Transmission line

COMMUNICATION SYSTEMS 1 (KNE334)

Transmission line Lab Report Submitted By: (50%) 143082 - Gabriella Tregurtha1 (50%) 176259 - Jiri Bednar2

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[email protected] [email protected]

Date of labs: 26 August 2015 9 September 2015 16 September 2015 Transmission line: #7 Date of submission: 23 September 2015

Submitted To: Peter Watt

KNE334 Transmission line Lab Report

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Contents 1 Introduction

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2 Theory

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3 Lab 3.1 Observation of step response . . . . . . . . . . . . 3.1.1 Unmatched source, open circuit . . . . . . 3.1.2 Unmatched source, short circuit . . . . . . 3.1.3 Measurement of travel time . . . . . . . . 3.1.4 Matching the R2 . . . . . . . . . . . . . . 3.1.5 Matching the RL . . . . . . . . . . . . . . 3.1.6 Deductions . . . . . . . . . . . . . . . . . 3.2 Steady state AC measurements . . . . . . . . . . 3.2.1 Propagation constant . . . . . . . . . . . . 3.2.2 Deduction of Z0 . . . . . . . . . . . . . . . 3.2.3 Standing waves . . . . . . . . . . . . . . . 3.2.3.1 Short circuit termination . . . . . 3.2.3.2 12 Z0 termination . . . . . . . . . 3.2.3.3 Z0 (matched) termination . . . . 3.2.3.4 Termination with a mystery load 3.2.3.5 Capacitive termination −jZ0 . . 3.3 Quarter-wavelength Lines . . . . . . . . . . . . . 3.3.1 Resonance . . . . . . . . . . . . . . . . . . 3.3.2 Anti-resonance . . . . . . . . . . . . . . . 3.3.3 Deduction of α . . . . . . . . . . . . . . . 3.3.4 Quarter-wavelength transformer . . . . . . 3.4 Limitations of the lumped model . . . . . . . . . 4 Discussion

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3 3 3 3 3 5 5 5 9 9 10 10 10 11 11 12 13 13 13 14 14 14 14 15

Appendices 17 A MATLAB code for simulations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 B Raw datapoints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19


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2

Introduction

When considering small electronic circuits operating at low frequencies, standard Circuit Theory can be used to calculate all characteristics of the circuit. However, when the size of the circuit and/or the operating frequency increases, the wavelength of the signals can get smaller than the circuit, and the Transmission Line theory needs to be employed to calculate all characteristics accurately. In this lab, a simulated transmission line composed of lumped LC sections is used to investigate the basic properties of transmission lines.

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Theory

An ideal lossless transmission line can be modelled by constructing a series of inductors and capacitors as seen in figure 2.1, and letting δx approach zero. In this lab, a model using 12

Figure 2.1: Lumped model of a transmission line. A good approximation to an ideal transmission line can be obtained by letting δx approach zero. lumped LC sections was used. In this case the δx is non-zero and hence the number of sections is finite, therefore this model can be easily solved using state space analysis instead of solving partial differential equations. Let ~x be a vector of states, A a system matrix, and B an input matrix, so that ~x˙ = A~x + Beg :   s 0 0 ··· 0 − C1s 0 ··· 0 0 − RT1Cs − G Cs 1  s 0 ··· 0 − C1s · · · 0 0  0 −G     Cs Cs eC1   .. .. .. Gs 1   . 0 0 . . 0 0 − Cs · · ·  eC2  Cs    .    .. .. .. .. .. .. . .  .  1 . .  . . −C . . 0  . . . .  .  s     Gs 1 1   0 0 0 · · · − 0 0 · · · − eC12  Cs Cs Cs  ~x =  , A = 1 1 R  − Ls 0 ··· 0 − Lss 0 ··· 0 0   iL1  Ls     1 1 R   s  iL2  0 − · · · 0 0 − · · · 0 0   L L L s s s  .    . . . . . . .  ..  1 . . . . . . .   . . 0 0 . . . . . Ls     .. .. .. iL12 .. Rs 1 .  0 0 · · · − Ls 0  . . . − Ls s 0 0 0 · · · L1s 0 0 ··· 0 − ZLL+R s i h B = RAT TCs 0 · · · 0 where the Thevenin amplitude AT =

