Equivalent Circuits and Transfer Functions
Samantha R Summerson
14 September, 2009
1
Equivalent Circuits
𝑅𝑒𝑞
𝑖𝑠𝑐
+
+
𝑣𝑒𝑞
±
𝑣
−
−
Figure 1: Thévenin equivalent circuit.
+
𝑖𝑠𝑐
↑
𝑣𝑜𝑐
𝑅𝑒𝑞
−
Figure 2: Mayer-Norton equivalent circuit.
Recall the steps to finding the equivalent circuit values: 𝑣𝑜𝑐 , 𝑖𝑠𝑐 , and 𝑅𝑒𝑞 .
1. Consider the open circuit (𝑖 = 0). Find the voltage, 𝑣𝑒𝑞 = 𝑣𝑜𝑐 .
2. Consider the short circuit, setting the terminal voltage to zero (𝑣 = 0). Solve for 𝑖 = 𝑖𝑠𝑐 .
3. Zero the sources (all voltage sources short-circuited and current sources open-circuited) and find the
equivalent resistance, 𝑅𝑒𝑞 , looking back into the terminals.
𝑅𝑒𝑞 = −
1
𝑣𝑒𝑞
𝑖𝑠𝑐
We can find the equivalent circuit by performing any two of the three above steps. We can solve for
the third unknown quantity using the 𝑣 − 𝑖 relation.
Note that if we short-circuit our Thévenin equivalent circuit, the current traveling clockwise is 𝑖𝑠𝑐 .
Example 1. Find the Norton equivalent circuit. For sake of completeness, we will go through all three
100Ω 𝑖1 𝑖
+
𝑖2
15𝑉 ±
𝑣
50Ω
−
Figure 3: Example 1 - find the equivalent circuit values.
steps.
(a) Let 𝑖 = 0. Using our voltage divider rule,
𝑣𝑜𝑐
= 𝑣
50Ω
(15𝑉 )
50Ω + 100Ω
5𝑉
=
=
(b) Set 𝑣 = 0, i.e. put a wire across the terminal. The voltage across the 50Ω resistor is now zero,
meaning 𝑖2 must be zero. We are solving for 𝑖𝑠𝑐 = 𝑖. Using KCL,
𝑖1
=
𝑖,
=
𝑖𝑠𝑐 .
By KVL,
100𝑖1
=
15𝑉,
3
=
𝐴,
20
3
=
𝐴.
20
⇒ 𝑖1
⇒ 𝑖𝑠𝑐
(c) Zero out the voltage source, meaning short-circuit it (replace it with a wire). From the right-hand
side of the circuit, the two resistors are in parallel. Thus,
(
𝑅𝑒𝑞
=
=
=
1
1
+
100 50
(
)−1
3
100
100
Ω
3
Example 2. Find the equivalent circuit values.
2
)−1
100
3 Ω
+
5𝑉
±
𝑣
−
Figure 4: Example 1 - Thévenin equivalent circuit.
+
3
20 𝐴
100
3 Ω
↑
𝑣𝑜𝑐
−
Figure 5: Example 1 - Mayer-Norton equivalent circuit.
i. First, set 𝑖 = 0. This implies that the current through the 10Ω resistor, which is the resistor
we are measuring the output voltage across, is 𝑖4 . So,
𝑣𝑜𝑐 = 𝑣 = 10𝑖4 .
We only need to solve for 𝑖4 . Now we can write our KCL and KVL equations. By KVL, we
have
10
=
20𝑖𝑖 + 16𝑖4 ,
10
=
20𝑖1 + 5𝑖3 .
By KCL we have:
𝑖1
= 𝑖2 + 𝑖3 ,
𝑖2 + 1
=
𝑖4 ,
𝑖4
=
1 + 𝑖5 ,
𝑖3 + 𝑖5 = 𝑖1 .
First we will solve for 𝑖1 in terms of 𝑖4 .
