EECS 117 Homework Assignment 2 Spring

Fall 2004
2008
1(a). To find the voltage in each transmission line, we need to calculate the V1 and
propagate it along the transmission lines (figure (a).)

We first calculate the equivalent input impedance looking down from the end of
transmission line 1 (figure (b).)

Z cos β ' l '+ jZ 2 sin β ' l '

2 2
jZ 2 Z 2 Z
Z 2eq = Z2 L = Z2 = = 1
Z 2 cos β ' l '+ jZ L sin β ' l ' jZ L ZL 9Z L

R / jω C R 100Ω
= =
where ZL = R // 1/jωC = R + 1 / jωC 1 + jωRC 1 + j (2π * 10 Hz *100Ω *10 −11 F )
9

= 2.47 − 15.52 j

βl = π/3 β’l’ = 3π/2

Zg
+ + +
ZC = Z1 V2 Zc = Z2
V1 VL ZL
- - -

(a)

Zg
+ +
V1 ZC = Z1 V2 Z2eq
- -

(b)

Zg
+
V1 Z1eq
-

(c)
Figure (b) can be simplified to become figure (c), where Z1eq is the equivalent impedance
looking from the beginning of the transmission line 1.

Z 2 eq cos β l + jZ 1 sin βl Z 2eq / 2 + j 3 / 2Z 1 2

Z 1 + j 9 3Z 1 Z L
Z 1eq = Z 1 = Z1 = Z1
Z 1 cos βl + jZ 2eq sin βl Z 1 / 2 + j 3 / 2Z 2 eq 9 Z 1 Z L + j 3Z 1
2

= 55.3 − 425 j

Then,
Z 1eq
V1 = V g = 6.52 − 4.16 j
Z 1eq + Z g

To calculate the voltage, VTL1, along the transmission line 1, let’s define the origin of z at
Z2eq.

+ − +
VTL1 ( z ) = V1 e − jβz + V1 e jβz = V1 (e − jβz + Γ1e jβz )

and,

π +
V1 = VTL1 (− ) = V1 (e jπ / 3 + Γ1e − jπ / 3 )
3β

Γ1 is the reflection coefficient in the figure (b), and is equal to:

Z 2 eq − Z 1
Γ1 = = 0.567 + 0.680 j
Z 2eq + Z 1

Using the known information, we can get

+ V1
V1 = jπ / 3
= 2.50 − 4.33 j
e + Γ 1 e − jπ / 3

Thus, we have calculated the steady state voltage along the transmission line 1 (for z=0
at the end of the transmission line 1):

+
VTL1 ( z ) = V1 (e − jβz + Γ1e jβz ) = (2.50 − 4.33 j ) * ((e − jβz + (0.567 + 0.680 j ) * e jβz )

The corresponding current:

I TL1 ( z ) = (0.0083 − 0.0144 j ) * ((e − jβz − (0.567 + 0.680 j ) * e jβz )

For getting the steady state voltage along the transmission line 2, one can repeat the
above calculation using the corresponding parameters. For instance, V2 should replace
V1, the reflection coefficient Γ2 replaces Γ1, and Z2 replaces Z1.
 V2 = VTL1 (0) = V1+ (1 + Γ) = 6.86 − 5.09 j
Z − Z2
Γ2 = L = −0.908 − 0.289 j
ZL + Z2
+ V
V2 = j 3π / 2 2 − j 3π / 2 = 3.14 + 3.12 j
e +Γ 2 e

For z’ = 0 at the load,

+
VTL 2 ( z ' ) = V2 (e − jβz ' + Γ2 e jβz ' ) = (3.14 + 3.12 j ) * ((e − jβz ' − (0.908 + 0.289 j ) * e jβz ' )

I TL 2 ( z ' ) = (0.0314 + 0.0312 j ) * ((e − jβz ' + (0.908 + 0.289 j ) * e jβz ' )

1 + Γ1
(b) For line 1, VSWR = = 16.4
1 − Γ1

1 + Γ2
For line 2, VSWR = = 41.5
1 − Γ2

(c). The average power delivered to the load = Pav , L =

1
2
{ }
Re VL [I L ] , where VL and IL are
*

the voltage and current at the load, respectively.

Pav , L =
1  1
} {
ReVL [I L ] = Re VTL 2 (0)[I TL 2 (0)] = 9 mW
2 
*

2
*
}
(d). The average power delivered by the source to the transmission lines =

1   V1   V1
* 2
 1 
1 
2 
*
}
Pav , L = ReV1 [I 1 ] = ReV1
2   Z 1eq  
 =
2
Re  = 9 mW.
 Z 1eq 
  
This is the same as the one in part (c), which is expected as the transmission lines are
lossless.

