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UNIVERSITY OF TORONTO
LABORATORY OF ELECTROMAGNETIC FIELDS - ECE357S
RODOLFO SIQUEIRA
THOMAS SATTOLO
WAVES ON TRANSMISSION LINES
REPORT OF LABORATORY
TORONTO
2014
2
UNIVERSITY OF TORONTO
LABORATORY OF ELECTROMAGNETIC FIELDS - ECE357S
This report will demonstrate
everything that has been worked on
laboratory with base on the lectures.
RODOLFO SIQUEIRA
THOMAS SATTOLO
WAVES ON TRANSMISSION LINES
REPORT OF LABORATORY
TORONTO
2014
3
Contents
Contents .................................................................................................................................................... 3 Table of Figures ........................................................................................................................................ 4 Section A ................................................................................................................................................... 5
Section B ................................................................................................................................................... 6 Section C ................................................................................................................................................... 7 Section D ................................................................................................................................................. 10 Section E .................................................................................................................................................. 11 Section F ................................................................................................................................................. 18
Section G ................................................................................................................................................. 24
4
Table of Figures
Figure 1 β Expected Voltage Waveform at C ............................................................................................ 5 Figure 2β Waveform at point C (top) and D (bottom) .............................................................................. 7 Figure 3 β Waveform at point E (top) and F (bottom) .............................................................................. 7
Figure 4 β Waveform at t = T/2 = 100ns ................................................................................................... 8 Figure 5 β Waveform at t = 2T = 400ns .................................................................................................... 8 Figure 6 β (C,F) β RL = 0 ....................................................................................................................... 12 Figure 7 β (C,F) β RL = 20Ξ© ................................................................................................................... 12 Figure 8 β (C,F) β RL = 100Ξ© ................................................................................................................. 13
Figure 9 β (C,F) β RL = β ....................................................................................................................... 13 Figure 10 β (C,D) β C on top and D on the bottom ................................................................................ 14 Figure 11 β (E,F) β E on top and F on the bottom .................................................................................. 15 Figure 12 β Waveform at t = T/2 = 178 ns .............................................................................................. 15
Figure 13 β Waveform at t = 3T/2= 534 ns ............................................................................................. 16 Figure 14 β (C,D) β C on top and D on the bottom β pulse width = T ................................................... 18
Figure 15 β (E,F) β E on top and F on the bottom β pulse width = T ..................................................... 19 Figure 16 β (C,D) β C on top and D on the bottom β pulse width = 10T ............................................... 19 Figure 17 β (E,F) β E on top and F on the bottom β pulse width = 10T ................................................. 19
Figure 18 βTheoretical V vs. t at point F β pulse width = 10T ............................................................... 21 Figure 19 βTheoretical V vs. t at point D β pulse width = 10T .............................................................. 22
Figure 20 βTheoretical V vs. t at point F β pulse width = T ................................................................... 22 Figure 21 βTheoretical V vs. t at point D β pulse width = T .................................................................. 23 Figure 22 β format of wave for V(zβ) and I(zβ) ....................................................................................... 26
5
Section A
The value of characteristic impedance Zo is 50Ξ©. This was determined by trial and error by checking for the load impedance that produced no reflection. Below is the expected expected waveform at point C.
Figure 1 β Expected Voltage Waveform at C
0
20
40
60
80
100
120
0 50 100 150 200 250
Vo
latg
e (
mV
)
Time (ns)
6
Section B
This section shows a calculation of πΌ1(π‘, 0) from the source parameters and the characteristic impedance.
ππ = 100 ππ
π1(π‘, 0) = 100 ππ 0 β€ π‘ β€ 100
0 ππ‘βπππ€ππ π
π0 = 50Ξ©
π π πππππ = 50Ξ© + π0 = 50Ξ© + 50Ξ© = 100Ξ©
πΌ1(π‘, 0) = π1(π‘, 0)
π π πππππ =
1 ππ΄ 0 β€ π‘ β€ 100
0 ππ‘βπππ€ππ π
7
Section C
This section shows the different waveforms at points C, D, F and E for RL = 50Ξ©. It will be showed with two different types of plot: V vs. t, where we expect delayed waveform at positions further from the source; and V vs. d, where we expect to observe the pulse move along the line.
