Download Applications of the Laplace Transform

EE 422G Notes: Chapter 6
Instructor: Cheung
Applications of the Laplace Transform
Application in Circuit Analysis
1. Review of Resistive Network
1) Elements
2) Superposition
1
Page 6-1
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EE 422G Notes: Chapter 6
Instructor: Cheung
3) KVL and KCL – Select a node for ground. Watch out for signs!
4) Equivalent Circuits
Thevenin
Equivalent
Circuit
Norton
Equivalent
Circuit
Vs = VOC = Open Circuit Voltage
Rs = Equivalent Resistance
I s = I SC = Short Circuit Current
Rs = Same as before
Page 6-2
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EE 422G Notes: Chapter 6
Instructor: Cheung
5) Nodal Analysis and Mesh Analysis
(Use KCL)
Mesh analysis (use KVL)
VS 1 = R1 I1 + R2 ( I1 − I 2 ) + R3 ( I1 − I 2 )

VS 1 = R1 I1 + R4 I 2 + VS 2
Solve for I1 and I2.
2. Characteristics of Dynamic Networks
1) Inductor
d
iL ( t )
dt
1 t
or iL (t ) = ∫ v L (τ )dτ
L −∞
v L (t ) = L
2) Capacitor
d
vC ( t )
dt
1 t
or vC (t ) = ∫ iC (τ )dτ
C −∞
iC (t ) = C
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EE 422G Notes: Chapter 6
Instructor: Cheung
3) Operation Amplifier
Non-Inverting input
+
vi(t)
-
+
Inverting input
+
+
Ii(t)
vi(t)
≡
vo(t)
-
Rin
Avi(t)
-
+
+
-
vo(t)
-
A general op-amp model is described above.
In practice, the input resistance, Rin, is very large (> 1012 Ω) and the gain, A, is
very large (>105). Thus, we will use the ideal model in the analysis:
1. Input current Ii(t) = 0 (due to the large input impedence)
2. Input voltage difference vi(t) = 0 and output voltage vo(t) is dictated by
external circuit (due to the large gain)
Example:
v2(t)
+
-
v1(t) +Ra
io(t)
Rb
+
vo(t)
Based on the ideal op-amp model,
v2(t) = v1(t)
(1)
Also, as the op-amp does not have any input current, applying KCL at the inverting
port, we have
v2(t)/Ra= (vo(t)-v2(t))/Rb
vo(t)/v2(t) = 1+Rb/Ra
Plug in (1), we have
vo(t)/v1(t) = 1+Rb/Ra
This circuit is called Non-Inverting Amplifier.
Page 6-4
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EE 422G Notes: Chapter 6
Instructor: Cheung
4) Mutual Inductor – used in transformer.
Two separate circuits with coupling currents.
Make sure both i1 and i2 point either away or toward the polarity marks to
make the mutual inductance M positive.
To link the two circuits together, introduce a combined current term (i1+i2):
di
di
di
di
v1 (t ) = L1 1 − M 1 + M 1 + M 2
dt
dt
dt
dt
di
d
= ( L1 − M ) 1 + M (i1 + i2 )
dt
dt
di
di
di
di
v 2 (t ) = M 1 + M 2 + L2 2 − M 2
dt
dt
dt
dt
d
di
= M (i1 + i2 ) + ( L2 − M ) 2
dt
dt
Equivalent circuit:
Page 6-5
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EE 422G Notes: Chapter 6
Instructor: Cheung
Example : Apply mesh analysis to the following circuit
Using Laplace Transform
V ( s ) = VL ( s ) + VC ( s ) + VR ( s )
= L( sI ( s ) − i (0− )) +
= ( Ls ) I ( s ) +
1  I ( s ) vC (0) 
+
+ RI ( s )
C  s
s 
1
I ( s ) + RI ( s )
Cs
Define ‘Generalized Resistors’ (Impedances)
Z L ( s ) = Ls ⇒ VL ( s ) = I ( s ) Z L ( s )
1
⇒ VC ( s ) = I ( s ) Z C ( s )
Cs
Both capacitor and inductor behave exactly like a resistor!
ZC ( s) =
⇒ Vs ( s ) = Z L ( s ) I ( s ) + Z C ( s ) I ( s ) + RI ( s )
⇒ I ( s) =
Vs ( s )
Z L ( s) + Z C ( s) + R
Everything we know about resistive network can be applied to dynamic network in
Laplace domain:


