇒ 1 = s 2 + 2 s + 2 + ( Bs + C ) ( s + 1)
( s +1) ( s + 2s + 2 ) s +1 s + 2s + 2
2
⇒ 1 = ( B +1) s 2 + ( B + C + 2 ) s + C + 2
Equating coefficients:
s 2 : 0 = B + 1 ⇒ B = −1
s : 0 = B + C + 2 ⇒ C =−1
Finally
P 14.4-5
Y (s) =
1
s +1
−
⇒
s +1 ( s +1)2 +1
y ( t ) = ⎡⎣e− t − e − t cos t ⎤⎦ u ( t )
Find the inverse transform of
F (s) =
Solution:
F (s) =
2s + 6
( s + 1) ( s 2 + 2s + 5)
2( s +3)
( s +1) ( s 2 + 2 s +5)
=
−( s +1)
1
2
+
+
2
s +1 ( s +1) + 4 ( s +1)2 + 4
f ( t ) = ⎡⎣e −t − e− t cos ( 2t ) + e− t sin ( 2t ) ⎤⎦ u ( t )
4
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P 14.4-6
Find the inverse transform of
F (s) =
2s + 6
s ( s 2 + 3s + 2 )
Answer: f(t) = [3 – 4e–t+e–2t] u(t)
Solution:
F (s) =
where
A = sF ( s ) s =0 =
2( s+3)
A
B
C
= +
+
s( s+1) ( s + 2 ) s s +1 s + 2
2( s + 3 )
2( s + 3 )
= 3, B = ( s +1) F ( s ) s =−1 =
=−4
s ( s + 2 ) s =−1
( s +1) ( s + 2 ) s =0
and
( s + 2 ) F ( s ) s = −2 =
Finally
3 −4
1
F (s) = +
+
⇒
s s +1 s + 2
2( s + 3)
= C =1
s ( s +1) s =−2
f ( t ) = ( 3 − 4e− t + e −2t ) u ( t )
5
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P 14.4-7
Find the inverse transform of F(s) expressing f(t) in cosine and angle forms.
(a) F ( s ) =
8s − 3
2
s + 4s + 13
(b) F ( s ) =
3e− s
s 2 + 2s + 17
Answer: (a) f(t) = 10.2e–2t cos (3t + 38.4°), t ≥ 0
(b) f(t) = 34 e (
− t −1)
Solution:
(a)
F (s) =
sin [4(t–1)], t ≥ 1
8 s −3
1 2( 8s −3)
= ×
s + 4s +13 2 ( s + 2 )2 + 9
2
∴ a = 2, c =8, ω =3 & ca −ω d =−3 ⇒ d =
−3−( 8 )( 2 )
= 7.33
−3
2
2
⎛ 6.33 ⎞
°
∴ θ = tan −1 ⎜
⎟ =38.4 , m = ( 8 ) +( 6.33) =10.85
⎝ 8 ⎠
⇒ f ( t ) =10.85 e −2 t cos( 3 t + 42.5 ) u ( t )
(b)
Given F ( s ) =
( 2( 3) ) .
3e − s
3
1
= ×
, first consider F1 ( s ) = 2
2
s + 2s +17
s + 2s +17 2 ( s +1)2 +16
⎛ −3 4 ⎞
Identify a =1, c = 0, ω = 4 and −ω d =3 ⇒ d =−3 4. Then m =|d |=3 4, θ = tan −1 ⎜
⎟=−90°
⎝ 0 ⎠
So f1 (t ) = (3 4)e −t sin 4t u ( t ) . Next, F ( s ) = e− s F1 ( s ) ⇒ f ( t ) = f1 ( t −1) . Finally
∴ f ( t ) =(3 4)e − (t −1) sin ⎡⎣ 4( t −1) ⎤⎦ u ( t −1)
6
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Find the inverse transform of F(s).
P 14.4-8
(a) f ( s ) =
s2 − 5
s ( s + 1)
4s 2
(b) f ( s ) =
2
( s + 3)
3
Answer: (a) f(t) = –5 + 6e–t + 4te–t, t ≥ 0
(b) f(t) = 4e–3t – 24te–3t + 18t2e–3t, t ≥ 0
Solution:
(a)
F (s) =
where
A = sF ( s ) |s = 0 =
Multiply both sides by s ( s + 1)
s 2 −5
s ( s +1)
=
2
A B
C
+
+
s s +1 ( s +1)2
−5
1−5
2
= −5 and C = ( s +1) F ( s ) |s =−1 =
=4
−1
1
2
s 2 − 5 = −5 ( s +1) + Bs ( s +1) + 4 s ⇒
2
Then
F (s) =
Finally
4
−5 6
+
+
s s +1 ( s +1)2
f ( t ) = ( −5 + 6 e − t + 4 t e − t ) ,
(b)
F (s) =
4s 2
( s + 3)
3
B=6
=
t≥0
A
B
C
+
+
2
( s + 3 ) ( s + 3 ) ( s + 3 )3
where
A=
1 d2 ⎡
d ⎡
3
3
s + 3) F ( s ) ⎤
= 4, B =
( s +3) F ( s )⎤⎦ s =−3 = −24
2 ⎣(
⎦
⎣
s
=−
3
2 ds
ds
and
C = ( s +3) F ( s ) s =−3 = 36
3
Then
F (s) =
Finally
−24
4
36
+
+
2
( s + 3 ) ( s + 3 ) ( s + 3 )3
f ( t ) = ( 4 − 24 t + 18t 2 ) e−3t ,
t≥0
7
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Section 14-5: Initial and Final Value Theorems
A function of time is represented by
P 14.5-1
2 s 2 − 3s + 4
F (s) = 3
s + 3s 2 + 2s
(a) Find the initial value of f(t) at t = 0.
(b) Find the value of f(t) as t approaches infinity.
Solution:
(a)
2s 2 −3s + 4 2s 2
= 2 =2
f ( 0 ) = lim sF ( s ) = lim
2
s
s →∞
s →∞ s +3s + 2
(b)
4
f ( ∞ ) = lim sF ( s ) = = 2
2
s →0
Find the initial and final values of v(t) when
P 14.5-2
V (s) =
( s + 16 )
s + 4s + 12
2
Answer: v(0) = 1, v(∞) = 0 V
Solution:
Initial value:
s →∞
Final value:
s( s +16 )
s 2 + 16 s
=
lim
=1
s →∞ s 2 + 4 s + 12
s →∞ s 2 + 4 s + 12
v ( 0 ) = lim sV ( s ) = lim
⎛
⎞
s +16
s 2 + 16 s
=
=0
v ( ∞ ) = lim s ⎜ 2
lim
⎟
2
s→ 0
⎝ s + 4 s + 12 ⎠ s → 0 s + 4 s +12
(Check: V(s) is stable because Re { pi } < 0 since pi = − 2 ± 2.828 j . We
expect the final value to exist.)
14-1
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Find the initial and final values of v(t) when
P 14.5-3
V (s) =
( s + 10 )
(3s + 2s 2 + 1s )
2
Answer: v(0) = 0, v(∞) = 10 V
Solution:
Initial value:
s 2 +10 s
= 0
s →∞ 3 s 3 + 2 s 2 + s
v(0) = lim sV ( s ) = lim
s →∞
Final value:
v ( ∞ ) = lim sV ( s ) = lim
s →0
s →0
s ( s +10 )
s ( 3s 2 + 2s +1)
= 10
(Check: V(s) is stable because p = − 0.333 ± 0.471 j . We expect the
i
final value to exist.)
Find the initial and final values of f(t) when
P 14.5-4
F (s) =
−2 ( s + 7 )
s 2 − 2 s + 10
Answer: initial value = –2; final value does not exist
Solution:
Initial value:
s →∞
Final value:
−2 s 2 −14 s
= −2
s →∞ s 2 − 2 s + 10
f ( 0 ) = lim s F ( s ) = lim
F(s) is not stable because Re { pi } > 0 since pi = 1 ± 3 j . No final value of
f ( t ) exists.
14-2
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P14.5-5
Given that L ⎡⎣v ( t ) ⎤⎦ =
as+b
where v ( t ) is the voltage shown in Figure P14.5-5, determine the
s2 +8s
values of a and b.
Figure P14.5-5
Solution:
From the plot, v(0) = 4 V and lim v ( t ) = 12 V . From the final value theorem,
t →∞
lim v ( t ) = lim sV ( s ) = lim s
t →∞
Consequently, 12 =
s →0
s →0
as+b
as+b b
= lim
= .
2
s + 8 s s →0 s + 8 8
b
⇒ b = 96 . From the initial value theorem
8
as+b
as+b
lim v ( t ) = lim sV ( s ) = lim s 2
= lim
=a
t →0
s →∞
s →∞ s + 8 s
s →∞ s + 8
Consequently a = 4.
14-3
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P14.5-6
Given that L ⎡⎣v ( t ) ⎤⎦ =
as+b
where v ( t ) is the voltage shown in Figure P14.5-6, determine
2 s 2 + 40 s
the values of a and b.
Figure P14.5-6
Solution:
From the plot, v(0+) = 10 V and lim v ( t ) = 2 V . From the final value theorem,
t →∞
lim v ( t ) = lim sV ( s ) = lim s
t →∞
s →0
s →0
as+b
as+b
b
= lim
=
.
2
2 s + 40 s s→0 2 s + 40 40
b
⇒ b = 80 . From the initial value theorem
40
as+b
as+b a
= lim
=
lim v ( t ) = lim sV ( s ) = lim s 2
t →0 +
s →∞
s →∞ 2 s + 40 s
s →∞ 2 s + 40
2
a
⇒ a = 20 .
Consequently 10 =
2
Consequently, 2 =
14-4
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Section 14.6 Solution of Differential Equations Describing a Circuit
Figure P14.6-1
P14.6-1 The circuit shown in Figure P14.6-1 is at steady state before the switch closes at time t = 0.
Determine the inductor current, i(t), after the switch closes.
Solution:
The initial inductor current is
i ( 0) =
12
=2A
6
d
i (t ) + 4 i (t )
dt
Take the Laplace Transform of both sides of this equation:
Apply KVL after the switch closes
12 = 2
12
= 2 ⎡⎣ s I ( s ) − 2 ⎤⎦ + 4 I ( s )
s
I (s) =
Solve for I(s):
Taking the Inverse Laplace Transform:
2s +6 3
1
= −
s ( s + 2) s s + 2
i ( t ) = 3 − e −2t A
Figure P14.6-2
P14.6-2 The circuit shown in Figure P14.6-1 is represented by the differential equation
d 2v ( t )
d v (t )
+7
+ 10 v ( t ) = 120
2
dt
dt
after time t = 0. The initial conditions are
i(0) = 0 and v(0) = 4 V
Determine the capacitor, v(t), after time t = 0.
Solution:
Let’s take the Laplace Transform of both sides of the differential equation.
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⎡ d v (t ) ⎤
£⎢
⎥ = sV ( s ) − 4
⎣ dt ⎦
First, using Table 14.2-2
Next, notice that
Using Table 14.2-2 again
Now we have:
or:
d v (t )
d v ( 0) i ( 0)
⇒
=
=0
0.1
dt
dt
⎡ d 2v ( t ) ⎤
⎡ d v (t ) ⎤ d v ( 0)
£⎢
= s 2 V (s) − 4 s
⎥−
2 ⎥ = s£ ⎢
dt
⎣ dt ⎦
⎣ dt ⎦
i ( t ) = 0.1
s 2 V ( s ) − 4 s + 7 ( sV ( s ) − 4 ) + 10V ( s ) =
+ 28 + 4 s
( s + 7 s + 10 )V ( s ) = 120
s
2
40 16
4 s + 28 s + 120 3
1
12
V (s) =
= −
= + 3 + 3
s ( s + 2 )( s + 5 ) s s + 2 s s + 2 s + 5
−
2
Solve for V(s):
120
s
Taking the inverse Laplace transform: v ( t ) = 12 −
40 − 2 t 16 −5t
e + e A
3
3
Figure P14.6-3
P14.6-3 The circuit shown in Figure P14.6-3 is at steady state before time t = 0. The input to the circuit is
v s ( t ) = 2.4 u ( t ) V
Consequently, the initial conditions are i1(0)=0 and i2(0)=0. Determine the inductor current, i2(t), after
time t = 0.
Solution:
First, let’s simplify the circuit by replacing the 3 12-Ω resistors at the right of the circuit by an equivalent
resistor:
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vs (t ) = 2
Write the mesh equations
and
0=2
d i 2 (t )
dt
d i1 ( t )
dt
+ 12 ( i1 ( t ) − i 2 ( t ) )
+ 8 i 2 ( t ) − 12 ( i1 ( t ) − i 2 ( t ) ) = 2
d i2 (t )
dt
+ 20 i 2 ( t ) − 12 i1 ( t )
Taking the Laplace Transforms of these equations:
2.4
= 2 s I 1 ( s ) + 12 ( I 1 ( s ) − I 2 ( s ) )
s
and
0 = 2 s I 2 ( s ) + 20 I 2 ( s ) − 12 I 1 ( s )
1
5
5⎞
⎛1
s I 2 (s) + I 2 (s) = ⎜ s + ⎟ I 2 (s)
6
3
3⎠
⎝6
2.4
5⎞
⎛1
= ( 2 s + 12 ) I 1 ( s ) − 12 I 2 ( s ) = ( 2 s + 12 ) ⎜ s + ⎟ I 2 ( s ) − 12 I 2 ( s )
s
3⎠
⎝6
7.2
0.3 0.03794 0.33974
=
+
−
I 2 (s) =
2
s
s + 14.3 s + 1.68
2 s + 16 s + 24
Next, some algebra:
Solving for I2(s):
I1 (s) =
(
)
Taking the Inverse Laplace Transform gives i2(t) for t ≥ 0:
i 2 ( t ) = 300 + 39.74 e −14.3t − 339.74 e −1.68t A
Figure P14.6-4
P14.6-4 The circuit shown in Figure P14.6-4 is at steady state before the switch opens at time t = 0.
Determine the capacitor voltage, v(t), after the switch opens.
Solution:
v (0)
3
4
=− A
(12 ) = 4 V and i ( 0 ) = −
3
3
3+9
2
1 d v (t ) 3 d v (t )
Apply KVL after the switch opens 0 =
+
+ v (t )
2 dt 2
2 dt
The initial conditions are
v (0) =
d 2 v (t )
d v (t )
That is
0=
+3
+ 2 v (t )
2
dt
dt
Let’s take the Laplace Transform of both sides of the differential equation.
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⎡ d v (t ) ⎤
£⎢
⎥ = sV ( s ) − 4
⎣ dt ⎦
First, using Table 14.2-2
d v (t )
d v ( 0)
i ( 0) 8
⇒
=−
= A
dt
dt
0.5 3
⎡ d 2v ( t ) ⎤
⎡ d v (t ) ⎤ d v ( 0)
2
= s 2 V (s) − 4 s −
Using Table 14.2-2 again £ ⎢
⎥−
2 ⎥ = s£ ⎢
dt
3
⎣ dt ⎦
⎣ dt ⎦
Next, notice that
Now we have:
Solve for V(s):
i ( t ) = −0.5
8⎞
⎛ 2
⎜ s V ( s ) − 4 s − ⎟ + 3 ( sV ( s ) − 4 ) + 2V ( s ) = 0
3⎠
⎝
44
32
20
4s+
3
= 3 − 3
V (s) =
( s + 1)( s + 2 ) s + 1 s + 2
Taking the Inverse Laplace Transform: v ( t ) =
32 − t 20 − 2 t
e − e
V
3
3
Figure P14.6-5
P14.6-5 The circuit shown in Figure P14.6-5 is at steady state before the switch closes at time t = 0.
Determine the capacitor voltage, v(t), after the switch closes.
Solution:
v (0) =
The initial capacitor voltage is
40
(12 ) = 0.98 V
10 + 40
Write a node equation after the switch closes:
vs (t ) − v (t )
10 × 10
3
= ( 2 × 10− 6 )
1200 − 100 v ( t ) = 2
600 =
v (t )
d
v (t ) +
10 × 10 3
dt
d
v ( t ) + 100 v ( t )
dt
d
v ( t ) + 100 v ( t )
dt
Take the Laplace Transform of both sides of this equation:
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600
= ⎡⎣ sV ( s ) − 0.98⎤⎦ + 100V ( s )
s
Solve for V(s):
V (s) =
Taking the Inverse Laplace Transform:
600 s + 0.98 6
5.02
= −
s ( s + 100 )
s s + 100
v ( t ) = 6 − 5.02e −100 t V
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Section 14.7 Circuit Analysis Using Impedance and Initial Conditions
P14.7-1
Figure P14.7-1a shows a circuit represented in the time domain. Figure P14.7-1b shows the same circuit,
now represented in the complex frequency domain. Figure P14.7-1c shows a plot of the inductor current.
(b)
(a)
(c)
Figure P14-7-1
Determine the values of D and E used to represent the circuit in the complex frequency domain. Determine
the values of the resistance R 2 and the inductance L.
Solution:
First, we find the values of D and E used to represent the circuit in
the complex frequency domain:
L ⎡⎣12 − 6 u ( t ) ⎤⎦ = L ⎡⎣6 u ( t ) ⎤⎦ =
6
⇒ D = 6 A.
s
E is the initial inductor current = 8 A from the plot.
Next, we find the values of the resistance R 2 and the inductance L:
The circuit is at steady state before t = 0, so the inductor acts like a short circuit. Using current division,
⎛ 30 ⎞
8=⎜
12 ⇒ R 2 = 15 Ω . Similarly, the circuit will at steady state for t → ∞. Again, the inductor
⎜ 30 + R 2 ⎟⎟
⎝
⎠
acts like a short circuit. Using current division,
14.7-1
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⎛ 30 ⎞
4=⎜
6 ⇒ R 2 = 15 Ω .
⎜ 30 + R 2 ⎟⎟
⎝
⎠
The inductor current can be represented as v ( t ) = 4 + 4 e − at for t ≥ 0. From the plot,
then
⎛ 4.54 − 4 ⎞
ln ⎜
⎟
4 ⎠
4.54 = 4 + 4 e − a ( 0.5) so a = ⎝
= 4.005 ≅ 4 1/s .
−0.5
1
45
L
=τ =
⇒ L=
= 11.25 H .
4
15 + 30
4
As a check, apply KVL to the center mesh in the complex frequency domain to get
R D
E
D
L s + R1
Es+ 1
E
D
⎛
⎞
⎛
⎞
s
s =
L
R 2 I ( s ) + L s ⎜ I ( s ) − ⎟ + R1 ⎜ I ( s ) − ⎟ = 0 ⇒ I ( s ) =
+
R
R2 ⎞
s⎠
s⎠
L s + R1 + R 2
⎛
⎝
⎝
s⎜s + 1
⎟
L ⎠
⎝
8 s + 16 4
4
= +
s ( s + 4) s s + 4
Taking the inverse Laplace transform gives
Substituting values gives
I (s) =
4 ⎤
⎡4
i ( t ) = L−1 ⎡⎣ I ( s ) ⎤⎦ = L−1 ⎢ +
= 4 + 4 e−4 t A for t ≥ 0
⎥
⎣ s s + 4⎦
14.7-2
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P14.7-2
Figure P14.7-2a shows a circuit represented in the time domain. Figure P14.7-2b shows the same circuit,
now represented in the complex frequency domain. Figure P14.7-2c shows a plot of the inductor current.
