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Cauer introduced two forms of circuit realization from the network function. These are known as
The network function used for realization is such that of n>m,where n is the degree of numerator (P(s)) polynomial and m is the degree of the denominator polynomial. This leads to a pole at ω=0 producing inductor as the first element. This form of circuit realization produces ladder type of circuit realization with 1^{st} element as an inductor and the parallel element is a capacitor and this way a ladder is formed.
These driving point immitance for Cauer 1st form is:
In the 2nd form of Cauer, arranger the polynomial in ascending order.
when the degree of the lowest order term of denominator polynomial is higher than the degree of the lowest order term numerator polynomial, then the realization produces first element as series capacitor and then a series capacitor.
The driving point immitance according to cauer 1st Form is obtained as:
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Foster has given two type of network synthesis for oneport reactive network
Foster 1^{st} Form
When numerator polynomial is of higher degree than the denominator polynomial
Example: Obtain the first form of the Foster Network for the driving point impedance of LC network given as:
Z(s) = 10(s^{2}+4)(s^{2}+16) / s(s^{2}+9)
Solution:
It is observed from the function Z(s) that the numerator has one higher degree of s than the denominator, hence two poles exists one at ω =0 and another at ω=∞, therefore there will be the presence of first element as capacitor and the last element as inductor
The partial fraction expansion of the network function is:
Z(s) = A_{0}/s + A_{1}/(s+j3) + A_{1}*/(sj_{3}) + Hs
The values of residues is found to be as:
A_{0} at s=0 is = 640/9 = 71.11;
so we find C_{0} = 1/A_{0}
= 1 / 71.11 = 0.0141 Farad
A_{1} at s= j_{3} is= 350/18 = 19.45
Now we know Li and Ci of parallel network is calculated as
C_{1} = 1/(2A_{1})
= 1/(2*19.45)
= 0.0257 Farads
and
L_{1} = 2A_{1}/ω^{2}
= 2*19.45 / 9
= 4.322 Henry
The end element inductor is ‘H’ = 10 henry
Therefore the realization of the impedance function is:
Foster’s 2nd Form
2nd for is for synthesis of admittance function, and is:
As observed:
Given an admittance function:
i. The inductor is represented by the term B_{0}/s and this corresponds to a pole at origin
ii. The capacitor C_{∞ is represented by Hs and corresponds to a pole at infinity.}
iii. The series combination of L and C is determined from 2*B_{1} / (s^{2} + ω^{ 2})
iv. In case there is no pole at ω=0 or at ω=∞ or at both, signifies absence of end elements.
Example:
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Network analysis: You are given a network then you analyse the behaviour (output) by applying certain excitation.
Network Synthesis: You know the output response for certain excitation, then you are required to find appropriate Z(s) or Y(s) that will give that desired behaviour.
Let us consider the network equations in terms of impedance and admittance:
Process of network synthesis: let us first consider figure1(a)
Z(s) = Z_{1}(s) + Z_{2}(s)
Rearranging,
Z2(s) = Z(s) – Z_{1}(s)
That is we find Z_{2}(s) by removing Z_{1}(s) from Z(s).
This removal of Z_{1}(s) from Z(s) can be performed in following different ways:
Procedure of removal of a pole:
Z(s) = Hs + A(s) /B(s) where H is the quotient i.e. H= a_{n1}/b_{n}.
Comparing the polynomial Z(s) and the Z_{2}(s) reveals that:
= Z_{1}(s) + Z_{2}(s)
Z_{1}(s) term has ‘s’ in denominator indicating that removal of a pole at zero means removal of a capacitor (of value k0= a_{0}/b_{1}) from the network. If the original function was an admittance function, then it would mean removal of an inductor (of value b_{1}/a_{0} ) from the circuit.
If the Q(s) has roots of the form (s^{2}+ω_{1}^{2}) then roots will be complex conjugate and the poles (s+jω) and (sjω) at the imaginary axis. Since denominator factor is (s^{2}+ω_{1}^{2}) we have to find the residues.
