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Simple Signal Processing
Introduction
The circuits in this set of problems each have a single input and a single output. The input is
either the voltage of a voltage source or the current of a current source. The output is either the
voltage measured by a voltmeter or the current measured by an ammeter.
A variety of circuits are considered in this set of problems. Some can be analyzed easily using
voltage and current division. Circuits containing dependent sources can be analyzed using mesh
and node equations. Circuits containing more than one independent source can be analyzed using
superposition.
Voltage and current division are discussed in Sections 3.4 and 3.5 of Introduction to Electric
Circuits by R.C. Dorf and J.A Svoboda. Mesh and node equations are discussed in Chapter 4.
Superposition is described in Section 5.4.
Worked Examples
Example 1:
Consider the circuit shown in Figure 1. The input to this circuit is the current, Is, of the current
source. The output is the voltage, Vm, measured by the voltmeter. A plot showing the relationship
between the input and output of the circuit is shown in Figure 1. Find the value of the resistance
R that is required to cause the circuit to behave in the manner specified by the plot.
Figure 1 The circuit considered in Example 1.
1
Solution: First, we can represent the relationship between Vm and Is analytically by writing the
equation of the straight line shown in the plot.
Vm = 5 I s
(1)
Next, Figure 2 shows the circuit from Figure 1 after replacing the (ideal) voltmeter by the
equivalent open circuit and labeling the voltage measured by the voltmeter. Also, the current in
the 45-Ω resistor has been labeled as Ia. This current is related to the current source current by
current division


R
 R 
Ia = 
 Is = 
Is
 R + 60 
 R + ( 45 + 15 ) 
The current directed downward in the 15-Ω resistor is equal to Ia. Using Ohm’s law gives
Vm = 15 I a
Therefore
 R 
Vm = 15 
Is
 R + 60 
(2)
Equation 1 describes the plot from Figure 1 and equation 2 describes the circuit. To make the
circuit behave in the manner specified by the plot, we require
 R 
5 = 15 

 R + 60 
Solving this equation gives R = 30 Ω.
Figure 2 The circuit from in Figure 1.
2
Example 2:
Consider the circuit shown in Figure 3. The input to this circuit is the voltage, Vs, of the voltage
source. The output is the current, Im, measured by the ammeter. A plot showing the relationship
between the input and output of the circuit is shown in Figure 3. Find the value of A, the gain of
the CCCS, required to cause the circuit to behave in the manner specified by the plot.
Figure 3 The circuit considered in Example 2.
Solution: First, we can represent the relationship between Im and Vs analytically by writing the
equation of the straight line shown in the plot.
 100 × 10−3 
Im = 
 Vs = 0.05 Vs
2


(3)
Next, Figure 4 shows the circuit from Figure 3 after replacing the (ideal) ammeter by the
equivalent short circuit and labeling the current measured by the ammeter. Also, the mesh
currents have been labeled in Figure 4. Notice that one mesh current is equal to the controlling
current of the dependent source while the other mesh current is equal to the current measured by
the meter. Finally, the resistor voltages have been expressed as functions of the mesh currents.
Figure 4 The circuit from in Figure 3.
3
The current of the dependent source is related to the mesh currents by
A Ia = Ia + Im
⇒ I m = ( A − 1) I a
Apply KVL to the supermesh, i.e. the outside loop, to get
−50 I a + 30 I m − Vs = 0
Combining these equations gives
−50 I a + 30 ( A − 1) I a = Vs
Finally
⇒ Ia =
Vs
−50 + 30 ( A − 1)


( A − 1)
I m = ( A − 1) I a = 
 Vs
 −50 + 30 ( A − 1) 
(4)
Equation 3 describes the plot from Figure 3 and equation 4 describes the circuit. To make the
circuit behave in the manner specified by the plot, we require
0.05 =
( A − 1)
−50 + 30 ( A − 1)
⇒ − 2.5 + 1.5 ( A − 1) = ( A − 1)
Solving this equation gives A = 5 A/A.
4
Example 3:
Consider the circuit shown in Figure 5. The input to this circuit is the current, Is, of the current
source. The output is the voltage, Vm, measured by the voltmeter. A plot showing the relationship
between the input and output of the circuit is shown in Figure 5. Find the values of the
resistance, R, and voltage, Vs, required to cause the circuit to behave in the manner specified by
the plot.
Figure 5 The circuit considered in Example 3.
Solution: First, we can represent the relationship between Vm and Is analytically by writing the
equation of the straight line shown in the plot.
Vm = 5 I s + 2
(5)
Next, Figure 6 shows the circuit from Figure 5 after replacing the (ideal) voltmeter by the
equivalent open circuit and labeling the voltage measured by the voltmeter. Vm is the response to
both sources working together. This response can be calculated using superposition. Figure 7
shows circuits that can be used to find V1, the part of Vm due to the current source, and V2, the
part of Vm due to the voltage source. The voltage source voltage is set zero when we find the part
of Vm due to the current source. Consequently, we replace the voltage source by a short circuit in
the circuit used to calculate V1. Similarly, the current source current is set zero when we find the
part of Vm due to the voltage source. We replace the current source by an open circuit in the
circuit used to calculate V2.
Analyzing the circuits in Figure 7, we find
 R 
V1 = 15 
 Is
 R + 60 
 15 
and V2 = 
 Vs
 R + 60 
Using superposition
 R 
 15 
Vm = V1 + V2 = 15 
 Is +
 Vs
 R + 60 
 R + 60 
(6)
5
Figure 6 The circuit from in Figure 5.
Figure 7 Analyzing circuit from in Figure 6 using superposition.
Equation 5 describes the plot from Figure 5 and equation 6 describes the circuit. To make the
circuit behave in the manner specified by the plot, we require
 R 
 15 
5 I s + 2 = 15 
 Is +
 Vs
 R + 60 
 R + 60 
Equating coefficients gives first
 R 
5 I s = 15 
 Is
 R + 60 
⇒ 1=
3R
⇒ R = 30 Ω
R + 60
and then
Vs
 15 
 15 
2=
 Vs = 
 Vs =
6
 R + 60 
 30 + 60 
⇒ V s = 12 V
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