R1 R1 + R2

(2.1)


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and the Thevenin resistance

R2 Rg (2.2) R1 The MATLAB code used to produced to the full matrix A is provided in appendix A.2, and the code to solve the state space equation numerically using Euler’s method in appendix A.1. To implement reactive load, the ZL should be set to infinity (open circuit), and the values of Ls , Rs , Cs , or Gs in the last segment modified to suit the needs. RT = R2 + Rg +

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Lab

The transmission line model is composed of 12 inductors (assumed identical) and 12 capacitors (also assumed identical), connected as seen in figure 2.1. When the switch S1 is set to open (labeled as ”Setup”), one individual capacitor and inductor can be measured using LRC meter. The values have been found to be Ls = 0.713 mH, Cs = 6.79 nF, Gs = 0 S, and D = 0.055. The α coefficient can be calculated as α = tan−1 (D) = 0.055 Np, from which the resistance per segment Rs can be calculated as Rs = 2πf Ls α = 0.0246 Ω. The model also includes 2 potentiometers R1 and R2 for adjusting the internal impedance of the generator, with ranges R1 ∈ [0.1, 117.9] Ω and R2 ∈ [1.1, 487] Ω. From the information q above it can be inferred Rs +jωLs which is equal to that the characteristic impedance of the transmission line is Z0 = G s +jωCs Z0 = (325.5 − j6.2) Ω at ω = 2π100 Hz. However, because the Rs and Gs is very small, they q

can be neglected, yielding a simpler expression Z0 =

3.1

Ls Cs

= 324 Ω.

Observation of step response

During all measurements below the value of R1 is set to its maximal value of R1 = 117.9 Ω. 3.1.1

Unmatched source, open circuit

The potentiometer R2 is set to its lowest value of R2 = 1.1 Ω. According to (2.2), this results in the sending impedance of RT = 19.3 Ω. Therefore the sending reflection coefficient ks = 19.3−324 = −0.888. For open circuit, the receiving reflection coefficient is kr = +1. This results 19.3+324 in the voltage at the receiving end overshooting the input voltage and falling below it in an oscillatory manner, slowly converging towards the input voltage, as seen in figure 3.1. 3.1.2

Unmatched source, short circuit

Again, the R2 = 1.1 Ω and therefore ks = −0.888. The receiving coefficient for a short circuit is kr = −1. Therefore the output will stay at zero volts, and the voltage at the line input will decay asymptotically towards zero from the initial voltage, as seen in figure 3.2 3.1.3

Measurement of travel time

To measure the travel time, a unit step was sent into the transmission line and the reflections were observed on an oscilloscope. To achieve greater accuracy, the time between the initial step and the determined number of reflections was measured, and divided by the number of reflections to produce an average of the travel time. This has been observed to be T = 32813µs = 25.2 µs. The travel time then can be used to derive the length of the modelled transmission line. Assuming that the velocity of propagation is approximately the speed of light (vp = c =


(a) Observed waveform at the input (yellow) and the output (cyan) of the transmission line.

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(b) Theoretical waveform based on simulation. The bottom diagram shows propagation of the wave across the line, and reflections at both ends.

Figure 3.1: When the sending and receiving impedance doesn’t match the characteristic impedance of the transmission line, a reflection occurs which can be measured at both ends of the line. This diagram shows the input and output voltage when the termination is an open circuit.

(a) Observed waveform at the input (yellow) and the output (cyan) of the transmission line.

(b) Theoretical waveform based on simulation. The bottom diagram shows propagation of the wave across the line, and reflections at both ends.

Figure 3.2: When the sending and receiving impedance doesn’t match the characteristic impedance of the transmission line, a reflection occurs which can be measured at both ends of the line. This diagram shows the input and output voltage when the termination is a short circuit.


