DETAILED SOLUTIONS D07 JUNE 2003
1.a D To calculate Thevenin’s equivalent impedance value in a circuit, all independent voltage sources are shorted while all independent current sources are opened.
b. A A 26dBm output in watts equals to 0.4 W because
c A The characteristic impedance of a low pass filter in attenuation band is purely imaginary.
d C The real part of the propagation constant shows reduction in voltage, current values of signal amplitude.
e D The purpose of an Attenuator is to decrease value of signal strength.
f D In parallel resonance of R-L-C circuit having a R-L branch and ‘C’ forming parallel branch,
g A In a transmission line terminated by characteristic impedance, Zo, there is no reflection of the incident wave.
h B For a coil with inductance L and resistance R in series with a capacitor C has a resonance impedance R.
2.a) Millman’s theorem states that if n voltage sources V1, V2, ---,Vn having internal impedances Z1, Z2, ---, Zn respectively are connected in parallel, then these sources may be replaced by a single voltage source Vm having internal series impedance Zm where Vm and Zm are given by the equations
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where Y1, Y2, ---, Yn are the admittances corresponding to Z1, Z2, ---, Zn
Proof:
A voltage source V1
with series impedance Z1 can be replaced by a current source V1Y1
with shunt admittance Y1 = 1/Z1
\The network of Fig 2.1 can be replaced by its equivalent network as shown in Fig 2.2a, where
I1= V1Y1, I2= V2Y2, - - - In= VnYn ,
and Y1 = 1/Z1 , Y2 = 1/Z2 , - - -, Yn = 1/Zn ,
Network of Fig 2.2b is equivalent to that of Fig 2.2a where by
Im
= I1 + I2+ - - - + In
and Ym = Y1
+ Y2 + - - - + Yn
Reconverting the current source of
Fig 2.2b into an equivalent voltage source,
The equivalent circuit of 2.2c is obtained where
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2.b) Applications of Millman’s theorem:
· This theorem enables us to combine a number of voltage (current) sources to a single voltage (current) source.
· Any complicated network can be reduced to a simple one by using the Millman’s theorem.
· It can be used to determine the load current in a network of generators and impedances with two output terminals.
3.a) Consider a constant – K filter, in which the series and shunt impedances, Z1 and Z2 are connected by the relation
Z1 Z2 = R02
Where R0
is a real constant independent of frequency. R0 is often termed as
design impedance or nominal impedance of the constant – K filter.
Let Z1 = jwL and Z2 = 1/jwC
At cutoff-frequency fc
Given the values of R0 and wc, using the eq 3a.1 and 3a.2 the values of network elements L and C are given by the equations
3.b) Advantages of m - derived filters:
i) M - derived filters have a sharper cut-off characteristic with steeper rise at fc (cut-off frequency). The slope of the rise is adjustable by fixing the distance between fc and f.
ii) Characteristic impedance (Z0) of the filter is uniform within the pass band when m - derived half sections, having m = 0.6 are connected at the ends.
iii) M - derived filters are used to construct “composite filters” to have any desired attenuation/frequency characteristics.
4. The impedance of the series RLC circuit is given by
The circuit is at resonance when the imaginary part is zero,
i.e.
at w = w0,
X = 0
\To find the condition for resonance X = 0.
The current at any
instant in a series RLC circuit is given by
At resonance, X = 0
Z0 = R
The
Q-factor of an RLC series resonant circuit is given as the voltage
magnification that the circuit produces at resonance.
The ability to discriminate the different frequencies is called the Q factor of the circuit. The quality factor of the circuit determines the overall steepness of the response curve.
Higher the value of Q of a series resonant circuit, the smaller is the bandwidth and greater is the ability to select or reject a particular narrow band of frequencies.
5.i) COMPENSATION THEOREM: The theorem may be stated as “Any impedance linear or nonlinear, may be replaced by a voltage source of zero internal impedance and voltage source equal to the instantaneous potential difference produced across the replaced impedance by the current flowing through it”.
PROOF:
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fig 5.ii.(a) fig 5.ii.(b)
Consider a network of impedance and voltage source, together with the particular impedance Z1, that is to be replaced, considered as the load.
By Kirchhoff’s law to fig 5.ii.(b),
SI1Z1+Z I = SV1.
Where the summation extends over a number of unspecified impedances in the mesh shown in fig 5.ii.(b).
By Kirchhoff’s law to fig 5.ii.(a), the equation is the same as for fig 5.ii.(b) except the equation for the right hand side mesh.
SZ1I1 = -IZ1 + SV1.
\ All the equations are identical for the two networks, and so are the currents and voltages through out the two networks, that is networks are equivalent.
5.ii) Stub matching: When there are no reflected waves, the energy is transmitted efficiently along the transmission line. This occurs only when the terminating impedance is equal to the characteristic impedance of the line, which does not exist practically. Therefore, impedance matching is required. If the load impedance is complex, one of the ways of matching is to tune out the reactance and then match it to a quarter wave transformer. The input impedance of open or short circuited lossless line is purely reactive. Such a section is connected across the line at a convenient point and cancels the reactive part of the impedance at this point looking towards the load. Such sections are called impedance matching stubs. The stubs can be of any length but usually it is kept within quarter wavelength so that the stub is practically lossless at high frequencies. A short circuited stub of length less than l /4 offers inductive reactance at the input while an open circuited stub of length less than l /4 offers capacitive reactance at the input. The advantages of stub matching are:
· Length of the line remains unaltered.
· Characteristic impedance of the line remains constant.
· At higher frequencies, the stub can be made adjustable to suit variety of loads and to operate over a wide range of frequencies.
5.iii) Star delta conversion: At any one frequency, a star network can be interchanged to a delta network and vice-versa, provided certain relations are maintained.
Let Z1, Z2, Z3 be the three elements of the star network and ZA, ZB, ZC be the three elements of the delta network as shown in Fig 5.iii.a. and Fig 5.iii.b
The impedance between the terminal 1 and terminal 3 is
The impedance between the terminal 3 and terminal 4 is
The impedance between the terminal 1 and terminal 2 is
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Adding eq.1, eq.2 and subtracting eq.3
Adding eq.2, eq.3 and subtracting eq.1
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Adding eq.3, eq.1 and subtracting
eq.2
Consider Z1Z2 + Z2Z3 + Z3Z1 = SZ1Z2
From eq.4, eq.5, eq.6 we get
From eq .6
6.a) An attenuator is a four terminal resistive network connected between the source and load to provide a desired attenuation of the signal. An attenuator can be either symmetrical or asymmetrical in form. It also can be either a fixed type or a variable type. A fixed attenuator is known as pad.