2. (Two typos in the question: … desired to deliver 10W into ZL2 and the remaining
power into ZL3.)
VSWR − 1
Since VSWR = 0 dB is required for each line and Γ = , |Γ| = 0. Thus, ZL2=Z2,
VSWR + 1
ZL3=Z3.
Because it is a perfect match, the length of the transmission lines does not change the
impedance of the transmission lines seen from the junction downwards.

The voltages at the junction of the two transmission lines are the same, for they are
measured at the same junction.

Thus, the currents going into individual lines determine the amount of power flowing into
the respective loads.

PL 2 ( )
V1 / 2 Re{1 / Z 2 } Z 3 1
2

PL 3
=
( 2
) = =
V1 / 2 Re{1 / Z 3 } Z 2 4

So Z2 = 4 Z3 would make 10 W going into ZL2 and 40 W into ZL3.

VSWR − 1
3. Given VWSR = 3.2, Γ = = 0.524
VSWR + 1

At the position in the transmission line where the voltage is minimum, the current is
maximum, so the impedance and the impedance normalized to Z0 at that point is
minimum. The impedance is also real. Thus the normalized impedance of this position
has a point lying on the left hand part along the horizontal axis in the Smith chart.

Since this position is 0.23 wavelength away from the load, you need to move the point
counterclockwise 0.23λ to find the normalized impedance of the load. The new point has
an angle subtended at the origin of -14.40.

To find the distance of the new point from the origin of the chart, one needs to make use
of the fact that propagation along the transmission line just rotates the impedance point
about the origin. The radius of the circular locus does not change. At the position of the
line where the voltage is maximum, the normalized impedance is maximum and real, and
is equal to VWSR = 3.2. Thus, the point corresponding to the normalized load impedance
can be obtained by moving this point about the origin of the chart, by –14.40. This point
has a reading of 2.8 – 1.0j. Therefore, the load impedance is 196 – 70j Ω.
4. This is somewhat similar to the problem 1. We need to find the input impedance of the
transmission network.

Z L1eq cos π / 4 + jZ 1 sin π / 4

Z in = Z 1
Z 1 cos π / 4 + jZ L1eq sin π / 4
where Z L1eq = jX 1 + ( jX 2 // Z 2 eq ) = jZ 1 + ( jZ 1 // Z 2eq )
Z L cos 3π / 4 + jZ 1 sin 3π / 4 2 Z cos 3π / 4 + jZ 1 sin 3π / 4
Z 2eq = Z 1 = Z1 1
Z 1 cos 3π / 4 + jZ L sin 3π / 4 Z 1 cos 3π / 4 + j 2 Z 1 sin 3π / 4
Simplifying the above expressions,
 Z 2eq = (0.8 + 0.6 j ) Z 1
jZ 1 * Z 1 (0.8 + 0.6 j )
Z L1eq = jZ 1 + = (0.25 + 1.5 j ) Z 1
jZ 1 + Z 1 (0.8 + 0.6 j )
(0.25 + 1.5 j ) Z 1 cos π / 4 + jZ 1 sin π / 4
Z in = Z 1 = (1.6 − 4.2 j ) Z 1
Z 1 cos π / 4 + j (0.25 + 1.5 j ) Z 1 sin π / 4

Thus, voltage across the beginning of the line 1, V1 is equal to:

Z in
V1 = V g = 4.47 − 0.861 j
Z in + Z g

Since the transmission lines are lossless, the power delivered into the line is equal to the
power consumed by the load.

Hence, power transmitted to the load =

1   V1   V1
* 2
 1  0.820
Pav , L = ReV1  = Re = W
2   Z 1eq   2  Z 1eq  Z1
 

(0.25 + 1.5 j − 1)
Γ1 for transmission line 1 = = 0.344 + 0.787 j
(0.25 + 1.5 j + 1)

1 + Γ1
VSWR = = 13.2
1 − Γ1

(2 − 1)
Γ2 for transmission line 2 = = 1/ 3
(2 + 1)

1 + Γ2
VSWR = =2
1 − Γ2

(Using the Smith chart)

The normalized load impedance is 2, which is represented by point A in the figure below.
To find the normalized impedance at the beginning of line 2, one would move clockwise
through 2700 and get to point B. The reading of this point is (0.8+0.6j) Z1. This
impedance needs to combine with the load “jX2” in parallel and subsequently the load
“jX1” in series. The result is (0.25+1.5j) Z1. This impedance is shown as point C in the
figure. The input impedance of the transmission line network can be found by moving
clockwise through 900. The new point has a reading of (1.6-4.2j) Z1. This result is the
same as the one found above. Using the method indicated above, one can find the average
power delivered to the load.
The VSWR of a transmission line is equal to the maximum normalized impedance along
the line, i.e.,
Z
VSWR = max
Z0

It is a real number, and lies on the right side along the horizontal axis (u axis). So for
transmission line 1, we can find its VSWR by moving point C about a circle and reading
the intersection at the right u axis. The intersection is denoted by point D, and has a value
of ~ 13.