superposition

KVL and KCL


Equivalent circuit

Nodal analysis and mesh analysis
Generalized Ohms Law
Page 6-6
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EE 422G Notes: Chapter 6
Instructor: Cheung
3. Laplace transform models of circuit elements.
What if the initial conditions are not zero?
1) Capacitor
Alternatively, you can also represent it as an impedance and a parallel current
source (Norton equivalent circuit)
+
I(s)
ZC
Cv(0-)
-
V(s)
BE VERY CAREFUL ABOUT THE POLARITY OF VOLTAGE SOURCE AND
THE DIRECTION OF CURRENT SOURCE!!
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EE 422G Notes: Chapter 6
Instructor: Cheung
2) Inductor
Alternatively, you can also represent it as an impedance and a parallel current
source (Norton equivalent circuit)
+
I(s)
ZL
i(0-)/s
-
V(s)
3) Resistor V(s) = RI(s)
4) Voltage and Current Sources (Don’t forget to apply Laplace Transform on
them)
5) Op-Amp : same ideal model assumption
Page 6-8
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EE 422G Notes: Chapter 6
Instructor: Cheung
6) Mutual Inductance (Transformers)
⇓
Laplace transform model: Obtain it by using inductance model
Page 6-9
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EE 422G Notes: Chapter 6
Instructor: Cheung
Example: Find Complex Norton Equivalent circuit given vc (0 − ) = 0
Solution
1) Compute the Short-Circuit Current I s ( s) = I sc
Straightforward to see: I S ( s ) = −2 I ( s )
To compute I(s), apply mesh analysis on the left loop:
1
1
3
s+3
= I ( s ) × 1 Ω + 3I ( s ) = (1 + ) I ( s ) =
I ( s)
s
s
s
s
1
⇒ I (s) =
s+3
2
⇒ I s ( s ) = −2 I ( s ) = −
s+3
No need to do inverse Laplace transform as the equivalent circuit is in the sdomain.
Page 6-10
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EE 422G Notes: Chapter 6
Instructor: Cheung
2) Find the equivalent impedance Z s
Normally, we can just kill all the independent sources and combine the
impedances (using resistive combination rules). However, as there is a
dependent source, we need to drive it with a test voltage:
Itest(s)
Vtest(s)
Zs =
Zs
+
-
Vtest(s)
Vtest ( s) Vtest ( s)
=
I test ( s ) 2 I ( s )
Mesh analysis on the left loop:
3I ( s )
=0
s
 3
⇒ (1 Ω) I ( s ) = −  I ( s )
s
⇒ I ( s) = 0
I ( s) × 1 Ω +
So we got an interesting result: Z s =
Vtest ( s )
= ∞ ⇒ OPEN CIRCUIT
0
a
ZL
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EE 422G Notes: Chapter 6
Instructor: Cheung
Example: Find the transfer function H(s) = Vo(s)/Vi(s) of the following circuit.
Assume all initial conditions are zero.
This is called the Sallen-Key circuit, which we will see again in filter design.
Rewrite everything in Laplace domain, we have
We recognize the op-amp configuration as a non-inverting amplifier, so we have
K = 1+
Rb
Ra
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EE 422G Notes: Chapter 6
Instructor: Cheung
To find Vo, we need Vb which depends on Va.
All other nodal voltages are known. Thus, we need two nodal equations:
Applying KCL at node a, we have:
Va − Vi Va − Vb Va − KVb
+
+
=0
1
R1
R2
C1 s
 1

 1

1
1
⇒  +
+ C1 s Va − 
+ KC1 s Vb = V1
R1
 R1 R2

 R2

(1)
Applying KCL at node b, we have:
Vb − Va
+ C 2 sVb = 0
R2
⇒−
 1

1
Va + 
+ KC1 s Vb = 0
R2
 R2

(2)
Combining equations (1) and (2) by eliminating Va, we get:
Vb
1
=
2
Vi R1 R2 C1C 2 s + [R2 C 2 + R1C 2 + R1C1 (1 − K )]s + 1
Since Vo = KVb , we have
Vo
K
=
2
Vi R1 R2 C1C 2 s + [R2 C 2 + R1C 2 + R1C1 (1 − K )]s + 1
where K = 1 +
Rb
Ra
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