(a)
(b)
(c)
Figure P14-7-2
Determine the values of D and E used to represent the circuit in the complex frequency domain. Determine
the values of the resistance R 1 and the capacitance C.
Solution:
First, we find the values of D and E used to represent the circuit
in the complex frequency domain:
L ⎣⎡6 + 12 u ( t ) ⎦⎤ = L ⎣⎡18 u ( t ) ⎦⎤ =
18
⇒ D = 18 V.
s
E is the initial capacitor voltage = 4 V from the plot.
Next, we determine the values of the resistance R 1 and the
capacitance C:
The circuit is at steady state before t = 0, so the capacitor acts like an open circuit. Using voltage
⎛ 30 ⎞
6 ⇒ R1 = 15 Ω . Similarly, the circuit will at steady state for t → ∞. Again, the
division, 4 = ⎜
⎜ R1 + 30 ⎟⎟
⎝
⎠
capacitor acts like an open circuit. Using voltage division,
14.7-3
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⎛ 30 ⎞
12 = ⎜
18 ⇒ R1 = 15 Ω .
⎜ R1 + 30 ⎟⎟
⎝
⎠
The capacitor voltage can be represented as v ( t ) = 12 − 8 e − at for t ≥ 0. From the plot,
⎛ 11.6 − 12 ⎞
ln ⎜
⎟
−8 ⎠
= 7.9886 ≅ 8 1/s .
11.6 = 12 − 8 e − a ( 0.375) so a = ⎝
−0.375
1
1
then
= τ = (15 || 30 ) C = 10 C ⇒ C =
= 12.5 mF .
8
80
As a check, apply KCL at the top node of the 30 Ω resistor, R 2 to get
D
D
Es+
R1 C
s + V ( s ) + C s ⎛V ( s ) − E ⎞ = 0 ⇒ V ( s ) =
⎜
⎟
R1
R2
s⎠
⎛
R + R2 ⎞
⎝
s⎜s + 1
⎟
⎜
R1 R 2 C ⎟⎠
⎝
Substituting values and performing a partial fraction expansion gives
V (s) −
V (s) =
4 s + 96 12
8
= −
s ( s + 8) s s + 8
Taking the inverse Laplace transform gives
8 ⎤
⎡12
v ( t ) = L−1 ⎡⎣V ( s ) ⎤⎦ = L−1 ⎢ −
= 12 − 8 e− 8t V for t ≥ 0
⎥
⎣ s s + 8⎦
14.7-4
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P14.7-3
Figure P14.7-3a shows a circuit represented in the time domain. Figure P14.7-3b shows the same circuit,
now represented in the complex frequency domain. Determine the values of a, b and d used to represent the
circuit in the complex frequency domain.
(a)
(b)
Figure P14.7-3
Solution:
= −12 for t > 0 ฀L [ −12] =
24 − 36 u ( t ) ฀
−12
⇒ a = −12 V.
s
The circuit is at steady state before t = 0, and the input
is constant, so the capacitor acts like an open circuit
and the inductor acts like a short circuit.
24
⎛ 8 ⎞
v (0) = ⎜
=2A
⎟ 24 = 16 V and i ( 0 ) =
4+8
⎝ 4+8⎠
Consequently, b = v ( 0 ) = 16 V and d = L i ( 0 ) = ( 6 )( 2 ) = 12 .
14.7-5
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P 14.7-4
The input to the circuit shown in
t=0
Figure P 14.7-4 is the voltage of the voltage
source, 12 V. The output of this circuit is the
voltage, vo(t), across the capacitor. Determine
12 V
–
+
+
6Ω
6Ω
0.5 F
6Ω
vo(t) for t > 0.
–t/2
Answer: vo (t) = – (4 + 2e
vo(t)
–
)V for t > 0
Figure P 14.7-4
Solution:
t < 0
time domain
frequency domain
Mesh equations in the frequency domain:
6 I 1 ( s ) + 6 ( I 1 ( s ) − I 2 ( s )) + 6 I 1 ( s ) +
12
2
2
= 0 ⇒ I1 (s) = I 2 (s) −
s
3
3s
2
6
2⎞
6
⎛
I 2 ( s ) − − 6 ( I 1 ( s ) − I 2 ( s )) = 0 ⇒ ⎜ 6 + ⎟ I 2 ( s ) − 6 I 1 ( s ) =
s
s
s⎠
s
⎝
1
⎛2
2⎞
2 ⎞ 6
⎛
⇒ I 2 (s) = 2
Solving for I2(s):
⎜ 6 + ⎟ I 2 (s) − 6⎜ I 2 (s) − ⎟ =
1
3s ⎠ s
s⎠
⎝
⎝3
s+
2
⎛ 1 ⎞
−2
1
6 1⎜ 2 ⎟ 6
4
Calculate for Vo(s):
Vo ( s ) = I 2 ( s ) − = ⎜
− =
−
⎟
s 2⎜ s+ 1 ⎟ s s+ 1 s
2
2⎠
2
⎝
Take the Inverse Laplace transform:
(
vo ( t ) = − 4 + 2 e −t /2
) V for t > 0
(Checked using LNAP, 12/29/02)
14.7-6
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P 14.7-5
The input to the circuit shown in Figure
t=0
P 14.7-5 is the voltage of the voltage source, 12 V.
2Ω
The output of this circuit is the current, i(t), in the
12 V
inductor. Determine i(t) for t > 0.
–
+
2Ω
5H
Answer: i(t) = –3(1 + e–0.8t) A for t > 0
i(t)
Figure P 14.7-5
Solution:
t <0
frequency domain
time domain
Writing a mesh equation:
2⎞
⎛
⎛
⎞
−6 ⎜ s + ⎟
⎜
⎟
12
3
3
( 4 + 5 s ) I ( s ) + 30 + = 0 ⇒ I ( s ) = ⎝ 45 ⎠ = − ⎜ + 4 ⎟
s
s s+ ⎟
⎛
⎞
s⎜s + ⎟
⎜⎜
⎟
5⎠
⎝
5⎠
⎝
Take the Inverse Laplace transform:
(
i ( t ) = −3 1 + e −0.8t
) A for t > 0
(Checked using LNAP, 12/29/02)
14.7-7
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P 14.7-6 The input to the circuit shown in Figure P 14.7-6 is the voltage of the voltage source, 18 V. The
output of this circuit, the voltage across the capacitor, is given by
vo (t) = 6 + 12e–2t V when t > 0
Determine the value of the capacitance, C, and the value of the resistance, R.
Figure P 14.7-6
Solution:
Steady-state for t < 0:
Steady-state for t > 0:
From the equation for vo(t):
vo ( ∞ ) = 6 + 12 e
From the circuit:
Therefore:
6=
− 2 (∞)
vo ( ∞ ) =
=6 V
3
(18)
R+3
3
(18) ⇒ R = 6 Ω
R+3
⎛
1 ⎞ 18 6
−6
I (s)⎜ 2 +
⎟ + − = 0 ⇒ I (s) =
1
Cs⎠ s s
⎝
s+
2C
14.7-8
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⎛
⎞
1
18 1 ⎜ −6 ⎟ 18 −12
12
18
12
6
Vo ( s ) =
I (s) + =
+ =
+
+ =
+
⎜
⎟
1
Cs
s Cs ⎜ s + 1 ⎟ s
s
s s+ 1
s
s+
⎜
⎟
2C ⎠
2C
2C
⎝
Taking the inverse Laplace transform:
vo ( t ) = 6 + 12 e − t / 2C V for t > 0
Comparing this to the given equation for vo(t), we see that 2 =
1
2C
⇒ C = 0.25 F .
(Checked using LNAP, 12/29/02)
14.7-9
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P 14.7-7
The input to the circuit shown in Figure P 14.7-7 is the voltage source voltage
vs(t) = 3 – u(t)
V
The output is the voltage
vo(t) = 10 + 5e–100t V for t ≥ 0
Determine the values of R1 and R2.
Figure P 14.7-7
Solution:
We will determine Vo ( s ) , 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
Vo ( s ) =
10
5
+
s s + 100
Next, we determine Vo ( s ) from the circuit. For t ≥ 0 , 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.
When t < 0 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
3 − v (0 −)
= 0 ⇒ v (0 −) = 3 V
R1
The initial condition is
v (0 +) = v (0 −) = 3 V
Now we can represent the circuit in the frequency domain, using Laplace transforms.
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Apply KCL at the noninverting input of the op
am to get
2
3
−V ( s) V ( s) −
s
s
=
6
R1
10
s
Solving gives
106
3s + 2
R1
2
1
= +
V (s) =
⎛ 106 ⎞ s ⎛ 106 ⎞
s ⎜s+
⎟
⎜s+
⎟
⎜
⎜
R1 ⎟⎠
R1 ⎟⎠
⎝
⎝
Apply KCL at the inverting input of the op amp to get
⎛
⎞
⎜
⎟
Vo ( s) −V ( s) V ( s)
R2 ⎞
R2 ⎞⎜ 2
⎛
⎛
⎟
1
=
⇒ V o ( s ) = ⎜1 +
⎟V ( s ) = ⎜1 +
⎟⎜ +
⎟
R2
1000
⎝ 1000 ⎠
⎝ 1000 ⎠ ⎜ s ⎛ 106 ⎞ ⎟
⎜s+
⎟⎟
⎜
⎜
⎟
R
1
⎝
⎠⎠
⎝
The expressions for Vo(s) must be equal, so
⎛
⎞
⎜
⎟
R
⎛
⎞⎜
⎟
10
5
2
1
2
+
= ⎜1 +
⎟⎜ +
⎟
s s + 100 ⎝ 1000 ⎠ ⎜ s ⎛ 106 ⎞ ⎟
⎜s+
⎟
⎜
⎜
R1 ⎟⎠ ⎟
⎝
⎝
⎠
Equating coefficients gives
106
= 100 ⇒ R1 = 10 kΩ
= 5 ⇒ R 2 = 4 kΩ and
1+
1000
R1
R2
(checked using LNAPTR 7/31/04)
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P 14.7-8 Determine the inductor current, iL(t), in the circuit shown in Figure P 14.7-8 for each
of the following cases:
(a)
R = 2 Ω, L = 4.5 H, C = 1/9 F, A = 5 mA, B = – 2 mA
(b)
R = 1 Ω, L = 0.4 H, C = 0.1 F, A = 1 mA, B = – 2 mA
(c)
R = 1 Ω, L = 0.08 H, C = 0.1 F, A = 0.2 mA, B = – 2 mA
Figure P 14.7-8
Solution:
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 t = 0 − so
vC ( 0 − ) = 0 and i L ( 0 − ) = B
The capacitor voltage and inductor current are continuous so vC ( 0 + ) = vC ( 0 − ) and
iL (0 +) = iL (0 −) .
For t < 0, represent the circuit in the
frequency domain using the Laplace
transform as shown. V C ( s ) is the node
voltage at the top node of the circuit.
Writing a node equation gives
A + B VC ( s ) B VC ( s )
=
+ +
+ C sV C ( s )
s
R
s
Ls
so
A L s + R + R L C s2
=
VC ( s )
s
RLs
Then
A
AR L
C
=
VC ( s ) =
R L C s2 + L s + R s2 + 1 s + 1
RC
LC
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A
B
B
LC
and
+ =
+
I L (s) =
Ls
s
⎛
1
1 ⎞ s
s ⎜ s2 +
s+
⎟
RC
LC ⎠
⎝
1
a.) When R = 2 Ω, L = 4.5 H, C = F, A = 5 mA and B = −2 mA , then
9
5
40
−2 3
10
+
= + 7 − 7
I L (s) =
2
s s+4 s+ 1
s ( s + 4.5 s + 2 ) s
2
Taking the inverse Laplace transform gives
VC ( s )
5
5
i L ( t ) = 3 + e− 4 t − e − 0.5 t mA for t ≥ 0
7
7
b.) When R = 1 Ω, L = 0.4 H, C = 0.1 F, A = 1 mA and B = −2 mA , then
I L (s) =
⎛1
25
25
5
1 ⎞
−2
−2
+
=
+
=
−
+
+
⎜
⎟
2
⎜ s ( s + 5)2 s + 5 ⎟
s
s ( s 2 + 10 s + 25 ) s
s ( s + 5)
⎝
⎠
Taking the inverse Laplace transform gives
i L ( t ) = − (1 + 5 t e − 5 t − e− 5 t ) mA for t ≥ 0
c.) When R = 1 Ω, L = 0.08 H, C = 0.1 F, A = 0.2 mA and B = −2 mA , then
I L (s) =
25
10
s+5
−2 −1.8
−0.2 s − 2
−1.8
0.2
0.1
+
=
+
=
−
−
2
2
2
s
s ( s + 10 s + 125 ) s
( s + 5) + 102 s
( s + 5) + 102
( s + 5) + 102
2
Taking the inverse Laplace transform gives
i L ( t ) = −1.8 − e − 5 t ( 0.2 cos (10 t ) + 0.1sin (10 t ) ) mA for t ≥ 0
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P 14.7-9 Determine the capacitor current, ic(t), in the circuit shown in Figure P 14.7-9 for each
of the following cases:
(a)
R = 3 Ω, L = 2 H, C = 1/24 F, A = 12 V
(b)
R = 2 Ω, L = 2 H, C = 1/8 F, A = 12 V
(c)
R = 10 Ω, L = 2 H, C = 1/40 F, A = 12 V
Figure P 14.7-9
Solution:
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.
A
A
and vC ( t ) =
i (t ) =
2
2R
Consequently,
A
A
and vC ( 0 − ) =
i (0 −) =
2R
2
Also
iC (0 −) = 0
The capacitor voltage and inductor current are continuous so vC ( 0 + ) = vC ( 0 − ) and
iL (0 +) = iL (0 −) .
For t > 0, the voltage source voltage is
12 V. Represent the circuit in the
frequency domain using the Laplace
transform as shown.
I L ( s ) and I C ( s ) are mesh currents.
Writing a mesh equations gives
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AL
A
+ R ( I L ( s ) − I C ( s )) − = 0
2R
s
A
1
+ R ( I L ( s ) − I C ( s )) = 0
I C (s) −
Cs
2s
⎛ AL A⎞
+ ⎟
−R ⎞
⎛Ls+ R
⎜
I
s
⎛
⎞
R
s
2
(
)
L
⎜
⎟
⎟
=⎜
1
⎜
⎟
⎜
⎟
⎜ −R
⎟ I C (s) ⎜
R
+
A
⎟
⎜
⎠
C s ⎟⎠ ⎝
⎝
⎜ − 2s ⎟
⎝
⎠
L s I L (s) −
Or, in matrix form
⎛
I C (s) =
⎛ AL A⎞
A⎞
A
+ ⎟
⎟ + R⎜
2L
⎝ 2s ⎠
⎝ 2R s ⎠ =
1
1
⎛
1 ⎞
s2 +
s+
( L s + R ) ⎜ R + ⎟ − R2
RC
LC
Cs⎠
⎝
( L s + R)⎜ −
a.) When R = 3 Ω, L = 2 H, C =
1
F and A = 12 V ,
24
3
3
3
2
I C (s) = 2
=
= 4 − 4 .
s + 8 s + 12 ( s + 2 )( s + 6 ) s + 2 s + 8
Taking the inverse Laplace transform gives
3
⎛3
⎞
i C ( t ) = ⎜ e− 2t − e− 6t ⎟ u ( t ) A
4
⎝4
⎠
b.) R = 2 Ω, L = 2 H, C =
1
F and A = 12 V ,
8
3
3
I C (s) = 2
=
s + 4 s + 4 ( s + 2 )2
Taking the inverse Laplace transform gives
i C ( t ) = 3 t e− 2t u ( t ) A
1
F and A = 12 V
40
3
3
3
4
=
= ×
I C (s) = 2
2
s + 4 s + 20 ( s + 2 ) + 16 4 ( s + 2 )2 + 16
c.) R = 10 Ω, L = 2 H, C =
Taking the inverse Laplace transform gives
i C (t ) =
3 − 2t
e sin ( 4 t ) u ( t ) A
4
(checked using LNAP 4/11/01)
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P 14.7-10
The voltage source voltage in the circuit shown in Figure P 14.7-10 is
Determine v(t) for t ≥ 0.
vs(t) = 12 – 6u(t) V
Figure P 14.7-10
Solution:
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 − v ( t ) ⎞ v ( t ) ⎛ 3
⎞
−⎜
+ ⎜ − v (t ) ⎟ = 0
⎟+
8
4
⎝ 4
⎠
⎝
⎠
−12 + v ( t ) + 2 v ( t ) − 6 v ( t ) = 0 ⇒ v ( t ) = −4 V
v ( 0 + ) = v ( 0 − ) = −4 V
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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
V a ( s ) = −V ( s )
Writing a node equation gives
6
s + V ( s ) + 3 s ⎛ V ( s ) + 4 ⎞ = 0.75 V ( s )
⎜
⎟
s⎠
8
4
40 ⎝
V (s) −
Solving gives
−2
10
10
4
2
4
1 ⎞
⎛1
− 4 ⇒ V (s) =
−
=
+
−
= −2 ⎜ +
⎟
s
s ( s − 5) s − 5 s s − 5 s − 5
⎝ s s −5⎠
Taking the inverse Laplace transform gives
( s − 5)V ( s ) =
v ( t ) = −2 (1 + e5 t ) V for t ≥ 0
This voltage becomes very large as time goes on.
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P 14.7-11
Determine the output voltage, vo(t), in the circuit shown in Figure P 14.7-11.
Figure P 14.7-11
Solution:
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.
i (0 −) =
2−0
= 0.04 mA
10 × 103 + 40 × 103
and
vC ( 0 − ) = ( 40 × 103 )( 0.04 × 10−3 ) = 1.6 V
The capacitor voltage is continuous so vC ( 0 + ) = vC ( 0 − ) .
For t > 0, the voltage source voltage is
12 V. Represent the circuit in the
frequency domain using the Laplace
transform as shown.
V C ( s ) and V o ( s ) are node voltages.
Writing a node equation gives
12
1.6
VC ( s ) −
s +
s + V C ( s ) = 0 ⇒ 4 ⎛ V ( s ) − 12 ⎞ + 0.08 s ⎛ V ( s ) − 1.6 ⎞ + V ( s ) = 0
⎜ C
⎟
⎜ C
⎟ C
6
3
0.5 × 10
s ⎠
s ⎠
10 × 10
40 × 103
⎝
⎝
s
−8
48
80 s + 48
1.6 s + 600 9.6
+ 0.128 ⇒ V C ( s ) =
=
=
+
V C ( s )( 0.08 s + 5 ) =
s
s ( 0.08 s + 5 ) s ( s + 62.5 )
s s + 62.5
VC ( s ) −
Taking the inverse Laplace transform gives
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v C ( t ) = 9.6 − 8 e − 62.5 t V for t ≥ 0
The 40 kΩ resistor, 50 kΩ resistor and op amp comprise an inverting amplifier so
v o (t ) = −
50
50
v C ( t ) = − ( 9.6 − 8 e − 62.5 t ) = −12 + 10 e − 62.5 t V for t ≥ 0
40
40
so
⎧ −2 V for t ≤ 0
v o (t ) = ⎨
− 62.5 t
V for t ≥ 0
⎩ −12 + 10 e
(checked using LNAP 10/11/04)
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P 14.7-12
Determine the capacitor voltage, v(t), in the circuit shown in Figure P 14.7-12.
