Superposition
The response of a linear circuit to several inputs working together is equal to
the sum of the responses to each input working separately.
Circuit inputs and outputs
•
Inputs are usually the voltages of independent voltage sources and the currents of
independent current sources.
•
Outputs can be any current or voltage.
•
The circuit designer designates the input and output of the circuit.
•
A circuit can have more than on input or output.
Linearity
A linear circuit, without capacitors or inductors, is represented by an algebraic equation of the
form:
v o = a1 v1 + a 2 v 2 + + a n v n
where
•
v o is the output of the circuit (it could be a current instead of a voltage).
•
v1 , v 2 ,…, v n are the inputs to the circuit (any or all of the inputs could be currents
instead of voltages).
•
The coefficients a1 , a 2 ,…, a n are real constants, called gains.
We say that the output of the circuit is a linear combination of the inputs to the circuit. To
determine a1 , set v 2 , v 3 ,…, v n to zero. Then
a1 =
vo
v1
v 2 , v3 …vn = 0
The other gains are determined similarly.
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Example
This circuit has one output, v o , and three inputs, v1 ,
i 2 and v 3 . (As expected, the inputs are voltages of
independent voltage sources and the currents of
independent current sources.)
Express the output as a linear combination of the
inputs.
Solution
Let’s analyze the circuit using node equations. Label the node voltage at the top node of the
current source and identify the supernode corresponding to the horizontal voltage source.
Apply KCL to the supernode to get
v1 − ( v 3 + v o )
40
+ i2 =
vo
10
Multiply both sides of this equation by 40 to eliminate the fractions
v1 − ( v 3 + v o ) + 40 i 2 = 4 v o
⇒ v1 + 40 i 2 − v 3 = 5 v o
Dividing both sides by 5, expresses the output as a linear combination of the inputs:
vo =
v1
5
+ 8i2 −
v3
5
Also
a1 =
vo
v1
=
i2 , v3 = 0
vo
1
V/V , a 2 =
i2
5
= 8 V/A and a 3 =
v1 , v 3 = 0
vo
v3
=−
v1 , i 2 = 0
1
V/V
5
2
Alternate Solution
Consider the circuit itself when i 2 , v 3 = 0 . (A zero current source is equivalent to an open circuit
and a zero voltage source is equivalent to a short circuit.)
Using voltage division
vo =
10
1
v1 = v1
40 + 10
5
In other words
a1 =
vo
v1
=
i2 , v3 = 0
1
V/V
5
Next, consider the circuit itself when v 1 , v 3 = 0 .
The resistors are connected in parallel. Applying Ohm’s law to the equivalent resistance gives
vo =
40 × 10
i2 = 8 i2
40 + 10
In other words
3
a2 =
vo
i2
= 8 V/A
v1 , v 3 = 0
Finally, consider the circuit itself when i 2 , v 3 = 0 .
Using voltage division
vo =
10
1
−v 3 ) = − v3
(
40 + 10
5
In other words
a3 =
vo
v3
=−
v1 , i 2 = 0
1
V/V
5
Now the output can be expressed as a linear combination of the inputs
1
⎛ 1⎞
v o = a1 v1 + a 2 i 2 + a 3 v 3 = v1 + 8 i 2 + ⎜ − ⎟ v 3
5
⎝ 5⎠
as before.
Summary
When the voltages of all independent voltage sources and the currents of all independent current
sources are inputs to a linear circuit, then
The response of the linear circuit to several inputs working together is equal
to the sum of the responses to each input working separately.
To calculate the response to one input working separately, set all other inputs to zero. (A zero
current source is equivalent to an open circuit and a zero voltage source is equivalent to a short
circuit.)
4
Example
This circuit has one output, i o , and three inputs, v1 , i 2 and v 3 . Express the output as a linear
combination of the inputs.
Solution
When v 2 , i 3 = 0 :
Writing a node equation:
v1 − 40 i o1
40
=
40 i o1
40
+
40 i o1
10
= 5 i o1 ⇒ v1 − 40 i o1 = 200 i o1 ⇒ i o1 =
v1
240
When v1 , i 3 = 0 :
Writing a node equation:
5
v 2 − 40 i o2
10
40 i o2
=
40
+
40 i o2
40
= 2 i o2
⇒ v 2 − 40 i o2 = 20 i o2
⇒ i o2 =
v2
60
When v1 , v 2 = 0 :
Writing a node equation:
va
40
+
va
+
va
40 10
= 0 ⇒ v a = 0 ⇒ i o3 =
va
40
=0
The response of the linear circuit to several inputs working together is equal to the sum of the
responses to each input working separately. Consequently
i o = i o1 + i o2 = i o3 =
v1
240
+
v2
60
+0
Alternate Solution:
Let’s analyze the circuit using node equations. Label the node voltage at the top node of the
vertical resistor:
The node equation is
v a − v1
40
+
va
40
+
va − v2
10
= 0 ⇒ 6 v a = v1 + 4 v 2
⇒ va =
v1
2
+ v2
6 3
6
Using Ohm’s law, we get
io =
va
40
=
v1
240
+
v2
60
as before.
It’s easier to solve this particular problem without using superposition. That’s not always the
case.
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