Two port network is electrical equivalent

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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.

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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

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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)

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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 0.995003 -0.04351 0

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10 Hz 10.606 V 10.554V 0.995003 -0.04351 -0.001
100 Hz 10.606 V 10.553V 0.995003 -0.04351 -0.00999
1 kHz 10.606 V 10.502V 0.995003 -0.04351 -0.09961
10 kHz 10.606 V 7.473V 0.704601 -3.04113 -0.78519
100 kHz 10.606 V 1.053V 0.099283 -20.0625 -1.47156
1 MHz 10.606 V 0.1V 0.009429 -40.511 -1.56148
10 MHz 10.606 V 0.01V 0.000943 -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.

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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;
θActual
Stop =←0.78519 radians
(S5.10.6)The phase radians at the cut-off frequency is
θActual
Cut off =−0.78519 radians

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(S5.10.7) Fractional percentage error between the cut-off frequency value and the one
specified in the design was computed as illustrated below.
Ef = ∆ f c
f design
×100 % (14)
Substituting the values;
Ef =(12−10)kHz
10 kHz ×100 %=+20 % (15)
(S5.11) The circuit below was used to determine the transfer function of the passive RC low
pass filter.
The output voltage across Load resistance Rout of the circuit is equivalent to the voltage across the
capacitor.
V out = { ZC
Z R +ZC }V ¿ (16)
The transfer function is;
V out
V ¿
= ZC
ZR + ZC
(17)
The impedances of the circuit in the Laplace transform are as shown below.
ZC = 1
Cs (18)
And
ZR=Rin (19)

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The transfer function is therefore;
V out
V ¿
=H=
1
Cs
R+ 1
Cs
(20)
Algebraically rearranging the equation above
H= 1
RCs+1 (21)
Alternatively; by rewriting;
s= jω
Then the transfer function becomes;
H ( jω ) = 1
1+ jωRC (23)
Therefore, RC low pass filter is basically a first-order type filter
(S5.12) Deriving a two port representation for this circuit
Two port circuit of the network is generally represented by;
Basing on the Z-impedance model, general equations of two port network are as shown below.
V 1=Z11 I1 + Z12 I 2 (24)
V 2=Z21 I 1+Z22 I 2 (25)
Passive RC low pass filter is represented as shown below.

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The model parameters are determined by individually setting currents to zero.
Assuming I 2=0.
Z11= V 1
I 1
=R¿− j
ωC (26)
Substituting the values;
Z11= (5000− j
2 π ( 5000 Hz ×3.18 ×10−9 F ) )Ω (27)
Simplifying equation (27)
Z11= ( 5000− j 10010 ) Ω (28)
Z11= (5− j 10 ) kΩ (29)
Also, when I2=0
Z21= V 2
I1
= − j
2 πfC (30)
Substituting the values in the equation above;
Z21= − j
2 π ( 5000 Hz ×3.18 ×10−9 F ) (31)
Simplifying the equation;
Z21= ( − j 10010 ) Ω (32)
Z21=− j 10 kΩ

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Assuming I 1=0.
Z12= V 1
I 2
= − j
2 πfC (33)
Z12=− j10 kΩ (34)
Also
Z22= V 2
I 2
= − j
2 πfC (35)
Z22=− j 10 kΩ (36)
6 Passive LC low pass filter
S6.1. The circuit diagram of a passive pi-type LC filter is as shown in the figure below. The
filter is connected to the function generator via the source resistance RS and as well as
connected to the oscilloscope via load resistance RL .
S6.2 The design specification of the cut-off frequency for the LC passive low pass filter is
provided in this section.

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The cut-off frequency of this pi type LC passive low pass filter was chosen as;
f c=15 kHz
S6.3 Determination of the right values of components that are required to achieve the
desired specifications.
The combination of capacitive reactance of C1∧C2 gives shunt equivalent impedance as;
Z2 =Zc1+ Zc2 (37)
Where;
Zc1=Zc2=1
2 Zc (38)
The product of shunt and series impedance gives a constant Rk . For instance, therefore;
Z1 Z2= ( Rk )2. (39)
Where;
Z1 = jωL and (40)
Z2 = − j
ωC
Substituting equation (40) into equation (39);
( jωL ) ( − j
ωC ) = ( Rk ) 2 (41)
Simplifying equation (41) algebraically;
L
C = ( Rk )2 (42)
Therefore, source impedance is expressed as;
Rk = √ L
C (43)
The characteristic impedance of the circuit is given by;
ZOT = √ z1
2
4 +Z1 Z2 (44)
Substituting the equivalent expressions;

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ZOT = √ ( jωL ) 2
4 + L
C (45)
Algebraically simplifying the expression;
ZOT = √ L
C − ( ωL )2
4 (46a)
ZOT = √ L
C √1− ω2 CL
4 (46b)
At cut-off frequency;
1− ω2 CL
4 =0 (47)
From equation (47), the cut-off frequency is given by;
ω2 CL
4 =1 (49a)
ω= √ 4
LC (49b)
Since;
ω=2 πf (50)
Then by substitution; the cut-off frequency in Hertz is given by;
f c= 1
π √ LC (51)
According to the design;
f c=15 kHz
And therefore with an inductance of 3mH
L=3 mH
Then from equation (51), the capacitance is given by;
C= 1
( π f c )2 L (52)
Substituting the values;
C= 1
( π ×15000 Hz )2 × ( 0.003 H ) (53a)