Figure 2β Waveform at point C (top) and D (bottom)
Figure 3 β Waveform at point E (top) and F (bottom)
Note that the voltage scale is not the same across these first four plots: the first pair is scaled
8
to half the size of the second pair. Also there is a time delay of 236ns for the second pair compared to the first. For the following two plots the blue dots represent measured values a point A, B, C and D whereas the dashed line represents the expected values assuming a phase velocity of 2.47 Γ 108
m/s, T is the pulse width.
Figure 4 β Waveform at t = T/2 = 100ns
Figure 5 β Waveform at t = 2T = 400ns
Calculations to find expected V vs. d:
0
20
40
60
80
100
120
0 10 20 30 40 50 60 70 80 90
Vo
ltag
e (
mV
)
Distance from source (m)
0
20
40
60
80
100
120
0 10 20 30 40 50 60 70 80 90
Vo
ltag
e (
mV
)
Distance from source (m)
9
Figure 4:
Plotting at π‘ =π
2 < π therefore the end of the pulse will not yet have reached the source;
therefore the source is at Vg = 100 mV.
By π‘ =π
2 the front of the pulse will have reached
π = π£π Γπ
2= 2.47 Γ 108
π
π Γ 100 ππ = 24.7 π
therefore the line will be at 100 mV from the source to π = 24.7 π, and at 0 V, beyond there.
Figure 5:
Plotting at π‘ = 2π > π therefore the end of the pulse passed the source; therefore the source is at 0 V.
By π‘ = 2π the back of the pulse will have reached
π = π£π Γ (2π β π) = 2.47 Γ 108π
π Γ 200 ππ = 49.4 π
and the front of the pulse will have reached
π = π£π Γ 2π = 2.47 Γ 108π
π Γ 400 ππ = 98.8 π
which is passed the end of the line, therefore the line will be at 100 mV from π =49.4 π to the end of the line and at 0 V before that.
10
Section D
In section D is proposed to complete a table with the values of the delay time and the phase velocity in which point C, D, E and F. To evaluate the delay time we just need to compare the delay between the point and the point of reference z=0, the point C. The table follows above with the calculations. LCC = 0 LCD = 30m LCE = 60m LCF = 90m
π£ππΆ =πΏπΆπΆ
π‘π·=
0π
128ππ = 0 π/π
π£ππ· =πΏπΆπ·
π‘π·=
30π
128ππ = 2.35π₯108π/π
π£ππΈ =πΏπΆπΈ
π‘π·=
60π
236ππ = 2.54π₯108π/π
π£ππΉ =πΏπΆπΉ
π‘π·=
90π
356ππ = 2.53π₯108π/π
z (m) t (ns) v x 10^8 (m/s)
C 0 0
D 128 2.35
E 236 2.54
F 356 2.53
π£ππ΄ππΈ =(π£ππ· + π£ππΈ + π£ππΉ)
3= 2.47π₯108π/π
π£ππ΄ππΈ = 1/βΒ΅β° , considering Β΅ = Β΅o and β° = β°o.β°, π£ππ΄ππΈ =1
βΒ΅β°πβ°πβ΄ β°π =
1
π£ππ΄ππΈ2 β°πΒ΅0
= 1.47
11
Section E
This section ask to us compare in a table the values of ΞL measured and ΞL calculated from a pulse with width approximately equal to the total delay of the transmission line. The ΞL is measured in accord with the values of the voltage reflected and the voltage incident. Those values can be extracted from the plot V vs t. Those plots will follow above the table, C will be the top (yellow) wave and F the bottom (green) wave. The ΞL can be calculated with the values of RL and Ro.
