7.1
The equation
r r
r
r V
dt dx d C dtq I d
I −1− = = 0 at the limit of dx→0 becomes:
dt C dV dx dI
= 0 (7.1.1)
The equation
1 0 Ir =Vr −Vr+
dt dx d L
at the limit of dx→0 becomes:
t L I x V
∂
= ∂
∂
∂
0 (7.1.2)
The derivative of equation (7.1.1) on t gives:
2
2 0 2
t C V t x
I
∂
= ∂
∂
∂
∂
(7.1.3)
The derivative of equation (7.1.2) on x gives:
t x L I x
V
∂
∂
= ∂
∂
∂ 2
2 0 2
(7.1.4)
Equation (7.1.3) and (7.1.4) give:
2 2 0 2 0
2
t C V x L
V
∂
= ∂
∂
∂
The derivative of equation (7.1.1) on x gives:
t x C V x
I
∂
∂
= ∂
∂
∂ 2
2 0 2
(7.1.5)
The derivative of equation (7.1.2) on t gives:
2
2 0 2
t L I t x
V
∂
= ∂
∂
∂
∂
(7.1.6)
Equation (7.1.5) and (7.1.6) give:
2 2 0 2 0
2
t C I x L
I
∂
= ∂
∂
∂
7.2
Fig Q.7.2.1
A pair of parallel wires of circular cross section and radius a are separated at a distance 2d
between their centres.
To find the inductance per unit length we close the circuit by joining the sides of a section of length l.
The self inductance of this circuit is the magnetic flux through the circuit when a current of 1 amp flows around it.
If the current is 1 amp the field outside the wire at a distance r from the centre is μ0I 2πr, where μ0 is the permeability of free space. For a clockwise current in the circuit (Fig Q.7.2.1) both wires contribute to the magnetic flux B which points downwards into the page and the total flux through the circuit is given by:
∫
ad− a = ⎜⎝⎛ a ⎟⎠⎞ d lI rl 2 2 0Idr 0 2 2 ln
2 π
μ π
μ for d >>a
Hence the self inductance per unit length is:
⎟⎠
⎜ ⎞
⎝
= ⎛
a
L 2d
0 ln π μ
To find the capacitance per unit length of such a pair of wires we first find the electrostatic potential at a distance r from a single wire and proceed to find the potential from a pair of wires via the principle of electrostatic images.
If the radius of the wire is a and it carries a charge of λ per unit length then the electrostatic flux E per unit length of the cylindrical surface is: 2πrE(r)=λ ε0 , where ε0 is the permittivity of free space. Thus E(r)=λ 2πε0r for r>a and we have the potential:
I l
a 2
d 2
⎟⎠
⎜ ⎞
⎝
= ⎛ +
−
= r
r b
r ln
constant 2 )
2 ln(
) (
0
0 πε
λ πε
φ λ for φ =0 at r =b.
Fig Q.7.2.2
The conducting wires are now represented in the image system of Fig Q.7.2.2. The equipotential surfaces will be seen to be cylindrical but not coaxial with the wires. Neither the electric field nor the charge density is uniform on the conducting surface.
The surface charge is collapsed onto two line carrying charges ±λ per unit length. The y axis represents an equipotential plane.
The conducting wires, of radius a, are centred a distance d from the origin ο(x= y=0). The distances ± p can be chosen of the line charges so that the conducting surfaces lie on the equipotentials of the image charge. Choosing the potential to be zero at ∞ on the y axis, the potential at point p in the xy plane is given by:
⎟⎟⎠
⎜⎜ ⎞
⎝
= ⎛
⎟⎟⎠
⎜⎜ ⎞
⎝
− ⎛
⎟⎟⎠
⎜⎜ ⎞
⎝
= ⎛ 2
1 2 2 0 2
0 1
0
4 ln ln 1
2 ln 1
2 r
r r
p r πε
λ πε
λ πε
φ λ
In Fig Q.7.2.2:
δ βγ
δ αγ
+
= +
= 2 2
2 1
2 2
r r
where α =(d+ p); β =(d−p); γ =(d+rcosθ) and δ =(r2+p2−d2). If the position of the image charge is such that p2 =(d2−a2), then, at r=a:
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛
−
= +
p d
p
a ln d
) 4 (
πε0
φ λ , independent of θ, and the right-hand conductor is an equipotential of
the image charge. Symmetry requires that the potential at the surface of the other conductor is
a θ
r r r
λ λ +
−
o
p p
d d
p
x y
) φ(a
− and the potential difference between the conductors is:
⎟⎟⎠
Gauss’s theorem applied to one of the equipotentials surrounding each conductor proves that the surface charge on each conductor is equal to the image charge.