Applications of Attenuators:
i) Resistive attenuators are used as volume controls in broadcasting stations.
ii) Variable attenuators are used in laboratories, when it is necessary to obtain small value of voltage or current for testing purposes.
iii) Resistive attenuators can also be used for matching between circuits of different resistive impedances.
6.b) Given Zoc = 900Ð-300
Zsc = 400Ð-100
The characteristic impedance of the line is given by
7.a) The transmission properties of the line are
improved by satisfying the condition
Where L is the inductance of the line, R is the resistance, C is the capacitance and G is the capacitance of the line per unit length. The above condition is satisfied either by increasing L or decreasing C. C cannot be reduced since it depends on the construction. The process of increasing the value of L to satisfy the condition
so as to reduce attenuation and distortions of the line is known as “loading of the line”. It is done in two ways. i) Continuous loading ii) Lumped loading.
· Continuous loading: Continuous loading is done by introducing the distributed inductance throughout the length of the line. Here one type of iron or some other material as mu-metal is wound around the conductor to be loaded thus increasing the permeability of the surrounding medium. Here the attenuation increases uniformly with increase in frequency. It is used in submarine cables. This type of loading is costly.
· Lumped loading: Lumped loading is done by introducing the lumped inductances in series with the line at suitable intervals. A lumped loaded line behaves as a low pass filter. The lumped loading is usually provided in open wire lines and telephone cables. The a.c resistance of the loading coil varies with frequency due to hysteresis and eddy current losses and hence a transmission line is never free from distortions.
7.b In a hybrid parameter model, the voltage of the input port and the current of the output port are expressed in terms the current of the input port and the voltage of the output port. The equations are given by
V1 = h11I1 + h12V2
I2 = h21I1 + h22V2
When the output terminal is short circuited, V2 = 0, we can determine h11 and h21,
where h11 is the input impedance expressed in ohms and h21 is the forward current gain.
When the input terminal is open circuited, I1 = 0, we can determine h12 and h22,
where h12 is the reverse voltage gain and h22 is the output admittance expressed in mhos.
The equivalent circuit of the hybrid parameter representation is shown in Fig 7.b
h12V2 is the controlled voltage source.
h21I1 is the controlled current source.
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h- parameters are widely used in modeling of electronic devices and circuits particularly transistors where both the open circuit and short circuit conditions are utilized.
8.
Given Ro
= 600W, D = 40dB
We know that
R2R3 = R12 = R02
Þ R1 = R0 = 600W
\ R1 =
600W
R3 = R0(N -1)
= 600(100 – 1) = 59.4W
9.a) The attenuation in decibel (dB) is given by
1 dB = 20 x log10(N)
where
The attenuation in Neper (Nep) is given by
1 Nep = loge(N)
The relation between decibel and neper is
dB = 20 x log10(N)
= 20 x loge(N) x log10(e)
= 20 x loge(N) x 0.434 (\log10(e) = 0.434)
= 8.686 loge(N)
\ Attenuation in decibel = 8.686 x attenuation in Neper.
\ Attenuation in Neper = 0.1151 x attenuation in decibel.
9.b Image impedance is that impedance, which when connected across the appropriate pair of terminals of the network, the other is presented by the other pair of terminals.
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VSWR (Voltage Standing Wave Ratio) is defined as the ratio of maximum and minimum magnitudes of voltage on a line having standing waves.
VSWR is always greater than 1. When VSWR is equal to 1, the line is correctly terminated and there is no reflection.
9.c At the points of voltage maxima,
|Vmax| = |VI| + |VR|
where VI is the r.m.s value of the incident voltage.
where VR is the r.m.s value of the reflected voltage.
Here the incident voltages and reflected voltages are in phase and add up.
At the points of voltage minima,
|Vmin| = |VI| - |VR|
Here the incident voltages and reflected voltages are out of phase and will have opposite sign.
VSWR (Voltage Standing Wave Ratio) is defined as the ratio of maximum and minimum magnitudes of voltage on a line having standing waves.
The voltage reflection coefficient, k is defined as the ratio of the reflected voltage to incident voltage.
10.a The network function N(s) can be written in
the form of ratio of polynomials in (s).
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The equation P(s) = 0, has n roots say a1, a2, ---an.
The equation Q(s) = 0, has n roots say b1, b2, ---bn.
Then both P(s) and Q(s) are written as
product of linear factors
s
Where H = a0/b0 is constant called scale factor.
If the value of s is equal to any of the values a1, a2, ---an, the network function N(s) = 0 as P(s) = 0. Therefore the complex frequencies a1, a2, ---an, are called zeros of the network function.
If the value of s is equal to any of the values b1, b2, ---bn, the network function N(s) = ¥ as Q(s) = 0. Therefore the complex frequencies b1, b2, ---bn, are called poles of the network function. In a pole-zero configuration or pole-zero plot, zeros are conventionally denoted by circle O, while poles are denoted by the cross X as shown in Fig 10.a.
If the poles and zeros are not repeated, then the network function has simple poles and zeros. On the other hand, if the poles and zeros are repeated, then the network function has multiple poles and zeros.
10.b Consider an anti-resonant RLC circuit as shown in Fig 10.b.i
When
the capacitor is perfect and there is no leakage and dielectric loss.
i.e. RC = 0 and let RL = R as shown in Fig 10.b.ii.
The admittance
At
resonance frequency wo, the
susceptance is zero.
If R is negligible
It is possible to have parallel resonance as long as
to have fo or wo to be real.
11.a The unit step function is defined as
The Laplace transform is given by
The
ramp function is given by
f(t) = t, t ³ 0
The Laplace transform is given by
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The unit impulse function is given by
Which is nothing but the derivative of the unit step function.
d(t) has the value zero for t > 0 and unity at t = 0.
Let g(t) = 1 - e-at
g(t) approaches d(t) when a is very large.
On applying Laplace transform
\F(s) = 1 for impulse function.
11.b Consider the two functions f1(t) and f2(t) which are zero for t < 0.
The convolution of f1(t) and f2(t) in time domain is normally denoted by f1(t) * f2(t) and is given by
and
Where t is a dummy variable part.
The Laplace transform of convolution of two time domain functions is given by convolution theorem. The convolution theorem states that the Laplace transform of the convolution of f1(t) and f2(t) is the product of individual Laplace transforms.
L [ f1(t) * f2(t) ] = F1(s) * F2(s).
