Unit-5
Two-port network and Filters
Short circuit Admittance parameters
Y-parameter:-
I1 = Y11V1 + Y12V2
I2 = Y21V1 + Y22V2
Y11 = V2=0
Y12 = V1=0
Y21 = V2=0
Y22 = V1=0
V1& V2 should be independent
Equivalent circuit: -
Fig 1 Equivalent Y-parameter circuit
Symmetrical two-port N/W: -
I2 = 0 = I1=0
Y12 = Y22
Reciprocal two-port N/W: -
I1=0 =I2 = 0
Y12 = Y21
Que 1. Find overall Y-parameter?
Solution:V1 – I1R – V2 = 0
V1 – V2 = I1R
I1 = V1 - V2
V2 = I2R + V1
I2 = - V1 + V2
Y11 =
Y12 = Y21 =
Y22 =
Que 2. Find overall Y-parameter
Solution: Y-parameter does not exist as V1 = V2
Que 3. Find overall Y-parameter?
Solution: I1 = V1Ya + (V1 – V2)Yc
I1 = (Ya + Yc)V1 - YcV2
I2 = V2Yb + (V2 – V1)Yc
I1 = (Yb + Yc)V2 - YcV1
Y11 = Yb + Yc
Y12 = Y21 = - Yc
Y22 = Yb + Yc
Que 4. Find overall Y-paramter?
Solution:
Que 5:
Find Y-parameter?
Solution:
-I1 + – 2V2 + + 2V1 = 0
I1= V1 + V1 - 3V2 + 2V1
I1= 4V1 - 3V2V
V2 + 2V2- 2V1 = 2(I2 + 2V1)
- 2V1 + 3V2 = 2(I2 + 2V1)
3V2 - 2V1 – 4V1 = 2I2
I2 = -3V1 + V2
Y11 = 4
Y12 = -3
Y21 = -3
Y22 =
Open circuit impedance parameters
Z-parameter :-
v1 = Z11I1 + Z12I2
v2 = Z11I1 + Z22I2
=
Z11 = I2=0
Z12 = I1=0
Z21 = I2=0
Z22 = I1=0
I1 and I2 are excitations at ports 1 & 2 respectively.
V1 and V2 are the responses at ports 1 and 2 respectively.
Equivalent circuit:-
Fig 2 Equivalent z-parameter circuit
Symmetry:-
I2=0 =I1 = 0
Z11 = Z22
Reciprocal two-port N/W:-
I2=0= I1=0
Z12 = Z21
I1& I2 should be independent
Que 1.
Find Z-parameter
Solution:I1 = -I2
Current dependent so Z-parameter doesn’t exist
Que 2.
Find z-parameter
Solution:V1 =R (I1 + I2)
V2 = R (I1 + I2)
Z11 = Z12 = Z21 = Z22 = R
Que 3.
Solution: V1 = I1Za + I1Zc + I2Zc
= (Za + Zc)I1 + ZcI2
V2 = I2Zb + I2Zc + I1Zc
= (Zb + Zc)I1 + ZcI1
Z11 = (Za + Zc)
Z12 = Zc = Z21
Z22 = (Zb + Zc)
Que 4.
Solution:V1 = Za(I1 - I)
(I - I1)Za+ IZc+ Zb(I + I2) = 0
I(Za + Zb + Zc) – I1Za + I2Zb = 0
I =
V1 = ZaI1 - Za
= I1 + I2
V2 = Zb(I2 + I)
= ZbI2 + Zb
= I2 + I2
Z11 =
Z12 = Z21 =
Z22 =
Can be solved by Y-A conversion
Que 1.
Solution :
Z11 = I2=0
V1 - (Za + Zb) = 0
= Z11 =
Z21 = I2=0
V2 - Zb +Za = 0
=
Z12 =
Z22 =
Que 2.
Find Z21?
Solution: Z21 = I2=0
I1/2 =
= I1
V2 = I1/2
= × I1
= I1
Z21 = I2 = 0 = I1 Ω
Que 3.
Solution : + + 2V1 = 0
2Vx – 2V1 + Vx – V2 + 2V1 = 0
3Vx = V2
Vx =
I1 = V1 +
= 3V1 – 2Vx
= 3V1 – 2
V1 =
+ 2V2 = I2
3V2 - = I2
V2 = I2
V2 = I2
V1 =
=
Z11 =
Z12 =
Z21 = 0
Z22 =
Transmission parameter [ABCD]
V1 = AV2 - BI2
I1 = CV2 - DI2
A = I2=0
B = V2=0
C = -I2=0
D = V2=0
Symmetrical two-port N/W:-
I2 = 0 = I1=0
A = D
Reciprocal two-port N/W:-
I1=0 =I2 = 0
AD – BC = 1
Que 1. Find all the Transmission parameters?
