DIT, Kevin St.
Electric circuits
Waed 3
Chapter 10
Transmission Lines
A transmission line consists of two conductors. Each conductor has a constant
cross -section along its length. Examples of transmission lines are: telephone
lines, guitar lead, lead to a TV, power lead etc. We may predict the signal
behaviour on a transmission line by analysing the voltage and current signals
along the line.
I(z) - I(z+δz) I(z) - I(z+δz)
Figure 10.55 Cascaded network model of a Transmission line.
Figure 10.55 shows a cascaded network representation of discrete parameters for a
short length of line. In actual transmission lines, these parameters are distributed
evenly along its length. The model is accurate for a short length of line. The primary
parameters, defined per metre of line, are:
R: The loop resistance of both conductors.
G: The two conductors, separated by a dielectric material, has a leakage current
flowing across the dielectric between the two conductors. The insulating
properties of the dielectric determine the magnitude of the leakage current.
C: The capacitance formed between the two conductors and the dielectric.
L: The cross-sectional area of the conductor determines the line inductance.
For a given line length, these values are multiplied by the line length to find the total
resistance, inductance, conductance and capacitance. The continuous distribution of
these parameters is modelled as a cascaded network of elements, each element
having a very short length, δz.
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DIT, Kevin St.
Electric circuits
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Figure 10.56: Equivalent circuit for element of length δz.
The shunt current is:
I(z) - I(z + δ z) = Gδ zV(z) + Cδ z
∂
V(z) (10.1)
∂t
The series element voltage drop is:
V(z) - V(z + δ z) = Rδ zI(z + δ z) + Lδ z
∂
I(z + δ z) (10.2)
∂t
Each impedance in (10.1) and (10.2), is multiplied by δz to determine the actual
impedance for the small section. For a small δ z,
I(z + δ z) ≅ I(z) +
∂
I(z)δ z
∂z
(10.3)
From the example below, at some instant in time, the variation of I(z) with z is
shown. At point 'z' the slope of the line is:
I ( z) =
I(z + δz) - I(z)
δz
Using equation 10.3, (10.1):
∂
∂

I(z) = -  G + C V(z)
∂z
∂t 

(10.4)
And (10.2) is:
∂
∂

V (z ) = -  R + L  I(z + δz)
∂z
∂t 

∂
∂
V(z) = - ( R + L )
∂z
∂t
∂


 I ( z ) + ∂z I(z)δz 
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Assuming
Electric circuits
Waed 3
∂
I(z)δz is small compared with the other terms, then we can write:
∂z
∂
∂

(10.5)
V(z) = -  R + L  [I(z)]
∂z
∂t 

Apply a sinusoidal signal to the transmission line input and measure the response.
j t
j t
Defining V(z) = Ve ω and I(z) = Ie ω i.e. V(z) and I(z) are signals each represented
as a rotating phasor with an angular frequency of ω radians/sec. The magnitude of
each phasor V and I, vary with time and distance z. V and I represent the peak values
of the time varying signals.
∂
∂z
I(z) =
d
I
dz
And
∂
d
j t
j t
V(z) =
Ve ω = jωVe ω = jωV(z)
∂t
dt
We can separate the time and distance variations in the signals and replace partial
differentials with total differentials. Rewriting (10.4) as:
dI
= - (G + jωC )V
dz
(10.6)
dV
= -(R + jωL)I
dz
(10.7)
Similarly
Differentiating (10.7) gives:
d2 V
dz
And substituting for
2
= - (R + jωL)
dI
dz
dI
from (10.6):
dz
d2 V
dz
2
= - (R + jωL)(G + jωC)V
Expressed more conveniently as
d2 V
dz
2
=
2
γ V2
(10.8)
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Electric circuits
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The propagation constant, gamma, is:
2
γ = (R + jωL)(G + jωC)
Or
γ=
(R + jωL)(G + jωC) = α + jβ
Alpha is the attenuation coefficient and beta is the phase change coefficient. (10.8),
is a standard form of differential equation, and has a solution:
V = V1 e
-γ z
+ V2 e γ
z
V1 and V2, are determined by the initial conditions on the line. By similar analysis,
the current relationship to z on the line is:
I = I1 e
-γ z
Then
V = V1 e
+ I2 e γ
-(α + jβ) z
z
+ V2 e α
( + j β) z
V = V1 e α e β + V2 eα e β
The voltage on the line at any point z from the sending end is the sum of two
components:
- z
The forward or incident wave V1 e
-j z
z j z
-α z - jβ z
e
The applied voltage at z = 0 is attenuated as it travels down the line (e α reduces
-j z
with increasing z). The term e β has a magnitude of unity, which does not affect
the amplitude of the signal down the line but causes a phase shift in the transmitted
signal, which increases with z.
- z
The reverse or backward V2 e α e
z jβ z
This component increases in amplitude with increasing z. However, the amplitude of
the signal must reduce with distance z. z must be decreasing for this component in
order for its level to be reducing as it travels along the line. This component must be
travelling in the opposite direction to the forward wave. It is the backward or
reflected wave produced by a mismatch between the transmission line and the load.
The above equations include the terms α and β.
α: Attenuation coefficient: Both forward and reflected waves are attenuated
exponentially at a rate α with the distance travelled.
β: Phase constant: This factor produces a phase shift along the line because the
wave requires finite time to travel down the line. Consider the wave shown in figure
3.
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Electric circuits
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Figure 10.57: No reflected signal for a matched load i.e. ZL = Z0.
.
Assume the wave is travelling with a velocity v and angular frequency ω. The
distribution of voltage over a distance z = λ, corresponds to one full cycle and the
phase shift is therefore 2π radians.
Hence
βλ = 2π
Or
β = 2π/λ
Now v = fλ where f is the frequency of operation. Hence
v= f
2π
ω
=
m/s
β
β
Example 1
A 1 MHz signal is applied to the input of a transmission line of β = 0.2π radians/m.
6.
6
The velocity of propagation of the wave is 1.10 2π /0.2π = 10.10 m/s. The
wavelength of the signal is λ = 2π/β = 2π/0.2π = 10 metres. The line causes a phase
shift of 0.2π radians/m, taking 10 metres for the signal to be phase shifted by 2π
radians.
Example 2
A 1 kHz sine wave is applied to the input of a 40 m length of loss-less transmission
line. The phase constant β = π/10 radians/sec.
λ = 2π/β = 2π/(π/10) = 20 m.
Propagation velocity v = fλ = 1×20 = 20 m/s. Wave propagates 5 metres in 0.25
seconds.
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Electric circuits
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The above wave propagates at 20 m/s and the wavelength is also 20 metres. For a
transmission line with a given propagation velocity, the higher the input frequency
the lower the wavelength. More cycles of the input signal is seen in the same length
of line than would be seen for an input signal of lower frequency.
Characteristic Impedance:
The total voltage is
V = V1e
-γz
+ V2e γ
z
Thus
dV
z
- z
= γ(V2e γ - V1e γ )
dz
From before
dV
= -(R + jωL)I
dz
Therefore
I=
-1
dV
γ
- z
z
[V1e γ -V2 e γ ]
=
R + jωL dz
R + jω L
But
γ = (R + jωL)(G + jωC)
Substituting for γ gives
 G + jωC 
- z
z
I= 
[V1e γ -V2e γ ]

