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ECE 3144 Lecture 18
Dr. Rose Q. Hu
Electrical and Computer Engineering Department
Mississippi State University
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Linear system
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Linear element: a passive element that has a linear voltage-current
relationship- example: v(t) = Ri(t)
Linear dependent source: a dependent current or voltage source whose output
current or voltage is proportional only to the first power of a specified current
or voltage variable in the circuit
Linear circuit: a circuit composed entirely of independent sources, linear
dependent sources and linear elements.
All circuits we have analyzed are linear circuits
For a linear system, it satisfies both additivity and homogeneity.
x1(t)
x2(t)
x1(t)
Linear system
Linear system
Linear system
y1(t)
additivity
x1(t)+ x2(t)
Linear system
y1(t)+ y2(t)
y2(t)
y1(t)
homogeneity
Ax1(t)
Linear system
Ay1(t)
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Superposition
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The most important consequence of linearity is superposition.
– Assume linear system L yields an output y=L(x) with an input x. The superposition
says L(ax1(t)+bx2(t)) = aL(x1(t)) + bL(x2(t)).
Consider a very simple circuit v(t) = Ri(t)
v1(t)= Ri1(t), v2(t)=Ri2(t). If i(t) = Ai1(t)+Bi2(t)
v(t) = R(Ai1(t) + Bi2(t)) = ARi1(t)+BRi2(t) = Av1(t) + Bv2(t) => superposition holds.
Let us develop the general circuit superposition principle by considering the following
circuit:
The two equations for this mesh network are:
6ki1(t) – 3ki2(t) = v1(t)
-3ki1(t) + 9ki2(t) = -v2(t)
Solving equations for i1(t) yields
v1 (t ) v 2 (t )
i1 (t ) 

5k
15k
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The current i1(t) has a component due to v1(t) and a component due to v2(t). First we are
interested in finding what each source acting alone would contribute to i1(t). And second
if i1(t) is just equal to the sum of what each source contributes alone to i1(t), then
superposition principle holds.
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Superposition
Let us determine the part of i1(t) due to v1(t). This part
of i1(t) is denoted as i’1(t). Based on Ohm’s law
i1' (t ) 
v1 (t )
v (t )
 1
3k  3k // 6k
5k
Let us determine the part of i1(t) due v2(t). This
part of i1(t) is denoted as i’’1(t). Based on Ohm’s
law
v2 (t )
 2v2 (t )
i (t )  

6k  3k // 3k
15k
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Using current divider =>
i1'' (t ) 
 2v2 (t )
3k
 v (t )
(
) 2
15k
3k  3k
15k
Now we know the current values for i1(t) contributed by each current source. Now add
them up i ' (t )  i '' (t )  v1 (t )  v 2 (t ) , which is just equal to i1(t). So i1(t) have been
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1
5k
15k
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superposed into these two parts.
Superposition principle for linear circuits
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What we have demonstrated is true in general for linear circuits. It actually
provides a very good example for how to perform superposition on a
complicated problem so that it can reduce to several easier problems-each
only containing a single independent source.
The general circuit superposition principle states:
– In any linear circuit containing multiple independent sources, the current
or voltage at any point the network may be calculated as the algebraic
sum of individual contributions of each source acting alone, which is
Vo = A1 I 1+A2I2+…+AnIn + B1 V1+B2V2+…+BmVm
– When determining the contribution for a single independent source (the
coefficient Ai or Bi), all other independent voltage sources are replaced by
short circuits and all other independent current sources are replaced by
open circuits.
Thus if there are in total N independent sources we must perform N
experiments with each having only one of the independent sources active and
the others zeroed out.
Be careful: although superposition applies to the networks containing
dependent sources, it is not useful in this case => dependent sources must
remain active in every experiment.
Superposition can replace nodal analysis and loop analysis to solve the circuit
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problems!
Superposition example 1
Find Io in the network shown using superposition
6 V
4 k
3 k
2 mA
9 V
4 k
2 k
Io
Either nodal analysis and mesh analysis may be applied
here. But we want to show you how to use superposition
to solve the same problem. There are 3 independent
sources in the network => we need to solve three circuit
problems:
Problem 1: Assume network with only 9V independent voltage source
The 9-V source is connected directly across the output resistor.