Z(s) = A/(sjω_{1}) + A*/(s+jω_{1}) (since the residues are complex conjugate A=A* = say K; so, the impedance function:
Z(s) = 2Ks/(s^{2}+ω_{1}^{2}) + Z_{2}(s)
= Z_{1}(s) + Z_{2}(s)
Next we note that
Re[Z1(s)] = Re[2Ks/((jω)^{2}+ω_{1}^{2})]
= Re[2Ks/(ω^{2}+ω_{1}^{2})]
Thus, Re[Z2(jω)] becomes zero or +ve for any value of w since Z(s) is a PRF.
Now Z(s) = 2Ks / (s^{2}+ω_{1}^{2})
= 1/ {(s^{2}/2Ks) + w_{1}^{2}/2ks)}
= 1/(Y_{as} + Y_{bs})
Here Yas = s/2K represent a capacitor and Ybs = ω_{1}^{2}/2ks represent an inductor connected in parallel to form Z_{1}(s)
Similarly, if we write the Y(s) as the riginal function, then we can prove that Y_{1}(s) = 1/{(Z_{a}(s) + Z_{b}(s)}
Z_{a}(s)=s/2K representing an inductor of value 1/2K henry and a Z_{b}(s)= ω_{1}^{2}/2ks representing a capacitor of value ω_{1}^{2}/2K and both Za(s) and Zb(s) connected in series.
Example: Realize the network whose impedance is given as: Z_{1}(s) = (s^{4} + 10s^{2} + 7) / (s^{3} + 2s)
Solution:
Step1:
Z_{1}(s) = ((jω)^{4} + 10(jω)^{2} + 7) / ((jω)^{3} + 2(jω))
= (ω^{4} 10ω^{2} + 7) / (jω^{3} + 2jω)
Function reveal that for s=jω, Z_{1}(s) is not real, therefore it is inferred that Z_{1}(s) does not have any resistance element.
Step2: Since an inductor is present, find its value by long division as:
s^{3} + 2s ) s^{4} + 10s^{2} + 7 ( s
(s^{4} + 2s^{2} )
————–
8s^{2} + 7
Therefore Z1(s) becomes:
Z1(s) = s + (8s^{2} + 7) / s^{3} + 2s
= Z_{2}(s) + Z_{3}(s) ; where Z_{2}(s) = s representing an inductor of 1 henry
Step3
Z_{3}(s) = (8s^{2} + 7) / s^{3} + 2s
Inverting we get Y_{3}(s) = 1/Z_{3}(s)
= (s^{3} + 2s) / (8s^{2} + 7)
Again as we that in this admittance function, degree of numerator is one higher than the degree of the numerator, therefore it represents a capacitor in parallel
Step4:
Since the impedance Y_{3}(s) represent a capacitor, its value can be found by long division as:
8s^{2} + 7 ) s^{3} + 2s ( s/8
( s^{3} + 7s/8)
—————————
9s/8
Therefore we can write Y_{3}(s) as
Y_{3}(s) = s/8 + (9s/8) / (8s^{2} + 7)
= Y4(s)+ Y5(s)
The admittance Y_{4}(s) = s/8 represent a capacitor of 1/8 Farad
Step5
Convert the admittance Y_{5}(s) in to impedance Z_{5}(s) as
Z_{5}(s) = 1/Y_{5}(s)
= 8(8s^{2} + 7)/9s
= 64s^{2}/9s + 56/9s
= 64s/9 + 56/9s
= Z_{6}(s) + Z_{7}(s)
Where Z_{6}(s) = 64s/9 represent an inductor of value 64/9 Henry
Step6:
Invert Z_{7}(s) to convert it into admittance Y_{7}(s) as:
Y_{7}(s) = 1/ Z_{7}(s)
=9s/56
Y_{7}(s) represent a capacitive impedance of value 9/56 Farad
]]>We know that the driving point impedance (Z(s) and driving point admittance (Y(s))are of the following type:
N(s) = P(s) / Q(s)
The function is prf if:
Properties of prf
Given a transfer function N(s) = P(s)/Q(s)
How to test whether the polynomial is Hurwitz or not
a) All quotients must be +ve
b) The process should not abruptly end because of some common factor. If does so, then do D(s) /D’(s) where D(s) is the common polynomial which was the last divisor that resulted in zero remainder
RouthHurwitz Criterion of stability of network function
An example:
Q(s) = b_{0}s^{5} + b_{1}s^{4} + b_{2}s^{3} + b_{3}s^{2} + b_{4}s^{1} + b_{5}
Even coefficient terms b_{0},b_{2}, b_{4}s,
Odd coefficient terms b_{1},b_{3}, b_{5},
Writing the coefficients in an array form for m=5, the array will contain m+1 =6 elements
s  Coefficients 
S^{5}  b_{0} b_{2} b_{4} 
S^{4}  b_{1} b_{3} b_{5} 
S^{3}  c_{1} c_{2} 
S^{2}  d_{1} d_{2} 
S^{1}  e_{1} 
S^{0}  f_{1} 
Find other coefficients c_{1},c_{2},d_{1},d_{2},e_{1},f_{1} using
c1 = – (b_{0}b_{3}b_{1}b_{2})
c2 = – (b_{0}b_{5} – b_{1}b_{4})
d1 = – (b_{1}c_{3} – c_{1}b_{3})
d2 = – (b_{1}*0 – c_{1}b_{5})
e1 = (c_{1}d_{2} – d_{1}c_{2})
f1 = – (d_{1}*0 – e_{1}d_{2})
Note the pattern carefully
According to the Routh Hurwitz stability criterion, the system is said to be stable , if and only if, there are no change in sign of the first column of the array.