𝑖1
16𝑖4
16
𝑖4
⇒
5
⇒ 𝑖1
=
𝑖2 + 𝑖3
=
(𝑖4 − 1) + 𝑖3
=
5𝑖3 from KVL equations
=
𝑖3
=
𝑖4 − 1 +
=
21
𝑖4 − 1
5
3
16
𝑖4
5
20Ω
10𝑉 ±
𝑖1 𝑖2
𝑖3
𝑖
6Ω
𝑖4
+
↑ 1𝐴
5Ω
10Ω
𝑣
−
𝑖5
Figure 6: Example 2 - find the equivalent circuit values.
Using this expression for 𝑖1 , we plug it into our first KCL equation and solve for 𝑖4 .
21
𝑖4 − 1) + 16𝑖4
5
= 100𝑖4 − 20
3
=
𝐴
10
10
=
⇒ 𝑖4
20(
Now,
𝑣𝑜𝑐 = 10𝑖4 = 3𝑉.
ii. Put a wire across the terminals, i.e. set 𝑣 = 0. We are solving for 𝑖𝑠𝑐 = 𝑖. Due to the short
circuit, no current will go through the 10Ω resistor. Thus, 𝑖𝑠𝑐 = 𝑖 = 𝑖4 . From KVL, we have
These two equations together imply
10
=
20𝑖1 + 6𝑖4 ,
10
=
20𝑖1 + 5𝑖3 .
6
5 𝑖4
= 𝑖3 . From KCL, we have
𝑖1
= 𝑖2 + 𝑖3 ,
1 + 𝑖2
= 𝑖4 ,
𝑖
= 𝑖4 ,
𝑖
=
𝑖3 + 𝑖5
1 + 𝑖5 ,
= 𝑖1 .
Combining the first two equations, we have
(𝑖4 − 1) + 𝑖3 ,
6
= 𝑖4 − 1 + 𝑖4 , from the KVL equations
5
11
=
𝑖4 − 1.
5
Using this value for 𝑖1 in the first KVL equation,
(
)
11
𝑖4 − 1 + 6𝑖4 ,
10 = 20
5
= 50𝑖4 − 20,
3
⇒ 𝑖4 =
𝐴.
5
𝑖1
=
iii. By the 𝑣 − 𝑖 relation,
𝑅𝑒𝑞 =
4
𝑣𝑜𝑐
3
= 3 = 5Ω.
𝑖𝑠𝑐
5
2
2.1
Equivalent Capacitance and Inductance
Capacitors in Series and Parallel
Consider two capacitors in parallel. The voltage across the capacitors will be the same, 𝑣, though
the current will be different. By KCL, the current going into the node is the equal to the sum of
the current through both capacitors:
𝑖 = 𝑖1 + 𝑖2 .
Using our 𝑣 − 𝑖 relation for capacitors, we can re-write this.
𝑖
= 𝑖1 + 𝑖2 ,
𝑑𝑣
𝑑𝑣
+ 𝐶2 ,
= 𝐶1
𝑑𝑡
𝑑𝑡
𝑑𝑣
= (𝐶1 + 𝐶2 ) ,
𝑑𝑡
𝑑𝑣
= 𝐶𝑒𝑞 .
𝑑𝑡
Thus, the equivalent capacitance for two capacitors in parallel is
𝐶𝑒𝑞 = 𝐶1 + 𝐶2 .
This is analogous to the rule for resistors in series.
Consider two capacitors in series. In this case, the current through both is the same but the
voltages are different. However, the voltage across both has the relation
𝑣 = 𝑣1 + 𝑣2 .
Since differentiating is a linear operation, we can take the derivative of both sides and plug in the
𝑣 − 𝑖 relations.
𝑑𝑣
𝑑𝑡
=
=
=
=
𝑑𝑣1
𝑑𝑣2
+
,
𝑑𝑡
𝑑𝑡
𝑖
𝑖
+
,
𝐶
𝐶2
(1
)
1
1
+
𝑖,
𝐶1
𝐶2
𝑖
.
𝐶𝑒𝑞
Thus, the equivalent capacitance for two capacitors in series is
(
)−1
1
1
𝐶𝑒𝑞 =
+
.