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C=0.15 ×10−6 F or (53b)
C=0.15 μF
And
C1=C2 =1
2 C=0.075
The design parameters are as shown below.
L=3 mH
C=0.15 μF
S6.4 The designed circuit was built as shown in the figure below.
S6.5 The preferred (manufactured) values of the resistor and capacitor that you have used
that approximate the designed ones are;
L = 3mH
C = 0.068 μF
S6.6 The gain and phase of the circuit for a range of frequencies were measured and
recorded in the table below The gain of the LC low pass filter was computed for each
frequency and recorded in the same table.

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Frequency f I/P Volts Vi O/P Volts
Vo
Gain G 20log(Vo/Vi)¿ GdB Phaseɸ
radians
0 Hz 10.606V 10.605V 0.999906 -0.00082 0
10 Hz 10.606V 10.605V 0.999906 -0.00082 0
100 Hz 10.606V 10.605V 0.999906 -0.00082 0
1 kHz 10.606V 10.605V 0.999906 -0.00082 0
10 kHz 10.606V 10.603V 0.999717 -0.00246 -3.14251
15 kHz 10.606V 7.346 V 0.692627 -3.19001 -3.14251
20 kHz 10.606V 3.955 V 0.372902 -8.5681 -3.14268
100 kHz 10.606V 0.160V 0.015086 -36.4286 -3.14286
1 MHz 10.606V 0.109V 0.010277 -39.7625 -3.14286
S6.7 The plot of the magnitude response of the circuit is as shown in the figure below.
S6.8 The plot the phase response of the circuit is as shown in the logarithmic graph below.

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S6.9 Graph Measurements are highlighted below.
(S6.9.1) The cut-off frequency is;
f c
Actual=14 kHz
(S6.9.2) The per decade slope of the gain magnitude is
SActual
Decade=40 dB /decade
(S6.9.3) The per octave slope of the gain magnitude is
SActual
Octave=−12 dB
(S6.9.4) The passband phase in radians is;
θActual
Pass =0 ¿−3.14 radians
(S6.9.5) The stopband phase in radians is;
θActual
Stop =←3.14 radians
(S6.9.6) The phase in radians at the cut-off frequency is;
θActual
Cut off =−3.14 radians

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(S6.9.7) The fractional percentage error between the cut-off frequency value and the one
specified in the region is computed as shown below;
Ef = ∆ f c
f design
×100 % (54)
Substituting the values;
Ef =(14−15)kHz
15 kHz ×100 %=−6.7 % (55)
S6.10 The transfer function of the LC passive low pass filter was derived from the circuit
below.
The impedance of the circuit is given by;
Z=ZC 1 /¿ ( ZL1 + ZC 1 ) (56)
Simplifying equation (18);
Z= ZC 1 ( ZL 1+ ZC 1 )
2 ZC 1 +Z L1
(57)

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But
1
2 ZC =ZC 1 (58)
Substituting equation (58) into equation (57), the expression becomes;
Z=
1
2 ZC (Z L+ 1
2 ZC )
ZC +Z L
(59a)
Z= ( 1
2 ZC Z L+ 1
4 ZC )
ZC + ZL
(59b)
Total current in the circuit is given by;
I = ZC +Z L
1
2 ZC (ZL+ 1
2 ZC ) V ¿ (60)
Using current divider rule, the current through output capacitor is given by;
IC 1= ZC+ Z L
1
2 ZC (Z L+ 1
2 ZC ) ( 1
2 ZC
ZC+ Z L )V ¿ (61)
The output voltage is equivalent to the voltage across output capacitor
V out = 1
(Z L+ 1
2 ZC ) V ¿ (62)
The transfer function is the ratio of output against input voltage.
H= V out
V ¿
= 1
(ZL + 1
2 ZC ) (63)
But in Laplace transform
ZL=sL (64)
And
ZC = 1
sC

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Substituting into equation (63);
H ( s)= 1
sL+ 1
2 sC
(65)
Simplifying;
H (s)= 2 sC
2 s2 LC+1 (66)
Substituting s= jω ;
H ( jω)= 2 jωC
1−2 ω2 LC (67)
S6.11 The two port representation of the circuit is as shown in the figure below.
Basing on the Z-impedance model, general equations of two port network are as shown below.
V 1=Z11 I1 + Z12 I 2 (68)
V 2=Z21 I 1+ Z22 I 2 (69)
The Z parameters can be determined using mesh analysis of the circuit below.