This gives the roots with –ve real parts and hence gives the condition for stability.
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N(s) = P(s) / Q(s)
Above equation consist of factors of general form where p_{r} and p_{n} are complex frequencies. Their difference is also a complex frequency which can be written as:
Where M_{nr} is the magnitude of the phasor and Φ_{nr} is the phase angle of the same phasor
Thus
All Ms and phase angles can be calculated using the pole zero plot on splane
Procedure for calculation of M and Φ
Given a polynomial N(s) = P(s) / Q(s)
By factorising the numerator and denominator polynomials, we can easily show that the polynomial becomes zero when the ‘s’ terms in the numerator polynomial have the values s=0, z1,z2…….zn, thus the roots of numerator define the zeros. Also the roots of the denominator polynomial define the poles of the function. Zeros are marked by ‘O’ on the splane while a pole is indicated by a X
The splane is shown in figure below
Unit step function F(s) =1/s  F(cos(w)) = s/(s^{2}+ω^{2})  F(sin(w)) = w/(s^{2}+ω^{2}) 
Function has a only one pole at s=0 
The function has a zero at s=0 and two poles at +/ jω 
There are two poles at +/jω 
In all the three functions we see that the poles and zeros are on the imaginary axis only 
Damped cosine function
F(s) = s+σ_{0}/{(s+ σ_{0})2 +ω_{0}^{2}}

Exponentials f1(t) =Me^{σ1t} and f2(t) = Me^{σ1t}
F1(s) =1/(s+ σ_{1}) and F2(s) = 1/(s+ σ_{2}) 
Sinusoidal
F1(s) = ω_{1}/(s^{2}+ω_{1}^{2}) F1(s) = ω_{2}/(s^{2}+ω_{2}^{2}) 
Again we observe that the poles and zero in all these three cases are either on imaginary axis or on the left hand splane 
A system which has poles on the right hand of the splane is basically unstable because such system gives rise to exponentially increasing transient response.
Ex1: check for the stability using polezero concept
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Network: A network is an interconnection of passive elements and dependent sources. There are no active or independent sources inside the box. The network is treated to be part of the square box shown in figure.
Terminal: A terminal is the end point of the conductor which is connected to the node of the network which is brought out of the designated square box representing the network
Port: port is a pair of terminal which is used to connect to the sources of energy i.e. independent sources and or the excitation.
A network may be oneport in which there is only one driving force or multiport network to which multiple sources of energy are connected. We consider a 2port network for our study.
In a multiport network of specifically in our study a 2port network, one port may be connected to driving force and other port may be used for measurement, for connecting to load or for interconnecting to the input of some other networks.
Driving Point: The ports which are used for connecting sources or load and for interconnecting networks is called driving points of a network.
Network functions:
Transform impedance: It is written as Z(s) = V(s) /I(s) is defined as the ratio of the voltage transform and the current transform at the same port. They are also called as driving point impedance.
Transform admittance: It is written as Y(s) and equals the inverse of transform impedance i.e. Y(s) = 1/Z(s) is defined as the ration of current transform to the voltage transform at the same port.
