Laplace Transforms Circuit Analysis • Passive element equivalents ...
Transcript of Laplace Transforms Circuit Analysis • Passive element equivalents ...
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Laplace Transforms Circuit Analysis
• Passive element equivalents
• Review of ECE 221 methods in s domain
• Many examples
J. McNames Portland State University ECE 222 Laplace Circuits Ver. 1.63 1
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Example 1: Circuit Analysis
We can use the Laplace transform for circuit analysis if we can definethe circuit behavior in terms of a linear ODE.
For example, solve for v(t). Check your answer using the initial andfinal value theorems and the methods discussed in Chapter 7.
10 u(t) v(t)
-
+
5 mH
i(0-) = -2 mA
5 kΩ
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Example 1:Workspace
Hint: � [r,p,k] = residue([-2e-3 2e3],[1 1e6 0])
r = -0.0040, 0.0020,
p = -1000000, 0
k = []
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Example 1:Workspace
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Laplace Transform Circuit Analysis Overview
• LPT is useful for circuit analysis because it transforms differentialequations into an algebra problem
• Our approach will be similar to the phasor transform
1. Solve for the initial conditions– Current flowing through each inductor
– Voltage across each capacitor
2. Transform all of the circuit elements to the s domain
3. Solve for the s domain voltages and currents of interest
4. Apply the inverse Laplace transform to find time domainexpressions
• How do we know this will work?
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Kirchhoff’s Laws
N∑k=1
vk(t) = 0N∑
k=1
Vk(s) = 0
M∑k=1
ik(t) = 0M∑
k=1
Ik(s) = 0
• Kirchhoff’s laws are the foundation of circuit analysis
– KVL: The sum of voltages around a closed path is zero
– KCL: The sum of currents entering a node is equal to the sumof currents leaving a node
• If Kirchhoff’s laws apply in the s domain, we can use the sametechniques that you learned last term (ECE 221)
• Apply the LPT to both sides of the time domain expression forthese laws
• The laws hold in the s domain
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Defining s Domain Equations: Resistors
R
v(t) -+
i(t) R
V(s) -+
I(s)
v(t) = R i(t) V (s) = R I(s)
• Generalization of Ohm’s Law
• As with KCL & KVL, the relationship is the same in the s domainas in the time domain
• Note that we used the linearity property of the LPT for bothOhm’s law and Kirchhoff’s laws
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Defining s Domain Equations: Inductors
v(t) -+
i(t) L
V(s) -+
I(s) LsL I0
V(s) -+
I(s) Ls
I0s
v(t) = Ldi(t)dt
i(t) =1L
∫ t
0-
v(τ) dτ + I0
V (s) = L [sI(s) − I0] I(s) =1sL
V (s) +1sI0
V (s) = sLI(s) − LI0
Where I0 � i(0-)
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Defining s Domain Equations: Capacitors
v(t) -+
i(t)
V(s) -+
I(s)
V(s) -+
I(s)
CV0C
1sC
V0s 1
sC
i(t) = Cdv(t)dt
v(t) =1C
∫ t
0-
i(τ) dτ + V0
I(s) = C [sV (s) − V0] V (s) =1C
[1sI(s)
]+
1sV0
I(s) = sCV (s) − CV0 V (s) =1
sCI(s) +
V0
s
Where V0 � v(0-)
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s Domain Impedance and Admittance
Impedance: Z(s) =V (s)I(s)
Admittance: Y (s) =I(s)V (s)
• The s domain impedance of a circuit element is defined for zeroinitial conditions
• This is also true for the s domain admittance
• We will see that circuit s domain circuit analysis is easier when wecan assume zero initial conditions
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s Domain Circuit Element Summary
Resistor V (s) = RI(s) V = RI R
Inductor V (s) = sLI(s) V = sLI Ls
Capacitor V (s) = 1sC I(s) V = 1
sC I1
sC
• All of these are in the form V (s) = ZI(s)
• Note similarity to phasor transform
• Identical if s = jω
• Will discuss further later
• Equations only hold for zero initial conditions
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Example 2: Circuit Analysis
10 u(t) v(t)
-
+
5 mH
i(0-) = -2 mA
5 kΩ
Solve for v(t) using s-domain circuit analysis.
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Example 2: Workspace
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Example 3: Circuit Analysis
t = 0
vo
-
+1 kΩ
sin(1000t) 1 μF
Given vo(0) = 0, solve for vo(t) for t ≥ 0.