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3 × 108 m s−1 ), then the length can be calculated as l = vp T = 7.57 km. Therefore, this length can be divided between 12 sections, making each 630 m long. 3.1.4

Matching the R2

The source is matched with the transmission line if its internal impedance is equal to line’s characteristic impedance (RT = Z0 ). Therefore using (2.2) it can be calculated that R2 = 262 Ω. There are two methods for ascertaining the optimum value of R2 which results in matched sending end of the transmission line. First method uses the oscilloscope in the X-T mode and matching is deduced by eye, where the target is to minimise the second reflection. The second method is to use the oscilloscope in the X-Y mode where the input of the transmission line is on the horizontal axis and the output on the vertical axis. This way the matched value of R2 can be found by trying to make the resulting Lissajous figure thin. Figure 3.3 shows the effect of tuning the value of R2 around the matched value for the open circuit termination. The matched value has been found to be R2 = 269 Ω for the open circuit termination, which according to (2.2) results in sending impedance of Zs = 328 Ω, which is very close to the theoretical characteristic impedance of the transmission line. The same process applies to the short circuit termination. Figure 3.4 shows the effect of tuning the value of R2 around the matched value for the short circuit termination. The matched value has been found to be R2 = 259 Ω for the short circuit termination, which according to (2.2) results in sending impedance Zs = 317 Ω, which is also very close to the theoretical characteristic impedance of the transmission line. 3.1.5

Matching the RL

Setting R2 to the optimum value found in the previous step, the effect of adjusting RL can be investigated without interference from waves reflected from the source. The method is similar to the one used to find the optimum value of R2 , and figure 3.5 shows the effect of varying the RL . By applying the method, the optimum value has been found to be RL = 340 Ω, which is very close to the theoretical characteristic impedance of the transmission line. 3.1.6

Deductions

Since both sending and receiving impedances matched the theoretical characteristic impedance of the transmission line in all cases where no reflections have been observed, it can be deduced that the theoretical characteristic impedance is very similar to the one observed during matching. As such, the transmission line theory previously discussed is consistent with practical observations. The model of the transmission line therefore exhibits all major characteristics of a real transmission line.


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Voltage

Voltage

Voltage

1.4 Input Output

Input Output

0.8

0.8

0.6

0.6

e(t) [V]

1

0.6

Input Output

1

0.8

e(t) [V]

e(t) [V]

1.2

1

0.4

0.4

0.4 0.2

0.2

0.2 0

0 1.9

1.95

2

2.05

2.1

2.15

2.2

2.25

2.3

2.35

t [s]

×10

1.9

2.4

2

2.05

2.1

2.15

2.2

2.25

2.3

t [s]

Lissajous figure

1.4

0 1.95

-3

2.35 ×10

2.4

1.9

-3

2

2.05

1.2

2.1

2.15

2.2

2.25

2.3

t [s]

2.35

2.4

×10 -3

Lissajous figure

0.9

1.2

0.8

1 1

0.7 0.8

0.6 0.4 0.2 0

0.6

Output e(t) [V]

Output e(t) [V]

0.8

Output e(t) [V]

1.95

Lissajous figure

0.6

0.4

0.5 0.4 0.3 0.2

0.2

-0.2

0.1 0

-0.4 -0.6 -0.2

0 0

0.2

0.4

0.6

0.8

1

1.2

Input e(t) [V]

(a) Negative reflection coefficient, R2 = 0.5Z0 . Oscillatory behaviour is observed.

-0.2 -0.2

0

0.2

0.4

0.6

0.8

1

Input e(t) [V]

(b) Zero reflection coefficient ks = 0, matched sending impedance R2 = Z0 .

-0.1 -0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

Input e(t) [V]

(c) Positive reflection coefficient, R2 = 1.5Z0 . Asymptotic behaviour is observed.

Figure 3.3: The matched value of R2 can be found by trying to align the reflected voltage with the sent voltage when the termination is open circuit (receiving reflection coefficient kr = +1), hence eliminating the reflection at the sending end of the transmission line. When the sending impedance is too low (figure 3.3a), the sending reflection coefficient is negative and the reflected wave subtracts, resulting in oscillatory behaviour. When the sending impedance is too high (figure 3.3c), the reflected wave adds, resulting in the voltage asymptotically approaching the final value in steps. When the sending impedance is matched (figure 3.3b), the reflected wave is absorbed entirely, so the voltage stabilises in just one reflection.


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Voltage

Voltage

0.7

Voltage

0.5

0.5

Input Output

0.6

Input Output

Input Output

0.4

0.4 0.5

0.3 0.3

0.3

e(t) [V]

e(t) [V]

e(t) [V]

0.4 0.2

0.2

0.1

0.2 0.1 0

0.1 0

-0.1

0 -0.1 1.9

1.95

2

2.05

2.1

2.15

t [s]

2.2

2.25

2.3

2.35

2.4

×10 -3

(a) Negative reflection coefficient, R2 = 0.5Z0 . Asymptotic behaviour is observed.