Transform /Driving Point  Port1 voltage and current  Port2 voltage and current 
Impedance  Z_{11} = V_{1}(s) / I_{1}(s)  Z_{22} = V_{2}(s) / I_{2}(s) 
Admittance  Y_{11} = I_{1}/V_{1}  Y_{22} = I_{2 }/V_{2} 
Transfer Function: Transfer function is defined as the ratio of the voltage or current transform at the output to the transform of voltage or current at input port:
Transfer function ( Z, Y, G, α)


Transfer impedance (Z_{21})  Ratio of Voltage transform at output port (V_{2}(s) ) to the transform of current (I_{1}(s))at the input port  Z_{21} = V_{2}(s) / I_{1}(s) 
Transfer admittance (Y_{21 })  Ratio of current transform (I_{2}(s)) at output port to the transform of Voltage (V_{1}(s) ) at the input port  Y_{21} = I_{2}(s) / V_{1}(s) 
Voltage Transfer Ratio (G_{21})  Ratio of Voltage transform at output port (V_{2}(s) ) to the voltage transform (V_{1}(s)) at the input port  G_{21} = V_{2}(s) / V_{1}(s) 
Current Transfer Ratio (α_{21})  Ratio of Current transform at output port (I_{2}(s) ) to the current transform (I_{1}(s)) at the input port  α_{21} = I_{2}(s) / I_{1}(s) 
When solving for the transfer function, the degree of the polynomial may be found to be greater than the degree of the denominator polynomial, in which case we can reduce by simple division to make the degree of the numerator less than or equal to degree of denominator polynomial.
For determining the above quantities, we should be well conversant with the series and parallel combination of the components. Using the transform of the R, L and C components it is convenient and easy to determine as the transform allow us find these quantities as if we are calculating the values for series and parallel combination of resistance values. It should be clear from the following examples.
Example1: For the following oneport network find the transform impedance and admittance
Solution:
In figure(a) applying transform to the components R_{1} –> R_{1} L_{1} –> sL_{1} The network is a series connection of R1 and L1. The driving point impedance of the network is: Z(s) = R_{1} + sL_{1} The driving point admittance is: Y(s) = 1/Z(s) = 1/( R_{1} + sL_{1}) 
Solution:
From figure(b), applying the transform to the network components. R2 –> R2 L2 –> sL2 Impedance of parallel network = (Z_{A}Z_{B})/(Z_{A} + Z_{B}); therefore we get the driving point impedance of the parallel network : = (R_{2}.sL_{2}) / (R_{2} + sL_{2}) The driving point admittance = 1 / Z = (R_{2} + sL_{2}) / (R_{2}.sL_{2}) 
Practice Problem:
Examples of determining the driving point impedance, driving point admittance and the transfer function of a 2port network.
Ex2 Determine the transfer impedance of the network
Solution: The transfer impedance of the network is V_{2}(s) / V_{1}(s)
Writing the network components in the transform form R_{1} –> R_{1}; R_{2} –> R_{2}; and L –> sL; V_{1} –> V_{1}(s) and V_{2} –> V_{2}(s) and I_{1} –> I_{1}(s)
We can write down the KVL equation with the port2 as open circuited. With port2 open the current I2=0. Therefore the two loop equations are:
V_{1}(s) = I_{1}(s)(R_{1} + R_{2})
V_{2}(s) = I_{1}(s)R_{2}
The transfer impedance