4 k
I o1 
3 k
9V
4 k
2 k
9V
 2.25mA
4k
Io1
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Superposition example 1: cont’d
Problem 2: Assume network with only 6V independent voltage source
The 6-V source is also connected directly across the
output resistor =>
6V
4 k
3 k
Io2 
4 k
2 k
6V
 1.5mA
4k
Io2
Problem 3: Assume network with only 2mA independent current source
A short circuit is connected directly across the output
resistor =>
4 k
I o3  0
3 k
2 mA
I o  I o1  I o 2  I o3  2.25  1.5  0.75mA
4 k
2 k
Io3
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Superposition conclusions
• The example tells superposition sometimes (not always) can
simplify your problem a lot!
• In a network containing multiple independent sources, each
source can be applied independently with the remaining sources
turned off.
• To turn off a voltage source, replace it with a short circuit; to
turn off a current source, replace it with an open circuit.
• With the network containing individual source, all the circuit
laws and techniques we have learned (Ohm's Law,
current/voltage divider, series/parallel combinations) or will
learn soon can be applied.
• The results obtained by applying each source independently are
then added up algebraically to get the solution.
• Superposition applies to the current and voltage in a linear
circuit. However it can not be used to determine power (Why?).
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Practical sources vs. ideal sources
•Up to now we have been working exclusively with ideal voltage source and current sources
•Ideal voltage source: No matter what is the current through the voltage source, the output
voltage does not change.
•Ideal current source: No matter what is the voltage across the current source, the output
current does not change.
•The above two ideal cases result in infinite power (v*i)supplies to the external loads,
which is not possible in reality.
•It is now time to take a step closer to reality by considering practical sources.
•When a real battery is connected to a load (light bulb, radio receiver,...) the voltage across the
battery terminals decreases in proportion to the current flowing from the battery. Thus a more
accurate model can be represented by combining an ideal voltage source in series with a resistor
(called internal resistor or output resistor).
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Based on Ohm’s law, the actual voltage vo on the
load RL is:
vo = vS - ioRS,
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The equation can be expressed by the figure in the
next slide.
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Ideal source
Practical source
Practical sources vs. ideal sources
vo
Ideal source
vo=vS
vo = vS - ioRS,
Practical
source
io= vS/RS
io
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For a practical voltage source vo = vS – ioRS
– If RL = , no current is being drawn by the load. => vo = vS
– If RL = 0, the load terminal is a short circuit =>vo = 0. In practice, such as
experiment would probably result in the destruction of the voltage source!
– It is evident that the load voltage vo and the ideal source voltage vS are
approximately equal if load resistance RL are large compared with internal
resistance RS.
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Practical sources vs. ideal sources
•A practical current source is considered the same way. When a current source is connected
to a load, the current out of the current source decreases in proportional to the voltage
across the current source. Thus a more accurate model can be represented by combining an
ideal current source in parallel with a resistor
vo  RS (iS  io )
vo
vo= iSRS
Ideal source
Practical
source
Ideal source
Practical source
iS
io
–If RL =  , no current is being drawn by the load. => vo = iSRS,, io = 0.
–If RL = 0, the load terminal is a short circuit =>vo = 0, io = iS.
–It is evident that the load current io and the ideal source current iS are approximately
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equal if load resistance RL are small compared with internal resistance RS.
Source transformations
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Recall voltage source formula vo = vS – ioRS and current source formula vo = RS (iS – io). The
voltage source circuit is equivalent to the current source circuit if they have the same i-v
characteristics to the load resistor, i.e., vS = RSiS
vo
vo=vS
Ideal source
vo
vo= iSRS
Ideal source
Practical
source
Practical
source
iS
io= vS/RS
io
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io
After all, one circuit can't see another's structure, count the number of its components, notice
differences in color, smell or weight. All one circuit can "see" about another is its i-v
characteristic.
To another circuit, two circuits are equivalent (even if they are entirely different inside) provided
only that the two circuits have the same relationship between voltage and current at
corresponding load terminals (that is, have the same i-v characteristic).
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Homework for Lecture 18
• Problems 4.6, 4.14, 4.12, 4.16, 4.18
• Due Feb 25
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