Two port network is electrical equivalent

Names:.................................................... IDs:
............................................................................ 1
NAME:.............................................................................
STUDENT ID:......................................................................
STUDENT ID of partner:........................................................
Introduction
Two port network is electrical equivalent representation of complex network by using terminal
characteristics which are voltages and currents. Two port networks seeks to simplify analysis of
the complex network by an equivalent simpler model. General representation of two port
network is as shown in the figure below (Electrical4U, 2020).
Low pass passive filters can be easily analyzed using two port network model. In this report, two
types of passive low pass filters namely; RC low pass filter and LC low pass filters are dealt
with. These filters are as shown in figures below.
Passive RC low pass filter Passive pi LC low pass filter
The characteristics of passive low pass filters are embedded in their ability to select and allow
certain range of frequencies to be transmitted and attenuates any other frequency components in
the electrical signals. Attenuated frequencies are electronically termed as unwanted frequencies
that are filtered out or rejected so as to output the desired signal.
The objective of this experiment entailed deriving two port representation of these filters. The
filters were terminated to the supply and output using source resistance and load resistance
respectively.

Names:.................................................... IDs:
............................................................................ 2
5. Passive RC Low Pass Filter.
S5.1. The drawn circuit diagram of a passive RC filter that is connected to a function
generator with RS source impedance and an oscilloscope with load impedance RL.
S5.2. The specification of the source resistance and load resistance is as shown below.
Source Resistance
RS=50 Ω
Load Resistance
RL=1 MΩ
S5.3. Specification of the cut-off frequency (for the RC low pass filter);
The cut-off frequency for the RC low pass filter design is’
f C=10 kHz

Names:.................................................... IDs:
............................................................................ 3
S5.4. Determining the component values that are required to achieve the design specification
of the filter.
At the cut-off frequency, the real part of the impedance is equal to the imaginary part. Therefore,
the resistance of the resistor is equivalent to the capacitive reactance. Mathematically, the
following expression is used.
XC =R¿ (1)
Where;
R¿−¿Series combination of Source Resistance and Low pass filter resistance.
Thus; capacitive reactance is given by;
XC = 1
2 π f c C (2)
Therefore;
1
2 π f c C =R¿ (3)
In the design, the equivalent input resistance value was selected as;
R¿=5 kΩ (4)
From equation (3), making the capacitance the subject, then capacitance value is determined as;
C= 1
2 π f c R¿
(5)
Substituting the values in equation (5);
C= 1
2 π ( 10 ×103 Hz ) ( 5000Ω ) (6)
Simplifying the equation;
C=3.18 ×10−9 F (7)
Or
C=3.18 nF (8)
The RC resistor
R=R¿−RS (9)
Substituting the values;
R= ( 5000−50 ) Ω (10)

Names:.................................................... IDs:
............................................................................ 4
R=4950 Ω∨4.95 kΩ (11)
The design values are
R=4.95 kΩ (12)
C=3.18 nF (13)
S5.5. The circuit was built with the design parameters as shown in the figure below.
S5.6. The preferred as from the manufacturer which are approximately equal to the designed
parameters of resistor and capacitor is highlighted below.
R= 4.87kΩ
C = 3.3nF
S5.7. The gain and phase of the circuit was determined over range of frequencies. The
values were recorded in the table below.
Freq f I /P Volts Vi O/ PVolts Vo Gain G GdB=20 log(Vo/Vi) Phase ɸ
In radians
0 Hz 10.606 V 10.553V 0.9950 -0.04351 0

Names:.................................................... IDs:
............................................................................ 5
03
10 Hz 10.606 V 10.554V 0.9950
03 -0.04351 -0.001
100 Hz 10.606 V 10.553V 0.9950
03 -0.04351 -0.00999
1 kHz 10.606 V 10.502V 0.9950
03 -0.04351 -0.09961
10 kHz 10.606 V 7.473V 0.7046
01 -3.04113 -0.78519
100 kHz 10.606 V 1.053V 0.0992
83 -20.0625 -1.47156
1 MHz 10.606 V 0.1V 0.0094
29 -40.511 -1.56148
10 MHz 10.606 V 0.01V 0.0009
43 -60.511 -1.57038
S5.8. The magnitude response of the designed circuit was plotted on the logarithmic graph
as shown in the figure below.

Names:.................................................... IDs:
............................................................................ 6
S5.9. The phase response of the designed circuit was plotted in the logarithmic graph as
shown in the figure below.
S5.10. Graph measurements are shown below
(S5.10.1) The cut-off frequency is;
f c
Actual=12 kHz
(S5.10.2) The per decade slope of the gain magnitude is;
SActual
Decade=−20 dB/decade
(S5.10.3) The per octave slope of the gain magnitude is;
SActual
Octave =−6 dB
(S5.10.4) The passband phase in radians is;
θActual
Pass =0 ¿−0.78519 radians
(S5.10.5)The stopband phase in radians is;