 R + jω L 
(10.9)
1/2
Zo =[(R + jωL)/(G + jωC)] =, is the characteristic impedance of the line. Its value
at any frequency is determined by the line parameters R, L, G and C.
Z0 =
R + jω L
G + jω C
(10.10)
A transmission line is matched, if terminated in an impedance or load equal to
the characteristic impedance of the line. When the line is matched, the forward
wave is totally absorbed by the load and there is no reflected wave. (V2 = 0). Any
value of load, not equal to the characteristic impedance of the line, will produce
reflections back down the line and interacts with the forward wave. The amount of
reflection depends on the difference between the load impedance and the
characteristic impedance of the line. Consider the case of the load impedance less
than the characteristic impedance shown in figure 10.58.
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Electric circuits
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Figure 10.58: Condition where ZL < Z0.
Consider the case shown in figure 10.59, where the load impedance is greater than
the characteristic impedance.
Figure 10.59: Condition where ZL > Z0.
The amount of reflection produced by the mismatched load, is defined in terms of
the voltage reflection coefficient ρ. This is defined as the ratio of the reflected
voltage to the incident voltage. The total voltage at the load is:
V = V1 e
V1 e γ and V2 e
Thus
- l
γl
-γ l
+ V2 eγ
l
are the incident and reflected voltages respectively at the load.
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Electric circuits
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-γ l
-2γl
γl
ρ = V2 e /V1 e = (V2 /V1)e
(j-2 l)
Expressed in polar form as ρ = ρ e γ is the angle of the reflection coefficient.
Using 10.10 the load current is
IL =
V1 -γ l V2
l
e eγ
ZL
ZL
The load impedance is ZL = VL/IL
1 + (V2 /V1 )e 2γl 
ZL = 

IL


Using 10.9
1 + (V2 /V1 )e 2γl 
Z L = Zo 
2 γl 
 1 - (V2 /V1 )e 
1 + ρ 
Z L = Z 0 

1 - ρ 
(10.11a)
(10.12)
Consider three load values:
1. Short circuit load, ZL = 0.
ρ = -1 so ρ = 1 and θ = π.
2. Open circuit load, ZL = ∞.
ρ = 1 so ρ = 1 and θ = 0.
3. ZL = 150 + j100 Ω
o
ρ = (100 + j100)/(200 + j100) = 0.6 + j0.2 so ρ = 0.63 and θ = 18.43 .
Z − Zo
ρ= L
(10.13)
Z L + Zo
Sending End Impedance:
It is of interest to determine the input impedance, which the transmission line and
load offers, at a point on the line. The impedance looking into the line, from the
source, is the sending end impedance. The impedance at some point A, a distance z
from the source is
 -γz V2 γz 
γz
-γ z
 e +V e 
V 1 e + V 2 e 
VA
1
ZA =
= Zo
= Zo 

γz 
-γ z
V
e
e
γ
γz 
z
2
V
V
IA
e 2
 1

e


V1
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 -γz V2 γz 
 e +V e 
V 1 e -γz + V 2 e γz 
VA
1

= Zo 
= Zo 
-γz
γz 
V
e
e


γ
z
γ
z
IA
V2
2
 V1

 e -V e 
1


2γ l
Using ρ = (V2/V1)e we can rewrite the above as
 e - γ ( l − z ) + ρe γ ( l − z ) 
 e-γz + ρe-2γl eγz 
=
Z
ZA = Zo  -γz
o
 -γ ( l − z )

- 2 γl γz 
- ρe γ ( l − z ) 
 e - ρe e 
e
But x = l – z
 e γx + ρ e - γx 
ZA = Zo  xγ
-γx 
e − ρ e 
Substituting for ρ = (ZL - Zo)/(ZL + Zo)
 Z L [e γx + e -γx ] + Z o [e γx + e -γx ] 
 ( Z L + Z o )eγx + ( Z L - Z o )e-γx 
Z A = Zo 
Z A = Zo 
γx
- γx 
γx
γx
-γx
-γx 
 ( Z L + Z o )e - ( Z L - Z o )e 
Zo [ e - e ] + ZL [ e + e ] 
Replacing exponential terms with hyperbolic functions:
 Z coshγx + Z o sinh γx 
 Z L + Z o tanh γx 
Z A = Zo  L
 = Zo 