𝐶1
𝐶2
This is analogous to the rule for resistors in parallel.
2.2
Inductors in Series and Parallel
We can do similar analysis for inductors. If we consider two inductors in parallel, the voltage
across both will be the same,𝑣, though the current will be different. Thus current going into the
node connecting the two inductors has the following relation:
𝑖 = 𝑖1 + 𝑖2 .
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We can take the derivative of both sides and use our 𝑣 − 𝑖 relations.
𝑑𝑖
𝑑𝑡
𝑑𝑖1
𝑑𝑖2
+
,
𝑑𝑡
𝑑𝑡
𝑣
𝑣
=
+
,
𝐿
𝐿2
(1
)
1
1
=
+
𝑣,
𝐿1
𝐿2
𝑣
=
.
𝐿𝑒𝑞
=
The equivalent inductance for two inductors in parallel is
(
𝐿𝑒𝑞 =
1
1
+
𝐿1
𝐿2
)−1
.
This is analogous to the rule for resistors in parallel.
Finally, consider two inductors in series. The current through both inductors will be the same, 𝑖.
The voltage across the two is equal to the sum of the voltages,
𝑣 = 𝑣1 + 𝑣2 .
Using our 𝑣 − 𝑖 relations, we have
𝑣
=
=
=
=
𝑣1 + 𝑣2 ,
𝑑𝑖
𝑑𝑖
𝐿1 + 𝐿2 ,
𝑑𝑡
𝑑𝑡
𝑑𝑖
(𝐿1 + 𝐿2 ) ,
𝑑𝑡
𝑑𝑖
𝐿𝑒𝑞 .
𝑑𝑡
The equivalent inductance for two inductors in series is
𝐿𝑒𝑞 = 𝐿1 + 𝐿2 .
This is analogous to the rule for resistors in series. The rules for inductors are similar to those
for resistors, whereas the rules for capacitors are the opposite.
3
Transfer Function
For an LTI system, for any sinusoidal input, the output will be a sinusoid with the same frequency
but possibly different amplitude and phase. Consider an input voltage
𝑣𝑖𝑛 (𝑡) = 𝑉𝑖𝑛 𝑒𝑗2𝜋𝑓 𝑡 .
The output voltage for the system will have the form
𝑣𝑜𝑢𝑡 (𝑡) = 𝑉𝑜𝑢𝑡 𝑒𝑗(2𝜋𝑓 𝑡+𝜙) .
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We can write this output voltage in terms of the input voltage.
𝑣𝑜𝑢𝑡 (𝑡)
𝑉𝑜𝑢𝑡
𝑉𝑖𝑛 𝑒𝑗2𝜋𝑓 𝑡 𝑒𝑗𝜙 ,
𝑉𝑖𝑛
𝑉𝑜𝑢𝑡 𝑗𝜙
𝑒 𝑉𝑖𝑛 𝑒𝑗2𝜋𝑓 𝑡 ,
𝑉𝑖𝑛
𝑉𝑜𝑢𝑡 𝑗𝜙
𝑒 𝑣𝑖𝑛 (𝑡).
𝑉𝑖𝑛
=
=
=
The term scaling the input voltage is called the transfer function, 𝐻.
Definition 1. The transfer function for an LTI system is defined as
𝐻(𝑓 ) =
𝑣𝑜𝑢𝑡
.
𝑣𝑖𝑛
The transfer function defines the response of the system to any complex exponential input. It
essentially defines the system.
Example 3. Consider a circuit/system where
𝑣𝑜𝑢𝑡 (𝑡) = 𝑣𝑖𝑛 (𝑡 − 𝑀 ).
𝑀 is a constant. Find 𝐻(𝑓 ). First of all, we need to check if the system is LTI. Since it is a
simple delay system, it is. Now we consider a complex exponential input,
𝑣𝑖𝑛 = 𝑒𝑗2𝜋𝑓 𝑡 .
The corresponding output is
𝑣𝑜𝑢𝑡 = 𝑒𝑗2𝜋𝑓 (𝑡−𝑀 ) .