DETAILED SOLUTIONS D07 DEC 2003
1.a. D Laplace transform of a unit Impulse function is 1
b. D Millman's theorem is applicable during determination of Load current in a network with more than one voltage source.
c. D Asymmetrical two port networks have ZOC1 ¹ ZOC2 and ZSC1 ¹ ZSC2
d. A An attenuator is a R's network.
e. C A pure resistance, RL when connected at the load end of a loss-less 100 W line produces a VSWR of 2. Then RL is 50 W or 200 W, as follows:
f. A The reflection coefficient of a transmission line with a short-circuited load is 0.
g. D All pass filter, passes all frequencies without attenuation but phase change.
h. B A series resonant circuit is inductive at f = 1000 Hz. The circuit will be capacitive some where at f < 1000 Hz.
PART I
2.a. State and prove the superposition theorem with the help of a suitable network.
Superposition theorem: It states that “if a network of linear impedances contains more than one generator, the current which flows at any point is the vector sum of all currents which would flow at that point if each generator was considered separately and all other generators are replaced at that time by impedance equal to their internal impedances”
I1 = I1’ + I1”
I2 = I2’ + I2”
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Let the currents due to V1 alone be I1’ and I2’ and currents due to V2 alone be I1’’ and I2’’ and the currents due to V1 and V2 acting together be I1 and I2.
Applying kirchoff’s voltage law
Z3 I1 I2
When V1 is acting alone, as in Fig 2.a.1, then
Z1
I1’(Z1 + Z2) - I2’Z2
= V1 ----- (1)
Z2
I2’(Z2 + Z3) - I1’Z2
= 0 ----- (2)
When V2 is acting alone, as in Fig 2.a.2, then
I1”(Z1 + Z3)
- I2”Z2 = 0 ----- (3)
I2”(Z2
+ Z3) - I1”Z2 = -V2 ----- (4)
When
both V1, and V2
are acting, as in Fig 2.a.3, then
I1(Z1 + Z2)
- I2Z2 = V1 ----- (5)
Fig 2.a.3
I2(Z2 + Z3) - I1Z2
= -V2 -----
(6)
Adding equations (1) and (3),
(I1’ + I1”) (Z1 + Z2) – (I2’+ I2“)Z2 = V1 ----- (7)
Adding equations (2) and (4),
(I2’ + I2”) (Z1 + Z2) – (I1’+ I1“)Z2 = -V2 ----- (8)
Comparing equations (5), (7) and (6), (8)
I1 = I1’ + I1”
I2 = I2’ + I2”
which proves the superposition theorem.
b. Find the power dissipated in 8W resistor in the circuit shown in Fig 2.b.1 using Thevenin's theorem.
To find RTH, open circuiting the 8W resistor and short-circuiting the voltage sources
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RTH = 5 W
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To find VOC,
Let the potential at x be V1.
On applying kirchoff’s current law at point x
Þ 8V1 + 70 = 0
Þ
V1 = - 70 / 8 = - 8.75V
\Current through 5 W resistor is
\Drop across 5 W resistor is
Current through 10 W resistor left to point x is
Drop across
10 W resistor is
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\VOC = - 15 - 1.25 + 3.75 = - 12.5 V
Power loss in 8W resistor = (0.96) 2 x 8 = 7.37 Watts
3.a. Derive the expression for characteristic impedance of a symmetrical Bridged -T -network.
Consider the symmetrical Bridged -T network shown in Fig 3.a
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Let the terminals c-d be open circuited as shown in fig 3.a.1
Let the terminals c-d be short circuited as shown in Fig 3.a.2
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Since the characteristic impedance is given by
b. Design an asymmetrical T -network shown below in Fig 3.b having ZOC1 = 1000 W, ZOC2 = 1200 W and ZSC1= 700 W.
Given ZOC1 = 1000 W, ZOC2 = 1200 W and ZSC1= 700 W.
From
the Fig 3.b ZOC1 = R1
+ R3 = 1000 W,
R1 = 1000 - R3
ZOC2 = R2 + R3
= 1200 W,
R2 = 1200 – R3
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Þ R32 = 360000.
\ R3 = 600W.
we know that R1 + R3 = 1000 W.
\ R1 = 400W.
we know that R2 + R3 = 1200 W.
\ R2 = 600W.
4.a Define the h-parameters of a two port network. Draw the h-parameter equivalent circuit. Where are the h-parameters used mostly?
In a hybrid parameter model, the voltage of the input port and the current of the output port are expressed in terms the current of the input port and the voltage of the output port. The equations are given by
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1 2
V1 = h11I1
+ h12V2
I2 = h21I1 + h22V2
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When the output terminal is short circuited, V2 = 0
V1 = h11I1
I2 = h21I1
Where h11 is the input impedance expressed in ohms and h21 is the forward current gain.
When the input terminal is open circuited, I1 = 0
V1 = h12V2
I2 = h22V2
Where h12 is the reverse voltage gain and h22 is the output admittance expressed in mhos.
The equivalent circuit of the hybrid parameter representation is shown in Fig 4.a
h12V2 is the controlled voltage source and h21I1 is the controlled current source.
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h- parameters are widely used in modeling of electronic components and circuits particularly transistors where both the open circuit and circuit conditions are utilized.
b. Calculate the transmission parameters of the network shown below. Also verify the reciprocity & symmetricity of the network.
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On open circuiting the terminals 2-2’ as in Fig 4.b.2
Applying Kirchoff’s voltage law (KVL) for the first loop
V1 = I1 + 3(I1- I3) = 4I1- 3I3 --- (1)
Applying KVL for the second loop
7I3 = I1
---(2)
From (1) and (2)
On short circuiting the terminals 2-2’ as in Fig 4.b.3
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Þ
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Applying Kirchoff’s voltage law (KVL) for the first loop of Fig 4.b.4
V1 = I1 + 3(I1- I3) = 4I1- 3I2 --- (5)
I2 = -I3
Applying KVL for the second loop
0 = 3(I3 - I1) + 2I3
5I3 = 3I1 and
also 
.
Hence are get
From (5) and (6)
Þ A ¹ D
\The circuit is neither reciprocal nor symmetrical.
5.a. Design a symmetrical bridged -T attenuator shown below. Use necessary assumptions for simplification.
where Ro is the characteristic impedance and N = ea, a is the attenuation constant.
Consider
the loop abcde of the circuit in Fig 5.a
Es = I1RA + IRRO + ISRO ------(i)
Consider the loop afcde in Fig.5a
---------(ii)
Consider the loop abfcde of the circuit
in Fig 5.a
Es = (IS - I1)R1 + (IR - I1)R1 + IRRO + ISRO
-------(iii)
Solving (i) and (ii) & (iii) we obtain
Usually the load and arm impedances are selected in such a way that
R12 = RO2 = R2RA
5.b In a symmetrical T-network, if the ratio of input and output power is 6.76. Calculate the attenuation in Neper & dB. Also design this attenuator operating between source and load resistances of 1000W.