 Z o + Z L tanh γx 
 Z o sinh γx + Z L coshγx 
The sending-end impedance is obtained by setting x = l
 Z L + Z o tanh γl 
Z in = Z o 

 Z L tanh γl + Z o 
Or
Z in
Zo
=
(Z L /Z o ) + tanh γl
1+ ( Z L /Z o )tanh γl
It is useful if the input is expressed in terms of the characteristic impedance of the
line (i.e. normalised form). Normalised input impedance ZinN = Zin/Zo and
ZL
= Z LN
Zo
Thus
Z L N + tanhγl
ZinN =
1+ Z L N tanh γl
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Loss-less Lines:
In many cases it may be assumed that the line is loss-less. The reduction in accuracy
is not significant if the line is short and operating at high frequencies. We had γ = α
+ jβ. For a loss-less line α = 0 so γ = jβ. Using the relationship tanh(jθ) = jtan(θ)
we can express the sending end impedance as
 Z L N + jZ o tan βl 
Zin = Zo 

 Z o + jZ L N tan βl 
(10.15)
And in normalised form (divide by Zo)
Z in  Z L N + jtanβl 
=

Z 0 1 + jZ L N tanβl 
(10.16)
Stub matching:
Short lengths of line, called stubs, are used to adjust or match the impedance along a
line. Stubs are terminated in a short circuit or open circuit. Assuming a loss-less
stub, we can use (10.16) to examine the input impedance of the stub for each case. If
the stub is terminated in a short circuit, then ZL = 0 Ω and Zin = jtanβl. If the stub is
terminated in an open circuit then ZL = ∞ Ω and Zin = -jcotβl. Note when l = λ/4 then
βl = (2π/λ)(λ/4) = π/2
And
tan βl = ∞
Standing Waves:
The total voltage on a loss-less line at some point z from the sending end is
V = V1 e
We had ρ = (V2/V1)e
- jβ z
+ V2 e
jβ z
2γ l
2γ l
= > V2 = ρV1 e
In this case since we assume the line to be loss-less γ = β and
= > V2 = ρV1e
-2β l
And
V = V1e
V = V1[e
-jβ z
-jβ l
+ ρV1e
e
jβ x
-2β l jβ z
+ρe
e
-2β l
e
= V1[e
jβ l
e
-jβ(l -x)
-jβ x
+ ρe
] = V1e
-jβ l
-2βl jβ(l - x)
e
[e
j βx
]
+ ρe
-jβ x
]
This two-component equation represents a voltage stationary wave or standing
wave. One of the components is in the forward direction and one in the reverse
direction.
The shape of the wave depends on the value of the complex reflection coefficient ρ.
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When the load is formed from a short-circuit or an open-circuit, ρ is either -1 or +1.
For a short circuit
V = V1 e β [e β – e β ] = j2V1e β sin(βx)
The standing wave is measured by finding the real part of the modulus of this signal
-j l
i.e. Re[2V1 e β sinβx]. Minima will occur when βx = (n -1)π where π is a positive
integer. Adjacent minima are separated by βx = π, or x = π/β or since β = 2π/λ
then x = π/(2π/λ) = λ/2. Similarly for an open circuit at the load, the detected
-j l
voltage will be V = Re[2V1e β cosβx] whose minima are separated by λ/2.
-j l
j x
-j x
-j l
For other values of ρ e.g. ρ = 0.6 + j0.3, the minima do not fall to zero and the
maxima do not increase to 2V1 but are still λ/2 apart. Similarly, current standing
waves also exist along the line. Consider open and short-circuited situations. For the
short-circuited case, minima occur when the incident and reflected waves always
cancel. Maxima occur when, at various points in time, the incident and reflected
waves add to produce a maximum signal level.
Even though the level of the combined signal at t1 in a distance of λ/2 is zero,
it is non -zero at time t2. This means that the signal at a distance of λ/2 is increasing
and reducing in magnitude. It is the RMS level which must be measured when
looking for a standing wave pattern. This is similar to obtaining the magnitude of the
phasor as in the previous analysis.
Voltage Standing Wave Ratio (VSWR):
Vmax and Vmin is measured moving a detector along a slotted line ( A special line for
measuring standing waves). The ratio of these two values is called the VSWR and is
referred to as S.
V
S = max and has a range: 1 ≤ S ≤ ∞.
V min
Using V = V1 e β [e β + ρe
ρ e jψ ,we may write
-j l
j x
-jβ x
V = V1e
] and substituting the complex form for ρ i.e. ρ =
-jβ(l-x)
(j(ψ-2βx)
[1 + ρ e
]
Then
Vmax = V1[1 + ρ ] when (ψ - 2βx) = 2(m-1)π
m = 1, 2, 3......
And
Vmin = V1[1 − ρ ] when (ψ - 2βx) = (m-1)π
m = 1, 2, 3......
S=
Vmax
Vmin
And
ρ =
S -1
S + 1
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Electric circuits
Note this last equation can be reworked into a form S =
Waed 3
1+ ρ
1− ρ
Impedance at voltage minimum and voltage maximum:
We had V = V1e
-jβ(l -x)
[1 + ρ e
j(ψ -2βx)
]
A voltage minimum occurs when e ψ β = e π. At the first voltage minimum
from the load, when x = xmin, ψ = π + 2βxmin.
j(
-2 x)
j
At xmin, Z = Zmin = (V/I)xmin.. Putting x = xmin into the equation for V,
V = V1e
From equation 10.9, I =
-jβ(l -xmin)
]
V1e −νZ − V2 eνZ
Z0
Substituting γ = jβ, z = l - x, and V2 = ρV1e
I=
j(ψ -2βxmin
[1 + ρ e
-2γl
, gives
1
1
-j (l -x)
-2j l j (l -x)
-j l
j x
-j x
[V1e β
V1e β [e β -ρe β ]
-ρV1e β e β ] =
Zo
Zo
Now substituting ρ = ρ e ψ,
j
I=
1
1
-j l j x
j( - x)
-j (l -x)
j( -2 x)
V1e β [e β - ρ e ψ β ] = V1e β [1 - ρ e ψ β ]
Zo
Zo
And finally substituting x = xmin,
1
-j (l -xmin)
j( -2 xmin)
V1e β
[1 - ρ e ψ β
]
Zo
Then
I=
Z x min =
1 + ρ e (jπ )
V V e-jβ [1 + ρ ]
= - jβ
=
Z
= Zo
o
I
Ie [1 - e j( ) ]
1 - ρ e (jπ )
(10.17)
Or
Zmin = 1/S
(10.18)
Maximum impedance occurs for voltage maximum, which occurs for:
e
j(ψ -2βx)
=e
j2π
Thus
Zmax = (V/I)max = Zo
1 + ρe j2π
= Zo
1 - ρe j2π
(10.19)
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Electric circuits
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Load impedance on a loss-less line:
An unknown load impedance may be found by assuming a loss-less line and
measuring the VSWR, the wavelength, and the distance from the load to the nearest
voltage minimum.
We had
V = V1e
-jβ(l - x)
j(ψ -2βx)
[1 + ρ e
] Zmax =
j(ψ - 2βx)
As noted in the last section, this is a minimum when e
i.e. when ψ - 2βx = (2m -1)π, for m = 1, 2, 3.......
(10.20)
= -1.
The solution of interest is for the smallest value of x i.e. xmin.
ψ - 2βxmin = π or ψ = π + 2βxmin
The load impedance and the reflection coefficient are related by:
 1+ ρe jψ
1 + ρ 
ZL = Zo 
 = Z o 
jψ
1
ρ


