Đzmir University of Economics
CE 205 Fundamentals of Electrical Circuits
Supplementary Notes
Frequency Transfer Function
1. Circuits with Sinosoidal Inputs
When a sinusoidal voltage is applied to the input of an RC circuit (Fig. 1), at steady
state, all the voltage and current waveforms are also sinusoidal with different
amplitudes and phases.
iR + vR −
R
vS
iC
C
+
vC
vS (t) = V cos ω t volts
Fig. 1 First Order RC Circuit with Sinusoidal Input
The differential equation describing the capacitor voltage variation can be obtained
using the KVL equation
vS = vR + vC
Since
vR = R iR and iC = iR
Then
vR = RC
dvC
dt
Therefore it is obtained that
RC
dvC
dt
+ vC = vS
The natural and forced solutions may be obtained as follows:
i) Natural solution is
vC (t) = Ae- ⁄
to the natural part
RC
where
dvC
dt
+ vC = 0
τ = RC
and A is a parameter to be determined using the initial condition on the total
solution.
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ii) Forced solution for the input vS (t) = V cos ω t volts, i.e., the solution of the
equation
RC
dvC
dt
+ vC = V cos ω t
may be obtained by assuming a suitable solution to the system.
Assume the solution is in the form of
vC(forced) = α cos ω0 t + β sin ω0 t
Upon substituting this solution into the above differential eqation, it is obtained
that
RC #- αω0 sin ω0 t + βω0 cos ω0 t $ + (α cos ω0 t + β sin ω0 t ) = V cos ω t
Upon rearranging
(β ω0 R C + α) cos ω0 t + (- α ω0 RC + β) sin ω0 t = VS cos ω0 t
it yields,
(β ω0 R C + α) cos ω0 t + (- α ω0 RC + β) sin ω0 t = VS cos ω0 t
Equating the coefficients of cosine and sine terms to each other,
β ω0 R C + α = VS
- α ω0 RC + β = 0
Solving for a and b,
V
V (ω RC)
S 0
α = 1+(ω SRC)2 , β = 1+(ω
RC)2
0
0
Then the forced solution is
VS
vC(forced) =
=
1+(ω0
VS
RC)2
(1+(ω0 RC)2
cos ω0 t +
)
1
VS (ω0 RC)
1+(ω0 RC)2
(1+(ω0 RC)2
sin ω0 t
cos ω0 t +
ω0 RC
(1+(ω0 RC)2
sin ω0 t *
It can be put into the form of
vC(forced) (t) = VCP cos (ω0 t + θ) volts
where
VCP =
VSP
(1+(ω0 RC)2
and
θ = - tan-1 ( ω0 RC)
The total solution will be then
vC (t) = A e-t⁄τ + VCP cos (ω0 t + θ) volts
The parameter is the to be determined using the initial condition. For example for
the initial condition vC (t) = 0 V, A is obtained as
vC (0) = A + VCP cos θ = 0 and hence A = - VCP cos θ
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Another approach to determine amplitude and phase for a sinusoidal input is
“phasor” approach.
2. Phasor Approach for Circuits with Sinusoidal Inputs
A sinusoidal function may be expressed in a general form as
f(t) = A cos (ω0 t + θ)
Then
f(t) = Re .A ej(ω0t + θ) / = Re 0Aejθ ejω0t 2 = Re 0F ejω0t 2
where
F = Aejθ
is called phasor.
It is possible to express all sinusoidal signals in the circuit using phasors.
2.1. KVL for Phasors
Consider the circuit given below. The voltages v1, v2 and v3 can be related using the
Kichhoff’s Voltage Law (KVL).
+ v2 +
v1
-
+
v3
Fig. 2
Assume the input voltage vS is a sinusoidal specified as
vS (t) = VS cos (ωt + θS )
Then the voltages in the circuit will be sinusoidal of the same frequency but with
different amplitude and phase. So the voltages v1(t), v2(t) and v3(t) may be expressed
as:
and
and
v1 (t) = V1 cos (ωt + θ1 ) = Re { V1 ej(ωt + θ1) } = Re { V1 ejωt }
v2 (t) = V2 cos (ωt + θ2 ) = Re { V2 ej(ωt + θ2) } = Re { V2 ejωt }
v3 (t) = V3 cos (ωt + θ3 ) = Re { V3 ej(ωt + θ3) } = Re 0 V3 ejωt 2
Question: Can we establish a relationship between the phasors of vS, v1 and v2?