For the line 2, one may note that point A is already the intersection on the right u axis. So
VSWR for this line is 2.
5(a). In general, voltage and current along a transmission line have forms:

V ( z ) = V + e −γz + V − e γz
I ( z) =
Z0
[
1 + − γz
V e − V − e γz ]
where γ = ZY
Z
Z0 =
Y
Z = R'+ jωL' , Y = G '+ jωC '

In this case, because R’>>ωL’, Z ~ R’. Y ~ j ωC’. Thus, the propagation constant
γ = jωR' C ' = ωR' C ' / 2 (1 + j ) . Since γ = α + jβ , β = ωR'C ' / 2 , and propagation
velocity = ω / β = 2ω /( R ' C ' ) .

(b) Using the voltage and current expression given above, the input impedance is equal to
be (z = -l) :
V + e γl + V − e −γl 1 + Γe −2γl Z cosh γl + Z 0 sinh γl
Z i (l ) = Z 0 + γl = Z0 = Z0 L
V e −V e − −γl
1 − Γe − 2γl
Z 0 cosh γl + Z L sinh γl

where ZL is the load impedance, Z 0 = R' /( jωC ' ) = R' /(2ωC ' ) (1 − j ) , and l is the
distance of the point of interest to the load.

(c) For a shorted termination,

Z L cosh γl + Z 0 sinh γl  
Z i (l ) = Z 0 = Z 0 tanh γl =
R'
(1 − j ) tanh  ωR' C ' (1 + j )l 
Z 0 cosh γl + Z L sinh γl 2ωC '  2 

Expansion of Z0 around ω = 0 gives:

jR '
Z0 = − + O(ω )
ωC '

O(ω) represents the higher-order terms in the expansion.

Expansion of tanh(γl) around ω = 0 gives:

1
tanh (γl ) = jωR ' C 'l − jRC jR ' C 'l 3ω 3 / 2 + O(ω )
3

So,
 1
Z i ~ R − j ωR 2 C
3
R and C are the sums of the resistance and capacitance of the components in the
transmission model. At low frequency, the shorted transmission line acts like a resistor.
This is true until the frequency is so large that the imaginary part, which is the reactive
component of the impedance, is comparable to the real part. In particular, the parts
become equal when

R = ωR 2 C / 3
or ω = 3 / RC

Thus the pole frequency is 3/(2πRC).

1-section lumped approximation to the line is a resistor of resistance R in parallel with a

capacitor of capacitance C. Expansion of the impedance of the lumped circuit gives:

R
Z= ~ R + jωRC
1 + jωRC

Using the values given in the question, plots of the input impedance of the transmission
line and the impedance of the lumped circuit are shown below.

40

35 Z

(Ω)
30

Zi

25

20
1. ´ 10 1 0 2. ´ 10 1 0 5. ´ 10 1 0 1. ´ 10 1 1 2. ´ 10 1 1 5. ´ 10 1 1 1. ´ 10 1 2
frequency (Hz)

(d) For an open termination,

 Z L cosh γl + Z 0 sinh γl  
Z i (l ) = Z 0 = Z 0 tanh γl =
R'
(1 − j ) tanh  ωR' C ' (1 + j )l 
Z 0 cosh γl + Z L sinh γl 2ωC '  2 

Just like in part (c), expansion of Z0 around ω = 0 gives:

jR '
Z0 = − + O(ω )
ωC '

Expansion of 1/tanh(γl) around ω = 0 gives:

j ωR' C ' (ωR' C ') 2 3

3
1
1 tanh (γl ) = + l+ l + O(ω )
jωR' C 'l 3 j 45 j

Combining the two results gives:

R  1 ωR 2 C 
Zi ~ − + j  + 
3  ωC 45 

At low frequency, (ωC ) >> ωR 2 C because of the values of R and C in this problem.
−1

Thus,

R 1
Zi ~ − + j
3 ωC

It is a capacitor of capacitance C in series to a resistor of resistance R/3. An 1-section

lumped approximation to the line gives R in series with C, i.e., Z i = R − j /(ωC ) .

(e) The voltage along the line is equal to:

V (l ) = V + (e γl − e −γl ) = V + sinh(γl )

So for R = 106Ω/500µm = 2*109 Ω, C = 100*10-15F/500 µm = 2*1010 F, we can plot the

V/V+ vs. the distance from the load.
600

500

400

300

200

100

0.0001 0.0002 0.0003 0.0004 0.0005

Distance from the load (m)

For an ideal resistor, the plot would look like this:

0.8

0.6

0.4

0.2

0.0001 0.0002 0.0003 0.0004 0.0005

Distance from the load (m)

The plot for the transmission line is not linear like the one for a resistor because the
frequency is 3 orders of magnitude higher than the pole frequency.