 1 - ρe
Substituting ρ =
S -1
S +1



gives
 1 + [( S - 1)/( S + 1)]e jψ
ZL = Zo 
jψ
 1 - [( S - 1)/( S + 1)]e
 1 + [( S - 1)/( S + 1)]e j( π +2 βxmin ) 


 = Zo 
j( π + 2xmin ) 

 1 - [( S - 1)/( S + 1)]e

 S + 1+ ( S - 1)e jπ e j2 βxmin 
 S + 1 - ( S - 1)e j2 βxmin

ZL = Zo 
=
Z
o
j2 βxmin
jπ (j2 βxmin 
 S + 1 - ( S - 1)e e

 S + 1 + ( S - 1)e
(jπ) =
From e



-1
 S [1 - e j2 βxmin ] + [1 + e j2 βxmin ] 

ZL = Zo 
j2 βxmin
] + [1 - e j2 βxmin ] 
 S [1 + e
Dividing throughout by e
jβ xmin
, gives
 - jSsinβxmin + cos βxmin
ZL = Zo 
 Scosβxmin - jsinβxmin
And in normalised form

 1 - jStan( βxmin ) 
 = Zo 


 S - jtan( βxmin ) 
(10.21)
 1 - jStanβx min 

ZL = 
(10.22)
 S - jtanβx min 
The Smith Transmission Line Chart:
The following analysis shows the basis of the Smith chart. The relationship between
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Electric circuits
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the load impedance and the reflection coefficient is:
1+ ρ
ZL =
1- ρ
In general, both ZL and ρ are complex. Let ZL = r + jx and ρ = u + jv. Substituting
these into the above equation gives
r + jx =
1+ u + jv
( 1+ u + jv)( 1+ u + jv) 1 - u - v + 2jv
=
=
1 - (u + jv)
( 1 - u - jv)( 1+ u + jv)
( 1 - u )2 + v 2
Thus
1- u - v
( 1 - u )2 + v 2
r=
and x =
2v
( 1 - u )2 + v 2
Rearranging the equation for r:
2
2
2
2
[(1- u) + v ]r = 1 – u - v
2
2
2
[1 - 2u + u + v ]r = 1 – u – v
2
2
2
2
2
r - 2ur + u r + v r = 1- u – v
2
2
2
2
r - 2ur + u r + v r - 1 + u + v = 0
2
2
(u + v )(r + 1) + r - 1 - 2ur = 0
2
2
(u + v ) +
2
2
(u + v ) +
r
2ur
1
=
1+ r 1+ r
1+ r
2ur
r
r
1
r
r
r
+ =
−
+
1+ r 1+ r 1+ r
2
1+ r
1+ r
2
2ur
r
( 1 + r) - r( 1 + r) + r (1 + r) 2
+ =
1+ r
2
(1 + r) 2 2
2ur
r
2
2
Thus (u + v ) + =1
1+ r
2
Rearranging the equation for x:
2
2
(u + v ) -
2
2
2v = (1 - u) x + v x
2v
2
- v2 = (u - 1)
x
1 1
2 2v
(u -1) + v2 + =
x
2 2
Or
1
1
(u - 1) + (v - ) 2
x
2
=
1
x
(10.24)
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The centre of this circle is on a vertical line, which is 1 unit to the right of the origin.
If x is positive, the centre points are above v = 0, and if x is negative they will lie
below that point. The Smith Chart plots these two equations for values of r and x.
The result is a series of orthogonal circles. Resistance circles, having their centres on
the horizontal line and all passing through the point u = 1, v = 0, at the right -hand
edge of the chart. Reactance circles having their centres on the vertical line u = 1 and
their perimeters also tangential at this point.
Figure 10.60: Smith chart.
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Figure 10.60a
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Any value of impedance or admittance can be plotted on the Smith chart. The
horizontal line dividing the chart in half, is the locus of impedance or admittance,
which have no reactive or susceptive component. Along this line, the impedance z =
r + j0 and the corresponding admittance y = g + j0. See chart in figure 10.60. The
centre of the chart, where the circle representing unity normalised resistance or
conductance has co-ordinates 1, 0. The range of resistance values is from 0 on the
outer periphery to ∞ at the right hand side. The other set of arcs on the chart show
the imaginary component of the impedance. The top half of the circle represents
positive reactance and the lower half represents negative reactance.
The centre of these arcs, is on a vertical line, which runs through the right
hand end of the horizontal dividing line. An impedance is plotted and a circle is
drawn of centre 1, 0 and radius from 1, 0 to the point plotted. The corresponding
admittance is found, by drawing a line from this point to the opposite side of the
circle. Consider the impedance z = 0.5 + j1.0 shown on the chart as X. The point on
the other side of the circle is y = 0.4 - j0.8.
1/z =
(0.5 - j1)
0.5 - j1 0.5 - j1
=
=
= 0.4 - j0.8.
(0.5 + j1)(0.5 - j1)
1.25
22
The relative impedance and admittance are at opposite ends of a diameter of the
circle with the centre 1, 0 passing through them. The inner scale around the chart is
marked "Angle of reflection coefficient in degrees". For the previous example, the
0
reflection coefficient is 84 . The magnitude of the reflection coefficient ρ is related
to the distance between 1, 0 and X. Below the chart are eight radial scales and the
uppermost one is the "Voltage reflection coefficient ". ρ is found by measuring the
distance from 1, 0 to X along the scale. Another important radial scale is the lower
scale marked "Voltage Standing Wave Ratio" which is related to ρ.
S=
Vmax 1 + ρ
=
Vmin
1- ρ
If ρ is constant, S is also constant and is on a circle. The value of S is read from an
appropriate radial scale.
Problem 1
(a) Plot the following impedance values on a Smith chart
(i) A = 0.2 + j 0.9 Ω,
(ii) B = 1.25 + j0 Ω,
(iii) C = 0 – j 1.1 Ω, and
(iv) D = 0.175 – 0.625 Ω
(b) Convert the above, normalised impedance values, to actual impedance values if
the transmission line characteristic impedance is 50 Ω.
(c) Convert A, B, C, D, to normalised admittance values and then to actual
admittance values for Zo = 50 Ω. Read the points as before but
A=
(0.2 + j 0.9)
S = 0.004 + j 0.018 S
50
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Solution to 1 (i)
Figure 10.61: Problem 1.
Problem 2
Plot the following load impedance on a Smith chart.
(i)
ZL = 12.35 + j17.59 Ω on a 50 Ω line
Solution
A=
(12.35 + j17.59)
Ω = 0.247 + j 0.3518
50
Problem 3
A 50 Ω Tx line is terminated in an impedance ZL = 5 – j20 Ω. Determine the
impedance 0.2 λ from the load.
Solution
(i)
Normalise the load,
(ii)
(iii)
(iv)
(v)
(vi)
(i)
(ii)
Z L (5 − j 20)
=
Ω = 0.1 − j 0.4 Ω ,
Zo
50
Plot this point A on the chart, and draw a line, from the centre through this
point right out passed the outer circle,
Read the wavelength on the outer circle as 0.439 λ
Using a compass, draw a circle with radius A, This circle represents the
impedance on the line and also represents the VSWR for a loss-line (See
point D VSWR = 12),
Point B is located = 0.439 λ + 0.2 λ - 0.5 λ = 0.139 λ
Draw an arc from 0.439 λ to the point = 0.139 λ
From this point, draw a line to the centre,
The point where this line intersects the drawn inner circle is the read 0.2 +
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j1.18. The impedance is therefore
50(0.2 + j1.18) = 10 + j 59 Ω
Figure 10.62: Problem 3.
Problem 4
A load ZL = 450 + j 275 Ω is connected to a 300 Ω Tx line. If the line is excited by
a 100 MHz signal and the line length is 78 cm, find the input impedance and the
VSWR. (Assume the line is loss-less).
Solution
(i) Normalise the load,
Z L ( 450 − j 275 )
=
Ω = 1 − j 0.91 Ω ,
Zo
300
Plot this point A on the chart and draw a line from the centre
λ=
vp
=
300.10 6
=3m
100.10 6
f
The electrical length is therefore
0.78
= 0.26λ
3
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Figure 10.63: Problem 4.
Problem 5
A load Z = 100 - j50 Ω is connected to a transmission line whose characteristic
impedance is 75 Ω. Determine, with the aid of a Smith chart, the characteristic
impedance of a quarter wave transmission line to achieve correct matching. You
may assume the transformer is inserted at a point nearest the load. Figure 10.64a
illustrates the technique.
Figure 10.64a: Quarter – wave transformer.
Solution
(i)The normalised load impedance is ZL =
100 + j50
= 1.333 – j0.67 is plotted on
75
the chart l. Point A,
(ii)Draw the S circle through ZL Read the VSWR value,
(iii)Where the VSWR circle intersects the horizontal axis at point B = 0.53, is the
point where the line in purely resistive with a value 0.53 x 75 Ω = 39.8 Ω. The
characteristic impedance of the λ/4 section is calculated using the expression for
input impedance of a line:
 ZR
' 
+ jZ o 

tanβ l

Z in = Z ' o 
 Z o'