Solution:
-3V1 – I1 + + = 0
+ = I1
V1 – V2 = I1
V1 = V2 + I1
V1= V2- I1----------------(1)
I2 = 3V1 + V2 + V2 – V1
I2 = 2V1 + 2V2
2V1 = I2 - 2V2
2V1 = - 2V2 + I2
V1 = -V2 + I2 ----------------(2)
A = -1
B =
From (1) & (2)
-V2 + I2 = I1 - V2
V2 - V2 + I2 = I1
I1 = V2 + V2 - I2
I1 = V2 - I2
C =
D =
Que 2. Find all the Transmission parameters?
Solution : V1 = RI1 + V2 ----------------(1)
I2R = V2 – V1
I2 = V2 - V1
I2R = V2 – V1
V1 = I2R - V2 --------------------(2)
A = 1
B = R
From (2) in (1)
V2 - I2R = V2 + RI1
I2 = -I1
C = 0
D = 1
Que 3. Find all the Transmission parameters?
Solution: V1 = R(I1 + I2)
V2 = R(I1 + I2)
V1 = V2 + 0I2
A = 1
B = 0
V2 = RI1 + RI2
RI1= V2 - RI2
I1 = V2 – I2
C = , D = 1
Hybrid Parameter
H-parameter :-
V1 = h11I1 + h12V2
I2 = h21I1 + h22V2
Equivalent circuit for H-parameter:-
Fig 4 Equivalent circuit for H-parameter
h11 = V2=0
h12 = I1=0
h21 = V2=0
h22 = I1=0
Symmetrical two-port N/W: -
I2 = 0 = I1=0
∆h = 0
Reciprocal two-port N/W: -
I1=0 =I2 = 0
h12 = h21
Que 4. Find all h-parameter?
Solution:
+ = I1
V1 - = I1
V1 = + I1
V1 = + I1
- + 3I1 = I2
3I1 + V2 = I2
From (1)
I2 = 3I1 + V2 – [ I1 + V2]
I2 = I1 + V2
h11 =
h12 =
h21 =
h22 =
Relationship of two-port variables
Fig5 2 Port network
Fig 6 1 port network
Fig 7 3 port network
z-parameter open circuit impedance
y-parameter short circuit admittance
h-parameter hybrid parameter
g-parameter inverse hybrid parameter
ABCD-parameter transmission parameter
A’B’C’D’ parameter inverse transmission
s-parameter scattering
used for very high-frequency application
Relationship of Two-Part Variables
Condition for Reciprocity and Symmetry
Sr No. | Parameter | Reciprocity | Symmetry |
1. | Z Parameter | Z12 = Z21 | Z11= Z22 |
2. | Y Parameter | Y12 = Y21 | Y12 = Y21 |
3. | h Parameter | h12 = h21 | Δ = 1 |
4. | ABCD Parameter | AD-BC = 1 | A = D |
5. | Inverse h Parameter | g12 = g21 | Δ = 1 |
6. | Inverse Transmission | A1D1 – B1 C1= 1 | A1 = D1 |
Key takeaway
Relation between Two- Port Parameters :
Δ = X11 X22 – X12 X21 , ΔT = AD – BC
| [z] | [y] | [h] | [T] |
[z] | Z11 Z12 | y22 – y12 Δy Δy
| Δh – h12 h22 h22 | A – ΔT C C |
Z21 Z22 | y21 – y11 Δy Δy
| -h21 h h22 h22
| 1 D C C | |
|
|
|
|
|
[y] | Z22 -Z12 Δz Δz
| y11 y12 | 1 -h12 h11 h11
| D – ΔT B B |
-Z21 Z11 Δz Δz | y21 y22 | h21 Δn h11 h11 | -1 A B B
| |
|
|
|
|
|
[h] | Δz Z12 Z22 Z22
| 1 -y12 y11 y11
| h11 h12 | B ΔT D D |
-Z21 -1 Z22 Z22 | y21 Δy y11 y11 | H21 h22 | -1 C D D
| |
[T] | Z21 ΔZ Z21 Z21 | -y22 -1 y21 y21 | -Δh -h11 h21 h21
| A B |
1 Z22 Z21 Z21
| -Δy -y11 y21 y21
| -h22 -1 h21 h21 | C D |
Filters are those electrical circuits that perform signal processing functions, to remove unwanted frequency components.