Let Pin be the input power, Pout be the output power and N be the attenuation in Nepers.
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Attenuation D = 20 log10(N)
= 20 log10(2.6)
= 8.299dB
Load Resistance, RO = 1000W (given)
6.a. What are the disadvantages of the prototype filters? How are they removed in composite filters?
The two disadvantages of a prototype filters are:
· The attenuation does not increase rapidly beyond cutoff frequencies (fc), that is in the stop band.
· The characteristic impedance varies widely in the pass band from the required value.
In m-derived sections attenuation reaches a very high value at a frequency (f¥) very close to cut off frequency but decreases for frequencies beyond (f¥) and the characteristic impedance is more uniform within the pass band.
Composite filters are used to overcome the two disadvantages. A composite filter consists of
· One or more constant K sections to produce a specific cut off frequency.
· One or more m-derived sections to provide infinite attenuation at a frequency close to (fc).
· Two terminating half sections (m-derived) that provide almost constant input and output impedances.
In a composite filter, the attenuation rises very rapidly with frequency in the range of fc to f¥, and falls only marginally with frequencies after f¥.
b. Determine the Laplace transform of the function f(t) = (1 - e-at) sinat
where a is a constant.
f(t) = (1 - e-at) sinat
L(f(t)) = L((1-e-at) sinat)
i.e.F(s) = L(sinat) - L(e-at sinat)
F(s) = F1(s) – F2(s)
F1(s) = L(sinat)
F2(s) = L(e-at sinat)
PART II
7.a. Draw the equivalent circuit of a section of transmission line. Explain primary and secondary parameters.
A uniform transmission line consists of series resistance (R), series inductance (L), shunt capacitance (C), and shunt conductance (G). The series resistance is due to the conductors and depends on the resistivity and diameter. The inductance is due to the magnetic field of each of the conductor carrying current. The inductance is in series with resistance since the effect of inductance is to oppose the flow of the current. The shunt capacitance is due to the two conductors placed parallel separated by a dielectric. The dielectric is not perfect and hence a small leakage current flows in between the wires, which results in shunt conductance. All the four parameters are uniformly distributed over the length of the line.
The series impedance Z per unit length is Z = R + j w L ohms/unit length
The shunt admittance Y per unit length is Y = G + j w C seimens/unit length
Where Z ¹ 1/Y.
The parameters R, L, G, C are normally constant for a particular transmission line and are known as primary constants of a transmission line.
The characteristic impedance, Zo and propagation constant, g are the secondary constants of a transmission line and indicate the electrical properties of a line. Characteristic impedance, Zo is the input impedance of a infinite length line. Propagation constant, g is defined as the natural logarithm of the ratio of the input to the output current.
The
equivalent T section of transmission line of length Dx is shown in the Fig 7.a.
7.b Derive the expressions for characteristic impedance and propagation constant of a transmission line.
Consider a transmission line of length (Dx) units.
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Where Z = R + jwL (series impedance per unit length)
and Y = G + jwC (admittance per unit length)
When the line is terminated in ZO,
as Dx ® 0
From the theory of exponential series
8.a. Describe various types of losses in a transmission line. How these losses are reduced?
The three types of losses are
· Radiation loss: The radiation loss is due to the electromagnetic field around the conductors. The loss of energy is proportional to the square of the frequency and also depends on the spacing between the conductors. These losses are more in open wire lines than that in co-axial cables. The radiation loss increases with frequency and is more evident in high frequency cables. Radiation losses can be reduced by decreasing the spacing between the conductors and allowing only low frequency signals to pass through.
· Dielectric loss: Air acts as a dielectric in transmission line and chemical compounds in a coaxial cable. The dielectric medium possesses finite conductivity and there is leakage of current and loss of energy between the conductors. This loss is due to the imperfect dielectric medium. Dielectric loss is proportional to the voltage across the dielectric and inversely proportional to the characteristic impedance of the dielectric medium. Dielectric loss increases with frequency. Dielectric losses can be reduced by choosing a perfect dielectric or by using air as dielectric.
· Copper loss (thermal loss or conductor heating loss): Copper loss is the energy loss in the form of heat dissipated in the surrounding medium by the conductors. This loss is due the existence of the resistance in the line conductors. It is expressed as I2R loss, where I is the current through the conductor and R is the resistance.
8.b Find the image impedances of an asymmetrical-p(pi) network.
Let Y1, Y2 and Y3
be the admittances and Yi1 and Yi2 be the image
admittances of the asymmetric p network.
From Fig 8.b.1
From Fig 8.b.2
Adding equations (1) and (2)
Subtracting equation (1) from equation (2)
9.a A low-1oss coaxial cable of characteristic impedance of 100W is terminated in a resistive load of 150W. The peak voltage across the load is found to be 30 volts. Calculate,
(i) The reflection coefficient of the load,
(ii) The amplitude of the forward and reflected voltage waves and current waves.
(iii) and V.S.W.R
Given ZO = 100W and ZR = 150W
Þ VR = - 0.2 Vi
VR + Vi = 30 (given)
Vi - 0.2 Vi = 30
Let IR and Ii be the reflected and forward current respectively
But IR + Ii = 0.2
Þ 0.2 Ii + Ii = 0.2
b. In Laplace domain a function is given by
Where a,b, q & M are constants. Show by initial value theorem
As per the initial value theorem
10.a Three series connected coupled coils are shown below in Fig
Calculate
(i) The total inductance of these coils.
(ii) The coefficient of coupling between coils (1) & (2), coils (2) & (3) and coils (3) & (1).
It is given that L1 =1.0 H, L2 = 3 H, L3 = 7 H,
M12 = 0.5 H, M23 =
1 H, M13 = 1 H
The inductance of the coils is given by
For coil 1 = L1 – M12 + M13 = 1.0 – 0.5 + 1.0 = 1.5 H
For coil 2 = L2 – M23 - M12 = 3.0 – 1.0 – 0.5 = 1.5 H
For coil 3 = L3 + M13 - M32 = 7.0 – 1.0 + 1.0 = 7 H
The total inductance of the coils = 1.5 + 1.5 + 7 = 10 H
The coefficient of coupling between coils (1) and (2) is
The coefficient of coupling between coils (2) and (3) is
The coefficient of coupling between coils (3) and (1) is
b. It is required to match 300W load to a 400W transmission line, to reduce the VSWR along the line to 1.0. Design a quarter-wave transformer at 100 MHz.
When the line is made l/4 long, the input impedance becomes,
At any value of s
But when standing wave ratio s is equal to 1, then ZO = ZR
Hence, the characteristic impedance of a quarter-wave transformer should be 400W.