Figure P 14.7-12
Solution:
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
v (0 −) =
and
32
30
5−
5 = 0.25 V
32 + 96
120 + 30
v ( 0 + ) = v ( 0 − ) = 0.25 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.
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Using voltage division twice
⎛ 32 ⎞ 20 ⎛ 30 ⎞ 20 5 − 4 1
=
= V
V oc ( s ) = ⎜
⎟ −⎜
⎟
s
s
⎝ 32 + 96 ⎠ s ⎝ 120 + 30 ⎠ s
Z t = ( 96 || 32 ) + (120 || 30 ) = 24 + 24 = 48 Ω
After replacing the part of the circuit to the left of
terminals a-b by its Thevenin equivalent circuit as shown
1 0.25
−
s
s = 0.75
I (s) =
80 48 s + 80
48 +
s
V (s) =
80
0.25 ⎛ 80 ⎞ 0.75
0.25
I (s) +
=⎜ ⎟
+
s
s
s
⎝ s ⎠ 48 s + 80
V (s) =
60
0.25
1.25
0.25 0.75 −0.75 0.25 1 −0.75
+
=
+
=
+
+
= +
s ( 48 s + 80 )
s
s ( s + 1.67 )
s
s
s + 1.67
s
s s + 1.67
Taking the inverse Laplace transform gives
v ( t ) = 1 − 0.75e −1.67 t V for t ≥ 0
Then
⎧0.25 V for t ≤ 0
v (t ) = ⎨
−1.67 t
V for t ≥ 0
⎩ 1 − 0.75e
(checked using LNAP 7/1/04)
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P 14.7-13
Determine the voltage vo(t) for t ≥ 0 for the circuit of Figure P 14.7-13.
Hint: vC(0) = 4 V
Answer: vo(t) = 24e0.75t u(t) V (This circuit is unstable.)
Figure P 14.7-13
Solution:
Mesh Equations:
4 1
4 ⎛
1 ⎞
− − I C (s) − 6 I (s) + I C (s) = 0 ⇒ − = ⎜ 6 + ⎟ I C (s) + 6 I (s)
2s ⎠
s 2s
s ⎝
10
6 ( I ( s ) − I C ( s )) + 3 I ( s ) + 4 I C ( s ) = 0 ⇒ I ( s ) = − I C ( s )
9
Solving for I C(s):
4 ⎛ 2 1 ⎞
6
− = ⎜ − + ⎟ I C (s) ⇒ I C (s) =
3
s ⎝ 3 2s ⎠
s−
4
(
)
Vo ( s ) = 4 I C ( s ) =
So Vo(s) is
Back in the time domain:
24
3
s−
4
v o ( t ) = 24 e0.75t u (t ) V for t ≥ 0
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P 14.7-14
Determine the current iL(t) for t ≥ 0 for the circuit of Figure P 14.7-14.
Hint: vC(0) = 8 V and iL(0) = 1 A
1
⎛
⎞
Answer: iL ( t ) = ⎜ e − t cos 2t + e− t sin 2t ⎟ u ( t ) A
2
⎝
⎠
Figure P 14.7-14
Solution:
KVL:
8
⎛ 20
⎞
+ 4 = ⎜ + 8 + 4s ⎟ I L ( s )
s
⎝ s
⎠
so
I L ( s) =
( s + 1) + 1
2+ s
=
s + 2 s + 5 ( s + 1) 2 + 4
2
Taking the inverse Laplace transform:
1
⎛
⎞
i L ( t ) = ⎜ e − t cos 2 t + e − t sin 2 t ⎟ u ( t ) A
2
⎝
⎠
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P14.7-15
The circuit shown in Figure P14.7-23 is at steady
state before the switch opens at time
t = 0. Determine the inductor voltage v(t)
for t > 0.
Figure P14.7-15
Solution:
The circuit shown in Figure P14.7-23 is at steady
state before the switch opens at time
t = 0. Determine the inductor voltage v(t)
for t > 0.
Determine the initial conditions, i.e. the inductor
current and capacitor voltage at t = 0, as shown in
the circuit on the left below. Use those initial
conditions to represent the circuit in the s-domain
as s shown in the circuit on the left below.
Figure P14.7-15
Analysis of the s-domain circuit shows that
⎛
⎞
⎜
⎟⎛ 4
16 ( s + 2 )
4
⎞ 16 ( s + 2 )
V (s) = ⎜
+ 2⎟ = 2
=
⎟
⎜
2
8
⎠ s + 8 s + 16 ( s + 4 )
⎜⎜ 0.5 s + 4 + ⎟⎟ ⎝ s
s⎠
⎝
Performing the partial fraction expansion, we get
16 ( s + 2 )
( s + 4)
Finally
2
=
−32
k
+
s + 4 ( s + 4 )2
⇒ 16 ( s + 2 ) = k ( s + 2 ) − 32 ⇒ k = 16
V (s) =
−32
16
+
s + 4 ( s + 4 )2
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Now use linearity, e − at f ( t ) ↔ F ( s + a ) and t ↔
1
to find the inverse Laplace transform
s2
1 2
v ( t ) = 16 e −4 t ℒ -1[ − 2 ] = 16 (1 − 2 t ) e − 4 t for t > 0
s s
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P14.7-16
The circuit shown in Figure P14.7-16 is at
steady state before time t = 0. Determine the
resistor voltage v(t) for t > 0.
Figure P14.7-16
Solution:
The circuit shown in Figure 14.7-16 is at
steady state before time t = 0. Determine the
resistor voltage v(t) for t > 0.
Determine the initial conditions, i.e. the
inductor current and capacitor voltage at t =
0, as shown in the circuit on the left below.
Use those initial conditions to represent the
circuit in the s-domain as s shown in the
circuit on the left below.
Figure P14.7-16
Analysis of the s-domain circuit shows that
V (s) V (s) 2 ⎛ s ⎞
+
+ + ⎜ ⎟V ( s ) = 0 ⇒
5
6s
s ⎝ 30 ⎠
( 6 s + 5 + s )V ( s ) = −60 ⇒ V ( s ) = s +−660s + 5
2
Performing the partial fraction expansion, we get
V (s) =
−60
−15 15
=
+
s + 6 s + 5 s +1 s + 5
2
The inverse Laplace transform is
v ( t ) = 15 ( e −5t − e − t ) V for t > 0
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P14.7.17
The input to the circuit shown in Figure P14.7-17
is the voltage source voltage
⎧10 V when t < 0
v i ( t ) = 10 + 5 u ( t ) V = ⎨
⎩15 V when t > 0
Determine the response, vo(t). Assume that the
circuit is at steady state when t < 0. Sketch vo(t) as
a function of t.
Figure P14.7-17
Solution:
⎧10 V when t < 0
v i ( t ) = 10 + 5 u ( t ) V = ⎨
⎩15 V when t > 0
The circuit is at steady state when t < 0 so the capacitors
act like open circuits. Since the current in the resistor is
0 A, vo(0−) = 0 V. Then KVL gives v1(0−) = 10 V.
Use the initial conditions to represent the circuit for t>0
in the s-domain:
Calculate
5 ||
500
500
=
s
s + 100
and
500
0.8 s
s
=
125
500 s + 20
+ 5 ||
s
s
5 ||
Use voltage division to write
500
4
⎛ 15 10 ⎞ 0.8 s ⎛ 5 ⎞
s
Vo ( s ) =
− ⎟=
=
⎜
⎜
⎟
125
500 ⎝ s
s ⎠ s + 20 ⎝ s ⎠ s + 20
+ 5 ||
s
s
Taking the inverse Laplace transforms gives v o ( t ) = 4 e − 20 t V when t > 0 .
5 ||
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In summary
when t < 0
⎧ 0V
v o ( t ) = ⎨ − 20 t
V when t > 0
⎩4 e
Let t = 0.05 s and calculate
v o ( 0.05 ) = 4 e − 20 ( 0.05) =1.47 V
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P14.7-18
The input to the circuit shown in Figure P14.7-18
is the current source current
⎧25 mA when t < 0
i ( t ) = 25 − 15 u ( t ) mA = ⎨
⎩10 mA when t > 0
Determine the response, i2(t). Assume that the
circuit is at steady state when t < 0. Sketch i2(t) as
a function of t.
Figure P14.7-18
Solution:
⎧25 mA when t < 0
i ( t ) = 25 − 15 u ( t ) mA = ⎨
⎩10 mA when t > 0
The circuit is at steady state when t < 0 so the
inductors act like short circuits. Since the voltage
across the resistor is 0 V, i2(0−) = 0 V. Then KCL
gives i1(0−) = 25 mA.
Use the initial conditions to represent the circuit for
t>0 in the s-domain:
Use current division to write
I2 ( s ) =
1.25 s
⎛ 0.01 0.025 ⎞ 0.2 s ⎛ 0.015 ⎞ −0.003
−
⎜
⎟=
⎜−
⎟=
s ⎠ s+4⎝
s ⎠ s + 20
1.25 s + ( 25 + 5 s ) ⎝ s
Taking the inverse Laplace transforms gives i 2 ( t ) = 3 e − 4 t mA when t > 0 .
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In summary
⎧ 0 mA when t < 0
i 2 ( t ) = ⎨ − 4t
mA when t > 0
⎩3 e
Let t = 0.1 s and calculate
i 2 ( 0.1) = 3 e − 4 ( 0.1) = 2.01 mA
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P 14.7-19 All new homes are required to install a device called a ground fault circuit
interrupter (GFCI) that will provide protection from shock. By monitoring the current going to
and returning from a receptacle, a GFCI senses when normal flow is interrupted and switches off
the power in 1/40 second. This is particularly important if you are holding an appliance shorted
through your body to ground. A circuit model of the GFCI acting to interrupt a short is shown in
Figure P 14.7-19. Find the current flowing through the person and the appliance, i(t), for t ≥ 0
when the short is initiated at t = 0. Assume v = 160 cos 400t and the capacitor is intially
uncharged.
FIGURE P 14.7-19
Circuit model of person and appliance shorted to ground
Solution:
We are given v ( t ) = 160 cos 400 t .
The capacitor is initially uncharged, so
v C ( 0 ) = 0 V . Then
i ( 0) =
10−3
KCL yields
dvC
dt
+
vC
100
160 cos ( 400 × 0 ) − 0
= 160 A
1
=i
Apply Ohm’s law to the 1 Ω resistor to get
v −v C
i=
⇒ vC = v− i
1
di
+ 1010 i = 1600 cos 400t − ( 6.4 ×104 ) sin 400t
Solving yields
dt
Taking the Laplace transform yields
s I ( s ) − i (0) + (1010 ) I ( s ) =
so
I (s) =
s 2 + ( 400 )
( 6.4×10 ) ( 400 )
−
2
1600s
2
s 2 + ( 400 )
2
160
1600s − 2.5×107
+
s + 1010 ( s + 1010 ) ⎡⎣ s 2 + (400) 2 ⎤⎦
Next
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A
B
B*
1600s − 2.5×107
=
+
+
( s + 1010 ) ⎡⎣ s 2 + (400)2 ⎤⎦ s + 1010 s + j 400 s − j 400
where
A =
B =
1600 s − 2.5 x 107
( s +1010 ) ( s − j 400 )
1600 s − 2.5×107
s 2 + ( 400 )
=
s = − j 400
Then
I (s) =
= − 23.1 ,
2
s = −1010
2.56 × 107 ∠1.4°
8.69 × 10 ∠68.4
5
°
= 11.5 − j 27.2 and B* = 11.5 + j 27.2
136.9
11.5− j 27.2 11.5 + j 27.2
+
+
s + 1010
s + j 400
s − j 400
Finally
i ( t ) = 136.9e−1010t + 2 (11.5 ) cos 400t − 2 ( 27.2 ) sin 400t for t > 0
= 136.9e−1010t + 23.0 cos 400t − 54.4sin 400t for t > 0
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P 14.7-20 Using the Laplace transform, find vc(t) for t > 0 for the circuit shown in Figure P 14.720. The initial conditions are zero.
Hint:Use a source transformation to obtain a single mesh circuit.
Answer: vc =–5e–2t + 5 (cos 2t + sin 2t) V
Figure P 14.7-20.
Solution:
vC (0) = 0
vc +15×103 i = 10 cos 2t ⎫
⎪
⎬ ⇒
⎛ 1
−3 ⎞ d vc
i = ⎜ ×10 ⎟
⎪
⎝ 30
⎠ dt ⎭
d vc
+ 2 vc = 20 cos 2t
dt
Taking the Laplace Transform yields:
20s
s
A
B
B*
⇒ VC ( s ) =
=
+
+
sVC ( s ) − vC ( 0 ) + 2VC ( s ) = 20 2
s +4
( s + 2 )( s 2 + 4 ) s + 2 s + j 2 s − j 2
where
A=
20 s
20s
5
5
5
5
5
−40
=
= −5, B =
=
= + j and B* = − j
2
8
2
2
s + 4 s = −2
( s + 2 )( s − j 2 ) s = − j 2 1− j 2 2
Then
5 5 5 5
+j
−j
−5 2 2 2 2
VC ( s ) =
+
+
s+2 s+ j2 s− j2
⇒ vC ( t ) = −5e−2t + 5 ( cos 2t + sin 2t ) V
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P 14.7-21
Determine the inductor current, i(t), in the circuit shown in Figure P 14.7-21.
Figure P 14.7-21
Solution:
After the switch opens, apply KCL and KVL to
get
d
⎛
⎞
R1 ⎜ i ( t ) + C v ( t ) ⎟ + v ( t ) = Vs
dt
⎝
⎠
Apply KVL to get
d
i (t ) + R2 i (t )
dt
Substituting v ( t ) into the first equation gives
v (t ) = L
d⎛ d
d
⎛
⎞⎞
R1 ⎜ i ( t ) + C ⎜ L i ( t ) + R 2 i ( t ) ⎟ ⎟ + L i ( t ) + R 2 i ( t ) = Vs
dt ⎝ dt
dt
⎠⎠
⎝
2
d
d
then
R 1 C L 2 i ( t ) + R1 C R 2 + L
i ( t ) + R1 + R 2 i ( t ) = Vs
dt
dt
Dividing by R1 C L :
(
)
(
)
⎛ R1 C R 2 + L ⎞ d
⎛ R1 + R 2 ⎞
Vs
i
t
i
t
+
+
⎜
⎟
⎜
⎟ i (t ) =
(
)
(
)
2
⎜
⎟
⎜
⎟
R1 C L
dt
⎝ R1 C L ⎠ dt
⎝ R1 C L ⎠
d2
With the given values:
d2
dt
2
i ( t ) + 25
d
i ( t ) + 156.25 i ( t ) = 125
dt
Taking the Laplace transform:
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125
⎡ 2
⎛d
⎞⎤
⎢ s I ( s ) − ⎜ dt i ( 0 + ) + s i ( 0 + ) ⎟ ⎥ + 25 ⎡⎣ s I ( s ) − i ( 0 + ) ⎤⎦ + 156.25 I ( s ) = 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
v (0 −) =
9
20 = 14.754 V
9 + (16 || 4 )
Then, using current division
⎛ 4 ⎞ v (0 −)
i (0 −) = ⎜
= 0.328 A
⎟
⎝ 16 + 4 ⎠ 9
The capacitor voltage and inductor current are continuous so v ( 0 + ) = v ( 0 − ) and i ( 0 + ) = i ( 0 − ) .
After the switch opens
v (t ) = L
d
i (t ) + R2 i (t ) ⇒
dt
v ( 0 + ) 9 i ( 0 + ) 14.754 9 ( 0.328 )
d
i (0 +) =
+
=
+
= 29.508
0.4
0.4
0.4
0.4
dt
Substituting these initial conditions into the Laplace transformed differential equation gives
⎡ s 2 I ( s ) − ( 29.508 + 0.328 s ) ⎤ + 25 ⎡ s I ( s ) − 0.328⎤ + 156.25 I ( s ) = 125
⎣
⎦
⎣
⎦
s
( s2 + 25 s + 156.25) I ( s ) = 125s + ( 29.508 + 0.328 s ) + 25 ( 0.328)
so
I (s) =
=
0.328 s 2 + ( 29.508 + 25 ( 0.328 ) ) + 125
(
s s 2 + 25 s + 156.25
)
0.328 s 2 + ( 29.508 + 25 ( 0.328 ) ) + 125
s ( s + 12.5 )
2
=
23.6
0.8
−0.471
+
+
2
s + 12.5 ( s + 12.5 )
s
Taking the inverse Laplace transform
i ( t ) = 0.8 + e −12.5 t ( 23.6 t − 0.471) A for t ≥ 0
So
⎧0.328 A for t ≤ 0
i (t ) = ⎨
−12.5 t
( 23.6 t − 0.471) A for t ≥ 0
⎩ 0.8 + e
(checked using LNAP 10/11/04)
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P 14.7-22
Find v2(t) for the circuit of Figure P 14.7-22 for t ≥ 0.
Hint: Write the node equations at a and b in terms of v1 and v2. The initial conditions are v1(0) =
10 V and v2(0) = 25 V. The source is vs = 50 cos 2t u(t) V.
Answer: v2 ( t ) =
23 − t 16 −4t
e + e + 12 cos 2t + 12 sin 2t V t ≥ 0
3
3
Figure P 14.7-22
Solution:
Apply KCL at node a to get
1 d v1 v 2 − v1
=
48 dt
24
⇒ 2 v1 +
d v1
dt
= 2 v2
Apply KCL at node b to get
v 2 − 50 cos 2 t
20
+
v 2 − v1
24
+
v2
30
+
d v2
1 d v2
= 0 ⇒ − v1 + 3 v 2 +
= 60 cos 2 t
24 dt
dt
Take the Laplace transforms of these equations, using v1 (0) = 10 V and v2 (0) = 25 V , to get
( 2+ s ) V1 ( s) − 2V2 ( s) = 10 and − V1 ( s) + ( 3+ s ) V2 ( s) =
25s 2 + 60s +100
s2 + 4
Solve these equations using Cramer’s rule to get
⎛ 25s 2 + 60s +100 ⎞
( 2+ s ) ⎜
⎟ +10 ( 2+ s ) ( 25s 2 + 60 s +100 )+10 ( s 2 + 4 )
s2 +4
⎝
⎠
=
V2 ( s ) =
( 2+ s ) (3+ s)− 2
( s 2 + 4 ) ( s +1)( s + 4 )
=
25s 3 +120s 2 + 220s + 240
( s 2 + 4 ) ( s +1)( s + 4 )
Next, partial fraction expansion gives
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V2 ( s ) =
A
A*
B
C
+
+
+
s + j 2 s − j 2 s +1 s + 4
where
A =
25s 3 +120 s 2 + 220 s + 240
−240 − j 240
=
= 6 + j6
−40
( s +1) ( s + 4 ) ( s − j 2 ) s =− j 2
A* = 6 − j 6
25s 3 +120 s 2 + 220 s + 240
115 23
B =
=
=
2
( s + 4 ) ( s + 4 ) s=−1 15 3
25s 3 +120 s 2 + 220 s + 240
−320 16
C =
=
=
2
s =−4
3
−60
( s + 4 ) ( s +1)
Then
V2 ( s ) =
6+ j 6 6− j 6 23 3 16 3
+
+
+
s + j 2 s − j 2 s +1 s + 4
Finally
v2 (t ) = 12 cos 2 t + 12sin 2 t +
23 − t 16 −4t
e + e V t≥0
3
3
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P 14.7-23 The motor circuit for driving the snorkel shown in Figure P 14.7-23a is shown in
Figure P 14.7-23b. Find the motor current I2(s) when the initial conditions are i1(0 –) = 2 A and
i2(0 –) = 3 A. Determine i2(t) and sketch it for 10 s. Does the motor current smoothly drive the
snorkel?