RL(Ξ©) ΞL(meas)=Vreflect/Vincident ΞL(calculated) Error (%)
0 -86.875/104.375 = -0.8323 -1
16.77
20 -40/100.625 = -0.3975 -0.4286
7.26
100 29.375/101.25 = 0.2901 0.3333
12.96
β 92.5/98.75 = 0.9367 1
6.33
For RL = 0
ΞπΏ =ππΏ β ππ
ππΏ + ππ=
0 β 50
0 + 50= β1
For RL = 20
ΞπΏ =ππΏ β ππ
ππΏ + ππ=
20 β 50
20 + 50= β0.4286
For RL = 100
ΞπΏ =ππΏ β ππ
ππΏ + ππ=
100 β 50
100 + 50= 0.3333
For RL = β
ΞπΏ =ππΏ β ππ
ππΏ + ππ=
β β 50
β + 50= πππππ‘πππππππ‘π
RL = β means that the load is an open-circuit. In that way, we can interpret it as the wave that goes through the transmission line is the same that come back. In other words:
ΞπΏ =ππ
βππΎπ
ππ+πβπΎπ
= +1
12
Figure 6 β (C,F) β RL = 0
Figure 7 β (C,F) β RL = 20Ξ©
13
Figure 8 β (C,F) β RL = 100Ξ©
Figure 9 β (C,F) β RL = β
On the plots we can clearly observe the results of the table. The errors can occur because of imprecision, connexions and the waste of the equipment, like the internal resistors.
14
When RL = 0 we can see that there is no pulse at point F. This occurs because the pulses cancel themselves out at that point, since ΞL = -1. When RL = 20Ξ©, we can observe a small signal at point F. As this transmission line is lossless, this is the sum of the first and second wave at the point C. Since ΞL is negative and has a magnitude less than 1, the second pulse at point C is negative and smaller that the original. When RL = 100Ξ©, the wave at point F becomes bigger than the original one, because the ΞL is positive, and the sum of the incident and reflected pulses will be bigger than the incident pulse. Finally, when RL = β, we see no significant loss between the pulses at point C. The sum of those two waves is the wave F, the biggest value possible for this point. Below are the graphs V vs t for (C,D) and (E,F) pair, for RL = 20Ξ©.
Figure 10 β (C,D) β C on top and D on the bottom
15
Figure 11 β (E,F) β E on top and F on the bottom
Below are the graphs V vs. d at time T/2 and 3T/2, where T is the pulse width as well as the total delay of the line (i.e. they are the same). As before the blue dots are measured values (read of the V vs. t graphs) and the dashed lines are expected values.
Figure 12 β Waveform at t = T/2 = 178 ns
0
20
40
60
80
100
120
0 10 20 30 40 50 60 70 80 90
Vo
ltag
e (
mV
)
Distance from source (m)
16
Figure 13 β Waveform at t = 3T/2= 534 ns
Calculations to find expected V vs. d:
Figure 12:
Plotting at π‘ =π
2 < π therefore the end of the pulse will not yet have reached the source;
therefore the source is at Vg = 100 mV.
By π‘ =π
2 the front of the pulse will have reached half way across the line (as T is the
total delay of the line); therefore the line will be at 100 mV from the source to π = 45 π, and at 0 V, beyond there.
Figure 13:
Plotting at π‘ =3π
2> π therefore the end of the pulse passed the source. Also
3π
2< 2π
therefore the reflected pulse has yet to reach the source; therefore the source is at 0 V.
By π‘ =3π
2 the back of the pulse will have reached half way across the line. The
reflected pulse will have reached half way across the line as well. Therefore, one half of the line (the one closer to that source) will be at 0 V and the voltage on the other half will be the sum of the pulse and the reflected pulse:
π = ππ + π€πΏππ = 100 β 0.4286 Γ 100 = 57.14 π
The preceding plots demonstrate wave propagation with a mismatched load. This
0
10
20
30
40
50
60
70
0 10 20 30 40 50 60 70 80 90
Vo
ltag
e (
mV
)
Distance from source (m)
17
mismatching causes part of the incident pulse to be reflected. Without the reflection the
boundary conditions at the load (i.e. ππΏ = πΌπΏππΏ) cannot be satisfied. Since the source is matched the wave is only reflected once.
18
Section F
This section examines reflection at the load and the source (multiple reflections), with the RSOURCE = (50+100)Ξ© = 150 Ξ© and RL = 20Ξ©. T represents the total delay of the line (356 ns).
Figure 14 β (C,D) β C on top and D on the bottom β pulse width = T
19
Figure 15 β (E,F) β E on top and F on the bottom β pulse width = T
Figure 16 β (C,D) β C on top and D on the bottom β pulse width = 10T
Figure 17 β (E,F) β E on top and F on the bottom β pulse width = 10T
20
We can calculate the reflection at the load and source considering RL = 20Ξ©, RS = 150Ξ© and RO = 50Ξ©.