The capacitance per unit length is now given by:
⎟⎠
The integral of magnetic energy over the last quarter wavelength is given by:
2
The integral of electric energy over the last quarter wavelength is given by:
( )
The maximum of the magnetic energy is given by:
2
The maximum of the electric energy is given by:
( )
0 02The instantaneous value of the two energies over the last quarter wavelength is given by:
2
For a real transmission line with a propagation constant γ , the forward current wave Ix+ at position x is given by: Therefore the impedance seen from position x is given by:
x
Noting that:
If the transmission line of Problem 7.6 is short-circuited, i.e. ZL =0, The expression of input impedance in Problem 7.6 gives:
l
If the transmission line of Problem 7.6 is open-circuited, i.e. ZL =∞, The expression of input impedance in Problem 7.6 gives:
l
By taking the product of these two impedances we have:
2
Z0
Z
Zsc oc = , i.e. Z0 = ZscZoc
which shows the characteristic impedance of the line can be obtained by measuring the impedances of short-circuited line and open-circuited line separately and then taking the square root of the product of the two values.
7.8
The forward and reflected voltage waves at the end of the line are given by:
ikl l
l V V e
V+ =− − = 0+ −
where V0+ is the forward voltage at the beginning of the line. So the reflected voltage wave at the beginning of the line is given by:
kl
The forward and reflected current waves at the end of the line are given by:
ikl l
l l
ikl ikl
l l
e I Z V Z V I
e I Z e V Z V I
− + +
−
−
− +
− + +
+
=
=
−
=
=
=
=
0 0 0
0 0 0
0
where I0+ is the forward current at the beginning of the line. So the reflected current wave at the beginning of the line is given by:
kl i ikl
l e I e
I
I0− = − − = 0+ −2 Therefore the input impedance of the line is given by:
λ πl C
i L kl iZ kl
e e I
e e V e
I
e V I I
V
Z V ikl ikl
ikl ikl kl
i kl i i
tan2 cos
sin )
(
) (
) 1
(
) 1
(
0 0 0
0 0 2
0
2 0
0 0
0
0 = =
+
= − +
= − +
= + −
+ + −
− + + −
− +
− +
The variation of the ratio Zi L0 C0 with l is shown in the figure below:
7.9
The boundary condition at Z0Zm junction gives:
− +
−
++ 0 = 0 + 0
0 V Vm Vm
V
− +
−
++ 0 = 0 + 0
0 I Im Im
I
where V0+, V0− are the voltages of forward and backward waves on Z0 side of Z0Zm
junction; I0+, I0− are the currents of forward and backward waves on Z0 side of Z0Zm
junction; Vm0+, Vm0− are the voltages of forward and backward waves on Zm side of Z0Zm
…
…
0
0 C
L Zi
0 l
λ 2
λ 4 3λ 4 λ 4
− λ 2
− 4 3λ
−
junction; Im0+, Im0− are the currents of forward and backward waves on Zm side of Z0Zm
junction;
The boundary condition at ZmZL junction gives:
L mL
mL V V
V ++ − =
L mL
mL I I
I ++ − =
where VmL+, VmL− are the voltages of forward and backward waves on Zm side of ZmZL
junction; ImL+, ImL− are the currents of forward and backward waves on Zm side of ZmZL
junction; VL, IL are the voltage and current across the load.