11.a. Check, if the driving point impedance Z (s), given below, can represent a
passive one port network.
Also specify proper reasons in support of your answer.
The given function is not suitable to represent the impedance of one port network due to following reasons :
· In the numerator, one coefficient is missing.
· In the denominator, one coefficient is negative.
The given function is not suitable for representing the driving point impedance due to following reasons.
· In the numerator, one coefficient is negative.
· The degree of numerator is 4, while that of denominator is 1. Then a difference of 3 exists between the degree of numerator and denominator and is not permitted.
· The numerator gives s2(s2-s+2) that is double zero, at s = 0,
This not permitted.
· The term of the lowest degree in numerator is 2 while that in the denominator is zero.
b. A network function is given below
Obtain the pole-zero diagram (use graph paper).
The Scale factor H = 2.
On factorization of the denominator
. we get (s + 2)(2s + s2 + 2) = (s + 2)(s + 1 – j)(s + 1 + j)
The poles are situated at s = - 2, s = (-1 + j), s = (-1 – j)
The zeroes are situated at s = 0.
The pole-zero diagram is shown below.
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DETAILED
SOLUTIONS D07 JUNE 2004
1.a. C Compensation theorem is applicable to linear and non-linear networks.
b. C Laplace transform of a damped
sine wave e-at sin(qt) u (t) is
c. D A network function is said to have simple pole or simple zero if the poles and zeroes are not repeated.
d. D Symmetrical attenuators have attenuation ‘a’ given by
e. A The velocity factor of a transmission line is governed by the relative permitivity of the dielectric.
h. B If ‘a’ is attenuation in nepers then attenuation in dB = 8.686 a.
g. D For a constant K high pass p-filter, characteristic impedance Zo for f < fc is inductive or capacitive.
h. B A delta connection contains three impedances of 60 W each. The impedances of equivalent star connection will be 20 W each.
PART A
2.a) State the Millman theorem and prove its validity by taking a suitable example.
Millman’s theorem states that if n voltage sources V1, V2, ---,Vn having internal impedances Z1, Z2, ---, Zn, respectively, are connected in parallel, then these sources may be replaced by a single voltage source Vm having internal series impedance Zm where Vm and Zm are given by the equations
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where Y1, Y2, ---, Yn are the admittances corresponding to Z1, Z2, ---, Zn
Proof:
A voltage source V1
with series impedance Z1 can be replaced by a current source V1Y1
with shunt admittance Y1 = 1/Z1
\The network of Fig 2.a.1 can be replaced by its equivalent network as shown in Fig 2.a.2, where
I1= V1Y1, I2= V2Y2, - - - In= VnYn , and Y1 = 1/Z1 , Y2 = 1/Z2 , - - -, Yn = 1/Zn ,
Network of Fig 2.a.2 is equivalent to that of Fig 2.a.3 where by
Im = I1 + I2+ - - - + In and Ym = Y1 + Y2 + - - - + Yn
Reconverting the current source of Fig 2.a.3 into an equivalent voltage source,
The equivalent circuit of 2.a.4 is obtained where
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To find the current in the resistor R3 in the Fig 2.a.5 using Millmans theorem
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The two voltage sources V1 and V2 with series resistances R1 and R2 are combined into one voltage source Vm with series resistances Rm.
According to the Millman theorem,
Hence the current through R3 is given by
2.b For the network of Fig 2.b.1, replace the parallel combination of impedances with a compensation source.
The equivalent impedance of the parallel combination is given by
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The total impedance of the circuit,
The current I,
The compensation source,
The replacement of the parallel combination of impedances with a compensation source VC is shown in the Fig 2.b.2
3.a State the advantages of using Laplace transform in networks. Give the ‘s’ domain representations for resistance, inductance and capacitance.
The advantages of using Laplace transform in networks are:
· The solution is easy and simple.
· It gives the total solution i.e. complementary function and particular integral as a single entity.
· The initial conditions are automatically included as one of the steps rather that at the end in the solution.
· It provides the direct solution of non-homogenous differential equations.
The ‘s’ domain representations for resistance, inductance and
capacitance are
3.b Determine the elements of a ’T’ – section which is equivalent to a p – section.
At any one frequency, a p network can be interchanged to a T network and vice-versa, provided certain relations are maintained.
Let Z1,
Z2, Z3 be the three elements of the T network and ZA,
ZB, ZC be the three elements
of the p network as shown in Fig.3b.1 and Fig 3.b.2
The impedance between the terminal 1 and terminal 3 is
The impedance between the terminal 3 and terminal 4 is
The impedance between the terminal 1 and terminal 2 is
Adding eq.1, eq.2 and subtracting eq.3
Adding eq.2, eq.3 and subtracting eq.1
Adding eq.3, eq.1 and subtracting eq.2
4.a. Discuss the characteristics of a filter.
Ideal filters should have the following characteristics:
· Filters should transmit pass band frequencies without any attenuation.
· Filters should provide infinite attenuation and hence, completely suppress all frequencies in the attenuation band.
· Characteristic impedance of the filter should match with the circuit to which it is connected throughout the pass band, which prevents the reflection loss. Since the power is to be transmitted in the pass band, the characteristic impedance ZO of the filter within the pass band should be real and imaginary outside the pass band (i.e., within stop band) as the power has to be suppressed.
· The transition region between the stop and pass band is very small. The critical frequencies where the filter passes from pass band to a stop band are called the cut off frequencies. The cut off frequency is denoted by the letter fc and is also termed as nominal frequency because the practical filter does not cut off abruptly at that point. Since ZO is real in the pass band and imaginary in an attenuation band, fc is the frequency at which ZO changes from being real to being imaginary. This is the ideal requirement. Practically, it is not possible to realize such an abrupt change of impedance at fc.
4.b.
It is given that
Hence
5.a Find the sinusoidal steady state solution (iss) for a series RL circuit.
The driving voltage is given by
Considering voltage source Vejwt / 2 and applying Kirchoffs voltage law
--- Eq - 2
The steady state current is given by iss1 = A ejwt where A is the undetermined coefficient.
From Eq - 1 and Eq - 2
Considering voltage source Ve-jwt / 2and applying Kirchoffs voltage law
--- Eq - 3
The steady state current is given by iss2 = B e-jwt where B is the undetermined coefficient.
From Eq -1 and Eq - 3
On applying superposition principle, the total steady state current iss is the summation of the currents iss1 and iss2.