Classification of Passive Filters
Passive filters are made up of passive components such as resistors, capacitors, and inductors, and have no amplifying elements such as transistors, op-amps, etc. so have no signal gain. Therefore, the output level is always less than the input.
Filters are so named according to the frequency rangke of signals that they allow passing.
(a) Low Pass Filter:-These filters reject all frequencies above a specified value called the cut-off frequency.
(b) High Pass Filter:-These filters reject all frequencies below the cut off frequency.
(c) Band Pass Filter:-These filters allow signals falling within a certain frequency band and blocks all the frequencies outside this band.
(d). Band Stop Filter:-These filters reject transmission of a limited band of frequencies but allow the transmission of all other frequencies.
The figure below gives the ideal gain v/s frequency response of above all mentioned types of filters.
Fig 8 Ideal gain v/s frequency response of above-mentioned Filters
A BPF is usually obtained by series connections of LPF with an HPF, where FL of HPF < FH of LPF.
A BSP may be obtained by parallel connections of LPF with an HPF, where FH of HPF > FL of LPF.
Q –A LPF circuit consisting of a resistor of 40KΩ in series with capacitor 47nF across a 10V sinusoidal supply. Calculate VO at frequency 1Kz?
Soln:
R = 40KΩ
C = 47nF
Vi = 10V
F = 1000 Hz
For LPF,
Capacitive Reactance is XC
The output voltage is given as
VO = Vi
XC = =
XC = 3.386KΩ
VO =
=10 x
= 10 x
VO = 0.843 V
(b) RL LPF
FC = R / 2nt
VO = Vi
The characteristic impedance of low pass
A Basic LPF circuit can be designed using passive elements R, L, and C. The circuit is having a single ‘ R ’ in series with capacitor ‘C’.
Fig 9 Characteristic impedance of LPF
For RC LPF
VO = = Vi x fC = 1 / 2∏RC
But Z = Phase shift φ = -cot(2∏FC)
And XC =
VO – output voltage
Vi – input voltage
Z – Impedance
XC – Capacitive Reactance
FC – cut off Frequency
For RC LPF the reactance of ‘ C ’ is very high at a low frequency so, the capacitor acts as an open circuit and blocks i/p signal Vi until the cut-off frequency is reached.
Cut-off frequency for an LPF is that frequency at which the voltage at output equals 70.7% of input voltage( -3 dB point ).
LPF is used to filter noise from the circuit.
Q –A LPF circuit consisting of a resistor of 40KΩ in series with capacitor 47nF across a 10V sinusoidal supply. Calculate VO at frequency 1Kz?
Soln:
R = 40KΩ
C = 47nF
Vi = 10V
F = 1000 Hz
For LPF,
Capacitive Reactance is XC
The output voltage is given as
VO = Vi
XC = =
XC = 3.386KΩ
VO =
=10 x
= 10 x
VO = 0.843 V
(b) RL LPF
FC = R / 2nt
VO = Vi
Key takeaway
Comparison of RL – RC circuits
Q – An RL LPF consists of a 5.6mH coil a 3.3 KΩ resistor. The output voltage is taken across the resistor. Calculate the critical frequency?
Soln :Given:
L = 5.6 x 10-3 H
R = 3.3 x 103Ω
For RL LPF
F ==
F = 93.78 KHz
Q – A sinusoidal voltage with a peak to the peak value of 10 V is applied to an RC LPF. If reactance at the input is zero, find output voltage?
Soln :
VO =
XC =
F = XC = 0, F ->∞
Hence, VO = 0V.
Q . An RC – LPF consists of a 120 Ω resistor and 0.02 µF capacitor. The output is taken through the capacitor. Calculate the critical frequency?
For RC LPF FC is given as
FC==
FC = 66.31 KHz
High pass
(a). The RC – HPF circuits are shown below, which is having a capacitor at the input and resistor at the output.
The reactance of the capacitor is very high at low frequency.
XC =
XC when F ( C acts as open circuit )
XC when F ( C acts as a short circuit )
Fig 10 Ideal gain and Phase response of HPF
FC =
Phase shift = Cot
VO =Vi
VO = Vi
At low F :XC ->∞ VO = 0
At high F :XC -> 0 VO = Vi
Q . An RC HPF with input capacitor 10nf and the output resistor of 10K Ω. Find the cut off frequency.