Solution:
Here are the equations describing the coupled coils:
di1
di
+M 2
dt
dt
di
di
v2 (t ) = L2 2 + M 1
dt
dt
v1 (t ) = L1
⇒ V1 ( s) = 3 ( s I1 ( s) − 2 ) + ( sI 2 ( s) − 3) = 3s I1 ( s) + sI 2 ( s ) − 9
⇒ V2 ( s) = s( I1 ( s ) − 2 ) + 2( sI 2 ( s) −3) = sI1 ( s) + 2sI 2 ( s ) − 8
Writing mesh equations:
5
5
= 2 ( I1 ( s ) + I 2 ( s ) ) + V1 = 2 ( I1 ( s ) + I 2 ( s ) ) + 3s I1 ( s ) + sI 2 ( s ) − 9 ⇒ ( 3s + 2 ) I1 + ( s + 2 ) I 2 = 9 +
s
s
V1 ( s ) = V2 ( s ) + 1I 2 ( s ) ⇒ 3s I1 ( s ) + sI 2 ( s ) − 9 = sI1 ( s ) + 2 sI 2 ( s ) − 8+ I 2 ( s ) ⇒ 2s I1 − ( s +1) I 2 =1
Solving the mesh equations for I2(s):
I2 ( s ) =
15s + 8
3s + 1.6
0.64
2.36
=
=
+
2
s + 0.26
s + 1.54
5s + 9s + 2 ( s + 0.26 )( s +1.54 )
Taking the inverse Laplace transform:
i2 (t ) = 0.64e −0.26t + 2.36e −1.54t A for t > 0
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P14.7-24
Using Laplace transforms, find vo(t) for t > 0 for the circuit shown in Figure 14.7-24.
Figure 14.7-24
Solution:
Find vo(t) for t > 0 for this circuit:
For t < 0, with the circuit at steady state, we have
v ( 0 ) = 2 V and i ( 0 ) =
For t > 0 we have
Apply KVL to the left mesh to get
⎛ 1 d
⎞
10 ⎜
v C (t ) ⎟ + v C (t ) − 8 = 0 ⇒
⎝ 20 dt
⎠
1 d
vC (t ) + vC (t ) = 8
2 dt
Apply KVL to the right mesh to get
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1
A
2
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4
d
d
i L ( t ) + 12 i L ( t ) − 3 v C ( t ) = 0 ⇒ 4 i L ( t ) + 12 i L ( t ) = 3 v C ( t )
dt
dt
Take the Laplace transform of these differential equations to get
1
8
⎡⎣ sVC ( s ) − v C ( 0 ) ⎤⎦ + VC ( s ) =
⇒
2
s
1
8
2 s + 16
⎡⎣ sVC ( s ) − 2 ⎤⎦ + VC ( s ) =
⇒ VC ( s ) =
2
s
s ( s + 2)
and
1⎤
⎡
4 ⎡⎣ s I L ( s ) − i L ( 0 ) ⎤⎦ + 12 I L ( s ) = 3VC ( s ) ⇒ 4 ⎢ s I L ( s ) − ⎥ + 12 I L ( s ) = 3VC ( s )
2⎦
⎣
3
1
⇒ ( s + 3) I L ( s ) = VC ( s ) +
4
2
3 ⎛ 2 s + 16 ⎞ 1
⇒ ( s + 3) I L ( s ) = ⎜⎜
⎟+
4 ⎝ s ( s + 2 ) ⎟⎠ 2
1 2 5
s + s + 12
2
⇒ IL ( s ) = 2
s ( s + 2 )( s + 3)
9
−
2
3
⇒ IL ( s ) = + 2 +
s s+2 s+3
Taking the inverse Laplace transform gives
Finally
9
⎛
⎞
i L ( t ) = ⎜ 2 − e − 2 t + 3 e − 3t ⎟ u ( t ) A
2
⎝
⎠
−2t
v o ( t ) = 12 i L ( t ) = 24 − 54 e + 36 e − 3t u ( t ) V
(
)
(checked with LNAP 2/28/05)
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P14.7-25
The circuit shown in Figure P14.7-25 is at steady
state before the switch opens at time t = 0.
Determine the inductor voltage v(t) for t > 0.
Figure P14.7-25
Solution:
Determine the initial conditions, i.e. the inductor current and capacitor voltage at t = 0, as shown
in the circuit on the left below. Use those initial conditions to represent the circuit in the sdomain as shown in the circuit on the left below.
Analysis of the s-domain circuit shows that
⎛
1000 ⎞
1000 ⎞
⎛
⎟ (1.5 ) ⎜ 40 s +
⎟
3.846 s ⎠
12 s + 78
3.846 ⎠
⎝
⎝
V (s) = −
=
= 2
1000
1000
s + 8 s + 52
5 s + 40 +
5 s 2 + 40 s +
3.846 s
3.846
The denominator does not factor any further in the real numbers. Let’s complete the square in the
denominator
12 ( s + 4 ) + 30
12 ( s + 4 )
5 ( 6)
12 s + 78
12 s + 78
12 s + 78
= 2
=
=
=
+
V (s) = 2
2
2
2
2
s + 8 s + 52 ( s + 8 s + 16 ) + 36 ( s + 4 ) + 36 ( s + 4 ) + 36 ( s + 4 ) + 6 ( s + 4 )2 + 62
( −1.5) ⎜ 40 +
Now use the property e − at f ( t ) ↔ F ( s + a ) and the Laplace transform pairs
sin ωt for t ≥ 0 ↔
ω
s + ω2
2
and cos ωt for t ≥ 0 ↔
s
s + ω2
2
to find the inverse Laplace transform
5 ( 6)
12 s
v ( t ) = e − 4 t ℒ-1[ 2
+ 2
] = e − 4t [ 12 cos(6t) + 5 sin (6t) ] for t > 0
2
2
s +6 s +6
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P14.7-26
The circuit shown in Figure P14.7-26 is at
steady state before the switch opens at time
t = 0. Determine the inductor voltage v(t)
for t > 0.
Figure P14.7-26
Solution
The circuit shown in Figure P14.7-22 is at
steady state before the switch opens at time
t = 0. Determine the inductor voltage v(t)
for t > 0.
Determine the initial conditions, i.e. the
inductor current and capacitor voltage at t = 0,
as shown in the circuit on the left below. Use
those initial conditions to represent the circuit
in the s-domain as s shown in the circuit on the
left below.
Figure P14.7-26
Analysis of the s-domain circuit shows that
⎛
⎞
⎜
⎟ ⎛ 12 ⎞
−12
−144
−60
⎟⎜ ⎟ =
= 2
V (s) = ⎜
⎜ 2.4 s + 12 + 1000 ⎟ ⎝ s ⎠ 2.4 s 2 + 12 s + 1000 s + 5 s + 48.5
⎜
8.59 s ⎟⎠
8.59
⎝
The denominator does not factor any further in the real numbers. Let’s complete the square in the
denominator
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V (s) =
−9.23 ( 6.5 )
−60
−60
−60
=
=
=
2
2
2
2
s + 5 s + 48.5 ( s + 2.5 ) + 42.25 ( s + 2.5 ) + 6.5
( s + 2.5) + 6.5 2
2
Now use e − at f ( t ) ↔ F ( s + a ) and sin ωt for t > 0 ↔
transform
v ( t ) = e − 2.5t ℒ-1[
−9.23 ( 6.5 )
( s + 2.5) + 6.5
2
2
ω
to find the inverse Laplace
s + ω2
2
] = −9.23 e − 2.5t sin (6.5 t) for t > 0
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Section 14.8 Transfer Functions
P14.8‐1 The input to the circuit shown in
Figure P14.8‐1 is the voltage v i ( t ) and
the output is the voltage v o ( t ) .
Determine the values of L, C, k, R1 and R2
that cause the step response of this
circuit to be:
(
)
v o ( t ) = 5 + 20 e−5000 t − 25 e−4000t u ( t ) V
Figure 14.8‐1
Answer: One solution is R1 = 400 Ω, L = 0.1 H, k = 5 V/V, C = 0.1 μF, R1 = 2 kΩ.
Solution: First, determine the transfer function from the circuit. To do so, represent the circuit in the s‐
domain as shown below.
Applying voltage division twice
R1
Va ( s ) =
R1
R1 + L s
Vi ( s ) =
L V (s)
R i
s+ 1
L
k
R 2C
and V o ( s ) =
k Va ( s ) =
V (s)
1
1 a
R2 +
s+
Cs
R 2C
1
Cs
Substituting Va(s) from the first equation into the second, the transfer function of this circuit is
H (s) =
Vo ( s )
Vi ( s )
k R1
=
R 2C L
R1 ⎞ ⎛
⎛
1 ⎞
⎟
⎜ s + ⎟ ⎜⎜ s +
L ⎠⎝
R 2C ⎟⎠
⎝
Next, determine the transfer function from the step response.
H (s)
5
20
25
10 8
.
= L ⎡⎣5 + 20 e −5000 t − 25 e−4000 t ⎤⎦ = +
−
=
s
s s + 5000 s + 4000 s ( s + 5000 )( s + 4000 )
Performing partial fraction expansion:
k R1
R 2C L
10 8
= H (s) =
( s + 5000 )( s + 4000 )
R1 ⎞ ⎛
⎛
1 ⎞
⎟
⎜ s + ⎟ ⎜⎜ s +
L ⎠⎝
R 2C ⎟⎠
⎝
Equality requires
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R1
L
and
= 4000 and
1
= 5000 or
R 2C
R1
L
= 5000 and
1
= 5000
R 2C
108 = k ( 4000 )( 5000 ) ⇒ k = 5 V/V
The solution is not unique. One solution is R1 = 400 Ω, L = 0.1 H, k = 5 V/V, C = 0.1 μF, R1 = 2 kΩ.
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P14.8‐2 The input to the circuit shown in Figure P14.8‐2 is
the voltage v i ( t ) and the output is the voltage v o ( t ) .
Determine the step response of this circuit.
Figure 14.8‐2
Solution:
The transfer function of this circuit is
1
C s ⎛ R2 ⎞
H (s) =
⎜1 +
⎟
1 ⎜
R1 ⎟⎠
⎝
R+ Ls+
Cs
1
10 s 2 +
L C ⎛ R2 ⎞
=
⎜1 +
⎟⎟
R
1 ⎜
R
2
1
⎝
⎠
s + s+
L
LC
Ls +
s 2 + 500, 000
= 2
s + 200 s + 50, 000
⎡ 10 s 2 + 500, 000 ⎤
⎡ H (s) ⎤
−1
⎥.
The step response is v o ( t ) = L ⎢
⎥=L ⎢
2
s
200
50,
000
+
+
s
s
s
⎢
⎥⎦
⎣
⎦
⎣
−1
(
)
Let’s do some algebra:
−2000
10 s 2 + 500, 000
10
= + 2
2
s s + 200 s + 50, 000
s s + 200 s + 50, 000
(
)
=
−2000
10
+
s ( s + 100 )2 + 2002
=
10
200
− 10
2
s
( s + 100 ) + 2002
The step response is
⎡10
⎤
200
v o ( t ) = L −1 ⎢ − 10
⎥ = 10 − e−100 t ⎡⎣10sin ( 200 t ) ⎤⎦ = ⎡⎣10 + 10 e−100 t cos ( 200 t + 90° ) ⎤⎦ u ( t ) V
2
2
( s + 100 ) + 200 ⎦⎥
⎣⎢ s
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P14.8‐3 The input to the circuit shown in Figure P14.8‐3
is the voltage v i ( t ) and the output is the voltage v o ( t ) .
Determine the impulse response of this circuit.
Figure 14.8‐3
Solution:
Represent the circuit in the s‐domain:
The transfer function of this circuit is
Vo ( s )
s2
5s
H (s) =
=
=
V i ( s ) 25 + 20 + 5 s s 2 + 5 s + 4
s
⎡
⎤
s2
The impulse response is h ( t ) = L −1 ⎣⎡ H ( s ) ⎦⎤ = L −1 ⎢ 2
⎥.
⎣s + 5s + 4⎦
Let’s do some algebra:
−1 ⎞
⎛ −16
⎜ −3
⎟
5s + 4
5s + 4
s2
= 1− 2
= 1−
= 1− ⎜
+ 3 ⎟
2
s +5s + 4
s +5s + 4
( s + 4 )( s + 1)
⎜ s + 4 s +1⎟
⎝
⎠
The impulse response is
16
1 ⎤
⎡
⎢
⎥
1 ⎞
⎛ 16
h ( t ) = L −1 ⎢1 − 3 + 3 ⎥ = δ ( t ) + ⎜ − e −4 t + et ⎟ u ( t ) V
3 ⎠
⎝ 3
⎢ s + 4 s + 1⎥
⎣
⎦
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P14.8‐4 The input to the circuit shown in Figure P14.8‐4
is the voltage v i ( t ) and the output is the voltage v o ( t ) .
Determine the step response of this circuit.
Answer: step response = 5 − ( 5 + 20 t ) e −4 t u ( t )
(
)
Figure 14.8‐4
Solution:
Represent the circuit in the s‐domain as
shown.
Using voltage division
10
16
s
Va ( s ) =
Vi ( s ) = 2
Vi ( s )
20 5
+
8
s
+
s
16
+ s+5
s 8
Recognizing the noninverting amplifier:
80
⎛ 20 ⎞
V o ( s ) = ⎜1 + ⎟ V a ( s ) = 2
Vi ( s )
s + 8 s + 16
5 ⎠
⎝
The transfer function of this circuit is
H (s) =
Vo ( s )
Vi ( s )
=
80
80
=
s + 8 s + 16 ( s + 4 ) 2
2
⎡ 80 ⎤
⎡ H (s) ⎤
−1
step response = L −1 ⎢
⎥=L ⎢
2⎥.
⎣ s ⎦
⎣⎢ s ( s + 4 ) ⎦⎥
Performing partial fraction expansion:
80
80
80
A
16 +
+ −4 2
2 =
s s + 4 ( s + 4)
s ( s + 4)
The step response is
Multiplying both sides by s ( s + 4 ) and equating coefficients of like powers of s:
2
80 = 5 ( s + 4 ) + A s ( s + 4 ) + ( −20 ) s ⇒
2
A = −5
The step response is
⎡ 5 −5
−20 ⎤
−4 t
+
u (t ) V
step response = L −1 ⎢ +
2 ⎥ = 5 − ( 5 + 20 t ) e
⎣⎢ s s + 4 ( s + 4 ) ⎦⎥
(
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P14.8-5 The input to the circuit shown in Figure P14.8-5 is the voltage, vi(t), of the independent voltage
source. The output is the voltage, vo(t), across the 5-kΩ resistor. Specify values of the resistance, R, the
capacitance, C, and the inductance, L, such that the transfer function of this circuit is given by
H (s) =
Vo ( s )
Vi ( s )
=
15 × 106
( s + 2000 )( s + 5000 )
Answer: R = 5k Ω, C = 0.5 μF, and L = 1 H (one possible solution)
Figure P14.8‐5
Solution:
The transfer function can also be calculated from 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
⎛
R ⎞⎟
Va ( s ) = ⎜⎜1 +
V s
⎜⎝ 10000 ⎠⎟⎟ c ( )
Now, writing node equations,
Vc ( s ) −Vi ( s )
Vo ( s ) −Va ( s ) Vo ( s )
+ CsVc ( s ) = 0 and
+
=0
Ls
1000
5000
Solving these node equations gives
1 ⎛⎜
R ⎞⎟ 5000
⎟
⎜⎜⎝1 +
1000 C
10000 ⎠⎟ L
H (s) =
⎛
⎞⎛
⎞
⎜⎜ s + 1 ⎟⎟ ⎜⎜ s + 5000 ⎟⎟
⎟⎜
⎜⎝ 1000C ⎠⎝
L ⎠⎟
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Comparing these two equations for the transfer function gives
⎛
⎞
⎛
⎞
⎜⎜ s + 1 ⎟⎟ = ( s + 2000) or ⎜⎜ s + 1 ⎟⎟ = ( s + 5000)
⎝⎜ 1000C ⎠⎟
⎝⎜ 1000C ⎠⎟
⎛
⎛
5000 ⎟⎞
5000 ⎞⎟
⎟⎟ = ( s + 2000) or ⎜⎜⎜ s +
⎟ = ( s + 5000)
⎜⎝⎜⎜ s +
⎝
L ⎠
L ⎠⎟
1 ⎛
R ⎞ 5000
= 15 × 10 6
⎜1 +
⎟
1000C ⎝ 10000 ⎠ L
The solution isn’t unique, but there are only two possibilities. One of these possibilities is
⎛
⎞
⎜⎜ s + 1 ⎟⎟ = ( s + 2000) ⇒ C = 0.5 μ F
⎜⎝ 1000C ⎠⎟
5000 ⎞
⎛
⎟ = ( s + 5000) ⇒ L = 1 H
⎜s +
⎝
L ⎠
⎛
⎞
⎜⎜1 + R ⎟⎟ 5000 = 15×106 ⇒ R = 5 kΩ
⎟
⎜
1000 0.5×10 ⎝ 10000 ⎠ 1
(
1
6
)
(Checked using LNAP, 12/29/02)
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P14.8-6 The input to the circuit shown in Figure P14.8-6
is the voltage, vi(t), of the independent voltage source.
The output is the voltage, vo(t), across the 10-kΩ resistor.
Specify values of the resistances, R1 and R2, such that the
step response of this circuit is given by
vo(t) = –4(1 – e–250t)u(t) V
Answer: R1 = 10 kΩ and R2 = 40 kΩ
Figure P 14.8‐6
Solution:
The transfer function of the circuit is
R2
1
1+ R2 C s
R1 C
H (s) = −
=−
1
R1
s+
R2 C
The give step response is vo ( t ) = −4 (1 − e −250 t ) u ( t ) V . The correspond transfer function is calculated as
H (s)
−1000
−1000
4 ⎞
⎛4
= L −4 (1 − e − 250 t ) u ( t ) = − ⎜ −
⇒ H (s) =
⎟=
s
s + 250
⎝ s s + 250 ⎠ s ( s + 250 )
{
}
Comparing these results gives
1
1
1
= 250 ⇒ R 2 =
=
= 40 kΩ
250 C 250 ( 0.1× 10 − 6 )
R2 C
1
1
1
= 1000 ⇒ R1 =
=
= 10 kΩ
1000 C 1000 ( 0.1×10 − 6 )
R1 C
(Checked using LNAP, 12/29/02)
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P14.8‐7 The input to the circuit shown in Figure P14.8‐7 is
the voltage v i ( t ) and the output is the voltage v o ( t ) .
