İzmir University of Economics
EEE202 Electrical Circuits II
Title of the Experiment
Experiment – 7
Frequency Domain Transfer Function and Bode Plots
Name-Surname of the Student
Student Number
E-Mail
Grade for the Lab (To be filled in by the instructor.)
Task
Pre-Lab Task
In-Lab Task
Total
Score
Background
B.1. Frequency Transfer Function of an RC Circuit
When a sinusoidal voltage is applied to the input of an RC circuit (Figure 1), at steady state, all the
voltage and current waveforms are also sinusoidal with different amplitudes and phases.
R
+
vS
vC
C
vS (t) = VSP cos ω0 t volts
Figure 1. A First Order RC Circuit with Sinusoidal Input
At steady state, the capacitor voltage (output) is obtained as (see class notes)
vC (t) = VCP cos (ω0 t + θ)
volts
where
VCP =
VSP
and
√1+(ω0 RC)2
θ = - tan-1 ( ω0 RC)
Another approach to determine amplitude and phase for a sinusoidal input, is “phasor” approach.
Using the fact that each signal in the circuit has a general form of a sinusoidal as
vX (t) = Re { VXP e
j(ω0t + θ)
jθ
} = Re { VXP e ejω0t } = Re{ VX ejω0 t }
where
VX = VXP e
jθ
is called phasor, it is possible to express all signals in the circuit using phasors.
The relationships between phasors also follows the circuit equations.
VS = VR + VC
Using the relations
VR = R IR , VC =
1
jω0 C
IR (see class notes)
and using
IR = IC = I
it is obtained that,
VS = (R +
1
jω0 C
) I or I =
1
(R +
1
)
jω0 C
VS
Then the phasor of the output (capacitor) voltage
VC =
(R
1
jω0 C
1
+
)
jω0 C
VS =
1
(1+ jω0 RC )
VS
Then the ratio of the output and input phasors is
VC
VS
=
1
(1+ jω0 RC )
This ratio for general  is called the frequency transfer function of the system H(j), i.e.,
H(jω) =
1
1+ jωRC
B.2. 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|∠𝜃
where |∙| indicates the magnitude, θ represents the phase and ∠𝜃 represents the exponential term.
Then
β
2
2
c = α+ j β = √α2 + β ejθ = √α2 + β ∠𝜃 where θ = tan-1 ( )
α
Since the frequency transfer function is a complex number, the polar coordinate representation of
H(j) is
H(jω) = |H(jω)|∠H(jω) =
1
1+ jωRC
=
1
√1+(ωRC)2ejθ
=
e-jθ
√1+(ωRC)2
where θ = tan-1 ( ωRC)
The magnitude and the phase of H(j) are:
|H(jω)|=
1
√1+(RC)2
and
-1
∠H(jω) = - tan ( 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 Figure 2. Determine and plot the frequency transfer function.
R
+
VS
R = 7.96 kΩ
C = 10 nF
VC
C
VS (t) = 5 cos ω𝑡 volts
Figure 2. An RC Circuit
The transfer function as obtained above:
H(jω) =
1
1+ jωRC
or
H(jf) =
1
1+ j2πfRC
The magnitude of H(jf) is
|H(jf)|=
1
√1+(2πf RC)2
and its phase is
-1
∠H(jf) = - tan ( 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 Figure 3a. and Figure 3.b,
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
1
√2
of its maximum value. For the above RC circuit at the corner frequency
fC
|H(jfC )| = 1 or
1
√1+(2πf C RC)2
=
1
√2
Then
2πfC RC =1 or fC =
=
1
2πx 79.6x10−6
= 2 kHz
For this circuit at fC, the phase is
-1
-1
∠H(jfC ) = - tan ( 2πfC RC) = - tan ( 1) =
π
4
Magnitude Plot
Frequency (Hz)
(a)
Magnitude of H(j)
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
iii.
1
2πRC
Phase
Phase Plot
0
-10
-20
-30
-40
-50
-60
-70
-80
10000
3000
2000
1000
300
200
100
-90
Frequency (Hz)
(b)
Phase of H(j)
Figure 3
Pre-Lab Task 1
Consider the circuit given below in Figure 4. The input is a sinusoidal waveform. The output voltage
is the steady state voltage over the capacitor.
R
+
VS
C
VC
Figure 4
Determine the time constant and the corner frequency fC.
Implementation of Pre-Lab Task 1
R = 8.2 kΩ
C = 10 nF
In-Lab Task 1
Construct the circuit given in Figure 5.
R
+
VS
C
VC
R = 8.2 kΩ
C = 10 nF
Figure 5
i. Using the oscilloscope, measure peak-to-peak values of both input and the output capacitor
voltages, and the phase difference at the given frequencies. Fill Table 1.
Note that |H(jw)| =
VC(P-P)
VS(P-P)
Table 1
Frequency
VS(P-P) (volts)
VC(P-P) (volts)
|H(jw)| =
100
200
300
400
500
600
700
800
900
1000
2000
3000
4000
5000
6000
7000
8000
9000
10000
ii. Plot your magnitude and phase values on the following graphs.
VC(P-P)
VS(P-P)
∠H(jw) (degree)
Magnitude Plot
10000
3000
2000
1000
300
200
100
Magnitude
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
Frequency (Hz)
(a)
Magnitude of H(j)
Phase
Phase Plot
0
-10
-20
-30
-40
-50
-60
-70
-80
Frequency (Hz)
(b)
iii. What is the corner frequency?
Phase of H(j)
10000
3000
2000
1000
300
200
100
-90
Implementation of In-Lab Task
Student’s Evaluation of Learning Outcomes (Write in the following box what you have gained
from doing the Experiment VIII.)

The knowledge of ...

The skill of measuring/computing/calculating/determining/finding/simulating of...

The awareness/consciousness of...