Applications of the Laplace Transform - Vis Centercheung/courses/ee422g/lecture6.pdf · EE 422G...
Transcript of Applications of the Laplace Transform - Vis Centercheung/courses/ee422g/lecture6.pdf · EE 422G...
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EE 422G Notes: Chapter 6 Instructor: Cheung
Page 6-1
Applications of the Laplace Transform Application in Circuit Analysis 1. Review of Resistive Network
1) Elements
2) Superposition
1
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EE 422G Notes: Chapter 6 Instructor: Cheung
Page 6-2
3) KVL and KCL – Select a node for ground. Watch out for signs!
4) Equivalent Circuits
Open Circuit Voltages OCV V= =
Rs = Equivalent Resistance
Short Circuit Currents SCI I= =
Thevenin Equivalent Circuit
Norton Equivalent Circuit
Rs = Same as before
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EE 422G Notes: Chapter 6 Instructor: Cheung
Page 6-3
5) Nodal Analysis and Mesh Analysis
Mesh analysis (use KVL)
++=−+−+=
224111
213212111 )()(
SS
S
VIRIRV
IIRIIRIRV
Solve for I1 and I2. 2. Characteristics of Dynamic Networks 1) Inductor
2) Capacitor
∫ ∞−=
=
t
CC
CC
diC
tv
tvdt
dCti
ττ )(1
)(or
)()(
∫ ∞−=
=
t
LL
LL
dvL
ti
tidt
dLtv
ττ )(1
)(or
)()(
(Use KCL)
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EE 422G Notes: Chapter 6 Instructor: Cheung
Page 6-4
3) Operation Amplifier
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)
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.
-
+
≡
+
- +
-
vo(t)
vi(t)
+
-
vi(t)
-
+ - Avi(t)
Rin
I i(t)
Inverting input
Non-Inverting input
- +
+ - v1(t)
Ra Rb
vo(t) io(t)
v2(t) +
-
+
-
vo(t)
Example:
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EE 422G Notes: Chapter 6 Instructor: Cheung
Page 6-5
4) Mutual Inductor – used in transformer. Two separate circuits with coupling currents.
To link the two circuits together, introduce a combined current term (i1+i2):
dt
diMLii
dt
dM
dt
diM
dt
diL
dt
diM
dt
diMtv
iidt
dM
dt
diML
dt
diM
dt
diM
dt
diM
dt
diLtv
2221
222
212
211
1
211111
)()(
)(
)()(
)(
−++=
−++=
++−=
++−=
Equivalent circuit:
Make sure both i1 and i2 point either away or toward the polarity marks to make the mutual inductance M positive.
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EE 422G Notes: Chapter 6 Instructor: Cheung
Page 6-6
Example : Apply mesh analysis to the following circuit
Using Laplace Transform
)()(1
)()(
)()0()(1
))0()((
)()()()(
sRIsICs
sILs
sRIs
v
s
sI
CissIL
sVsVsVsV
C
RCL
++=
+
++−=
++=
−
Define ‘Generalized Resistors’ (Impedances)
)()()()( sZsIsVLssZ LLL =⇒=
)()()(1
)( sZsIsVCs
sZ CCC =⇒=
Both capacitor and inductor behave exactly like a resistor!
RsZsZ
sVsI
sRIsIsZsIsZsV
CL
s
CLs
++=⇒
++=⇒
)()(
)()(
)()()()()()(
Everything we know about resistive network can be applied to dynamic network in Laplace domain:
analysismesh and analysis Nodal
circuit Equivalent
KCL and KVL
ionsuperposit
Law Ohms dGeneralize
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EE 422G Notes: Chapter 6 Instructor: Cheung
Page 6-7
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)
BE VERY CAREFUL ABOUT THE POLARITY OF VOLTAGE SOURCE AND THE DIRECTION OF CURRENT SOURCE!!
ZC
Cv(0-) I(s)
+ -
V(s)
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EE 422G Notes: Chapter 6 Instructor: Cheung
Page 6-8
2) Inductor
Alternatively, you can also represent it as an impedance and a parallel current source (Norton equivalent circuit)
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
ZL
i(0-)/s I(s)
+ -
V(s)
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EE 422G Notes: Chapter 6 Instructor: Cheung
Page 6-9
6) Mutual Inductance (Transformers)
⇓ Laplace transform model: Obtain it by using inductance model
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EE 422G Notes: Chapter 6 Instructor: Cheung
Page 6-10
Example: Find Complex Norton Equivalent circuit given 0)0( =−cv
Solution 1) Compute the Short-Circuit Current scs IsI =)(
Straightforward to see: )(2)( sIsI S −= To compute I(s), apply mesh analysis on the left loop:
3
2)(2)(
3
1)(
)(3
)()3
1(1
)(3 1)(1
+−=−=⇒
+=⇒
+=+=+Ω×=
ssIsI
ssI
sIs
ssI
sssIsI
s
s
No need to do inverse Laplace transform as the equivalent circuit is in the s-domain.
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EE 422G Notes: Chapter 6 Instructor: Cheung
Page 6-11
2) Find the equivalent impedance sZ
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:
)(2
)(
)(
)(
sI
sV
sI
sVZ test
test
tests ==
Mesh analysis on the left loop:
0)(
)(3
)() 1(
0)(3
1)(
=⇒
−=Ω⇒
=+Ω×
sI
sIs
sI
s
sIsI
So we got an interesting result: CIRCUITOPEN 0
)(⇒∞== sV
Z tests
Vtest(s)
a
ZL
V test(s) +
-
Itest(s)
Zs
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EE 422G Notes: Chapter 6 Instructor: Cheung
Page 6-12
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
a
b
R
RK += 1
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EE 422G Notes: Chapter 6 Instructor: Cheung
Page 6-13
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:
01
121
=−+−+−
sC
KVV
R
VV
R
VV babaia
11
12
121
1111V
RVsKC
RVsC
RR ba =
+−
++⇒ (1)
Applying KCL at node b, we have:
022
=+−b
ab sVCR
VV
011
122
=
++−⇒ ba VsKC
RV
R (2)
Combining equations (1) and (2) by eliminating Va, we get:
[ ] 1)1(
1
1121222
2121 +−+++=
sKCRCRCRsCCRRV
V
i
b
Since bo KVV = , we have
[ ] 1)1(1121222
2121 +−+++=
sKCRCRCRsCCRR
K
V
V
i
o
where a
b
R
RK += 1
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