If the length of the matching line is l, we have:
ikl mL
m V e
V 0+ = +
ikl mL
m I e
I 0+ = +
ikl mL
m V e
V 0− = − −
ikl mL
m I e
I 0− = − − In addition, we have the relations:
L L
L Z
V =I
0 0
0 Z
V =I
m mL mL mL
mL m
m m
m Z
I V I
V I
V I
V =− = =− =
−
− +
+
−
− +
+
0 0 0
0
The above conditions yield:
ikl m
mL V e
V + = 0+ −
ikl m
mL I e
I + = 0+ −
+
− +
= − mL
m L
m L
mL V
Z Z
Z V Z
+
By dividing the above equations we have:
kl
Analysis in Problem 7.8 shows the impedance of a short-circuited loss-free line has an impedance given by:
so, if the length of the line is a quarter of one wavelength, we have:
∞
If this line is bridged across another transmission line, due to the infinite impedance, the transmission of fundamental wavelength λ will not be affected. However for the second harmonic wavelength λ 2, the impedance of the bridge line is given by:
0 4 tan
2
tan 2 0
0 = =
= λ π
λ
π iZ
iZ Zi
which shows the bridge line short circuits the second harmonic waves.
7.11
For Z0 to act as a high pass filter with zero attenuation, the frequency
LC 2
2 > 1
ω , where
C L Z0 = .
The exact physical length of Z0 is determined by ω. Choosing the frequency ω1 determines
1 1 =2π λ
k .
For a high frequency load ZL and a loss- free line, we have, for the input impedance:
⎟⎟⎠
⎜⎜ ⎞
⎝
⎛
+
= +
kl iZ kl Z
kl iZ kl Z Z
Z
L L
in cos sin
sin cos
0
0 0
For n even, we have:
1 2 cos
cos2
cos 1
1
1 = λ = π =
λ
π n n
l k
For n odd, we have:
1 2 cos
cos2
cos 1
1
1 = λ = π =−
λ
π n n
l k
The sine terms are zero.
So Zin =ZL for n odd or even, and the high frequency circuits, input and load, are uniquely matched at ω1 when the circuits are tuned to ω1.
7.12
The phase shift per section β should satisfy:
1 2 1 2
1 2 cos
2
2
1 LC
C i
L i Z
Z ω
ω
β = + = + ω = −
i.e.
cos 2 1
2LC β =ω
−
i.e.
2 sin 2
2
2
2 β ω LC
=
For a small β, β ≈sinβ , so the above equation becomes: where the phase velocity is given by v=1 LC and is independent of the frequency.
7.13
The propagation constant γ can be expanded as:
⎟⎟⎠
which is the Q value of this transmission line.
7.14
0 , where K is constant, the characteristic impedance of a lossless line is
given by:
which is a real value.
7.15
Try solution ψ =ψmeγx in wave equation:
0 ) 8 (
2 2 2 2
=
−
∂ +
∂ψ π ψ
V h E
m x
we have:
) 8 (
2 2
2 V E
h
m −
= π γ
For E >V (inside the potential well), the value of γ is given by:
) ( 2 2
V E h m
in =±i π −
γ
So the ψ has a standing wave expression given by:
x E V h m i x E V h m i
Be
Ae ( )
) 2
2 ( − − −
+
= π π
ψ
where A, Bare constants.
For E<V (outside the potential well), the value of γ is given by:
) ( 2 2
E V h m
out =± π −
γ
So the expression of ψ is given by:
x E V h m x
E V
h m Be
Ae ( )
) 2 2 (
−
−
− +
= π π
ψ
where A, Bare constants. i.e. the x dependence of ψ is e±γx, where )
( 2 2
E V
h m −
= π γ
7.16
Form the diffusion equation:
2
1 2
x H t
H
∂
= ∂
∂
∂
μσ
we know the diffusivity is given by: d =1 μσ. The time of decay of the field is approximately given by Einstein’s diffusivity relation:
μσ μσ
2 2 2
1L L
d
t = L = =
where L is the extent of the medium.
For a copper sphere of radius 1m, the time of decay of the field is approximately given by:
By comparing the above derivatives, A.7.17.1 and 2, we can find the solution
)2
satisfies the equation:
2