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Example 3: Workspace
Hint: � [r,p,k] = residue([1e6],conv([1 0 1e6],[1 1e3]))
r = [ 0.5000, -0.2500 - 0.2500i, -0.2500 + 0.2500i]
p = 1.0e+003 *[ -1.0000, 0.0000 + 1.0000i, 0.0000 - 1.0000i]
k = []
� [abs(r) angle(r)*180/pi]
ans = [ 0.5000 0, 0.3536 -135.0000, 0.3536 135.0000]
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Example 3: Workspace
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Example 3: Plot of Results
0 5 10 15 20 25−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1 TotalTransientSteady State
Time (ms)
vo(t
)(V
)
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Example 4: Circuit Analysis
40 V
10 mF
10 mH
t = 0
v
-
+
50 Ω
50 Ω
100 Ω
175 Ω
175 Ω
Solve for v(t).
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Example 4: Workspace
Hint: � [r,p,k] = residue([1e-3 20 0],[1 21.25e3 10e3])
r = [-1.2496, -0.0004]
p = [-21250,-0.4706]
k = [0.0010]
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Example 4: Workspace
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Example 5: Parallel RLC Circuits
C L R v(t)
-
+
i(t)
Find an expression for V (s). Assume zero initial conditions.
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Example 6: Circuit Analysis
8 H v
-
+
iL
20 kΩ0.125 μF
Given v(0) = 0 V and the current through the inductor isiL(0-) = −12.25 mA, solve for v(t).
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Example 6: Workspace
Hint: � [r,p,k] = residue([98e3],[1 400 1e6])
r = [ 0 -50.0104i, 0 +50.0104i]
p = 1.0e+002 * [ -2.0000 + 9.7980i, -2.0000 - 9.7980i]
k = []
� [abs(r) angle(r)*180/pi]
ans =[ 50.0104 -90.0000, 50.0104 90.0000]
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Example 6: Workspace
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Example 6: Plot of v(t)
0 5 10 15 20 25 30 35 40
−40
−20
0
20
40
60
80
Time (ms)
(vol
ts)
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Example 6: MATLAB Code
t = 0:0.01e-3:40e-3;v = 50*exp(-200*t).*sin(979.8*t);t = t*1000;h = plot(t,v,’b’);set(h,’LineWidth’,1.2);xlim([0 max(t)]);ylim([-23 40]);box off;xlabel(’Time (ms)’);ylabel(’(volts)’);title(’’);
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Example 7: Series RLC Circuits
R L
Cv(t)
vR(t) -+ vL(t) -+
vC(t)
-
+
Find an expression for VR(s), VL(s), and VC(s). Assume zero initialconditions.
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Example 7: Workspace
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Example 8: Circuit Analysis
50 nF
10 nF 2.5 nF
80 V
t = 0
v1(t)
-
+
v2(t)
-
+
20 kΩ
There is no energy stored in the circuit at t = 0. Solve for v2(t).
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Example 8: Workspace
Hint: � [r,p,k] = residue([320e3],[1 5e3 0])
r = [-64, 64]
p = [-5000,0]
k = []
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Example 8: Workspace
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Example 9: Circuit Analysis
0.25 v1
20 H
0.1 F600 u(t)
v1 v2
10 Ω
140 Ω
Solve for V2(s). Assume zero initial conditions.
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Example 9: Workspace
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Example 9: Workspace
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Example 10: Circuit Analysis
10 mF
9 i(t)
u(t)
a
b
i(t)
100 Ω
Find the Thevenin equivalent of the circuit above. Assume that thecapacitor is initially uncharged.
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Example 10: Workspace
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Example 10: Workspace
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Example 11: Circuit Analysis
v(t)
L
R vo(t)
-
+
iL
Find an expression for vo(t) given that v(t) = e−αtu(t) andiL(0) = I0 mA.
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Example 11: Workspace
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Example 11: Workspace
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Example 12: Circuit Analysis
100 mH
vs(t) vo(t)
-
+
200 Ω
1 kΩ1 μF
Find an expression for Vo(s) in terms of Vs(s). Assume there is noenergy stored in the circuit initially. What is vo(t) if vs(t) = u(t)?
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Example 12: Workspace
Hint: � [r,p,k] = residue([-0.2 0],conv([1 2e3],[1 1e3]))
r = [-0.4000, 0.2000]
p = [-2000, -1000]
k = []
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Example 12: Workspace
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Example 13: Circuit Analysis
vo(t)
-
+vs(t)
RL
RA
CA RB
CB
Find an expression for Vo(s) in terms of Vs(s). Assume there is noenergy stored in the circuit initially.
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Example 13: Workspace
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Example 13: Workspace
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Example 14: Circuit Analysis
vs(t)vo(t)
-
+
RL
C
R
Find an expression for Vo(s) in terms of Vs(s). Assume there is noenergy stored in the circuit initially.
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Example 14: Workspace
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