Determine the step response of this circuit.
Answer: v o ( t ) = ( 4 × 10 3 ) t u ( t ) V
Figure 14.8‐7
Solution: First, determine the transfer function from the
circuit. To do so, represent the circuit in the s‐domain as
shown.
Notice that Va(s) is the node voltage at both node a and
node b. Apply KCL at node a to get
Va ( s )
10, 000
+
Va ( s ) − Vo ( s )
30, 000
=0 ⇒
⎛ 30, 000 ⎞
V o ( s ) = ⎜1 +
⎟V a ( s ) = 4V a ( s )
⎝ 10, 000 ⎠
Apply KCL at node b to get
Vi ( s ) − Va ( s )
10, 000
=
Vi ( s )
Va ( s ) − Vo ( s ) Va ( s )
V a ( s ) V a ( s ) − 4V a ( s ) sV a ( s )
+
⇒
=
+
+
7
10
30, 000
10, 000 10, 000
30, 000
10 7
s
V i ( s ) sV a ( s ) sV o ( s )
⇒
=
=
10, 000
10 7
4 ×10 7
⇒ Vo ( s ) =
H (s) =
The transfer function is
The step response is
Vo ( s )
Vi ( s )
=
4 ×10 7
4 × 10 3
Vi ( s ) =
Vi ( s )
s
10, 000 s
4 × 10 3
s
3
⎡ H (s) ⎤
−1 ⎡ 4 × 10 ⎤
3
L ⎢
⎥=L ⎢
⎥ = ( 4 ×10 ) t u ( t ) V .
2
⎣ s
⎦
⎣ s ⎦
−1
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P14.8‐8 The input to the circuit shown in Figure P14.8‐8
is the voltage v i ( t ) and the output is the voltage
v o ( t ) . Determine the step response of this circuit.
⎡ ⎛4
2
⎞⎤
Answer: v o ( t ) = ⎢ 2 − ⎜ e −1000 t + e −4000 t ⎟ ⎥ u ( t ) V
3
⎠⎦
⎣ ⎝3
Figure 14.8‐8
Solution: First, determine the transfer function from
the circuit. To do so, represent the circuit in the s‐
domain as shown.
Recognizing the noninverting amplifier we
⎛ 10 7
⎛
10 7 ⎞
4 ⎞
||10
⎜
⎟
⎜
3 |⎟
10
s
s
+
V a ( s ) = ⎜1 +
⎟ V ( s ) = ⎜1 +
⎟V ( s )
10 4 ⎟ i
10 4 ⎟ i
⎜
⎜
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
⎛ s + 2000 ⎞
=⎜
⎟Vi ( s )
⎝ s + 1000 ⎠
(Alternately, this equation can obtained by applying
KCL at the inverting input node of the op amp.)
Use voltage division to write
Vo ( s ) =
The transfer function is
2000
4000
Va ( s ) =
Va ( s )
2000 + 0.5 s
s + 2000
H (s) =
Vo ( s )
Vi ( s )
=
4000 ( s + 2000 )
( s + 1000 )( s + 4000 )
The step response is
⎡ 4000 ( s + 2000 ) ⎤
⎡ H (s) ⎤
−4 3
−2 3 ⎤
−1
−1 ⎡ 2
+
v o ( t ) = L −1 ⎢
⎥=L ⎢ +
⎥=L ⎢
⎥⎦ .
s
s
s
1000
s
4000
s
s
1000
s
4000
+
+
+
+
(
)(
)
⎣
⎣
⎦
⎣
⎦
⎡ ⎛4
2
⎞⎤
v o ( t ) = ⎢ 2 − ⎜ e−1000t + e−4000 t ⎟ ⎥ u ( t ) V
3
⎠⎦
⎣ ⎝3
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P14.8‐9 The input to the circuit shown in Figure P14.8‐9 is the voltage, vi(t), of the independent voltage
source. The output is the voltage, vo(t). The step response of this circuit is
vo(t) = 0.5(1 + e–4t)u(t) V
Determine the values of the inductance, L, and the resistance, R.
Answer: L = 6 H and R = 12 Ω
Figure P14.8‐9
Solution:
R
s+
Vo ( s )
R+ Ls
L
=
=
H (s) =
12
Vi ( s ) 12 + R + L s s + + R
L
From the circuit:
From the given step response:
H (s)
s+2
0.5 0.5
= L ⎣⎡ 0.5 (1 + e −4 t ) u ( t ) ⎦⎤ =
+
=
⇒
s
s
s + 4 s ( s + 4)
H (s) =
s+2
s+4
Comparing these two forms of the transfer function gives:
R
⎫
=2 ⎪
12 + 2 L
⎪
L
= 4 ⇒ L = 6 H, R = 12 Ω
⎬ ⇒
12 + R
L
= 4⎪
⎪⎭
L
(Checked using LNAP, 12/29/02)
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P 14.8‐10 An electric microphone and its associated circuit can be represented by the circuit shown in
Figure 14.8‐10. Determine the transfer function H(s) = V0(s)/ V(s).
V (s)
RCs
Answer: o
=
V ( s ) ( R1Cs + 2 )( 2 RCs + 1) − 1
Figure 14.8-10 Microphone circuit
Solution:
Mesh equations:
1 1 ⎞
1
⎛
V ( s ) = ⎜ R1 + + ⎟ I1 ( s ) −
I2 ( s )
Cs Cs ⎠
Cs
⎝
1 ⎞
1
⎛
0 = ⎜ R+ R+ ⎟ I2 ( s ) −
I1 ( s )
Cs ⎠
Cs
⎝
Solving for I2(s):
Then Vo ( s ) = R I 2 ( s ) gives
H (s) =
⎛ 1 ⎞
V (s) ⎜ ⎟
⎝ Cs ⎠
I2 ( s ) =
2 ⎞⎛
1 ⎞
1
⎛
⎜ R1 + ⎟ ⎜ 2 R + ⎟ −
(Cs ) 2
Cs ⎠ ⎝
Cs ⎠
⎝
V0 ( s )
RCs
=
=
V (s)
[ R1Cs + 2][ 2 RCs +1] − 1
s
⎡
⎤
4 RC + R1C
1
2
⎢
⎥
2 R1C s +
s+
2 2⎥
⎢
2 RR1C 2
( 2RR1C ) ⎦
⎣
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P 14.8‐11 Engineers had avoided inductance in long‐distance circuits because it slows transmission.
Oliver Heaviside proved that the addition of inductance to a circuit could enable it to transmit without
distortion. George A. Campbell of the Bell Telephone Company designed the first practical inductance
loading coils, in which the induced field of each winding of wire reinforced that of its neighbors so that
the coil supplied proportionally more inductance than resistance. Each one of Campbell’s 300 test coils
added 0.11 H and 12 Ω at regular intervals along 35 miles of telephone wire (Nahin, 1990). The loading
coil balanced the effect of the leakage between the telephone wires represented by R and C in Figure P
14.8‐11. Determine the transfer function V2(s)/V1(s).
V (s)
R
Answer: 2
=
2
V1 ( s ) RCLs + ( L + Rx RC ) s + Rx + R
Figure P 14.8‐11 Telephone and load coil circuit
Solution:
Let
⎛ 1 ⎞
R⎜ ⎟
R
Cs
Z2 = ⎝ ⎠ =
1
RCs + 1
R+
Cs
Z1 = Rx + Lx s
Then
R
RCs + 1
V2
Z2
=
=
V1 Z1 + Z 2
Rx + Lx s +
V2
V1
R
RCs + 1
=
R
Lx RCs + ( Lx + Rx RC ) s + Rx + R
2
1
Lx C
=
( L + R RC ) s + Rx + R
s2 + x x
Lx RC
Lx RC
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P14.8‐12 The input to the circuit shown in Figure P14.8‐12
is the current i ( t ) and the output is the voltage v ( t ) .
Determine the impulse response of this circuit.
Figure P14.8‐12
Solution: First, determine the transfer function from the circuit. To do so, represent the circuit in the
s‐domain as shown below.
Applying current division
Then
The transfer function is
The impulse response is
I o (s) =
10
500
I (s) =
I (s)
10 + 40 + 0.02 s
s + 2500
V ( s ) = 0.02 s I o ( s ) =
H (s) =
10 s
I (s)
s + 2500
V (s)
10 s
25000
=
= 10 −
I ( s ) s + 2500
s + 2500
£ −1 ⎡⎣ H ( s ) ⎤⎦ = 10 δ ( t ) − 25000 e−2500t u ( t ) V
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P14.8‐13 The input to the circuit shown in Figure
P14.8‐13 is the current i ( t ) and the output is the
voltage v ( t ) . Determine the impulse response of this
circuit.
(
)
Answer: v ( t ) = 1.25 ×10 7 e−5000t − e−25000t u ( t ) V
Figure P14.8‐13
Solution: First, determine the transfer function from the circuit. To do so, represent the circuit in the s‐
domain as shown below.
Applying KCL at the inverting input node of the op amp
Va ( s ) − 0 Va ( s ) − 0
s ⎞
⎛ 1
⎛ s + 5000 ⎞
+
= 0 ⇒ I (s) = −⎜
+ 7 ⎟Va ( s ) = − ⎜
⎟V a ( s )
7
7
10
2000
⎝ 2000 10 ⎠
⎝ 10
⎠
s
5000
25000
V (s) =
Va ( s ) =
Va ( s )
Using voltage division
5000 + 0.2 s
s + 25000
I (s) +
Combining these equations:
7
⎞
⎛ 25000 ⎞ ⎛ 10
V (s) = ⎜
⎟ I (s)
⎟⎜
⎝ s + 25000 ⎠ ⎝ s + 5000 ⎠
The transfer function is
V (s)
25 × 10 10
=
H (s) =
I ( s ) ( s + 5000 )( s + 25, 000 )
Partial fraction expansion:
25 × 10 10
1.25 ×10 7 −1.25 × 10 7
=
+
( s + 5000 )( s + 25, 000 ) s + 5000 s + 25, 000
The impulse response is
£ −1 ⎡⎣ H ( s ) ⎤⎦ = 1.25 × 10 7 e −5000t − e −25000t u ( t ) V
(
)
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P14.8‐14 A series RLC circuit is shown in Figure P14.8‐14.
Figure P14.8-14
Determine (a) the transfer function H(s), (b) the impulse response, and (c) the step response for each set of
parameter values given in the table below.
L
C
R
a
2H
0.025 F
18 Ω
b
2H
0.025 F
8Ω
c
1H
0.391 F
4Ω
d
2H
0.125 F
8Ω
Solution:
1
Cs
1
V (s)
LC
=
=
H (s) = o
R
1
Vi ( s ) Ls + R + 1
2
s + s+
Cs
L
LC
L, H
C, F
R, Ω
H(s)
2
0.025
18
20
20
=
s + 9 s + 20 ( s + 4 )( s + 5 )
2
2
0.025
8
20
20
=
s + 4 s + 20 ( s + 2 )2 + 42
1
0.391
4
2.56
2.56
=
s + 4 s + 2.56 ( s + 0.8 )( s + 3.2 )
2
0.125
8
4
4
=
s + 4s + 4 ( s + 2 )2
2
2
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a) H ( s ) =
20
( s + 4 )( s + 5)
20
20
−
⇒ h ( t ) = ( 20e −4t − 20e −5t ) u (t )
s + 4 s +5
20
1 −5
4
H ( s)
L {step response} =
=
= +
+
⇒
s
s ( s + 4) ( s + 5) s s + 4 s +5
L {h(t )} = H ( s ) =
step response = (1+ 4e −5t −5e −4t ) u (t )
b) H ( s ) =
20
( s + 2) + 4
2
2
L {h( t )} = H (s) =
5(4)
⇒ h ( t ) = 5e −2t sin 4t u (t )
2
2
( s + 2) + 4
H ( s)
20
1
K s + K2
=
= + 21
L {step response} =
2
s
s ( s + 4s + 20) s s + 4s + 20
20 = s 2 + 4s + 20 + s ( K1s + K 2 ) = s 2 (1+ K1 ) + s ( 4+ K 2 ) + 20
⇒ K1 = −1, K 2 = − 4
1
− ( 4)
−( s + 2 )
1
2
+
L {step response} = +
s ( s + 2 )2 + 42 ( s + 2 ) + 42
1
⎛
⎛
⎞⎞
step response = ⎜1− e −2t ⎜ cos 4t + sin 4t ⎟ ⎟ u (t )
2
⎝
⎠⎠
⎝
c) H ( s ) =
2.56
( s + 0.8)( s + 3.2 )
L {h( t )} = H ( s ) =
1.07
1.07
−
⇒ h ( t ) = 1.07 ( e −.8t − e −3.2t ) u(t)
s + .8 s + 3.2
1
−4
2.56
1
H (s)
=
= + 3 + 3
L {step response} =
s
s ( s + .8) ( s + 3.2) s s + .8 s + 3.2
4
⎛ 1
⎞
step response = ⎜1+ e −3.2t − e −.8t ⎟ u (t )
3
⎝ 3
⎠
d) H ( s ) =
h( t ) = 4te −2t u (t )
4
( s + 2)
2
step response = (1−(1+ 2t )e −2t ) u (t )
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14.8‐15 A circuit is described by the transfer function
VO
9s + 18
= H (S ) = 3
V1
3s + 18s 2 + 39s
Find the step response and impulse response of the circuit.
Solution:
For an impulse response, take V1 ( s ) = 1 . Then
V0 ( s ) =
3( s + 2 )
A
B
B*
= +
+
s ( s +3− j 2 ) ( s +3+ j 2 ) s s +3− j 2 s + 3+ j 2
where
A = sV0 ( s )
s =0
= 0.462, B = (s + 3 − j 2) V0 ( s ) s =−3+ j 2 = 0.47∠ − 119.7° and B* = 0.47 ∠119.7°
Then
The impulse response is
V0 ( s ) =
0.462 0.47 ∠−119.7° 0.47 ∠119.7°
+
+
s
s +3− j 2
s +3+ j 2
v0 (t ) = ⎡⎣0.462 + 2(0.47)e −3t cos ( 2 t − 119.7o ) ⎤⎦ u ( t ) V
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P14.8‐16 The input to the circuit shown in Figure P14.8‐16 is the voltage of the voltage source, vi(t),
and the output is the voltage, vo(t), across the 15‐kΩ resistor.
(a)
Determine the steady-state response, vo(t), of this circuit when the input is vi(t) = 1.5 V.
(b)
Determine the steady-state response, vo(t), of this circuit when the input is vi(t) = 4 cos (100t +
30°) V.
(c)
Determine the step response, vo(t), of this circuit.
Figure P14.8‐16
Solution:
a.
A capacitor in a circuit that is at steady state
and has only constant inputs acts like an open
circuit. Then
vo ( t ) = −
10
(1.5 ) = −3.75 V
4
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
Vo (ω )
Vo (ω )
4∠30°
+
+
=0
4 ×103 − j 10 × 103 10 × 103
or
10∠30° + (1 + j ) Vo (ω ) = 0
Vo (ω ) = −
10∠30°
= 7.07∠165°
1+ j
Finally,
vo(t) = 7.07 cos(100t +165°) V.
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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
1
s + Vo ( s ) + Vo ( s ) = 0
4 ×103 1 ×106 10 × 103
s
3
10
+ ( s + 100 ) Vo ( s ) = 0
4s
250
−2.5
2.5
Vo ( s ) =
=
+
s ( s + 100 )
s
s + 100
Finally,
vo ( t ) = 2.5 ( e −100 t − 1) u ( t ) V
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P14.8-17 The input to the circuit shown in Figure P14.8-17 is the
voltage of the voltage source, vi(t), and the output is the capacitor
voltage, vo(t). Determine the step response of this circuit.
Figure P14.8‐17
Solution
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,
1
× ( R2 + L s )
R2 + L s
1
Cs
|| ( R 2 + L s ) =
=
2
1
Cs
+ ( R2 + L s ) C L s + C R2 s +1
Cs
Next, using voltage division,
V (s)
=
H (s) = o
Vi ( s )
R2 + L s
C L s2 + C R2 s + 1
R2 + L s
=
R2 + L s
R 2 + L s + R1 ( C L s 2 + C R 2 s + 1)
+
R
1
C L s2 + C R2 s + 1
R2
s
+
R1 C R1 L C
2s + 4
=
= 2
L + R1 R 2C
R + R 2 s + 4 s + 29
s2 +
s+ 1
R1 L C
R1 L C
1
gives
s
H (s)
2s +4
0.1379 −0.1379 s + 1.4483
=
=
+
Vo ( s ) =
2
s
s
s 2 + 4 s + 29
s ( s + 4 s + 29 )
Using Vi ( s ) =
=
0.1379 −0.1379 s + 1.4483
+
2
s
( s + 2 ) + 52
=
s+2
0.1379
5
− 0.1379
+ 0.3449
2
2
2
s
( s + 2) + 5
( s + 2 ) + 52
Taking the inverse Laplace transform
v o ( t ) = 0.1379 + e− 2 t ( −0.1379 cos ( 5 t ) + 0.3448sin ( 5 t ) )
= 0.1379 + 0.3713 e− 2 t cos ( 5 t − 111.8° ) V
(checked using LNAP 10/15/04)
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P14.8-18 The input to the circuit shown in
Figure P14.8-18 is the voltage of the voltage source, vi(t),
and the output is the resistor voltage, vo(t). Specify values
for L1, L2, R, and K that cause the step response of the
circuit to be
vo(t) = (1 + 0.667e–50t – 1.667e–20t)u(t) V
Figure P14.8‐18
P14.8-18
First, we determine the transfer function corresponding to the step response. Taking the Laplace
transform of the given step response
H (s)
1 0.667 1.667 ( s + 50 )( s + 20 ) + 0.667 s ( s + 20 ) − 1.667 s ( s + 50 )
= Vo ( s ) = +
−
=
s
s s + 50 s + 20
s ( s + 50 )( s + 20 )
=
Consequently,
H (s) =
1000
s ( s + 50 )( s + 20 )
Vo ( s )
1000
=
Vi ( s ) ( s + 50 )( s + 20 )
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
Vi ( s ) = L1 s I a ( s ) + K I a ( s ) ⇒ I a ( s ) =
Vi ( s )
K + L1 s
Next, using voltage division,
Vo ( s ) =
R
RK
K I a ( s ) ⇒ Vo ( s ) =
V (s)
L2 s + R
( L2 s + R )( K + L1 s ) i
Then, the transfer function of the circuit is
Vo ( s )
RK
L1 L 2
RK
=
Vi ( s ) ( L 2 s + R )( L1 s + K ) ⎛
R ⎞⎛
K⎞
⎜⎜ s +
⎟⎟ ⎜⎜ s + ⎟⎟
L2 ⎠ ⎝
L1 ⎠
⎝
Comparing the two transfer functions gives
RK
L1 L 2
1000
= H (s) =
⎛
( s + 50 )( s + 20 )
R ⎞⎛
K⎞
⎜⎜ s + ⎟⎟ ⎜⎜ s + ⎟⎟
L2 ⎠ ⎝
L1 ⎠
⎝
H (s) =
=
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We require 1000 =
RK
R
R
K
K
and 20 =
and 50 = . These equations
and either 50 =
or 20 =
L1 L 2
L2
L2
L1
L1
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)
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P14.8-19 The input to the circuit shown in Figure P14.8-19
is the voltage of the voltage source, vi(t), and the output is
the capacitor voltage, vo(t). Determine the step response of
this circuit.
