Example:
The input to this circuit is the voltage of the voltage
source, v i ( t ) . The output is the capacitor voltage, v o ( t ) .
Determine the transfer function, impulse response and step
response of this circuit.
Solution:
The transfer function is
R2
H (s) =
1
R1 C
R1 + R 2
R 2 C s +1
=
R2
R1 +
s+
R 2 C s +1
R1 R 2 C
Using R1 = 2 Ω, R 2 = 8 Ω and C = 5 F
gives
H (s) =
0.1
s + 0.125
The impulse response is h (t ) = L -1 ⎡⎣ H ( s )⎤⎦ = 0.1e−0.125 t u (t ) V .
The step response is
⎡
⎤
⎡ H ( s )⎤
0.1
⎥ = L -1 ⎢⎡ 0.8 − 0.8 ⎥⎤ = 0.8 1− e−0.125 t u (t ) V
⎥ = L -1 ⎢
L -1 ⎢
⎢
⎥
⎢ s ⎥
⎢⎣ s
s + 0.125 ⎥⎦
⎢⎣ s ( s + 0.125)⎥⎦
⎣
⎦
(
)
Example:
The input to a linear circuit is the voltage, v i ( t ) . The output is the voltage, v o ( t ) . The impulse
response of the circuit is
h (t ) = 12 t e− 4 t u (t ) V
Determine the step response of this circuit.
Solution:
The transfer function is:
1
H ( s ) = L ⎡⎣12 t e −4 t u ( t ) ⎤⎦ =
12
( s + 4)
2
=
12
s + 8s + 16
2
The Laplace transform of the step response is:
3
H (s)
−3
k
12
=
= 4+
+
2
2
s
s ( s + 4)
s+4
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 ⎠⎟⎠⎟
⎣
⎦
Example:
The input to this circuit is the voltage source
voltage, v i ( t ) . The output is the voltage,
v o ( t ) . Specify value of R, C and L that cause
the transfer function of this circuit to be
H (s) =
Vo ( s )
15 × 106
=
Vi ( s ) ( s + 2000 )( s + 5000 )
Solution:
The transfer function can also be calculated form the circuit itself. The circuit can be represented
in the frequency domain as
2
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
1000
5000
Ls
Solving these node equations gives
1 ⎛⎜
R ⎞⎟ 5000
⎟
⎜⎜⎝1 +
1000 C
10000 ⎠⎟ L
H (s) =
⎛
⎞⎛
⎞
⎜⎜ s + 1 ⎟⎟ ⎜⎜ s + 5000 ⎟⎟
⎟⎜
⎜⎝ 1000C ⎠⎝
L ⎠⎟
Comparing these two equations for the transfer function gives
⎛
⎞
⎛
⎞
⎜⎜ s + 1 ⎟⎟ = ( s + 2000) or ⎜⎜ s + 1 ⎟⎟ = ( s + 5000)
⎜⎝ 1000C ⎠⎟
⎝⎜ 1000C ⎠⎟
⎛
⎞
⎛
⎞
⎜⎜ s + 5000 ⎟⎟ = ( s + 2000) or ⎜⎜ s + 5000 ⎟⎟ = ( s + 5000)
⎜⎝
⎝⎜
L ⎠⎟
L ⎠⎟
R ⎞ 5000
1 ⎛
= 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 +
⎟ = ( s + 5000) ⇒
⎝
L ⎠
L=1H
⎛
⎞
⎜1 + R ⎟⎟ 5000 = 15×106 ⇒ R = 5 kΩ
⎜
⎟
1000 0.5×106 ⎜⎝ 10000 ⎠ 1
(
1
)
3
Example:
The input to this circuit is the voltage of the voltage source, v i ( t ) . The output is the capacitor
voltage, v o ( t ) . Determine the step response of this circuit.
Solution:
⎛ 4 ⎞
⎛ 2 ⎞
Va ( s ) = ⎜
⎟ Vi ( s ) = ⎜
⎟ Vi ( s )
⎝ s+2⎠
⎝ 4+ 2s ⎠
⎛ 12
⎜
Vo ( s ) = ⎜ s
12
⎜⎜ 6 +
s
⎝
⎞
⎟
⎛ 2 ⎞
⎛ 2 ⎞
⎛ 2 ⎞ ⎛ 2 ⎞
⎟ Vb ( s ) = ⎜
⎟ Vb ( s ) = ⎜
⎟ 5 Va ( s ) = ⎜
⎟5 ⎜
⎟ Vi ( s )
⎝ s+2⎠
⎝ s+2⎠
⎝ s+2⎠ ⎝ s+2⎠
⎟⎟
⎠
The transfer function is:
H (s) =
Vo ( s )
20
=
Vi ( s ) ( s + 2 )2
The Laplace transform of the step response is:
Vo ( s ) =
20
s ( s + 2)
2
=
5 −5
−10
+
+
s s + 2 ( s + 2 )2
Taking the inverse Laplace transform:
vo ( t ) = ⎡⎣5 − 5 e −2 t (1 + 2t ) ⎤⎦ u ( t ) V
4
Example:
The input to this circuit is the voltage source voltage, v i ( t ) . The output is the voltage, v o ( t ) .
Specify value of k, C and L that cause the step response of this circuit to be
vo ( t ) = ⎡⎣ 2 + 4 e −3t − 6 e −2 t ⎤⎦ u ( t ) V
P 14.9-8
From the circuit:
⎛ 1 ⎞
⎛ 1 ⎞ ⎛ 4 ⎞
⎜ 6C ⎟
⎜ Cs ⎟ ⎜ L ⎟
⎛ 4 ⎞
⎜
⎟
=
H (s) = ⎜
k
k
(
)
(
)
⎜
⎟ ⎜
⎟
⎟
1
1
4
⎝ 4 + Ls ⎠
⎜
⎟
⎜⎜ 6 +
⎟⎟ ⎜ s + ⎟
⎜ s + 6C ⎟
Cs ⎠ ⎝
L⎠
⎝
⎝
⎠
From the given step response:
H (s)
2
4
6
12
= L ⎡⎣( 2 + 4 e −3t − 6 e −2 t ) u ( t ) ⎤⎦ = +
−
=
s
s s + 3 s + 2 s ( s + 3)( s + 2 )
so
12
H (s) =
s ( s + 3)( s + 2 )
Comparing the two representations of the transfer functions let
1
1
=3 ⇒ C =
F,
6C
18
4
= 2 ⇒ L = 2 H and 2 × 3 × k = 12 ⇒ k = 2 V/V .
L
5