G_{21} = V_{2}(s) / V_{1}(s)
= I_{1}(s)R_{2 }/ I_{1}(s)(R_{1} + R_{2})
= R_{2 }/ (R_{1} + R_{2})
The transform / driving point impedance
Z_{11} = V_{1}(s) / I_{1}(s)
V_{1}(s) = I_{1}(s)(R_{1} + R_{2})
Z_{11 } = V_{1}(s) / I_{1}(s) = (R_{1} + R_{2})
Ex3: determine the transfer functions and the driving point impedance for the following network
Solution:
Writing the network components in transform form
L_{1 }–> sL_{1}; L_{2 }–> sL_{2}; C_{1} — > 1/sC_{1}; V_{1 }–> V_{1}(s);
V_{2} –> V_{2}(s); I_{1 } –> I_{1}(s)
With output port open (I_{2}=0) the loop equations gives:
V_{1}(s) = ( sL_{1} + R_{1} +1/sC_{1})I_{1}(s)
= {(s^{2}L_{1}C_{1} + sR_{1}C_{1} +1)/sC_{1}} I_{1}(s)
V2(s) = (R_{1}+1/sC_{1})I_{1}
_{ }={(1+sR_{1}C_{1})/sC_{1}}I_{1}
The transfer function G_{21}
= V2(s) /V1(s)
=(1+sR_{1}C_{1})/sC_{1}} / {(s^{2}L_{1}C_{1} + sR_{1}C_{1} +1)/sC_{1}}
=(1+sR_{1}C_{1}) / (s^{2}L_{1}C_{1} + sR_{1}C_{1} +1)
Z_{11} = V_{1}(s) /I_{1}(s)
= (s^{2}L_{1}C_{1} + sR_{1}C_{1} +1)/sC_{1}
Example4: Determine the driving point impedance Z_{11} and the transfer ratio G_{21 for the following network.}
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From the generalized ABCD parameter equation
V1=AV2 – BI2
I1=CV2 – DI2
V_{1}= AV_{2}; I_{1}= CV_{2}
Z_{1O} = V_{1}/I_{1}= A/C ——————————(i)
Applying the short circuit at port2; V_{2}=0, we get
V_{1}= – BI_{2}
I_{1}=DI_{2}
Or
Z_{1S} = V_{1}/I_{1} = B/D ————————–(ii)
Now looking from port and applying:
Open circuit port1, I_{1}=0
0 = CV_{2} – DI_{2}
CV_{2}= DI_{2}
Z_{2O} = V_{2}/I_{2} = D/C ————————–(iii)
Now short circuiting the port1, V_{1}=0, we get
0 = AV_{2} – BI_{2}
AV_{2} = BI_{2}
Z_{2s} = V_{2}/I_{2} = B/A —————————(iv)
Ratio of open circuit to short circuit ratio
Z1O/ Z1S = (B/D)(C/A) ————————–(v)
To find ABCD parameters
Z_{2O }– Z_{2S} = D/C – B/A = (AD –BC)/CA
But for the two network to be reciprocal, the condition is ADBC =1
Therefore:
Z_{2O }– Z_{2S} = 1/CA ———(vi)
CA = 1/ (Z_{2O }– Z_{2S} )
From eqn(i) Z_{1O} = A/C
Multiplying eqn(i) and eqn(v)
CA(A/C) = 1/ (Z_{2O }– Z_{2S} )* Z_{1O}
A^{2} = Z_{1O} / (Z_{2O }– Z_{2S} )
Or
A= √( Z_{1O} / (Z_{2O }– Z_{2S} ) ——————–(vii)
From eqn(i) i.e. Z_{1O} = A/C C = A/ Z_{1O}
Therefore from eqn (vii):
C = √( Z_{1O} / (Z_{2O }– Z_{2S} ) /Z_{1O}
C = √( 1/ Z_{1O} (Z_{2O }– Z_{2S} ) ———–(viii)
From short circuit equations (iv) Z_{2s} = B/A
B = AZ_{2s}
= √( Z_{1O} / (Z_{2O }– Z_{2S} )* Z_{2s}
From short circuit equations (ii) Z_{1s} = B/D
D = B/ Z_{1s}
= √( Z_{1O} / (Z_{2O }– Z_{2S} )* Z_{2s}/ Z_{1s}
= A* Z_{2s}/ Z_{1s }——————————–(ix)
= Z_{20}*√( 1/ Z_{1O} (Z_{2O }– Z_{2S} )——————(x)
Summery, the ABCD parameters in terms of open and short circuit impedance:
A = A= √( Z_{1O} / (Z_{2O }– Z_{2S} ) B = AZ_{2s} = √( Z_{1O} / (Z_{2O }– Z_{2S} )* Z_{2s}
C = √( 1/ Z_{1O} (Z_{2O }– Z_{2S} ) D = A* Z_{2s}/ Z_{1s }= Z_{20}*√( 1/ Z_{1O} (Z_{2O }– Z_{2S} )
Image Impedance in terms of ABCD parameters
In a two port network, if the impedance at input port with ouptput port impedance as Zi2 is Zi1 and simultaneously, the output impedance with input impedance being Zi1 is Zi2, then the impedances Zi1 and Zi2 are called as the image impedances.