Using the loop equation (KVL) it can be written that the voltage v1(t) satisfies
v1 (t) = v2 (t)+ v3 (t)
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The voltage v1(t) can be also be expressed as:
v1 (t) = V2 cos (ωt + θ2 ) + V3 cos (ωt + θ3 )
Then
v1 (t) = Re { V2 ejωt } + Re { V3 ejωt } = Re { V2 ejωt + V3 ejωt }
= Re {( V2 + V3 ) ejωt }
Therefore,
= Re 0 V1 ejωt 2
V1 = V2 + V3
Similarly for KCL (Fig. 3), the same kind of relationship hold for the phasors
I3 = I1 + I2
i3
i2
i1
Fig. 3
Fig.3
2.2. Voltage and Current Phasor Relationships for Circuit Components
Resistor
The i-v relationship for a resistor is given as
vR (t)= R iR (t)
i(t)
Assume
Then
iR (t) = IR cos (ωt + θR ) = Re { IR ejωt }
+
VR(t)
-
vR (t) = R Re {IIR ejωt } = Re {RIIR ejωt }
= Re {V
VR e
R
Fig. 4
jωt }
Therefore the relationship for the current and voltage phasors over a resistor is
expressed as
VR = R IR
Capacitor
The i-v relationship for a capacitor is given as
iC (t)= C
Assume
dvC (t)
dt
vC (t) = VC cos (ωt + θC ) = Re { VC ejωt }
iC(t)
+
vC(t)
-
C
Fig. 5
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Then
iC (t) = C
d
dt
d
[Re {VC ejωt }] = C Re { 67 ejωt } = Re {jωC 67 ejωt }
dt
= Re {IIC ejωt }
Therefore the relationship for the current and voltage phasors over a capacitor is
obtained as
IC = jωC 67
Inductor
For an inductor the i-v relationship is given as
vL (t)= L
Assume
Then
diL (t)
iL(t)
dt
iL (t) = IL cos (ωt + θL ) = Re { IL e
vL (t) = L
d
+
vL(t)
-
jωt }
d
[Re {IL ejωt }] = L Re { 9: ejωt } = Re {jωL 9: ejωt }
dt
dt
L
Fig. 6
= Re {V
VL ejωt }
Therefore the relationship for the current and voltage phasors over a capacitor is
obtained as
VL = jωL 9:
Table 1
Component
Resistor
Capacitor
Inductor
iR
Symbol
+ vR −
R
+ vC −
iR
iL
C
+ vL −
L
Time Domain
Phasor Domain
vR (t)= R iR (t)
VR = R IR
iC (t)= C
dvC (t)
dt
IC = jωC 67
vL (t)=L
diL (t)
dt
VL = jωL 9:
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3. Frequency Transfer Function
Consider the circuit given in Fig. 7. The input voltage is vi(t) and the output voltage is
vO(t) as shown on the figure.
iR + vR −
R
vi(t) = Vi cos (ωt+θ)
+
iC
C
vO(t)
-
Fig. 7 An RC Circuit with Sinusoidal Input
In phasor domain above circuit reduces to the circuit given in Fig. 8
I + VR −
R
R
Vi
IC
jωC
+
VO
-
Fig. 8 An RC Circuit with Sinusoidal Input
The input phasor Vi is equal to
Vi = VR + VO
Since
VR = R IR and VC =
1
jωC
IR
Since
IR = IC = I
it is obtained that,
Vi = (R +
1
jω0
) I or I =
C
1
(R +
1
)
jω0 C
Vi
Then the phasor of the output (capacitor) voltage
VO =
1
jω0 C
1
)R +
*
jω0 C
1
Vi = (<=
jω
0 RC )
Vi
Then the ratio of the output and input phasors is
VO
Vi
=
1
1+ jω0 RC
This ratio is called the frequency transfer function of the system H(jω), i.e.,
H(jω) =
1
1+ jωRC
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4. Magnitude and Phase Plots of Frequency Transfer Function
Any complex number c may be represented also in polar form as
c = α+ j β = |α+ j β|| ejθ = |α+ j β|| ∠A
where |∙| indicates the magnitude, θ represents the phase and ∠A represents the
exponential term. Then
β
c = α+ j β = (α2 + β2 ejθ = (α2 + β2 ∠A where θ = tan-1 ( α )
Since the frequency transfer function is a complex number, the polar coordinate
representation of H(jω) is
1
H(jω) = |H(jω)| ∠H(jω) = 1+ jωRC =
<
(1+(ωRC)2 ejθ
=
e-jθ
(1+(ωRC)2
where θ = tan-1 ( ωRC)
The magnitude and the phase of H(jω) are:
|H(jω)|=
1
J1+(ωRC)2
and
∠H(jω) = - tan-1 ( ωRC)
The magnitude and phase of the frequency transfer function are plotted separately to
specify H(jω).
Example
Consider the RC circuit given below in Fig. 9. The output is the voltage over the
capacitor. Determine and plot the frequency transfer function.
R
R = 7.96 kΩ
+
VS
C = 10 nF
VC
C
VS (t) = 5 cos ωC volts
-
Fig. 9 An RC Circuit with Sinusoidal Input
The transfer function as obtained above:
H(jω) =
1
jωC
D
R+
jωC
1
= 1 + jωRC
or
1
H(jf) = 1+ j2πfRC
The magnitude of H(jf) is
|H(jf)|=
1
(1+(2πf RC)2
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and its phase is
∠H(jf) = - tan-1 ( 2πf RC)
RC product is then
RC = (7.96 kΩ) (10 nF) = 79.6 µsec
For these values the magnitude and phase of H(jf) is plotted on the Fig. 10. and Fig. 11
respectively.
Important Notes:
i.
ii.
Note that since the frequency domain is large between 100 Hz – 10 kHz, usually a
logarithmic scale is used over the horizontal frequency axis.
There is a special frequency called corner frequency fC. At this corner frequency
the output amplitude drops its
at the corner frequency fC
|H(jfC )| = 1 or
1
√2
1
(1+(2πf C RC)2
of its maximum value. For the above RC circuit
=
1
√2
Then
2πfC RC =1 or fC =
1
2πRC
1
= 2πx 79.6x10−6 = 2 kHz
iii. For this circuit at fC, the phase is
π
∠H(jfC ) = - tan-1 ( 2πfC RC) = - tan-1 ( 1) = 4
10000
3000
2000
1000
300
200
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
100
Magnitude
Magnitude Plot
Frequency (Hz)
Fig. 10 Magnitude of H(jf)
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10000
3000
2000
1000
300
200
0
-10
-20
-30
-40
-50
-60
-70
-80
-90
100
Phase
Phase Plot
Frequency (Hz)
Fig. 11 Magnitude of H(jf)
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