+ jZ R 

 tanβ l

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The input impedance of the λ/4 line should have a value equal to the characteristic
impedance of the line to be matched. For a section of line which is l = λ/4, the total
phase angle is calculated:
βl = (2π/λ)(λ/4) = π/2
Hence
tan βl = ∞
Therefore the relationship between the characteristic impedance and the
impedance ZR is
 0 + jZ ' o  Z ' o 2
Z in = Z o = Z ' o 
=
 0 + jZ R  Z R
The input impedance of the λ/4 section of line should equal the characteristic
impedance of the main line Zo. Substituting values into this quarter-wave
transformer equation, yields a value for the characteristic impedance of the λ/4 line:
Z o' = Z 0 Z R = 75.39.8 = 54.5 Ω
This section of line is located at 0.184λ from the load
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Figure 10.64b: Quarter – wave transformer.
Problem 6
(a)Define the following transmission line terms: Voltage standing wave ratio and
voltage reflection coefficient.
(b) A transmission line, whose characteristic impedance Zo = 300 + j0 Ω, has an
antenna of impedance 225 - j175 Ω connected as a load. Matching by means of
a single stub connected, at a distance d metres from the load, is used. Estimate
the length in metres of the stub and the distance d if the operating frequency f =
500 MHz. Assume that the stub is formed from a section of the same air-spaced
transmission line.
Solution
a) Definitions
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b) The procedure for using a single stub as a matching,
i) Normalise the aerial impedance as Z = 0.75 - j0.5833 Ω,
ii) A circle of constant VSWR is then constructed through the point L, and
iii) Since we are connecting a short circuit stub across the line (i.e. in parallel)
then it makes sense to work with admittance. The admittance of the load is
iv) Going around the constant VSWR circle to the point where the intersection
with the constant conductance circle of 1(i.e. through the circle G/Y=1.0+j0.72).
v) A stub is attached whose admittance Ys = -j0.72s will cancel out the admittance
j0.72 s. The points M and B. which is read from the chart as M = 0.13 λ l towards
the generator and B = 0.153 λ
d = (0.153-0.131) λ
l = 0.022 λ metres towards the generator.
The relationship v = λ/.f gives us a way of calculating the distance d in metres i.e.
8
8
λ = v/f = 3.10 /500.10 = 0.006 m (or 60) cm so that the distance d is = 0.022 x
.006 cm = .0132 cm
vi) The length of the stub is found from the distance = 0.25-0.099 λ = 0.151 λ l so
that the length of the stub is = 0.151x60 cm = 9.06 cm
Length of cable = 10 km, frequency of operation =10 kHz.
Primary constants:
L = 700 mH per km,
C = 0.05 µF per km,
R = 28 Ω per km, and
G = 1 µS per km
The propagation constant is calculated
γ = ( R + jωL)(G + jωC ) = (280 + j 2π .10 4 7)(10 −5 + j 2π .10 4.0.5.10 −6 )
γ = (280 + j14π .104 )( 10−5 + jπ .10−1 ) =
0.12424 + j371.72
0.056125 + j117.55
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The attenuation constant is the real part and the phase change coefficient is the
imaginary part. The characteristic impedance of the line:
R + jω L
Z0 =
Ω
G + jω C
Substituting values
Zo =
( 280 + j14π .10 4 )
= 1183.2 - j0.3578 Ω
( 10 −5 + jπ .10 −1 )
Abs(Zo) = 1183 Ω
Problem 7
A loss-less transmission line of characteristic impedance 50 Ω is terminated in a
load of 150 + j75 Ω. Find the reflection coefficient, the load admittance, the VSWR,
and the distance between the load and the nearest voltage minimum to it. Find also
the input impedance, if the line is 92 cm long and the wavelength of the signal on
the line is 40 cm.
Solution
150 + j75
The normalised load impedance is ZL =
= 3 + j1.5 is plotted on the
50
chart l. Drawing the S circle through ZL we can read the VSWR as S = 3.8. The angle
o
of reflection coefficient is Ψ = 16 . ρ is measured on the appropriate scale as
(j0.28)
0.58. Thus ρ = 0.58e
. The load admittance is on the other side of the circle and
is yL = 0.27 - j0.13. The line is 92 cm long which is equivalent to (92/40)λ = 2.3λ.
To find the input impedance Zin, we must travel a distance 2.3λ from the load
towards the generator, on a constant S circle. The line from the centre through ZL
cuts the "Wavelengths towards generator" scale at 0.227λ. The input impedance is
therefore at (2.3 + 0.227) λ towards the generator. i.e. at 2.527λ.
One complete revolution on the S circle corresponds to 0.5λ (The impedance
along a line is periodic in λ/2) thus in 2.527λ there are 5 revolutions + 0.027λ.
Hence the input impedance is located at 0.027λ on the "Wavelengths towards
generator" scale and is Zin = 0.27 + j0.16 Ω. The normalised impedance at a voltage
minimum, which is 1/S must lie on the horizontal line to the left of 1, 0. Then xmin is
read on the distance from ZL to Vmin on the "Wavelengths towards generator" scale.
Vmin is 0.237λ from the load. All of these results may be calculated using the
equations derived earlier.
The reflection coefficient
ρ=
Z L − Zo
150 + j 75 - 50
100 + j 75
=
=
= 0.56 + j0.16
150 + j 75 + 50
200 + j 75
ZL+ Zo
ρ = (0.56 ) 2 + (0.16 ) 2 = 0.58
Angle of reflection coefficient = Ψ = arctan (0.16/0.56) = 0.28 radians. =
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o
0.28×180/π = 16 . The reflection coefficient is calculated, in normalised form, as:
Z / Z − 1 3 + j1.5 - 1
ρ = L o
=
= 0.56 + j 0.16
Z L / Z o + 1 3 + j1.5 + 1
Normalised admittance = 1/(3 + j1.5) = 0.27 - j0.13.
VSWR = S = (1 + ρ )/((1- ρ ) = (1 + 0.58)/(1 - 0.58) = 3.8
Normalised input impedance Zin =