\iss = iss1 + iss2
=iss1+ iss2 = A ejwt+ B e-jwt
5.b. Given two capacitors of 1 m F each and coil L of 10 mH, compute the following:
i) Cut-off frequency and characteristic impedance at infinite frequency for a HPF.
ii) Cut-off frequency and characteristic impedance at zero frequency for a LPF.
Draw the constructed sections of filters from these elements.
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i) For a HPF, Z2 = jwL and Z1 = 1/jwC (Fig 5.b.i)
At f = µ, ZOT = RO
The cutoff-frequency fc
ii) For a LPF, Z1 = jwL and Z2 = 1/jwC (Fig 5.b.ii)
At f = 0, ZOT = RO
The cutoff-frequency fc
6. Write short notes on any TWO if the following:-
i) Lumped and distributed elements.
ii) Criteria for stability from pole and zero plot.
iii) Transmission parameters.
i. When the circuit elements are lumped as single parameters like single resistance, inductance, capacitance then they are known as lumped elements.
When the elements are distributed throughout the entire line, which are not physically separable then they are known as distributed elements.
ii. The network function N(s) can be written in the form of ratio of polynomials in (s).
The equation P(s) = 0, has n roots say a1, a2, ---an.
The equation Q(s) = 0, has n roots say b1, b2, ---bn.
Then both P(s) and Q(s) are written as product of linear factors.
Where H = a0/b0 is constant called scale factor.
If the value of s is equal to any of the values a1, a2, ---an, the network function N(s) = 0 as P(s) = 0. Therefore the complex frequencies a1, a2, ---an, are called zeros of the network function.
If the value of s is equal to any of the values b1, b2, ---bn, the network function N(s) = ¥ as Q(s) = 0. Therefore the complex frequencies b1, b2, ---bn, are called poles of the network function. In a pole-zero configuration or pole-zero plot, zeros are conventionally denoted by circle O, while poles are denoted by the cross X as shown in Fig 6.ii.
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If the poles and zeros are not repeated, then the network function has simple poles and zeros. On the other hand, if the poles and zeros are repeated, then the network function has multiple poles and zeros.
Any active network or any general system is said to be stable if the transfer function has its poles confined to the left half of the s-plane and the imaginary axis.
iii. Transmission (ABCD) Parameters: The ABCD parameter equations are given by
V1 = AV2 – B I2 (1)
I1 = CV2 – D I2 (2)
When the output terminal is short circuited, V2 = 0
Where B is the impedance expressed in ohms.
V1 = -BI2
I1 = -DI2
When the output terminal is open circuited, I2 = 0
Where C is the admittance expressed in mhos.
V1 = AV2
I1 = CV2
PART B
7. Define input impedance of transmission line. Derive an expression for the Input impedance of a line and show that Zin for a lossless line is
Input impedance of transmission line is defined as the impedance measured across the input terminals. If Vs is the sending end voltage and Is is the sending end current then Input impedance, Zin is given by
Consider a transmission line of length 
terminated in an
impedance ZR. Let VR be the voltage cross ZR
and IR be the current flowing through it.
The voltage, V and current, I at a point distance x from the sending end of a transmission line are given by
At the sending end, x = 0, V = Vs and I = Is
For a
lossless line a = 0, g = jb
But tanh jbℓ = j tan bℓ
8.a Differentiate between attenuator and amplifier. List the practical applications of attenuators.
An amplifier is used to increase the signal level by a give amount, while an attenuator is used to reduce the signal level by a given amount. An amplifier consists of active elements like transistors. An attenuator is a four terminal resistive network connected between the source and load to provide a desired attenuation of the signal. An attenuator can be either symmetrical or asymmetrical in form. It also can be either a fixed type or a variable type. A fixed attenuator is known as pad.
Applications of Attenuators:
iv) Resistive attenuators are used as volume controls in broadcasting stations.
v) Variable attenuators are used in laboratories, when it is necessary to obtain small value of voltage or current for testing purposes.
vi) Resistive attenuators can also be used for matching between circuits of different resistive impedances when insertion losses can be tolerated.
8.b In a transmission line the VSWR is given as 2.5. The characteristic impedance is 50W and the line is to transmit a power of 25 Watts. Compute the magnitudes of the maximum and minimum voltage and current. Also determine the magnitude of the receiving end voltage when load is (100 – j80) W .
Given the standing wave ratio, S = 2.5
Characteristic impedance, Zo = 50W
Power, P = 25 Watts.
We know that
Since the powers at the sending end and receiving end are the same
9. Compute the values of resistance, inductance, capacitance of the series and shunt elements of a ’T’ network of 10Km line a characteristic impedance of 280Ð-300 and propagation constant of 0.08Ð400 per loop Km at a frequency of (5000 / 2p) Hz. Draw the ‘T’ network from the calculated values.
Given = 10km, Zo = 280Ð-300 , g = 0.08Ð400,
f = 2500/p, w , 2pf = 5000rad/sec
Z1/2
The T – network for the calculated values is shown in Fig. 9
10.a Explain the terms image impedance and Insertion loss.
Image impedance is that impedance, which when connected across the appropriate pair of terminals of the network, the other is presented by the other pair of terminals. If the driving point impedance at the input port with impedance Zi2 is Zi1 and if the driving point impedance at the output port with impedance Zi1 is Zi2, then Zi1 and Zi2 are the image impedances of the two-port network.
Insertion loss: If a network or a line is inserted between a generator and its load, in general, there is a reduction in the power received in the load, and the load current will decrease. The loss produced by the insertion of the network or line is known as the insertion loss. If the load current without the network is I1 and the load current with the network inserted is I2, then the insertion loss is given by
The value of the insertion loss depends on the values of the source and the load.
10.b. Explain the basis for construction of Smith chart. Illustrate as to how it can be used as an admittance chart.
The use of circle diagram is cumbersome, i.e. S and bℓ circles are not concentric, interpolation is difficult and only a limited range impedance values can be obtained in a chart of reasonable size. The resistive component, R and reactive component, X of an impedance are represented in a rectangular form while R and X of an impedance are represented in circular form in the Smith charts. Smith charts can be used as impedance charts and admittance charts.
If the normalized admittance is Y = Y/YO = g – jb
· Any complex admittance can be shown by a single point, (the point of intersection R/YO circle and j X/ZO) on the smith chart. Since the inductive resistance is negative susceptance, it lies in the region below the horizontal axis, and since capacitive reactance is positive susceptance, it lies in the region above the horizontal axis.
· The points of voltage maxima lie in the region 0 to 1 on the horizontal axis, since the conductance is equal to 1/S at such points. The points of voltage minima lie in the region 1 to 0 on the horizontal axis, since the conductance is equal to S at such points.