Figure P14.8‐19
Solution:
Represent the circuit in the frequency domain using the
Laplace transform as shown. (Set the initial conditions to
zero to calculate the step response.)
⎛
1 ⎞
R2 × ⎜ L s +
⎟
R 2 ( C L s 2 + 1)
Cs⎠
⎛
1 ⎞
⎝
=
First, R 2 || ⎜ L s +
⎟=
2
Cs⎠
⎛
1 ⎞ C L s + C R2 s +1
⎝
R2 + ⎜ L s +
⎟
Cs⎠
⎝
Next, using voltage division twice,
H (s) =
Vo ( s )
=
Vi ( s )
R 2 ( C L s 2 + 1)
C L s + C R2 s + 1
2
R 2 ( C L s 2 + 1)
C L s + C R2 s + 1
2
+ R1
×
1
Cs
Ls+
1
Cs
=
R2
(R + R )C L s + R R C s + R + R
2
1
2
1
2
1
2
R2
(R + R ) LC
=
1
s2 +
R1 R 2
(R + R ) L
1
Using Vi ( s ) =
2
s+
2
1
LC
=
8
s + 10 s + 16
2
1
gives
s
2
1
1
−
H (s)
8
8
Vo ( s ) =
=
=
=2+ 3 + 6
2
s
s ( s + 10 s + 16 ) s ( s + 2 )( s + 8 ) s s + 2 s + 8
Taking the inverse Laplace transform
1
⎛1 2
⎞
v o ( t ) = ⎜ − e− 2 t + e− 8t ⎟ u ( t ) V
6
⎝2 3
⎠
(checked using LNAP 10/15/04)
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P 14.8-20
The input to the circuit shown in
Figure P 14.8-20 is the voltage of the voltage source,
vi(t), and the output is the inductor current, io(t).
Specify values for L, C, and K that cause the step
response of the circuit to be
vo(t) = (3.2 – (3.2e–5t + 16te–5t))u(t) V
Figure P 14.8‐20
Solution:
First, we determine the transfer function corresponding to the step response. Taking the Laplace
transform of the given step response
2
H (s)
3.2 ⎛ 3.2
16 ⎞ 3.2 ( s + 5 ) − 3.2 s ( s + 5 ) + 16 s
80
= Io ( s ) =
−⎜
+
=
⎟=
2
2
2
s
s ⎝⎜ s + 5 ( s + 5 ) ⎠⎟
s ( s + 5)
s ( s + 5)
Consequently,
H (s) =
Io ( s )
80
=
Vi ( s ) ( s + 5 )2
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.)
1
R1
1
Cs
=
=
R1 ||
Cs R + 1
1 + R1 C s
1
Cs
R1 ×
First
Next, using voltage division,
R1
Va ( s ) =
1 + R1 C s
R1
Vi ( s ) =
Vi ( s )
R1
R1 + R 2 + R1 R 2 C s
+ R2
1 + R1 C s
K
R2 C L
K R1
K Va ( s )
⇒ Io ( s ) =
V (s) =
Vi ( s )
L s + R3
R3 ⎞ ⎛
R1 + R 2 ⎞
⎛
( L s + R3 )( R1 + R 2 + R1 R 2 C s ) i
⎟
⎜ s + ⎟ ⎜⎜ s +
L ⎠⎝
R1 R 2 C ⎟⎠
⎝
Then, the transfer function of the circuit is
Io ( s ) =
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K
R2 C L
Io ( s )
=
Vi ( s ) ⎛
R3 ⎞ ⎛
R1 + R 2 ⎞
⎟
⎜s+
⎟ ⎜⎜ s +
L ⎠⎝
R1 R 2 C ⎟⎠
⎝
Comparing the two transfer functions gives
K
R2 C L
80
= H (s) =
2
R3 ⎞ ⎛
R1 + R 2 ⎞
⎛
( s + 5)
⎟
⎜s+
⎟ ⎜⎜ s +
L ⎠⎝
R1 R 2 C ⎟⎠
⎝
R1 + R 2
40 + 10
=
⇒ C = 25 mF ,
5=
We require
R1 R 2 C ( 40 ×10 ) C
H (s) =
and
R3
20
⇒ L=4 H
L
L
K
K
80 =
=
⇒ K = 80 V/V .
R 2 C L 10 ( 0.025) 4
5=
=
(checked using LNAP 10/15/04)
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P 14.8-21
The input to a circuit is the voltage vi(t) and the output is the voltage vo(t). The impulse
response of the circuit is
vo(t) = 6.5e–2t cos(2t + 22.6°)u(t) V
Determine the step response of this circuit.
Solution:
First,
6.5cos ( 2 t + 22.6° ) = 6.5 ( cos 22.6° ) cos ( 2 t ) − 6.5 ( sin 22.6° ) sin ( 2 t ) = 6 cos ( 2 t ) − 2.5sin ( 2 t )
Consequently, the impulse response can be written as
v o ( t ) = e − 2 t ( 6 cos ( 2 t ) − 2.5sin ( 2 t ) ) u ( t ) V
The transfer function is
H (s) = 6
s+3
( s + 3) + 2
2
2
− 2.5
2
( s + 3) + 2
2
2
=
6 s + 13
( s + 3) + 2
2
2
=
6s + 13
s + 6s + 13
2
The Laplace transform of the step response is
H (s)
s
s
s+3
6s + 13
1
1
1
3
2
=
= − 2
= −
=
−
+
×
2
2
2
2
s
s ( s + 6s + 13) s s + 6s + 13 s ( s + 3) + 22 s ( s + 3) + 22 2 ( s + 3) + 22
Taking the inverse Laplace transform gives the step response:
(
)
v o ( t ) = 1 + e− 2 t (1.5sin ( 2 t ) − cos ( 2 t ) ) u ( t ) = (1 + 1.803 e− 2 t cos ( 2 t − 123.7° ) ) V
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P 14.8-22
The input to a circuit is the voltage vi(t) and the output is the voltage vo(t). The step response
of the circuit is
vo(t) = [1 – e–t(1 + 3t)]u(t) V
Determine the impulse response of this circuit.
Solution:
Taking the Laplace transform of the step response,
H (s) 1 ⎡ 3
s+6
1 ⎤ 1
9
= −⎢
+
=
⎥= −
2
2
2
s
s ⎢⎣ ( s + 3) s + 3 ⎥⎦ s ( s + 3)
s ( s + 3)
The transfer function is
H (s) =
9
( s + 3)
2
Taking the inverse Laplace transform gives the impulse response:
v o ( t ) = 9 t e − 3t u ( t ) V
(checked using LNAP 10/15/04)
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P 14.8-23
The input to the circuit shown in Figure
P14.8-23 is the voltage of the voltage source, vi(t),
and the output is the voltage, vo(t). Determine the
step response of the circuit.
Figure P14.8‐23
Solution:
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,
V (s)
Ia ( s ) = i
L s + R1
The equivalent impedance of the parallel capacitor and
inductor is
1
R2 ×
R2
1
Cs
=
=
R 2 ||
Cs R + 1
1+ R2 C s
2
Cs
Next, using voltage division,
K ( R3 + R 2 R3 C s )
R3
R3 + R 2 R3 C s
Vo ( s ) =
K Ia ( s ) =
K Ia ( s ) =
Vi ( s )
R2
R 2 + R3 + R 2 R3 C s
L s + R1 )( R 2 + R 3 + R 2 R 3 C s )
(
+ R3
1+ R2 C s
Then, the transfer function of the circuit is
H (s) =
Using Vi ( s ) =
1 ⎞
K⎛
⎜⎜ s +
⎟
L⎝
R 2 C ⎟⎠
5 ( s + 0.5 )
Vo ( s )
=
=
Vi ( s ) ⎛
R1 ⎞ ⎛
R 2 + R 3 ⎞ ( s + 5 )( s + 2.5 )
⎟
⎜ s + ⎟ ⎜⎜ s +
L ⎠⎝
R 2 R 3 C ⎟⎠
⎝
1
gives
s
5 ( s + 0.5 )
H (s)
0.2 −1.8
1.6
=
=
+
+
s
s ( s + 5 )( s + 2.5 )
s s + 5 s + 2.5
Taking the inverse Laplace transform
Vo ( s ) =
v o ( t ) = ( 0.2 − 1.8 e− 5t + 1.6 e − 2.5t ) u ( t ) V
(checked using LNAP 10/15/04)
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P14.8-24
The transfer function of a circuit is H ( s ) =
12
. Determine the step response of this circuit.
s + 8s + 16
2
P14.8-24
The Laplace transform of the step response is:
3
H ( s)
−3
k
12
12
=
=
= 4+
+
2
2
2
s
s ( s + 4)
s+4
s ( s + 8s + 16) s ( s + 4)
The constant k is evaluated by multiplying both sides of the last equation by s ( s + 4) .
2
⎛3
⎞
3
3
2
12 = ( s + 4) − 3s + ks ( s + 4) = ⎜⎜ + k ⎟⎟⎟ s 2 + (3 + 4k ) s + 12 ⇒ k = −
⎜⎝ 4
⎠
4
4
The step response is
⎡ H ( s ) ⎤ ⎛ 3 −4 t ⎛
⎞⎞
⎥ = ⎜⎜ − e ⎜⎜3 t + 3 ⎟⎟⎟⎟ u (t ) V
L −1 ⎢
⎢ s ⎥ ⎝⎜ 4
⎝⎜
4 ⎠⎟⎠⎟
⎣
⎦
P14.8-25
The transfer function of a circuit is H ( s ) =
80 s
s + 8 s + 25
2
. Determine the step response of this circuit.
P14.8-25
The step response is given by
⎡
⎤
⎡
⎤
⎡ H (s) ⎤
⎤
80 s
80
3
−1
−1 ⎡
−1 80
⎢
⎥
=
=
=
L
L
L
v o ( t ) = L−1 ⎢
×
⎢
⎥
⎥
⎢ s 2 + 8 s + 25 ⎥
2
2
2
⎣
⎦
⎣ s ⎦
⎣⎢ 3 ( s + 4 ) + 3 ⎦⎥
⎣⎢ s s + 8 s + 25 ⎦⎥
(
)
=
80 − 4 t
e sin ( 3 t ) u ( t ) V
3
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P14.8‐26 The input to the circuit shown in Figure P14.8‐26
is the current i ( t ) and the output is the current i o ( t ) .
Determine the impulse response of this circuit.
Figure P14.8‐26
Solution: First, determine the transfer function from the circuit. To do so, represent the circuit in the s‐
domain as shown below.
Applying current division
The transfer function is
I o (s) =
450
9000 s
I (s) = 2
I (s)
6
s 10
s + 9000 s + 20, 000, 000
450 + +
s
20
H (s) =
I o (s)
I (s)
=
9000 s
s + 9000 s + 20, 000, 000
2
Performing partial fraction expansion:
9000 ( −4000 ) 9000 ( −5000 )
9000 s
9000 s
−1000
1000
=
=
+
H (s) = 2
s + 9000 s + 20,000,000 ( s + 4000 )( s + 5000 )
s + 4000
s + 5000
=
−36000
45000
+
s + 4000 s + 5000
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Figure 14.8‐27
P14.8‐27 The input to the circuit shown in Figure P14.8‐29 is the voltage v i ( t ) and the output is the
voltage v o ( t ) . Determine the impulse response of this circuit.
(
)
Answer: h ( t ) = 10323 e −10,000 t − e−320,000 t u ( t ) V
Solution: First, determine the transfer function from the circuit. To do so, represent the circuit in the s‐
domain:
Recognizing the voltage follower, we use voltage division twice:
Va ( s ) =
10 9
2.5 s
4 ×10 4 +
9
10
2.5 s
Vi ( s ) =
The transfer function is
10, 000
8000
320, 000
Va ( s ) =
Va ( s )
V i ( s ) and V o ( s ) =
s + 10, 000
8000 + 0.025 s
s + 320, 000
H (s) =
Vo ( s )
Vi ( s )
=
−10323
3, 200, 000, 000
10323
=
+
( s + 10, 000 )( s + 320, 000 ) s + 10, 000 s + 320, 000
The impulse response is
10323 ⎤
⎡ −10323
+
= 10323 e−10,000 t − e−320,000 t u ( t ) V .
h ( t ) = L −1 ⎡⎣ H ( s ) ⎤⎦ = L −1 ⎢
⎥
⎣ s + 10, 000 s + 320, 000 ⎦
(
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P14.8‐28 The input to the circuit shown in Figure P14.8‐28 is the voltage v i ( t ) and the output is the
voltage v o ( t ) . Determine the impulse response of this circuit.
(
)
Answer: h ( t ) = 10323 e −320,000 t − 322.6e−10,000 t u ( t ) V
Figure 14.8‐28
Solution: First, determine the transfer function from the circuit. To do so, represent the circuit in the s‐
domain:
Recognizing the voltage follower, we use voltage division twice:
0.025 s
320, 000
s
4 × 10 4
Va ( s ) =
Va ( s )
Va ( s ) =
Vi ( s ) =
V i ( s ) and V o ( s ) =
9
10
s + 10, 000
s + 320, 000
8000 + 0.025 s
4
4 ×10 +
2.5 s
The transfer function is
H (s) =
Vo ( s )
Vi ( s )
=
320, 000 s
10323
−322.6
=
+
( s + 10, 000 )( s + 320, 000 ) s + 10, 000 s + 320, 000
The impulse response is
10323 ⎤
⎡ −322.6
h ( t ) = L −1 ⎡⎣ H ( s ) ⎤⎦ = L −1 ⎢
+
= 10323 e −320,000 t − 322.6e −10,000 t u ( t ) V .
⎥
⎣ s + 10, 000 s + 320, 000 ⎦
(
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P14.8‐29 The input to the circuit shown in Figure P14.8‐29 is the voltage v i ( t ) and the output is the
voltage v o ( t ) . Determine the impulse response of this circuit.
(
)
Answer: h ( t ) = δ ( t ) + 322.6e −10,000 t − 330323 e−320,000 t u ( t ) V
Figure 14.8‐29
Solution: First, determine the transfer function from the circuit. To do so, represent the circuit in the s‐
domain:
Recognizing the voltage follower, we use voltage division twice:
Va ( s ) =
0.025 s
s
s
4 × 10 4
Va ( s ) =
Va ( s )
Vi ( s ) =
V i ( s ) and V o ( s ) =
9
10
s + 320, 000
8000 + 0.025 s
s + 10, 000
4
4 ×10 +
2.5 s
The transfer function is
H (s) =
Vo ( s )
Vi ( s )
=
s2
( s + 10, 000 )( s + 320, 000 )
= 1−
−330323
330, 000 s + 3, 200, 000, 000
322.6
= 1+
+
s + 10, 000 s + 320, 000
( s + 10, 000 )( s + 320, 000 )
The impulse response is
−330323 ⎤
322.6
⎡
h ( t ) = L −1 ⎡⎣ H ( s ) ⎤⎦ = L −1 ⎢1 +
+
= δ ( t ) + 322.6e −10,000 t − 330323 e −320,000 t u ( t ) V .
⎥
⎣ s + 10, 000 s + 320, 000 ⎦
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Section 14-9: Convolution Theorem
P14.9-1 The input to the circuit shown in Figure P14.6-1a is the voltage vi(t) shown in Figure
P14.9-1. Plot the output, vo(t), of the circuit.
(a)
(b)
Figure P14.9-1
Solution:
To solve this problem using convolution, we first represent the input is by the function
0
t≤2
⎧
⎪ 10 t − 20
2≤t ≤3
⎪⎪
v i ( t ) = ⎨−7.5 t + 32.5 3 ≤ t ≤ 5
⎪ 5 t − 30
5≤t ≤6
⎪
0
6≤t
⎪⎩
Next, we obtain the impulse response. To do so, assume that the initial conditions are zero and
represent the circuit in the s-domain as
Using voltage division and equivalent impedance, the transfer function is
30
30
|| 5
Vo ( s )
5
1.25 1.25
s
H (s) =
=
= s+6 = 2
=
−
Vi ( s ) 6 s + 30 || 5 6 s + 30
s + 6 s + 5 s +1 s + 5
s
s+6
The impulse response is
1
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⎡1.25 1.25 ⎤
h ( t ) = £ −1 ⎡⎣ H ( s ) ⎤⎦ = £ −1 ⎢
−
= (1.25 e − t − 1.25 e −5t ) u ( t )
⎥
⎣ s +1 s + 5⎦
Edit the MATLAB script from Example 14.9.1 to obtain
% P14_9_1.m - plots the output for Problem 14.9-1
% -------------------------------------------------% Obtain a list of equally spaced instants of time
% -------------------------------------------------t0 = 0; % begin
tf = 12; % end
N = 5000; % number of points plotted
dt = (tf-t0)/N; % increment
t = t0:dt:tf; % time in seconds
% -------------------------------------------------% Obtain the input x(t) and the impulse response
h(t)
% -------------------------------------------------for k = 1 : length(t)
if t(k) < 2
x(k) = 0;
elseif t(k) < 3
x(k) = -20 + 10*t(k); %
elseif t(k) < 5
x(k) = +32.5 - 7.5*t(k); %
elseif t(k) < 6
x(k) = -30 + 5*t(k); %
else
x(k) = 0;
end
end
x=x*dt;
h=1.25*exp(-t)-1.25*exp(-5*t);
% -------------------------------------------------%
Perform the convolution
% -------------------------------------------------y=conv(x,h);
% -------------------------------------------------%
Plot the output y(t)
% -------------------------------------------------plot(t,y(1:length(t)))
axis([t0, tf, -5, 10])
xlabel('t')
ylabel('y(t)')
2
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Running this script produces the required plot of the output voltage:
3
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P14.9-2 The input to the circuit shown in Figure P14.9-2a is the voltage vi(t) shown in Figure
P14.9-2. (Perhaps vi(t) represents the binary sequence 1101 which, in turn, might represent the
decimal number 15.) Plot the output, vo(t), of the circuit.