Z + jtanβl
1 + jZtanβl
Where β = 2π/λ = 2π/(0.4) = 5π radians/m. and Z is the normalised load.
Thus
Zin =
3 + j1.5 + jtan(5π 0.92)
= 0.27 + j0.16.
1 + j(3 + j1.5)tan(5π 0.92)
Stub Matching
A reflected wave on the line, due to a mismatched load, can interfere with the
performance of the generator and the line. The voltage maximum for example could
break down a dielectric medium between the lines. It is advantageous to adjust the
VSWR to unity so there is no reflected wave at the generator. This is achieved using
stubs to match the line to the load at some point near the load.
Single stub matching:
Rotating around from the load towards the generator will, at some point, result in the
impedance of the line having a normalised real part of unity. If a stub is inserted at
this point, and has a reactance value, which is opposite to the reactance of the line
impedance, then the net impedance is changed to unity (reactance parts cancel). The
line is then matched on the generator side of the stub. There are two factors to be
determined; the distance from the load to the stub and the length of the stub itself.
The stub may be open or short-circuited and may be in series or parallel with the
line.
Problem 8
Consider a line terminated in an unknown load. Measurements show a VSWR of 3.5,
adjacent minima are 15 cm apart, and xmin = 1.2 cm. A short -circuit shunt stub is
used to match the line. The VSWR from the load to the stub is 3.5, and is drawn on
the Smith chart as a circle of radius 3.5. yin provides a match to the line i.e. yin = 1 +
j0. The stub can only add susceptance, jb and therefore
yin = y1 + jb
Or
1 + j0 = y1 + jb
And
y1 = 1 - jb
Hence y1 must lie on the locus of unity conductance i.e. the g = 1 circle. It must also
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lie on the S = 3.5 circle. These two circles intersect at points a and b shown on the
chart.
Solution: Consider point a:
The admittance of a is 1+ j1.3. The stub must provide a susceptance of -j1.3. The
load is not given, so its position on the chart is unknown, but the position of Vmin is
known (point X). This point is 0.42λ (0.42 × 30 = 12.6 cm) towards the load from
a. The voltage minimum is 1.2 cm from the load, making point a 13.8 cm from the
load. The short circuit stub produces a VSWR of ∞ or ρ = 1, which is the
outermost circle on the chart. The load admittance of the stub is yL = ∞ which is at
the right -hand end of the horizontal line (point m). The stub length is found by
moving from point m towards the generator on the outer circle to the point at
which the susceptance is -j1.31 (point n). The stub length is mn = 0.1λ = 3.1 cm.
Matching is obtained by placing a short -circuited stub 3.1 cm long, 13.8 cm from
the load.
Consider point b:
The susceptance at b is - j1.31 so the stub susceptance must be j1.31. The stub
length is given by the distance from m to the point on the S = ∞ circle where the
susceptance is j1.3 i.e. point p. So mp is 0.39λ = 11.8 cm. The stub position from
the voltage minimum to b is 0.08λ = 2.4 cm.
Double Stub Matching
A single-stub, matching device, may not be convenient in all cases because of the
need to adjust the position of the stub relative to the load if the load impedance is
changed. Using two stubs, the two variables needed to produce a match are the
lengths of the stubs and their positions can be fixed. Referring to figure 10.15. The
admittance at the stub nearest the generator, yin must match the line, so
yin = 1 + j0
st
The admittance on the load side of the 1 stub must be given by
yin = y1 + jb1
Or
y1 = 1 - jb1
Which lies somewhere on the unity conductance circle. The admittance on the
generator side of stub 2 (yd), must lie on the same VSWR circle as y1, since it belongs
to the same section of line, and it can be found by moving from y1 on a constant S
circle a distance d towards the load. However the precise position of y1 is not
known. All that we know is that it lies on the unity conductance circle so yd must lie
on the g = 1 circle, translated a distance d, on a constant S circle, towards the load.
This is called the L circle. See the Smith Chart enclosed. It can be seen quite clearly
from the Smith Chart that no matter where y1 is on the g = 1 circle, yd will be a
distance d away towards the load, which must be on this L circle. In practice, we
will be using this L circle and working backwards.
Designing a double stub matching network
1. Normalise the load impedance by dividing by the characteristic impedance
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2. Plot this point A on the chart. From this point through the centre of the circle
to the outer circle opposite, draw a straight line. Note yl the admittance of the
load. Read the wavelength on the outer circle. λ1. Point B
3. Travel from this point λ1(towards the generator or clockwise) a distance equal
to the distance from the load to the first stub. Note the wavelength reading λ2,
point C.
4. From this point, draw a line to the centre of the chart. Where it intercepts the
VSWR circle read the admittance value. Point D.
5. Rotate anti-clockwise, the unit conductance circle, a distance equal to the stub
separation
6. Where the conductance part of point D intercepts the swung conductance,
draw an inner circle. Note the susceptance value (the conductance is the same
at both points) at Point E.
7. Subtract the two susceptance values at D and E.
8. An opposite value to this difference susceptance will give the required
susceptance for stub1.