· The movement in the clockwise corresponds to traveling from the load towards generator and movement in the anti-clockwise corresponds to traveling from the generator towards load.
· Open circuited end will be point A and short circuited end will be B
11.a. What is resonance? Why is it required in certain electronic circuits? Explain in detail.
An a.c. circuit is said to be in resonance when the applied voltage and the circuit current are in phase. Thus at resonance the equivalent complex impedance of the circuit consists of only the resistance, the power factor of the circuit being unity. Resonant circuits are formed by the combination of inductance and capacitance, which may be connected in series or in parallel giving rise to series resonant and parallel resonant circuits, respectively. In other words property of cancellation of reactance when inductive and capacitive reactance are in series, or cancellation of susceptance when in parallel is called resonance. Such cancellation leads to operation of reactive circuit leading only to resistive circuit under unity power factor conditions, or with current and voltage in phase.
There are two types of resonance, series resonance and parallel resonance. Parallel resonance is normally referred to as anti-resonance. In a series resonant circuit, the impedance is purely resistive, the current is maximum at the resonant frequency and the current decreases on both sides of the resonant frequency. In a parallel resonant circuit, the impedance is maximum and purely resistive at the resonant frequency. The impedance decreases on both sides of the resonant frequency. It is easy to select frequencies around the resonant frequency and reject the other frequencies. The resonant circuits are also known as tuned circuits in particular parallel resonant circuit is also known as tank circuit. The resonant circuits or tuned circuits are used in electronic circuits to select a particular radio frequency signal for amplification.
11.b. Design an unbalanced p -attenuator with loss of 20dBs to operate between 200W and 500W. Draw the attenuator.
It is given that,
D = 20 dB.
Converting decibels into nepers, we have
We know that
Similarly
Likewise
And the desired p attenuator is as shown in Fig 11.b
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DETAILED
SOLUTIONS D07 DEC. 2004
1.a. B Which one of the following is a passive element? An Inductor
b. C Millman's theorem yields equivalent voltage or current source.
c. B The Z parameters of the T - network at Fig 1.1 are given by
13, 8, 8, 20
Z11 = Z1 + Z3 = 5 + 8 = 13, Z12 = Z3 = 8, Z21 = Z3 = 8, Z22 = Z2 + Z3 = 12 + 8 = 20
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d. C To a highly inductive circuit, a small capacitance is added in series. The angle between voltage and current will remain nearly the same.
e. A The equivalent inductance of Fig 1.2 at terminals 11’ is equal to
i. D The characteristic impedance Zo of a transmission line given by, ( where R, L, G, C are the unit length parameters
g. B The relation between R1 and R2 for the given symmetrical lattice attenuator shown in Fig 1.3 is
h.
D If Laplace transform of x(t) = X(s), then laplace transform
of x(t – t0) is given by
2a) Given i = 3e-t ´u(t) Amp.
We know that
The charge q that passes through the conductor is 3 Coloumbs.
2b) Given I = 10t Amp.
C = 10mF,
, Where Q is the charge across the condensor
The voltage across the condensor is given
by
When t = 1,
The voltage across the condensor is
V = 5 ´ 105 Volts
The charge across the condensor is given
by
At t = 1, the charge is given by
Q = 5 x 1 = 5 Coulombs
The energy stored in the capacitor is
given by
At t = 1, the energy stored in the capacitor is given by
E = 125 x 104 Joules.
3(a) (i) The Laplace transform of impulse function
\F(s) = 1
(ii) The Laplace transform of unit step function, u(t) is given by
(iii) The Laplace transform of tn eat, n + integer is given by
According to the theorems & replacement of parameters s by (s – b) where b is a constant.
3.b) (i)
According to the partial fraction method
ii)
According to the partial fraction method
4. On Applying Kirchoff’s voltage law,
Applying Laplace transformation to Eq. 1
At time t = 0+, current i(0+) must be the same as at time t = 0 – due to the presence of the inductor L.
\i(0+) = 0
At t = 0+, charge q(0+) across capacitor must be the same as at time t = 0 –
\q(0+) = 0
Substituting the initial conditions in Eq. 2
On inverse laplace transformation
The current i(t) is given by
5.a Given ZS = RS + j XS and ZL = RL + j XL
The power P in the load is IL2RL, where IL is the current flowing in the circuit, which is given by,
\Power to the load is P = IL2RL
for maximum power, we vary XL such that
This implies the reactance of the load impedance is of the opposite sign to that of the source impedance. Under this condition XL + XS = 0
The maximum power is
For maximum power transfer, now let us vary RL such that
\The necessary and sufficient condition for maximum power transfer from a voltage source, with source impedance ZS = RS + j XS to a load ZL = RL + j XL is that the load impedance should be a complex conjugate of that of the source impedance i.e. RL = RS, XL = - XS
The value of the
power transferred will be
5b) To find the short circuit current Isc, first find the equivalent resistance from Fig 5.b.2.
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To find the Norton’s equivalent resistance, short-circuit the voltage source as in Fig 5.b.3.
Therefore the equivalent circuit is shown in Fig. 5.b.4
6a) Bilateral elements: Network elements are said to be bilateral elements if the magnitude of the current remains the same even if the polarity of the applied voltage is changed. The bilateral elements offer the same impedances irrespective of the flow of current.
e.g. Resistors, Capacitors and Inductors.
Unilateral elements: Network elements are said to be unilateral elements if the magnitude of the current passing through and element is affected, when the polarity of the applied voltage is changed. The unilateral elements offer varying impedances with variations in the flow of current.
e.g. Diodes, Transistors.
6b)
We know that
V1 = AV2 – B I2
I1 = CV2 – D I2
On short circuiting port 2 as in Fig 6.b.2, V2 = 0
On open circuiting port 2 as in Fig 6.b.3, I2 = 0
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An element is a bilateral element if the impedance does not change or the magnitude of the current remains the same even if the polarity of the applied EMF if changed. Since the network consists of only resistive elements, the given network is a bilateral network.
7a) For a Two port Network, the Y-Parameter equations are given by
I1 = Y11 V1 + Y12V2 (1)
I2 = Y21 V1 + Y22V2 (2)
The ABCD parameter equations are given by
V1 = AV2 – B I2 (3)
I1 = CV2 – D I2 (4)
From Equations 1 & 2 Y21V1
= I2 – Y22V2
Comparing equations (3) & (5), (4) & (6)
In s domain the equivalent circuit is shown in Fig 7.a.2
When port 2 is short circuited as in Fig 7.a.3, V2 = 0
V1 = I1R
When Port 1 is short circuited as in Fig
7.a.4, V1 = 0
To determine ABCD parameters from Y parameters
7b) Poles and zeros provide useful information about the network functions.