Using the formula for the transfer function, we have
𝐻(𝑓 )
=
=
𝑒𝑗2𝜋𝑓 (𝑡−𝑀 )
,
𝑒𝑗2𝜋𝑓 𝑡
𝑒−𝑗2𝜋𝑓 𝑀 .
Since the magnitude of this filter is always one, it is an all-pass filter.
Example 4. Consider an LTI system with transfer function 𝐻(𝑓 ) and
𝑣𝑖𝑛 (𝑡) = 𝑐𝑜𝑠(2𝜋𝑓 𝑡).
What is 𝑣𝑜𝑢𝑡 in terms of 𝐻(𝑓 )? Like we do so many time, we re-write the cosine as a sum of
exponentials (Euler’s formula)
𝑣𝑖𝑛 (𝑡) =
)
1 ( 𝑗2𝜋𝑓 𝑡
𝑒
+ 𝑒−𝑗2𝜋𝑓 𝑡
2
Since
𝑒𝑗2𝜋𝑓 𝑡
𝑒−𝑗2𝜋𝑓 𝑡
→ 𝐻(𝑓 )𝑒𝑗2𝜋𝑓 𝑡 ,
→ 𝐻(−𝑓 )𝑒−𝑗2𝜋𝑓 𝑡 ,
and our system is LTI (i.e. superposition holds), we have
𝑣𝑜𝑢𝑡 (𝑡) =
1
1
𝐻(𝑓 )𝑒𝑗2𝜋𝑓 𝑡 + 𝐻(−𝑓 )𝑒−𝑗2𝜋𝑓 𝑡 .
2
2
7
From class we learned the two main properties of the magnitude and phase of the transfer function.
Namely,
∙ The magnitude of 𝐻 is an even function, i.e.
∣𝐻(𝑓 )∣ = ∣𝐻(−𝑓 )∣.
∙ The phase of 𝐻 is an odd function, i.e.
∠𝐻(𝑓 ) = −∠𝐻(−𝑓 ).
From class, we also saw that for a real input
𝑣𝑖𝑛 (𝑡) = 𝑐𝑜𝑠(2𝜋𝑓 𝑡),
the output always has the form
𝑣𝑜𝑢𝑡 (𝑡) = ∣𝐻(𝑓 )∣𝑐𝑜𝑠(2𝜋𝑓 𝑡 + ∠𝐻(𝑓 )).
Example 5. Consider an LTI system with input
𝑣𝑖𝑛 (𝑡) = 1 + 𝑐𝑜𝑠(𝜋𝑡) + 𝑐𝑜𝑠(2𝜋𝑡),
and transfer function
{
𝐻(𝑓 ) =
𝑒𝑗2𝜋𝑓 𝑡
0
∣2𝜋𝑓 ∣ < 4
.
otherwise
Find 𝑣𝑜𝑢𝑡 . First we note that the input is a sum of three cosine functions at frequences 𝑓 = 0, 21 , 1.
By supersition and our derivation in class,
(
( ))
1
1
𝑣𝑜𝑢𝑡 = ∣𝐻(0)∣𝑐𝑜𝑠(0) + ∣𝐻( )∣𝑐𝑜𝑠 𝜋𝑡 + ∠𝐻
+ ∣𝐻(1)∣𝑐𝑜𝑠(2𝜋𝑡 + ∠𝐻(1)).
2
2
By examination, we see that
𝐻(0) = 1,
( )
1
𝐻
= 𝑒𝑗𝜋 ,
2
𝐻(1) = 0.
Thus,
𝑣𝑜𝑢𝑡
=
1 + ∣𝑒𝑗𝜋 ∣𝑐𝑜𝑠(𝜋𝑡 + ∠𝑒𝑗𝜋 ) + 0,
=
1 + 𝑐𝑜𝑠(𝜋𝑡 + 𝜋).
We can see that the highest frequency component of the input, 𝑐𝑜𝑠(2𝜋𝑡), was filtered out (i.e.
that frequency does not appear in the output). This makes sense, as our filter is a low-pass filter.
8