elec1111 05 superposition bw
DESCRIPTION
.TRANSCRIPT
1
Electrical Engineering & Telecommunications
Lecturer:Ted Spooner
Superposition
Elec1111Elec1111
Rm 124A EE email: [email protected]
Superposition
If cause and effect are linearly related then:
Total effect of several causes actingsimultaneously is equal to the sum ofsum ofeffect of individual causeseffect of individual causes.
SuperpositionDeflecting Beam
F1 F2
SuperpositionDeflecting Beam
D1F1
F2
SuperpositionDeflecting Beam
D2F2
F1
SuperpositionDeflecting Beam
D1+ D2F1
F2
2
SuperpositionDeflecting Beam
D2F2
D1F1
D1+ D2F1
F2
SuperpositionDeflecting Beam
D3
D4
D3+ D4
F4
F3
F3 F4
Superposition - Linearity
Y
Xx1 x2 (x1+x2)
(Y1+Y2)
Y1
Y2
Y
X
Superposition in circuitsR1
R2 R3
V1
+
+
V2
I3
I31 = current in R3 caused by source V1 alone.
We can analyse this circuit by taking each independentsource separately and finding its effect on the parameter of interest. In this case the current through R3.
I3 = Current flowing in R3 due to both sources together.
I3 = I31 + I32
I32 = current in R3 caused by source V2 alone.
Superposition in circuitsR1
R2 R3
V1
+
+
V2
I3
R1
R2 R3
V1
+ I31
R1
R2 R3
+
I32
V2
I3 = sum of effects of eachindependent voltage sourcetaken individually.
I3 = I31 + I32
Replace independent voltage sources by a short circuit leaving only one voltage source in circuit at a time.
To find I31 and I32 we need to take one source at a time.
Superposition
• Independent Voltage sources replaced by short circuits.
• Independent Current sources replaced by open circuits.
• Dependent sources:CANNOT be replaced.
3
ixvs=3V +_
6Ω
9Ω is=2A
Example Example –– Independent SourcesIndependent Sources
3V+_
6Ω
9Ω
/xi
/ 3 0.26 9xi A= =+
// 6 2 0.86 9xi A= × =+
/ // 0.2 0.8 1x x xi i i A= + = + =
6Ω
9Ω 2A
//xi
Solve using superpositionSolve using superposition ixvs=3V +_
6Ω
9Ωis=2A
ixvs=3V +_
6Ω
9Ωis=2A
Solve using node analysisSolve using node analysis
3bv =
29 6a a bv v v−
= + 9av V=
⇒ 19a
xvi A= =
ab
For node b:
KCL for node a:
ix
10V +_
2Ω 1Ω
3A+_ 2ix
Example with a DEPENDENT SourceExample with a DEPENDENT Source
10V +_
2Ω 1Ω
+_
/xi
/2 xi
/ / /10 2 1 2 0x x xi i i− + + + =
/ 2xi A=⇒
Solve using superpositionSolve using superpositionix
10V+_
2Ω 1Ω
3A
+_ 2ix2Ω 1Ω
3A+_
a//xi
//2 xi
b2
av−
av−
For node b: b av v= −
For node a:1.2av V=
// 0.62
ax
vi A−= = −
ix
10V +_
2Ω 1Ω
3A+_ 2ix
312
=−
+ baa vvv
4
ix
10V +_
2Ω 1Ω
3A+_ 2ix
10V+_
2Ω 1Ω
+_
/xi
/2 xi
2Ω 1Ω
3A+_
a//xi
//2 xi
b
( )/ // 2 0.6 1.4x x xi i i A= + = + − =
1V
+_ 1Ω1V
+_ 1V
1Ω+_
+_ 1Ω1V
4W deliveredto resistor
1W deliveredto resistor
1W deliveredto resistor+≠
Superposition does not work for Superposition does not work for powerpower
Maximum power transferMaximum power transfer
vs
Rt+_ RL
+
_v
i
givenWhat value to getmaximum powerdelivered ?
vs
Rt+_ RL
+
_v
i
2
sL
t L
vp RR R
⎛ ⎞= ⎜ ⎟+⎝ ⎠
p is maximum when: t LR R=2
max 4s
L
vpR
=
Power delivered to RL as it varies w.r.t. Rt
Ref: Dorfp is maximum when: t LR R=
2
max 4t sR ip =
isRt
RL
+
_v
i
5
i
10Ω
10Ω10Ω30V
15V
3A
Final exam 2003: solve using superposition
i = ?
10Ω
10Ω10Ω30V
i1
Ai 11010
10)10||10(10
301 =
+×
⎭⎬⎫
⎩⎨⎧
+=
i2
10Ω
10Ω10Ω
15V
Ai 5.01010
10)10||10(10
152 =
+×
⎭⎬⎫
⎩⎨⎧
+=
i3
10Ω
10Ω10Ω3A
Apply current division:
3
110 3 11 1 1
10 10 10
i A= × =+ +
1 2 3 1 0.5 1 2.5i i i i A= + + = + + =
Superposition Example
20V
10Ω
Rradio=12Ω
5V
6Ω6Ω
+
+
20V
10Ω
Rradio=12Ω
5V
6Ω6Ω
+
+
6
20V
10Ω
Rradio=12Ω
5V
6Ω6Ω
+
+
20V
10Ω
Rradio=12Ω
5V
6Ω6Ω
+
+