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In mesh one
V 1=ZC 1 ( I1−I 3 ) (70)
Rearranging the equation;
V 1=ZC 1 ( I1 ) −¿ ZC 1 ( I3 ) (71)
In mesh two
ZL ( I3 ) +ZC 2 ( I2 +I 3 ) +ZC 1 ( I3 −I1 ) =0 (72)
Rearranging the equation;
( ZL+ ZC 2+ ZC 1 ) I 3= ( ZC 1 ) I 1− ( ZC 2 ) I 2 (73)
Making current I 3 the subject;
I 3= ( ZC 1 )
( Z L+ ZC 2 + ZC1 ) I 1− ( ZC 2 )
( ZL+ ZC 2+ZC 1 ) I 2 (74)
In mesh three
V 2=ZC 2 ( I 2+ I3 ) (75)
Rearranging the equation;
V 2=ZC 2 I2+ ZC 2 I3 (76)
Substituting equation (74) into equation (71)
V 1=ZC 1 ( I 1 )−¿ ZC 1 ( ( ZC 1 )
( ZL+ ZC 2+ ZC 1 ) I1− ( ZC 2 )
( Z L+ZC 2 + ZC 1 ) I 2 ) (77)

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Rearranging the equation;
V 1=ZC 1 ( I 1 )− ( ZC 1 ) 2
( ZL+ZC 2+ ZC 1 ) I1 + ( ZC 2 )2
( ZL+ ZC 2+ ZC 1 ) I2 (78)
But { ZC 2+ ZC 1=ZC }; therefore;
V 1=ZC 1 ( I 1 )− ( ZC 1 ) 2
( ZL+ZC ) I1+ ( ZC 2 )2
( ZL+ ZC ) I 2 (79)
Simplifying equation (79)
V 1= ( ZC 1− ( ZC 1 )
2
( ZL+ ZC ) ) ( I 1 )+ ( ZC 2 )
2
( ZL +ZC ) I 2 (80)
Also, substituting equation (74) into equation (76), the expression becomes;
V 2=ZC 2 I2+ ZC 2 ( ( ZC 1 )
( Z L+ ZC ) I1− ( ZC 2 )
( ZL+ ZC ) I 2 ) (81)
Rearranging the equation;
V 2=ZC 2 I 2+ ( ZC 1 )
2
( ZL+ZC ) I 1− ( ZC 2 ) 2
( Z L+ ZC ) I 2 (82)
Grouping like terms together;
V 2= ( ZC 1 )
2
( ZL+ ZC ) I 1+ ( ZC 2− ( ZC 2 )
2
( ZL +ZC ) ) I 2 (83)
Comparing equation (80) and equation (83) with two port impedance equations;
{ V 1=Z11 I1 +Z12 I 2
V 1=
(ZC 1− ( ZC 1 ) 2
( Z L+ZC ) ) ( I1 ) + ( ZC 2 )2
( ZL+ ZC ) I 2
(84)
And
{ V 2=Z21 I1 + Z22 I2
V 2= ( ZC1 )2
( Z L+ ZC ) I1 + (ZC 2− ( ZC 2 )2
( ZL+ZC ) ) I2
(85)
From equations (86) and (85), then in comparison;

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Z11=Z22=
(ZC 2− ( ZC 2 )2
( Z L+ ZC ) ) (86)
Substituting the real values;
Z11=Z22=
(ZC 2− ( ZC 2 )2
( Z L+ ZC ) ) (87)
Also;
Z12=Z21= ( ZC 1 )
2
( Z L+ZC ) (88)
Substituting the values at cut-off frequency
ZC = − j
ωC = − j
2 π (15× 103 ) ( 0.15 ×10−6 )
ZC =− j 70.74 Ω
And
ZC 1=0.5 ZC=− j35.37 Ω
Also
ZL= j 2 πfL= j 2 π (15000 Hz)(0.003 H )
ZL= j 282.74 Ω
Substituting in equation (87)
Z11=Z22=
( − j 35.37Ω− ( − j 35.37 Ω ) 2
( j 282.74 Ω− j70.74 Ω ) ) (89a)
Z11=Z22=− ( 5.901+ j 35.37 ) Ω
Substituting in equation (88);
Z12=Z21= ( − j35.37 Ω ) 2
( j282.74 Ω− j70.74 Ω ) (89b)
Z12=Z21=5.901Ω
7 Conclusion

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The experiment was implemented as discussed in the previous section of this report. However,
there were existence of slight deviations of the actual values plotted and the designed desired
values. The errors are attributed to the approximation of the components values basing on
designed and manufactures values. Also, since the inductor and capacitor have non-zero initial
conditions if not grounded resulted to errors. However; the errors are so small hence negligible.
The transfer function of the passive RC low pass filter has one pole and no zeros. Theoretically;
when the magnitude response is plotted, at the stop band frequency, the graph should drop with
20dB/decade. This was the similar trend practically obtained from the magnitude plot of the RC
low pass filter. On the other side, the transfer function of the LC low pass filter is of the second
order having two poles and one zero. The effects of two poles on the magnitude response results
to the 40dB/decade at the stop band frequency.
Reference.
Electrical4U. 2020. Two Port Network: Parameters And Examples | Electrical4u. [online]
Available at: <https://www.electrical4u.com/two-port-network/> [Accessed 23 March 2020].