1) In impedance functions, a pole of the function implies a zero current for a finite voltage i.e. on open circuit, while a zero of the function implies no voltage for a finite current i.e. a short circuit.
2) In admittance functions, a pole of the functions implies a zero voltage for a finite current, i.e. short circuit, while a zero of the function implies zero current for a finite value of voltage i.e. an open circuit.
3) The poles determine the variation of the response while zero determine the magnitude of the coefficients in the partial fraction expansion and hence determine the magnitude of the response.
Any active network or any general system is said to be stable if the transfer function has its poles confined to the left half of the s-plane.
A system will be stable if its polynomial roots have negative real parts.
The transform current is given by
From the function it is clear that the function has poles at – 1 and –2 and a zero at the origin. The plot of poles and zeros in shown below:
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(s+1) ad (s+2) are factors in the denominators, the time domain response is given by
To find the constants K1 and K2
From the pole zero plot
M01 = 1 and f01 = 1800
Q21 = 1 and q21 = 00
Where f01 and f02, q21 and q12 are the angles of the lines joining the given pole to other finite zeros and poles.
Where M01 and M 02 are the distances of the same poles from each of the zeros.
Q21 and Q12 are the distances of given poles from each of the other finite poles.
Similarly,
M02 = 2 and f02 = 1800
Q12 = 1 and q12 = 1800
Substituting the values of K1 and K2, the time domain response of the current is given by
8.i)
Consider
(1)
Let the steady state current be given by
(2)
From equation (1) and (2)
(3)
Consider
(4)
Let the steady state current be given by
(5)
from equation (4) and (5)
(6)
The total steady state current is given by
8.ii)
consider
(1)
Let the steady state current be given by
(2)
Now equation (1) and (2)
(3)
consider
(4)
Let the steady state current be given by
(5)
from equation (4) and (5)
(6)
The total steady state current is given by
9. Consider an anti-resonant RLC circuit as shown in Fig. 9
The admittance
At resonance, the susceptance is zero.
If R is negligible
ii) At resonance, the susceptance is zero
We know that
iii)
Half power bandwidth: It
is the band of frequencies, which lie on either side of the resonant frequency
where the impedance falls to1/Ö2 of its value at resonance.
When Qo is very large, RL is
very less then
The impedance of the parallel resonant circuit is
given by
Where Zo is the impedance at parallel resonance
At half power points
iv) The quality factor, Q of the circuit is given by
10.a(i) Consider a transmission line of length (Dx) units.
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Where Z = R + jwL (series impedance per unit length)
and Y = G + jwC (admittance per unit length)
When the line is terminated in ZO,
From the theory of exponential series
We know that
Equating the real parts
Adding equations (3) and (4)
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Subtracting equation (3) from (4)
10b) The standing wage minima points are the voltage minima points. The two consecutive Emin points are separated by 105Cms.
The first voltage minimum is given by
Given Zo = 300W and s = 2.3
The magnitude of the reflection coefficient K is given by
The reflection coefficient K in terms of ZR and Ro is given by
11.i) Consider a constant – K filter, in which the series and shunt impedances, Z1 and Z2 are connected by the relation
Z1 Z2 = R02
Where R0 is a real constant independent of frequency. R0 is often termed as design impedance or nominal impedance of the constant – K filter.
(Z1 = jwL, Z2 = 1/jwC) (Z1 = jwL, Z2 = 1/jwC)
Let Z1 = jwL and Z2 = 1/jwC
At cutoff-frequency fc
Given the values of R0 and wc, using the eq 3a.1 and 3a.2 the values of network elements L and C are given by the equations
ii)
T and p attenuators:
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Attenuators are of two types. They are symmetrical and asymmetrical attenuators. Symmetrical attenuators are placed between two equal impedances of value equal to the characteristic impedance of the attenuator. The characteristic impedance is resistive, Ro since resistive elements are only used in the attenuator. Depending on the placement of the elements, T and p configurations are shown in Fig 11.ii.1 and Fig 11.ii.2. The attenuator is driven at the input port by a voltage source V of internal impedance Ro and it feeds a resistor Ro at the output port.
The T network consists of a divided series arm and one central shunt arm.
The output current I2 is given by
Where N is the attenuation in Nepers.
The p network consists of a series arm and two shunt arms.
The output voltage V2 is given by
Asymmetrical attenuators are placed between two unequal impedances of value equal to the image impedances of the network. The image impedances are resistive, Ri1 and Ri2 since resistive elements are only used in the attenuator. Depending on the placement of the elements, asymmetrical T and p are shown in Fig 11.ii.3 and Fig 11.ii.4
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The attenuation N, in Nepers is given by
iii) Stub matching: When there are no reflected waves, the energy is transmitted efficiently along the transmission line. This occurs only when the terminating impedance is equal to the characteristic impedance of the line, which does not exist practically. Therefore impedance matching is required. If the load impedance is complex, one of the ways of matching is to tune out the reactance and then match it to a quarter wave transformer. The input impedance of open or short circuited lossless line is purely reactive. Such a section is connected across the line at a convenient point and cancels the reactive part of the impedance at this point looking towards the load. Such sections are called impedance matching stubs. The stubs can be of any length but usually it is kept within quarter wavelength so that the stub is practically lossless at high frequencies. A short circuited stub of length less than l/4 offers inductive reactance at the input while an open circuited stub of length less than l/4 offers capacitive reactance at the input. The advantages of stub matching are:
· Length of the line remains unaltered.
· Characteristic impedance of the line remains constant.
· At higher frequencies, the stub can be made adjustable to suit variety of loads and to operate over a wide range of frequencies.
11.(iv) In a hybrid parameter model, the voltage of the input port and the current of the output port are expressed in terms the current of the input port and the voltage of the output port. The equations are given by
V1 = h11I1 + h12V2
I2 = h21I1 + h22V2
When the output terminal is short circuited, V2 = 0
V1 = h11I1
I2 = h21I1
Where h11 is the input impedance expressed in ohms and h21 is the forward current gain.
When the input terminal is open circuited, I1 = 0
V1 = h12V2
I2 = h22V2
Where h12 is the reverse voltage gain and h22 is the output admittance expressed in mhos.
The Z parameter equations are given by
V1 = Z11I1 + Z 12 I 2
V2 = Z 21I1 + Z 22 I 2
The Y parameter equations are given by
I1 = Y11V 1 + Y12V2
I2 = Y21V 1 + Y22V2
The h parameters in terms of Z parameters are given by
Where DZ = Z11Z22 – Z12Z21
The h parameters in terms of Y parameters are given
by
Where DY = Y11Y 22 – Y12Y 21