Let us again write the general ABCD parameter equations
V_{1}= AV_{2} – BI_{2}
I_{1}= CV_{2} – DI_{2}
V_{1}/I_{1} = (AV_{2}BI_{2}) / (CV_{2} – DI_{2})
But V_{2}/(I_{2}) = Z_{i2} or V_{2} = Z_{i2}I_{2} —–(i)
Therefore:
Z_{i1} = V_{1}/I_{1} = (AZ_{i2} – BI_{2}) / (Z_{i2}C – DI_{2})
Or
Z_{i1} = (AZ_{i2} – B) / (Z_{i2}C – D) —————– (ii)
Looking inwards from port1 Z_{i1} = V_{1}/I_{1} Z_{i2} = V_{2}/(I_{2})
Looking inwrds from port2 Z_{i1} = V_{1}/(I_{1}) Z_{i2} = V_{2}/I_{2}
Now the general equations:
V_{1} = AV_{2} – BI_{2} and I_{1}= CV_{2} – DI_{2}
AV_{2} = V_{1} + BI_{2} and CV_{2} = I_{1} + DI_{2}
Rewriting by balancing, and adding
AV_{2} = Z_{i1}I_{1} + BI_{2}
Z_{i1}CV_{2} = Z_{i1}I_{1} + Z_{i1}DI_{2}
———————————————–
V_{2}(A+Z_{i1}C) = (B + Z_{i1}D)I_{2}
Or
Z_{i2} = V_{2}/I_{2} = (B + Z_{i1}D) / (A+Z_{i1}C) ————(iii)
Substituting the value of Zi2 from eqn(iii) in eqn(ii) and simplifying we get
Z_{i1} = (A((B + Z_{i1}D) / (A+Z_{i1}C)) – B) / (((B + Z_{i1}D) / (A+Z_{i1}C))C – D) —————– (ii)
or CDZ_{i1}^{2} = AB
Z_{i1} = √((AB) / (CD) ————————————(iv)
Similarily we can substitute Z_{i1} from eqn(ii) in eqn (iii) for Z_{i2} and get
Z_{i2} = √(BD)/ (AC) ————————————(v)
Image Transfer Parameters
Again looking at the general equation of ABCD parameter
I_{2} = – V_{2}/Z_{i2}
V_{1}= AV_{2} – BI_{2}
I_{1}= CV_{2} – DI_{2}
V_{1} = AV_{2} + BV_{2}/Z_{i2}
= (A + B/Z_{i2}) V_{2}
V_{1}/V_{2} = ( A + B√(AC/BD) because Zi2 = √(BD)/ (AC)
= ( A + √(ACBBD/BDD)
= ( A + √(ABCD)/D ) ———————————(vi)
Also Converting I_{1} equation only in terms of I_{2}
I_{1} = CV_{2} – DI_{2}
= Z_{i2}CI_{2} – DI_{2}
= ((√(BD)/ (AC) )C + D)I_{2}
= – (D + √(ABCCD)/ (AAC)I_{2}
= – (D + √(ABCD)/ A)I_{2}
I_{1}/I_{2} = (D + √(ABCD)/ A) ——————————–(vii)
Multiplying eqn vi and eqn vii, to get
V_{1}/V_{2}*I_{1}/I_{2} = (AD + √ABCD)^{2} / AD^{2}
= (√(AD) + √(BC))^{2}
√((V_{1}/V_{2})* (I_{1}/I_{2}) = √(AD) + √(BC) from reciprocity condition AD – BC = 1 or BC = AD1
√((V_{1}/V_{2})* (I_{1}/I_{2}) = √(AD) + √(AD – 1)
Now
√(AD) = coshθ θ = cosh^{1}√(AD)
√(BC) = √(AD – 1)=sinh θ
cosh θ + sinhθ = e^{θ} = √((V_{1}/V_{2})* (I_{1}/I_{2})
taking antilog
θ = log √((V_{1}/V_{2})* (I_{1}/I_{2})
= log(√((Z_{0}I_{1}/Z_{0}I_{2})* (I_{1}/I_{2})
= log (I_{1}/I_{2})
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