ΞπΏ =ππΏ β ππ
ππΏ + ππ=
20 β 50
20 + 50= β0.4286
Ξπ =ππ β ππ
ππ + ππ=
150 β 50
150 + 50= 0.5
This allows to find the theoretical V vs t plot at points D and F from 0 to 6T using a bounce diagram. For the plots with pulse width equal to 10T we can treat the pulse as a constant signal because the end of the pulse happens after the end of the plot At point F we have:
ππΉ1 = 0 π‘ < π
ππΉ2 = π1+ + π1
β = π1+(1 + ΞπΏ) =
ππ π 0
π π +π 0(1 + ΞπΏ) = 28.56 ππ π < π‘ < 3π
ππΉ3 = ππΉ2 + π2+ + π2
β = π1+(1 + ΞπΏ + ΞπΞπΏ + ΞπΞπΏ
2 ) = 22.44 ππ 3π < π‘ < 5π
ππΉ4 = ππΉ3 + π3+ + π3
β = π1+(1 + ΞπΏ + ΞπΞπΏ + ΞπΞπΏ
2 + Ξπ 2ΞπΏ
2 + Ξπ 2ΞπΏ
3 ) = 23.76 ππ 5π < π‘ < 7π
At point D we have:
ππ·1 = 0 π‘ <π
3
21
ππ·2 = π1+ =
ππ π 0
π π +π 0= 25 ππ
π
3< π‘ <
5π
3
ππ·3 = π1+ + π1
β = π1+(1 + ΞπΏ) = π1
+(1 + ΞπΏ) = 28.56 ππ 5π
3< π‘ <
7π
3
ππ·4 = ππ·3 + π2+ = π1
+(1 + ΞπΏ + ΞπΞπΏ ) = 17.86 ππ 7π
3< π‘ <
11π
3
ππ·5 = ππ·4 + π2β = π1
+(1 + ΞπΏ + ΞπΞπΏ + ΞπΞπΏ2 ) = 22.44 ππ
11π
3< π‘ <
13π
3
ππ·6 = ππ·5 + π3+ = π1
+(1 + ΞπΏ + ΞπΞπΏ + ΞπΞπΏ2 + Ξπ
2ΞπΏ2) = 24.74 ππ
13π
3< π‘ <
17π
3
ππ·7 = ππ·3 + π3β = π1
+(1 + ΞπΏ + ΞπΞπΏ + ΞπΞπΏ2 + Ξπ
2ΞπΏ2 + Ξπ
2ΞπΏ3 ) = 23.76 ππ
17π
3< π‘ <
19π
3
For the plots with pulse with equal to T we represent the pulse as two constant signals: one with amplitude Vg starting at t = 0 and another with amplitude -Vg starting at t = T. The first signal gives a V vs. t plot the same as the one calculated for pulse widths equal to 10T; the second gives the same plot once again but negative and shifted right by T. The sum of these two plots is the desired V vs. t plot for the pulse.
Figure 18 βTheoretical V vs. t at point F β pulse width = 10T
0
5
10
15
20
25
30
0 1 2 3 4 5 6
Vo
ltag
e (
mV
)
Time/T
22
Figure 19 βTheoretical V vs. t at point D β pulse width = 10T
Figure 20 βTheoretical V vs. t at point F β pulse width = T
0
10
20
30
40
50
60
0 1 2 3 4 5 6
Vo
ltag
e (
mV
)
Time/T
-10
-5
0
5
10
15
20
25
30
35
0 1 2 3 4 5 6
Vo
ltag
e (
mV
)
Time/T
23
Figure 21 βTheoretical V vs. t at point D β pulse width = T
The theoretical results are very similar to the experimentally measured ones. Of course the measured values are not perfectly precise, the lines on the oscilloscope are quite thick; but the shapes of the graphs as well as their amplitudes match the expected result
-40
-30
-20
-10
0
10
20
30
40
50
60
0 1 2 3 4 5 6
Vo
ltag
e (
mV
)
Time/T
24
Section G
In this section we vary the frequency, find the minimum of the voltage at the input of the line. First with the line terminated in a short circuit and then in a 0.01Β΅F capacitor. This can be done analysing the wave as a sinusoidal wave. We know that the length of the line (L) is 90 m and we know the wave speed as well as the
frequency (π) therefore we can find the line length is term of the wavelength for each case. π£π = ππ
π =π£π
π
πΏ
π=
πΏπ
π£π=
90π
2.47 Γ 108= 3.64 Γ 10β7π
πΏ
π1= 3.64 Γ 10β7 Γ 4.16 Γ 106 = 1.52
πΏ
π2= 3.64 Γ 10β7 Γ 6.949 Γ 106 = 2.53
πΏ
π3= 3.64 Γ 10β7 Γ 11.45 Γ 106 = 4.17
πΏ
π4= 3.64 Γ 10β7 Γ 4.18 Γ 106 = 1.52
πΏ
π5= 3.64 Γ 10β7 Γ 6.94 Γ 106 = 2.53
πΏ
π6= 3.64 Γ 10β7 Γ 11.09 Γ 106 = 4.04
Line terminated in short circuit.