Soln :
FC = =
FC = 15.915 Hz
Q . Calculate the breaking point (FC) for a simple passive HPF consisting of an 82pF capacitor connected in series with a 240 KΩ resistor?
Soln :
FC = =
FC = 8KHz
(b). RL – HPF :
The basic RL HPF is shown below. The inductors pass low-frequency signals with very little resistance and offer great resistance to high-frequency signals.
VO =
FC =
Q . For an RL HPF consists of a 470Ω resistor and 600 mH coil output is taken across the coil. Calculate the cut off frequencies.
Soln :
FC = == 125 Hz
M – derived filters :
There are few drawbacks of constant – k filters which are
(i). the attenuation rise near the cut off frequency is not sharp.
(ii). The characteristic impedance varies continuously in the passband and hence presents a serious problem regarding matching.
The m-derived filter meets these two objectives. They raise the attenuation near the cut off frequency. The m-derived LPF is shown below,
m – derived T and ∏ section network:
For m-derived T section let
Z1 = mZ1, Z0 = Z0, Z2 = ?
We already have ZOT = √Z1Z2(1 + Z1/4Z2)
√Z1Z2 + Z12/4 = √Z1Z2 + Z12/4
(mZ1)(Z2) + (mZ1/4)2 = Z1Z2 + Z12/4
mZ1Z2 = Z1Z2 + Z12/4 – (mZ1)2/4
Z2 = Z2/m + (1 – m2)Z1/4m - (23)
The m-derived T-network will now be
For m-derived ∏ section
Let Z2 = Z2/m, Z0 = Z0, Z1 = ?
For ∏ network we already know
Z0∏ = √Z1Z2/1 + Z1/4Z2
√Z1Z22/1 + Z1/4Z2 = √Z1Z2/1 + Z1/4Z2
Z1Z2/m / 1 + Z1m/4Z2 = Z1Z2 / 1 + Z1/4Z2
Solving for Z1 we have
Z1 = 4Z1Z2m / 4Z2 + Z1(1 – m2)
Multiplying and diving RHS by m
Z1 = mZ1(4m/1-m2)Z2 / mZ1 + (4m/1-m2)Z2 - (24)
(a). m-derived T and ∏ section LPF
F∞ = 1/∏√LC√1-m2 , F∞ = FC/√1-m2
FC = 1/∏√LC
m = √1 – (FC/F∞)2
(b). m-derived HPF :
F∞ = √1 - m2FC
m = √1 – (F∞/FC)2
FC = 1/4∏√LC
Q5) In a certain LPF circuit Fc = 4.5KHz.R=300 ohm Find a) Its passband b) If the input voltage is a 6v sine wave dc level calculate the output voltage
Sol: The passband of LPF is from 0 to fC.
The output voltage is given as
VO =
XC ===0.035mΩ
VO = =6x=0.012V
Q6) For RC HPF with 325Ω resistor. Calculate the value of the capacitor such that Xc is ten times less than the value of the resistor at input frequency 12KHz?
Sol: For RC HPF
FC =
XC =
Bur Xc is 10 times less than R. Hence, Xc=R/10=32.5Ω
Finding fC
fC ==1/(2x32.5)=4.89mHz
Now finding C
FC =
4.89x=
C=0.1F
Q7) For the given -section LPF calculate Ro & Fc
Sol: Given C/2=0.4F. The prototype -section LPF is given below
C=0.8F
L=200mH
FC = ==795.8Hz
R0 = ==500Ω
Key takeaway
(a). m-derived T and ∏ section LPF
F∞ = 1/∏√LC√1-m2 , F∞ = FC/√1-m2
FC = 1/∏√LC
m = √1 – (FC/F∞)2
(b). m-derived HPF :
F∞ = √1 - m2FC
Reference
[1] Network Analysis Third Edition by M. E. Van Valkenburg, Prentice Hall of India Private Limited.
[2] Network Analysis & Synthesis by G. K. Mittal, Khanna Publication.
[3] Network Analysis and Synthesis by Ravish R Singh, McGraw Hill.
[4] Introduction to Electric Circuits by Alexander & Sadiku, McGraw Hill.
[5] Introduction to Electric Circuits by S. Charkarboorty, Dhanpat Rai & Co.
[6] Fundamentals of Electrical Networks by B.R.Gupta & Vandana Singhal- S.Chand Publications
[7] Electrical Circuit Analysis 2nd Edition by P. Ramesh babu, Scitech Publication
India Pvt Ltd.