9. Plot this susceptance and draw a line from the centre through this point to the
outer circle and note the wavelength value λ3. Add 0.25 λ to this value to get
the first stub length in wavelengths.
10. From the Note the value 1+ jB where this inner circle cuts the unit
conductance circle point F. A – jB value will cancel this + jB. Locate this
value and plot a straight line from centre out to this point. Point F. Note the
wavelength value λ4. Add .25 λ to this value to get the stub length.
Problem 9
A loss -less transmission line, whose normalised admittance load is yL = 0.7 + j0.7.
Two short -circuited parallel stubs are to be used to match the line. Stub separation
is 0.12 λ and the last stub is 0.45 λ from the load. Find the length of the stubs.
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Figure 10.64: Problem 9.
Problem 10
A loss-less transmission line, of characteristic impedance 50 Ω, is terminated in a
load of 100 + j25 Ω. Two short -circuited series stubs are to be used to match the
line. One stub is 8 cm and the other, 32 cm from the load. The signal on the line is at
750 MHz. Find the length of the stubs.
Solution
Because series stubs are used, the problem is solved, using impedance values. 750
MHz is equivalent to a wavelength of 40 cm. So the 1st stub is 0.2 λ from the load
st
and the second stub is 0.6 λ from the 1 . The L circle is now drawn with its centre
0.6 λ (= 0.1 λ) towards the load from the x = 1 circle. The normalised load 2 + j0.5
is placed on the chart as zL and the VSWR circle is drawn. The value of z2 is found
by moving on the constant S circle a distance of 0.2 λ toward the generator from ZL.
The impedance of Z2 is 0.55 - j0.36.
The stub must add a reactance to Z2, which is sufficient to put Zd on the L
circle at the point shown. Moving on a constant resistance circle from Z2, Zd is on
the L circle at the point 0.55 + j0.25. The difference in reactance between Zd and Z2
is that provided by stub 2 and is j0.61. Z1 is now found by moving towards the
generator a distance d towards the generator on a constant S circle from Zd. i.e. by
moving on a constant S circle from Zd, on the L circle, to the corresponding point, d
wavelengths towards the generator, on the x = 1 circle.
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Then Z1 = 1 + j0.7 and stub 1 must provide -j0.7. The stubs are short circuited, so to find their length it is necessary to start at the point Z = 0, and go
round the S = ∞ circle towards the generator to the required reactance. So stub 1 is
16.1 cm long and stub 2 is 3.45 cm long. It is important to note that the distance d
between the stubs is not entirely arbitrary. A situation could arise as shown below
where it is not possible by adding positive or negative reactance to Z2 to reach the L
circle. If this is the case, the distance between the stubs must be changed.
Problem 11:
a) Standing waves were observed on a 50 Ω transmission line with a maximum
voltage Vmax = 10 mV. The minimum voltage Vmin = 2 mV was measured 50.8
mm from the load. If the wavelength for the operating frequency is 212 mm,
determine:
(i) The VSWR,
(ii) The magnitude of the reflection coefficient ρ,
(iii) The distance in wavelengths between the load and the first minimum, and
(iv) The mismatch loss in dB.
b) Determine the location and the length of a single stub (same characteristics as
the mismatched line) for correct matching.
c) Describe the technique of TDR for locating faults on transmission lines.
Solution
V
10
=5
(i). VSWR = max =
Vmin
2
VSWR − 1 5 − 1
=
= 0.67
VSWR + 1 5 + 1
(ii)
ρ=
(iii)
D min =
50.8
= 0.24λ
212
2
Mismatch loss =-10log[1-ρ ]=-5.1 dB
b) The procedure for using a single stub as a matching device
1) Locate the point Dmin = 0.24λ. From the centre of the chart to this point draw
a line. The intersection of this line and the circle of constant VSWR is read as,
the normalised impedance 4.5 - j1.4 Ω (point A). The load is thus 50(4.5 - j1.4
Ω) = 225 - j70 Ω.
2) Since we are connecting a short circuit stub across the line (i.e. in parallel) then
it makes sense to work with admittance. The admittance of the load is located
through the centre from A to the opposite side where it intersects the VSWR
circle Pt B is an equal distance from centre and is read as 0.2+ j0.07 S On the
outer circle read the wavelength measurement as 0.01λ.
3) Going around the constant VSWR circle to the point where the intersection with
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the constant conductance circle of 1 is the point C.
y =1+ j1.8 S.
5) A stub is attached across the line whose admittance is Ys = -j1.8s. This will
cancel out the admittance j1.8 s. The points M from the chart as = (0.183 –
0.01)λ=0.173 λ from the load l towards the generator and B = 0.153 λ. d =
(0.153 -0.131) λ
Figure 10.65: Problem 11.
This admittance – j1.8 S is obtained at 0.33 λ i.e. the length is 0.33λ -0.25λ =
0.08λ. This is a stub of length 0.08*212 mm = 17 mm. and is placed a distance
0.173λ*212 mm from the load = 36.7 mm
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Problem 12: TDR
Discussion of TDR.
3 (a) A 10 km length of screened telephone cable, operating at a frequency of 10
kHz, has the following primary transmission line parameters:
L = 700 mH per km,
C = 0.05 µF per km,
R = 28 Ω per km, and
G = 1 µS per km
Determine the characteristic impedance, phase and attenuation constants for
this cable.
(b) An expression for the input impedance of a transmission line, terminated in a
resistance ZR, = 35 Ω is:
 ZR