(a)
(b)
Figure P14.9-2
Solution:
To solve this problem using convolution, we first represent the input is by the function
⎧5 1 ≤ t ≤ 2 or 3 ≤ t ≤ 5 or 7 ≤ t ≤ 8
vi (t ) = ⎨
otherwise
⎩0
Next, we obtain the impulse response. To do so, assume that the initial conditions are zero and
represent the circuit in the s-domain as
Using voltage division, the transfer function is
50
Vo ( s )
2
s
H (s) =
=
=
50
Vi ( s )
+ 25 s + 2
s
The impulse response is
⎡ 2 ⎤
h ( t ) = £ −1 ⎡⎣ H ( s ) ⎤⎦ = £ −1 ⎢
= 2 e −2t u ( t )
⎥
⎣s + 2⎦
h ( t ) = 2 e −2 t
4
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Edit the MATLAB script from Example 14.9.1 to obtain
% P14_9_2.m - plots the output for Problem 14.9-2
% -------------------------------------------------% Obtain a list of equally spaced instants of time
% -------------------------------------------------t0 = 0; % begin
tf = 12; % end
N = 5000; % number of points plotted
dt = (tf-t0)/N; % increment
t = t0:dt:tf; % time in seconds
% -------------------------------------------------% Obtain the input x(t) and the impulse response
h(t)
% -------------------------------------------------for k = 1 : length(t)
x(k) = 0;
if (t(k)>1) & (t(k)<2)
x(k) = 5;
end
if (t(k)>3) & (t(k)<4)
x(k) = 5;
end
if (t(k)>7) & (t(k)<8)
x(k) = 5;
end
end
x=x*dt;
h=2*exp(-2*t);
% -------------------------------------------------%
Perform the convolution
% -------------------------------------------------y=conv(x,h);
% -------------------------------------------------%
Plot the output y(t)
% -------------------------------------------------plot(t,y(1:length(t)))
axis([t0, tf, -0.5, 5.5])
xlabel('t, sec')
ylabel('vo(t), V')
Running this script produces the required plot of the output voltage:
5
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6
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Section 14-10: Stability
P 14.10-1
5Ω
The input to the circuit shown in Figure
L
P 14.10-1 is the voltage, vi(t), of the independent
voltage source. The output is the voltage, vo(t), across
vi(t)
+
–
R
the resistor labeled R. The step response of this circuit
is
+
vo(t)
–
Figure P 14.10-1
vo(t) = (3/4)(1 – e
–100t
)u(t) V
(a)
Determine the value of the inductance, L, and the value of the resistance, R.
(b)
Determine the impulse response of this circuit.
(c)
Determine the steady-state response of the circuit when the input is vi(t) = 5 cos 100 t V.
Solution:
a. From the given step response:
H (s)
75
⎡3
⎤
= L ⎢ (1 − e −100 t ) u ( t ) ⎥ =
s
⎣4
⎦ s ( s + 100 )
From the circuit:
R
H (s) =
R + 5 + Ls
⇒
R
H (s)
L
=
R+5⎞
s
⎛
s⎜s +
⎟
L ⎠
⎝
Comparing gives
b. The impulse response is
R
⎫
= 75 ⎪
R = 15 Ω
⎪
L
⎬ ⇒
R+5
L = 0.2 H
= 100 ⎪
⎪⎭
L
⎡ 75 ⎤
h ( t ) = L −1 ⎢
= 75 e −100 t u ( t )
⎥
⎣ s + 100 ⎦
c.
H (ω ) ω =100 =
75
3
=
∠ − 45°
j 100 + 100 4 2
15
⎛ 3
⎞
Vo (ω ) = ⎜
∠45° ⎟ ( 5∠0° ) =
∠ − 45° V
4 2
⎝4 2
⎠
vo ( t ) = 2.652 cos (100 t − 45° ) V
(Checked using LNAP, 12/29/02)
14-1
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P 14.10-2 The input to the circuit shown in Figure P 14.10-2 is the voltage, vi(t), of the
independent voltage source. The output is the voltage, vo(t), across the capacitor. The step
response of this circuit is
vo (t) = [5–5e–2t(1 + 2t)]u(t) V
Determine the steady-state response of this circuit when the input is
vi(t) = 5 cos (2t + 45°) V
Answer: vo (t) = 12.5 cos (2t–45°) V
6Ω
L
vi(t)
+
–
4Ω
+
va(t)
–
+
+
–
vb(t) = k va(t)
C
vo(t)
–
Figure P 14.10-2
Solution:
The transfer function of this circuit is given by
H (s)
−10
5 −5
20
= L ⎣⎡( 5 − 5 e −2 t (1 + 2t ) ) u ( t ) ⎤⎦ = +
+
=
2
2
s
s s + 2 ( s + 2)
( s + 2)
⇒
H (s) =
20
( s + 2)
2
This transfer function is stable so we can determine the network function as
H (ω ) = H ( s ) s = j ω =
20
=
20
( s + 2) s= jω ( 2 + j ω )
2
2
The phasor of the output is
Vo (ω ) =
20
( 2 + j 2)
The steady-state response is
2
( 5∠45° ) =
20
( 2 2∠45°)
2
( 5∠45° ) = 12.5∠ − 45° V
vo ( t ) = 12.5cos ( 2 t − 45° ) V
(Checked using LNAP, 12/29/02)
14-2
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P 14.10-3 The input to a linear circuit is the voltage vi(t) and the response is the voltage vo(t).
The impulse response, h(t), of this circuit is
h(t) = 30te–5tu(t)V
Determine the steady-state response of this circuit when the input is
vi(t) = 10 cos(3t) V
Answer: vo(t) = 8.82 cos (3t – 62°) V
Solution:
The transfer function of the circuit is H ( s ) = L −1⎡⎣30 t e −5t u (t ) ⎤⎦ =
30
( s + 5)
2
. The circuit is stable
so we can determine the network function as
H (ω ) = H ( s ) s = j ω =
30
=
30
( s + 5) s = j ω ( 5 + j ω )
2
2
The phasor of the output is
Vo (ω ) =
30
( 5 + j 3)
The steady-state response is
2
(10∠0° ) =
30
( 5.83∠31° )
2
(10∠0° ) = 8.82∠ − 62° V
vo ( t ) = 8.82 cos ( 3 t − 62° ) V
14-3
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P 14.10-4 The input to a circuit is the voltage vs. The output is the voltage vo. The step response
of the circuit is
vo(t) = (40 + 1.03e–8t – 41e–320t)u(t)
Determine the network function
H (ω ) =
Vo (ω )
Vs (ω )
of the circuit and sketch the asymptotic magnitude Bode plot.
Solution:
H (s)
40 41.03 1.03
102400
= L ⎣⎡( 40 − 41.03 e − 8 t + 1.03 e − 320 t ) u ( t ) ⎦⎤ =
−
+
=
s
s
s + 8 s + 320 s ( s + 8 )( s + 320 )
so
H (s) =
102400
( s + 8)( s + 320 )
The poles of the transfer function are s1 = −8 rad/s and s 2 = −320 rad/s , so circuit is stable.
Consequently,
H (ω ) = H ( s ) s = j ω =
102400
40
=
( j ω + 8 ) ( j ω + 320 ) ⎛1 + j ω ⎞ ⎛1 + j ω ⎞
⎜
⎟⎜
⎟
8 ⎠⎝
320 ⎠
⎝
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
14-4
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P 14.10-5 The input to a circuit is the voltage vs. The output is the voltage vo. The step response
of the circuit is
vo(t) = 60(e–2t – e–6t)u(t)
Determine the network function
H (ω ) =
Vo (ω )
Vs (ω )
of the circuit and sketch the asymptotic magnitude Bode plot.
Solution:
H (s)
60
60
240
= L ⎣⎡60 ( e − 2 t − e − 6 t ) u ( t ) ⎤⎦ =
−
=
s
s + 2 s + 6 ( s + 2 )( s + 6 )
so
H (s) =
240 s
( s + 6 )( s + 2 )
The poles of the transfer function are s 1 = −2 rad/s and s 2 = −6 rad/s , so circuit is stable.
Consequently,
H (ω ) = H ( s ) s = j ω =
240 j ω
20 j ω
=
( j ω + 2 ) ( j ω + 6 ) ⎛1 + j ω ⎞ ⎛1 + j ω ⎞
⎜
⎟⎜
⎟
2 ⎠⎝
6⎠
⎝
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
14-5
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P 14.10-6 The input to a circuit is the voltage vs. The output is the voltage vo. The step response
of the circuit is
vo(t) = (4 + 32e–90t)u(t)
Determine the network function
H (ω ) =
Vo (ω )
Vs (ω )
of the circuit and sketch the asymptotic magnitude Bode plot.
Solution:
H (s)
4
32
36 s + 360 36 ( s + 10 )
= L ⎣⎡( 4 + 32 e − 90 t ) u ( t ) ⎦⎤ = +
=
=
s
s s + 90 s ( s + 90 )
s ( s + 90 )
so
H ( s ) = 36
( s + 10 )
( s + 90 )
The pole of the transfer function s 1 = −90 rad/s , so circuit is stable. Consequently,
ω⎞
⎛
1+ j ⎟
⎜
( j ω + 10 ) = 4 ⎝ 10 ⎠
H (ω ) = H ( s ) s = j ω = 36
( j ω + 90 ) ⎛1 + j ω ⎞
⎜
⎟
90 ⎠
⎝
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
14-6
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P 14.10-7 The input to a circuit is the voltage vs. The output is the voltage vo. The step response
of the circuit is
5
vo ( t ) = ( e −5t − e−20t ) u ( t ) V
3
Determine the steady-state response of the circuit when the input is
vs(t) = 12 cos (30t) V
Solution:
H (s)
25
⎡5
⎤ 1.67 1.67
= L ⎢ ( e − 5 t − e − 20 t ) u ( t ) ⎥ =
−
=
s
⎣3
⎦ s + 5 s + 20 ( s + 5 )( s + 20 )
so
H (s) =
25 s
( s + 5)( s + 20 )
The poles of the transfer function are s 1 = −5 rad/s and s 2 = −20 rad/s , so the circuit is stable.
Consequently the network function of the circuit is,
25 j ω
0.25 j ω
=
( j ω + 5 ) ( j ω + 20 ) ⎛1 + j ω ⎞ ⎛1 + j ω ⎞
⎜
⎟⎜
⎟
5 ⎠⎝
20 ⎠
⎝
Using Vo (ω ) = H (ω ) Vs (ω ) at ω = 30 rad/s gives
H (ω ) = H ( s ) s = j ω =
Vo (ω ) =
0.25 ( j 30 )
j 90
12∠0 ) =
= 8.2∠ − 47° V
(
30 ⎞⎛
30 ⎞
1 + j 6 )(1 + j1.5 )
⎛
(
⎜1 + j ⎟⎜1 + j ⎟
5 ⎠⎝
20 ⎠
⎝
Back in the time domain, the steady state response is
v o ( t ) = 8.2 cos ( 30 t − 47° ) V
(checked using LNAP 10/12/04)
14-7
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P 14.10-8 The input to a circuit is the voltage vs. The output is the voltage vo. The impluse
response of the circuit is
vo(t) = e–5t(10 – 50t)u(t) V
Determine the steady-state response of the circuit when the input is
vs(t) = 12 cos (10t) V
Solution:
H ( s ) = L ⎡⎣e − 5 t (10 − 50 t ) u ( t ) ⎤⎦ =
10 ( s + 5 ) − 50
10
50
10 s
−
=
=
2
2
2
s + 5 ( s + 5)
( s + 5)
( s + 5)
The poles of the transfer function are s 1 = −5 rad/s and s 2 = −5 rad/s , so the circuit is stable.
Consequently the network function of the circuit is,
H (ω ) = H ( s ) s = j ω =
10 j ω
( j ω + 5)
2
=
Using Vo (ω ) = H (ω ) Vs (ω ) at ω = 10 rad/s gives
Vo (ω ) =
0.4 ( j 10 )
10 ⎞
⎛
⎜1 + j ⎟
5⎠
⎝
2
(12∠0 ) =
j 48
(1 + j 2 )
2
0.4 j ω
ω⎞
⎛
⎜1 + j ⎟
5⎠
⎝
2
= 9.6∠ − 37° V
Back in the time domain, the steady state response is
v o ( t ) = 9.6 cos (10 t − 37° ) V
(checked using LNAP 10/12/04)
14-8
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P14.10-9 The input to a circuit is the voltage vs. The output is the voltage vo. The step response
of the circuit is
v o ( t ) = 1 − e − 20 t ( cos ( 4 t ) + 0.5sin ( 4 t ) ) u ( t ) V
(
)
Determine the steady state response of the circuit when the input is
v s ( t ) = 12 cos ( 4 t ) V
Solution:
H (s)
0.5 ( 4 ) ⎤
1 ⎡ ( s + 2)
= L ⎡ 1 − e − 2 t ( cos ( 4 t ) + 0.5sin ( 4 t ) ) u ( t ) ⎤ = − ⎢
+
⎣
⎦ s ⎢ ( s + 2 )2 + 42 ( s + 2 )2 + 42 ⎥⎥
s
⎣
⎦
(
)
1
s+4
= − 2
s s + 4 s + 20
=
s 2 + 4 s + 20 − s ( s + 4 )
s ( s + 4 s + 20 )
2
=
20
s ( s + 4 s + 20 )
2
so
H (s) =
20
s + 4 s + 20
2
−4 ± 16 − 80
= −2 ± j 4 , so the circuit is stable.
2
Consequently the network function of the circuit is,
The poles of the transfer function are s 1,2 =
H (ω ) = H ( s ) s = j ω =
20
20 − ω 2 + 4 j ω
Using Vo (ω ) = H (ω ) Vs (ω ) at ω = 4 rad/s gives
Vo (ω ) =
20
240
= 14.6∠ − 76° V
(12∠0 ) =
20 − 4 + 4 ( j 4 )
4 + j16
2
Back in the time domain, the steady state response is
v o ( t ) = 14.6 cos ( 4 t − 76° ) V
(checked using LNAP 10/12/04)
14-9
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P14.10_10
20
. When the input to this circuit is sinusoidal, the
s +8
output is also sinusoidal. Let ω 1 be the frequency at which the output sinusoid is twice as large
as the input sinusoid and let ω 2 be the frequency at which output sinusoid is delayed by one
tenth period with respect to the input sinusoid. Determine the values of ω 1 and ω 2.
The transfer function of a circuit is H ( s ) =
Solution:
The circuit is stable so H (ω ) = H ( s ) s ← j ω =
20
.
8+ jω
2
⎛ 20 ⎞
and ω1 = ⎜ ⎟ − 8 2 = 6 rad/s .
The gain is 2 at the frequency ω 1 so 2 =
2
2
⎝ 2 ⎠
8 + ω1
20
When the frequency is ω 2, the period is
2π
ω2
. Also a delay t o corresponds to a phase shift −ω 2 t o.
⎛ 2π ⎞
⎛ ω2 ⎞
In this case, t 0 = 0.1⎜
so the phase shift is -0.2 π. Then −0.2 π = − tan −1 ⎜
⎟
⎟ so
⎜ ω2 ⎟
⎝ 8 ⎠
⎝
⎠
ω 2 = 8 tan ( 0.2 π ) = 5.8123 rad/s .
14-10
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P14.10-11
The input to a linear circuit is the voltage, v i . The output is
the voltage, v o . The transfer function of the circuit is
H (s) =
Vo ( s )
Vi ( s )
The poles and zeros of H ( s ) are shown on the pole-zero
diagram in Figure P14.10-11. (There are no zeros.) The dc
gain of the circuit is
H ( 0) = 5
Figure P14.10-11
Determine the step response of the circuit.
Solution:
The transfer function of the circuit is
H (s) =
a
( s + 2 )( s + 5)
where a is a constant to be determined. The circuit is stable because all its poles lie in the right
half of the s-plane. Consequently,
a
10
H (ω ) = H ( s ) s = j ω =
=
ω
( 2 + j ω )( 5 + j ω ) ⎛1 + j ⎞ ⎛1 + j ω ⎞
⎜
⎟⎜
⎟
2 ⎠⎝
5⎠
⎝
a
At dc (ω = 0)
5 = H ( 0) =
a
⇒ a = 50
10
The step response is given by
10
25 ⎤
⎡
⎡
⎤
⎢
H
s
⎡
⎤
( ) = L−1
50
−1 5
3 − 3 ⎥ = ⎛ 5 + 10 e− 5t − 25 e− 2 t ⎞ u ( t ) V
v o ( t ) = L−1 ⎢
⎢
⎥=L ⎢ +
⎥ ⎜
⎥
⎟
3
3
⎠
⎣ s ⎦
⎣ s ( s + 2 )( s + 5 ) ⎦
⎢ s s + 5 s + 2⎥ ⎝
⎣
⎦
14-11
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P14.10-12
The input to a linear circuit is the voltage, v i . The output is
the voltage, v o . The transfer function of the circuit is
H (s) =
Vo ( s )
Vi ( s )
The poles and zeros of H ( s ) are shown on the pole-zero
diagram in Figure 14.10-12. At ω = 5 rad/s the gain of the
circuit is
H ( 5 ) = 10
Figure P14.10-12
Determine the step response of the circuit.
Solution:
The transfer function of the circuit is
H (s) =
a ( s − 0)
as
= 2
( s + 4 − j 3)( s + 4 + j 3) s + 8 s + 25
where a is a constant to be determined. The circuit is stable because all its poles lie in the right
half of the s-plane. Consequently,
H (ω ) = H ( s ) s = j ω =
j aω
( j ω ) + j 8 ω + 25
2
=
(
j aω
25 − ω 2 + j 8 ω
)
At ω = 5 rad/s
10 =
(
j a5
a
=
2
25 − 5 + j 8 ( 5 ) 8
)
⇒ a = 80
The step response is given by
⎡
⎤
⎡
⎤
⎡ H (s) ⎤
⎤
80 s
80
3
−1
−1 ⎡
−1 80
⎢
⎥
v o ( t ) = L−1 ⎢
=
=
=
L
L
L
×
⎢
⎥
⎥
⎢ 2
⎥
2
2
⎢⎣ s s 2 + 8 s + 25 ⎦⎥
⎢⎣ 3 ( s + 4 ) + 3 ⎥⎦
⎣ s + 8 s + 25 ⎦
⎣ s ⎦
(
)
=
80 − 4 t
e sin ( 3 t ) u ( t ) V
3
14-12
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P14.10-13
The input to a linear circuit is the voltage, v i . The output is
the voltage, v o . The transfer function of the circuit is
H (s) =
Vo ( s )
Vi ( s )
The poles and zeros of H ( s ) are shown on the pole-zero
diagram in Figure P14.10-13. (There is a double pole at s =
− 4.) The dc gain of the circuit is
H ( 0) = 5
Figure P14.10-13
Determine the step response of the circuit.