Frequency (MHz) Length (in terms of Ξ) V1 Vg
4.16 1.52 4.43 mV 100 mV
6.949 2.53 4.675 mV 100 mV
11.45 4.17 5.75 mV 100 mV
Line terminated in 0.01Β΅F.
Frequency (MHz) Length (in terms of Ξ) V1 Vg
4.18 1.52 4.75 mV 100 mV
6.94 2.53 5.125 mV 100 mV
11.09 4.04 6.125 mV 100 mV
25
To analyse the results we use the following equation from the course notes:
ππΏ = π πππ
π π + π ππβππ½π (
1 + ΞπΏ
1 β ΞπΞπΏπβ2ππ½π)
From which we derive
|ππΏ| = π πππ
π π + π π
β(1 + ΞπΏ
1 β ΞπΞπΏπβ2ππ½π) (
1 + ΞπΏβ
1 β ΞπΞπΏβπ2ππ½π
)
|ππΏ| = π πππ
π π + π π
β(1 + 2π πΞπΏ + |ΞπΏ|2
1 β 2π πΞπΞπΏπβ2ππ½π + |ΞπΞπΏ|2)
|ππΏ| = π πππ
π π + π π
β(1 + 2π πΞπΏ + |ΞπΏ|2
1 β 2Ξπ(π πΞπΏcos(2ππ½π) + πΌπΞπΏsin(2ππ½π)) + Ξπ2|ΞπΏ|2
)
From this equation we can see that |ππΏ| is greatest (as a function of π) when
2Ξπ (ΞπΏπcos(2ππ½π) + ΞπΏπ
sin(2ππ½π)) greatest. Since πΌπΞπΏ = 0 for a short circuit termination, this
maximum occur where cos(2ππ½π) = 1, where π is an integer multiple of π
2. A result that fits well
with what we observe for the short terminated line. For a capacitor terminated line πΌπΞπΏ β 0
nor is negligible compared to π πΞπΏ since the impedance of the capacitor and the characteristic
impedance are of the same order of magnitude (1
2 πΓ4.18Γ106Γ0.01Γ10β6 = 3.81 ~ 50). Therefore the
maximum voltages for the capacitor vs. short terminated lines should be different. The results are not different; this may indicate that the capacitor used to terminate the line was defective. The maximum current occurs when the voltage has a minimum value. We can see it simply analysing two formulas:
|π(π§β²)| =|πΌπΏ|
2|ππΏ + π π|β[1 + |ΞπΏ
2| + 2|ΞπΏ| cos(ππ β 2π½π§β²)] (1)
|πΌ(π§β²)| =|πΌπΏ|
2π π|ππΏ + π π|β[1 + |ΞπΏ
2| β 2|ΞπΏ| cos(ππ β 2π½π§β²)] (2)
Since those two formulas are in function of zβ, the (1) will reach the maximum when
cos(ππ β 2π½π§β²) is maximum. But the (2) will be a minimum when cos(ππ β 2π½π§β²) is maximum. Furthermore, when cos(ππ β 2π½π§β²) is minimum, the formula (2) will reach the maximum, and the formula (1) will reach his minimum, as we expected.
26
Figure 22 β format of wave for V(zβ) and I(zβ)
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