 tanβ l + jZ o 

Z in = Z o 
 Z o + jZ 
R 
 tanβ l


Where β is the phase change coefficient and Zo is the characteristic impedance
equal to 75 Ω. Determine the input impedance of a section of line whose
length is λ/4 metres.
(c) A transmission cable is terminated in an impedance, which has a value twice
the characteristic impedance of the line. Calculate the position and length, of a
short -circuited stub to achieve matched conditions on the line. The stub is
constructed from the same cable type as the line. The frequency of operation is
1 MHz and the velocity of propagation is 0.67 c(c is the velocity of light).
Solution
(a)
The propagation constant is calculated
γ = ( R + jωL)(G + jωC ) = (280 + j 2π .1047)(10−5 + j 2π .104.5.10−6 )
γ = (280 + j14π .104 )( 10−5 + jπ .10−1 ) = 0.12424 + j371.72
The attenuation constant is the real part and the phase change coefficient is the
imaginary part. The characteristic impedance of the line is:
R + jω L
G + jω C
Abs(Zo) = 1183 Ω
Z0 =
=
( 280 + j14π .10 4 )
=1183.2 – j 0.3578 Ω
( 10 −5 + jπ .10 −1 )
(b)
The input impedance of the line is:1
Copyright: Paul Tobin School of Electronics and Comms. Eng.
109
DIT, Kevin St.
Electric circuits
Waed 3
 ZR

 tanβ l + jZ o 

Z in = Z o 
 Z o + jZ 
R 
 tanβ l


. For l = λ/4, the total phase angle is calculated:
βl = (2π/λ)(λ/4) = π/2
And
tan βl = ∞
 0 + jZ o  Z o 2 75 2
Z in = Z o 
=
= 150 Ω
=
 0 + jZ R  Z R 37.5
Copyright: Paul Tobin School of Electronics and Comms. Eng.
110