Solution:
The transfer function of the circuit is
H (s) =
a ( s + 2)
( s + 4)
2
where a is a constant to be determined. The circuit is stable because all its poles lie in the right
half of the s-plane. Consequently,
H (ω ) = H ( s ) s = j ω =
a ( j ω + 2)
( j ω + 4)
2
=
a⎛
ω⎞
⎜1 + j ⎟
8⎝
2⎠
ω⎞
⎛
⎜1 + j ⎟
4⎠
⎝
2
At dc (ω = 0)
5 = H ( 0) =
a
8
⇒ a = 40
The step response is given by
⎡
⎤
⎡
⎡ H (s) ⎤
5
20 ⎤
−1 40 ( s + 2 )
−1 5
− 4t
v o ( t ) = L−1 ⎢
u (t ) V
=
L
=
L
−
+
⎢
⎥
⎢
⎥
2
2 ⎥ = 5 + ( 20 t − 5 ) e
s
s
s
+
4
⎢⎣ s ( s + 4 ) ⎥⎦
⎢⎣
( s + 4 ) ⎥⎦
⎣
⎦
(
)
14-13
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P14.10-14
The input to a circuit is the voltage, v i . The step response of the circuit is
v o = 5 e − 4 t sin(2 t ) u ( t ) V
Sketch the pole-zero diagram for this circuit.
Solution:
H (s)
2
10
= L ⎡⎣5 e − 4 t sin(2 t ) u ( t ) ⎤⎦ = 5 × 2
=
2
s
s + 2 s ← s + 4 ( s + 4 )2 + 22
=
10
10
=
s + 8s + 20 ( s + 4 − j 2 )( s + 4 + j 2 )
2
Consequently
H (s) =
10 s
( s + 4 − j 2 )( s + 4 + j 2 )
P14.10-15
The input to a circuit is the voltage, v i . The step response of the circuit is
v o = 5 t e− 4t u ( t ) V
Sketch the pole-zero diagram for this circuit.
Solution:
H (s)
1
5
= L ⎣⎡5 t e − 4 t u ( t ) V ⎦⎤ = 5 × 2
=
s
s s←s + 4 ( s + 4 )2
Consequently
H (s) =
5s
( s + 4)
2
14-14
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Figure 14.10-16
P14.10-16 The input to the circuit shown in Figure P14.10-16 is the voltage, v i , of the voltage
source. The output is the voltage, v o , across resistor R 3 . The transfer function of this circuit is
H (s) =
120 s
s + 24 s + 208
2
a. Determine values of circuit parameters A, R, R2, R3, L and C that cause the circuit to have
the specified transfer function.
b. Determine the step response of this circuit.
c. Determine the steady state response of the circuit to the input
v i ( t ) = 3.2 cos (10 t + 30° ) V .
Solution:
The input is v i , the output is
v o ,and the transfer function is
H (s) =
120 s
s + 24 s + 208
2
a. Using voltage division twice we see that the transfer function is:
⎛
⎞
1
L s ||
⎜
⎟⎛
⎞
Cs
⎜
⎟ ⎜ A R3 ⎟V ( s )
Vo ( s ) =
i
⎜
⎛
1 ⎞ ⎟ ⎜⎝ R 2 + R 3 ⎟⎠
⎜⎜ R + ⎜ L s ||
⎟⎟
C s ⎠ ⎟⎠
⎝
⎝
where
14-15
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s
C
1
1
1
s
s
L s ||
s2 +
s
1
Cs
LC
RC
C
L
=
=
L s ||
=
× =
and
s
1
1
s
1
Cs Ls+ 1
⎛
⎞
2
2
1
+
s
s
s +
+
R + ⎜ L s ||
⎟
C
RC
LC
Cs L
LC
Cs⎠ R+
⎝
1
2
s +
LC
L
C
⎛ A R3 ⎞ 1
s
⎜
⎟
V o ( s ) ⎜⎝ R 2 + R 3 ⎟⎠ R C
120 s
H (s) =
=
= 2
Vi ( s ) s2 + 1 s + 1
s + 24 s + 208
RC
LC
Equating coefficients gives:
A R3
1
1
= 208 ,
= 24 and
=5
R 2 + R3
LC
RC
The solution of these three equations is not unique as they involve five unknowns. After some
1
trial and error we see that C =
F = 15.625 mF is a convenient value of capacitance. The we
64
64
64
H = 307.7 mH and R =
Ω = 2.67 Ω . Suppose we choose R 2 = R 3 = 5 Ω
calculate L =
208
24
A
then = 5 is required so we chose A = 10 V/V.
2
b. The step response is
⎡ H (s) ⎤
step response = L−1 ⎢
⎥
⎣ s ⎦
Consider
15 ( 8 )
H (s)
120 s
120
120
⎡ 8 ⎤
=
= 2
=
=
= 15 ⎢ 2
2
2
2
2
2
s
s + 24 s + 208 ( s + 12 ) + 64 ( s + 12 ) + 8
s s + 24 s + 208
⎣ s + 8 ⎥⎦ s ← s +12
(
)
Here the notation ⎡⎣ F ( s ) ⎤⎦ s ←s + a indicates that we replace each s in F(s) by s + a. The result is
written as F(s+a) so
F ( s + a ) = ⎡⎣ F ( s ) ⎤⎦ s ←s + a
The Laplace transform pair
sin (ω t ) ↔
ω
s + ω2
2
and the Laplace transform property
14-16
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e−a t f ( t ) ↔ F ( s + a )
are now used to determine the step response
⎡ H (s) ⎤
−12 t
sin ( 8 t ) V for t > 0
step response = L−1 ⎢
⎥ = 15 e
s
⎣
⎦
Since we expect the step response to be 0 for t < 0, we can write
⎡ H (s) ⎤
−12 t
sin ( 8 t ) u ( t ) V
step response = L−1 ⎢
⎥ = 15 e
⎣ s ⎦
c. The poles of this circuit are the roots of s 2 + 24 s + 208 :
s 1,2 =
−24 ± 242 − 4 (1)( 208 )
2
= −12 ± j 8
The real part of all the poles are negative so the circuit is stable. Consequently
120 ( j ω )
j 120 ω
H (ω ) = H ( s ) s = j ω =
=
2
( j ω ) + 24 ( j ω ) + 208 208 − ω 2 + j 24 ω
When ω = 10 rad/s
H (10 ) =
j 1200
= 4.56∠24°
108 + j 240
When the input is v i ( t ) = 3.2 cos (10 t + 30° ) V , the steady state response is
v o ( t ) = 3.2 ( 4.56 ) cos (10 t + 30° + 24° ) = 14.59 cos (10 t + 54° ) V .
14-17
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Section 14.11 Partial Fraction Expansion Using MATLAB
P14.11-1
11.6 s 2 + 91.83 s + 186.525
Find the inverse Laplace transform of V ( s ) = 3
s + 10.95 s 2 + 35.525 s + 29.25
Solution:
Using MATLAB:
>> num = [11.6 91.83 186.525];
>> den = [1 10.95 35.525 29.25];
>> [r,p]=residue(num,den)
r =
8.2000
-3.6000
7.0000
p =
-5.2000
-4.5000
-1.2500
Consequently
V (s) =
and
8.2
−3.6
7
8.2
−3.6
7
+
+
=
+
+
s − ( −5.2 ) s − ( −4.5 ) s − ( −1.25 ) s + 5.2 s + 4.5 s + 1.25
v ( t ) = 8.2 e −5.2 t − 3.6 e −4.5t + 7 e −1.25t
for t > 0
1
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P14.11-2
Find the inverse Laplace transform of V ( s ) =
8 s 3 + 139 s 2 + 774 s + 1471
s 4 + 12 s 3 + 77 s 2 + 296 s + 464
Solution:
Using MATLAB:
>> num = [8 139 774 1471];
>> den = [1 12 77 296 464];
>> [r,p]=residue(num,den)
r =
3.0000 - 6.0000i
3.0000 + 6.0000i
2.0000
3.0000
p =
-2.0000 + 5.0000i
-2.0000 - 5.0000i
-4.0000
-4.0000
Consequently
V (s) =
3− j6
3+ j6
2
3
+
+
+
s − ( −2 + j 5 ) s − ( −2 − j 5 ) s − ( −4 ) ( s − ( −4 ) )2
Using the Laplace transform pair
ect ⎡⎣ 2 a cos ( d t ) − 2b sin ( d t ) ⎤⎦ ↔
a + jb
a− jb
+
s − (c + j d ) s − (c − j d )
with a = 3, b = −6, c = −2 and d = 5 we have
v ( t ) = e −2 t ( 6 cos ( 5 t ) + 12sin ( 5 t ) ) + e −4 t ( 2 + 3 t ) for t > 0
2
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P14.11-3
Find the inverse Laplace transform of V ( s ) =
s 2 + 6 s + 11
s 2 + 6 s + 11
=
3
s 3 + 12 s 2 + 48 s + 64
( s + 4)
Solution:
Using MATLAB:
>> num = [1 6 11];
>> den = [1 12 48 64];
>> [r,p]=residue(num,den)
r =
1.0000
-2.0000
3.0000
p =
-4.0000
-4.0000
-4.0000
Consequently
V (s) =
and
1
3
−2
+
+
2
s − ( −4 ) ( s − ( −4 ) ) ( s − ( −4 ) )3
v ( t ) = e −4 t (1 − 2 t + 3 t 2 ) for t > 0
3
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P14.11-4
Find the inverse Laplace transform of V ( s ) =
−60
s + 5 s + 48.5
2
Solution:
The denominator does not factor any further in the real numbers. Let’s complete the square in the
denominator
−9.23 ( 6.5 )
−60
−60
−60
=
=
=
V (s) = 2
2
2
s + 5 s + 48.5 ( s + 2.5 ) + 42.25 ( s + 2.5 ) + 6.5 2 ( s + 2.5 )2 + 6.5 2
Now use e − at f ( t ) ↔ F ( s + a ) and sin ωt for t > 0 ↔
transform
v ( t ) = e − 2.5t ℒ-1[
−9.23 ( 6.5 )
( s + 2.5) + 6.5
2
2
ω
to find the inverse Laplace
s + ω2
2
] = −9.23 e − 2.5t sin (6.5 t) for t > 0
Using MATLAB:
>> num = [-60];
>> den = [1 5 48.5];
>> [r,p]=residue(num,den)
r =
0 + 4.6154i
0 - 4.6154i
p =
-2.5000 + 6.5000i
-2.5000 - 6.5000i
Using the Laplace transform pair
ect ⎡⎣ 2 a cos ( d t ) − 2b sin ( d t ) ⎤⎦ ↔
a + jb
a− jb
+
s − (c + j d ) s − (c − j d )
with a = 0, b = 4.6154, c = −2.5 and d = 6.5 we have
v ( t ) = −9.2308 e −2.5t sin ( 6.5 t ) for t > 0
4
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P14.11-5
Find the inverse Laplace transform of V ( s ) =
−30
s − 25
2
Solution:
Using MATLAB:
>> num = [-30];
>> den = [1 -25];
>> [r,p]=residue(num,den)
r =
-3
3
p =
5
-5
Consequently
V (s) =
and
−3
3
−3
3
+
=
+
s − 5 s − ( −5 ) s − 5 s + 5
v ( t ) = −3 e5t + 3 e −5t
for t > 0
5
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P14.11-6 The input to the circuit shown in Figure P14.11-6 is the voltage v i ( t ) and the output is
the voltage v o ( t ) . Determine the output when the input
v i ( t ) = 5cos ( 4000 t ) u ( t ) mV
Figure 14.11-6
Solution:
The transfer function of this circuit is
1
⎛ R2 ⎞
Cs
H (s) =
⎜1 +
⎟
1 ⎜
R1 ⎟⎠
⎝
R+ Ls+
Cs
1
⎛ R2 ⎞
LC
=
⎜1 +
⎟⎟
1 ⎜
R
R
2
1
⎝
⎠
s + s+
L
LC
=
100 ×106
s 2 + 500 s + 25 ×106
The Laplace transform of the input is
Vi ( s ) = L ⎡⎣5cos ( 4000 t ) u ( t ) ⎤⎦ =
5s
s + 40002
2
⎡
100 × 106
5s
⎛
⎞⎤
. We can use MATLAB to
The output is v o ( t ) = L −1 ⎢ 2
6 ⎜ 2
2 ⎟⎥
⎣ s + 500 s + 25 × 10 ⎝ s + 4000 ⎠ ⎦
calculate this inverse transform:
>> den=conv([1 500 25e6],[1 0 16e6]);
>> num=[500e6 0];
>> [r,p]=residue(num,den)
r =
6
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-26.4706 + 6.0370i
-26.4706 - 6.0370i
26.4706 - 5.8824i
26.4706 + 5.8824i
p =
1.0e+003 *
-0.2500 + 4.9937i
-0.2500 - 4.9937i
-0.0000 + 4.0000i
-0.0000 - 4.0000i
Consequently
Vo ( s ) =
−26.4706 + j 6.0370
−26.4706 − j 6.0370 −26.4706 + j 5882.4 −26.4706 − j 5882.4
+
+
+
s − (−250 + j 4993.7) s − (−250 + j 4993.7)
s − ( j 4000)
s − ( j 4000)
Using the Laplace transform pair
ect ⎡⎣ 2 a cos ( d t ) − 2b sin ( d t ) ⎤⎦ ↔
a + jb
a− jb
+
s − (c + j d ) s − (c − j d )
with a = −26.4706, b = 6.0370, c = −250 and d = 4993.7 we have
⎡ −26.4706 + j 6.0370 −26.4706 − j 6.0370 ⎤
−250 t
L −1 ⎢
+
⎡⎣ −52.9 cos ( 4993.7 t ) − 12.073sin ( 4993.7 t ) ⎤⎦
⎥=e
−
−
+
−
−
+
250
4993.7
250
4993.7
s
j
s
j
(
)
(
)
⎣
⎦
= 54.26 e −250 t cos ( 4993.7 t + 167° )
Using the Laplace transform pair
a + jb
a− jb
+
ect ⎡⎣ 2 a cos ( d t ) − 2b sin ( d t ) ⎤⎦ ↔
s − (c + j d ) s − (c − j d )
with a = 26.4706, b = 5.8824, c = 0 and d = 4000 we have
⎡ 26.4706 + j 5882.4 26.4706 − j 5882.4 ⎤ 0
L −1 ⎢
+
⎥ = e ⎡⎣52.9 cos ( 4000 t ) − 11.765sin ( 4000 t ) ⎤⎦
s − ( j 4000 )
s − ( j 4000 ) ⎦
⎣
= 54.2 cos ( 4000 t − 12.5° )
Finally
v o ( t ) = ⎣⎡54.26 e −250 t cos ( 4993.7 t + 167° ) + 54.2 cos ( 4000 t − 12.5° ) ⎦⎤ u ( t ) mV
7
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Section 14.12 How Can We Check…?
P 14.12-1
Computer analysis of the circuit of Figure P14.12-1 indicates that
vC (t) = 6 + 3.3e–2.1t + 2.7e–15.9t V
iL(t) = 2 + 0.96e–2.1t + 0.04e–15.9t A
and
after the switch opens at time t = 0. Verify that this analysis is correct by checking that (a) KVL
is satisfied for the mesh consisting of the voltage source, inductor, and 12-Ω resistor and (b)
KCL is satisfied at node b.
Hint: Use the given expressions for iL(t) and vC(t) to determine expressions for vL(t), iC(t), vR1(t),
iR2(t), and iR3(t).
Figure P14.12-1
Solution:
v L (t ) = 3
iC (t ) =
d
i L ( t ) = −6 e − 2.1t − 2 e −15.9 t
dt
1 d
v C ( t ) = −0.092 e − 2.1t − 0.575 e −15.9 t
75 dt
v R1 ( t ) = 12 − v L ( t ) = 12 + 6 e − 2.1t + 2 e −15.9 t
i R2 (t ) =
12 − ( v L ( t ) + v C ( t ) )
i R3 (t ) =
Thus,
6
vC (t )
6
= 1 + 0.456 e − 2.1t − 0.123 e −15.9 t
= 1 + 0.548 e − 2.1t + 0.452 e −15.9 t
−12 + v L ( t ) + v R1 ( t ) = 0 and i R 2 ( t ) = i C ( t ) +i R 3 ( t )
as required. The analysis is correct.
1
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P 14.12-2
Analysis of the circuit of Figure P 14.12-2 when vC(0) =–12 V indicates that
i1(t) = 18e0.75t A and i2(t) = 20e0.75t A
after t = 0. Verify that this analysis is correct by representing this circuit, including i1(t) and i2(t),
in the frequency domain using Laplace transforms. Use I1(s) and I2(s) to calculate the element
voltages and verify that these voltages satisfy KVL for both meshes.
Figure P 14.12-2
Solution:
18
20
and I 2 ( s ) =
3
3
s−
s−
4
4
⎛
⎞ ⎛
⎞
12 1 ⎜ 18 ⎟ ⎜ 18
20 ⎟
+ ⎜
−
⎟ + 6⎜
⎟ = 0 (ok)
s 2s ⎜ s − 3 ⎟ ⎜ s − 3 s − 3 ⎟
4⎠ ⎝
4
4⎠
⎝
⎛
⎞ ⎛
⎞ ⎛
⎞
⎜ 18
20 ⎟ ⎜ 20 ⎟ ⎜ 18 ⎟
−6 ⎜
−
+3
−4
= 0 (ok)
3
3⎟ ⎜
3⎟ ⎜
3⎟
⎜s−
s− ⎟ ⎜s− ⎟ ⎜s− ⎟
4
4⎠ ⎝
4⎠ ⎝
4⎠
⎝
I1 (s) =
KVL for left mesh:
KVL for right mesh:
The analysis is correct.
2
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P 14.12-3
Figure P 14.12-3 shows a circuit represented in (a) the time domain and (b) the
frequency domain using Laplace transforms. An incorrect analysis of this circuit indicates that
IL ( s) =
−20 ( s + 2 )
s+2
and VC ( s ) =
s +s+5
s ( s 2 + s + 5)
2
(a) Use the initial and final value theorems to identify the error in the analysis. (b) Correct the
error.
Hint: Apparently the error occurred as V C(s) was calculated from IL(s).
Answer: VC ( s ) = −
20 ⎛ s + 2 ⎞ 8
⎜
⎟+
s ⎝ s2 + s + 5 ⎠ s
Figure P14.12-3
Solution:
Initial value of IL (s):
lim
s+2
s 2
= 1 (ok)
s→∞
s +s+5
Final value of IL (s):
lim
s+2
s 2
= 0 (ok)
s→0
s +s+5
Initial value of VC (s):
lim
−20 ( s + 2 )
= 0 (not ok)
s
s→∞
s ( s 2 + s + 5)
3
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Final value of VC (s):
lim
−20 ( s + 2 )
= −8 (not ok)
s
s→0
s ( s 2 + s + 5)
Apparently the error occurred as VC (s) was calculated from IL (s). Indeed, it appears that VC (s)
was calculated as −
VC ( s ) = −
20
20
8
I L ( s ) instead of − I L ( s ) + . After correcting this error
s
s
s
20 ⎛ s + 2 ⎞ 8
⎜
⎟+ .
s ⎝ s2 + s+5⎠ s
Initial value of VC (s):
⎛ −20 ( s + 2 ) 8 ⎞
s⎜
+ ⎟ = 8 (ok)
s → ∞ ⎜ s ( s 2 + s + 5) s ⎟
⎝
⎠
Final value of VC (s):
⎛ −20 ( s + 2 ) 8 ⎞
s⎜
+ ⎟ = 0 (ok)
s → 0 ⎜ s ( s 2 + s + 5) s ⎟
⎝
⎠
lim
lim
4
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