-0.1 2.9

2.95

3

3.05

3.1

3.15

t [s]

3.2

3.25

3.3

3.35

3.4

×10 -3

(b) Zero reflection coefficient ks = 0, matched sending impedance R2 = Z0 .

-0.2 2.9

2.95

3

3.05

3.1

3.15

t [s]

3.2

3.25

3.3

3.35

3.4

×10 -3

(c) Positive reflection coefficient, R2 = 2Z0 . Oscillatory behaviour is observed.

Figure 3.4: The matched value of R2 can be found by having a short circuit termination (receiving reflection coefficient kr = −1), and trying to get the reflected voltage cancel the sent voltage completely, without any over- or under-shoots, ending up with stable zero voltage in just one reflection. When the R2 is too small (figure 3.4a), the sending reflection coefficient is negative, which causes the reflected wave to subtract, resulting in asymptotic decrease of input voltage towards zero in steps. When the R2 is too large (figure 3.4c), the reflection coefficient is positive, which causes the reflected wave to add to the sent voltage, resulting in decaying oscillation around zero voltage. When the sending impedance is matched, the first reflection will cancel the sent voltage entirely, resulting in stable zero input voltage.


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Voltage

Voltage

Voltage Input Output

0.5

Input Output

0.7

Input Output

0.5

0.6 0.4

0.4

0.3

0.3

0.2

0.2

0.1

0.1

0

0

e(t) [V]

e(t) [V]

e(t) [V]

0.5 0.4 0.3 0.2 0.1

2.9

2.95

3

3.05

3.1

3.15

3.2

3.25

3.3

3.35

t [s]

2.95

3

3.05

3.1

3.15

3.2

3.25

3.3

t [s]

×10 -3

Lissajous figure

0.35

0

2.9

3.4

3.35

2.9

3.4

0.6

2.95

3

3.05

3.1

3.15

3.2

3.25

3.3

3.35

t [s]

×10 -3

Lissajous figure

3.4

×10 -3

Lissajous figure

0.7 0.6

0.3

0.5 0.5

0.25

0.15 0.1

Output e(t) [V]

Output e(t) [V]

Output e(t) [V]

0.4 0.2

0.3

0.2

0.3 0.2 0.1

0.05

0.1 0

0 0

-0.1

-0.05 -0.1 -0.2

0.4

-0.1

0

0.1

0.2

0.3

0.4

0.5

Input e(t) [V]

(a) RT = Z0 , RL = 0.5Z0

-0.1 -0.1

0

0.1

0.2

0.3

0.4

Input e(t) [V]

(b) RT = Z0 , RL = Z0

0.5

-0.2 -0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

Input e(t) [V]

(c) RT = Z0 , RL = 2Z0

Figure 3.5: The matched value of RL can be found by trying to align the reflected voltage with the sent voltage, hence eliminating the reflection at the receiving end of the transmission line.


3.2

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Steady state AC measurements

By using the method described in previous section, both sending and load impedances are matched with the transmission line characteristic impedance Z0 . 3.2.1

Propagation constant

To find the propagation constant γ, consider the following equation: p γ = α + jβ = (R + jωL)(G + jωC)

(3.1)

Substituting the values of Ls , and Cs measured in Section 3, and assuming the line is loss-less (Rs = 0, Gs = 0) p ∴ γ = jω Ls Cs (3.2) p rad rad rad ∴ β = ω Ls Cs = 2.2 · 10−6 ω = 26.4 · 10−6 ω = 3.48 · 10−9 ω (3.3) section line m To find the frequency at which the phase shift is 360◦ , the equation above can be solved for ω at β = 2π: 2π = 26.4 · 10−6 ω ω = 37.8 kHz ∴f = 2π

(3.4) (3.5)

To find the frequency at which the 360◦ phase shift occurs, the oscilloscope is switched into X-Y mode, the input of the transmission line is displayed on the horizontal axis, and the output on the vertical axis. When the resulting Lissajous figure is a diagonal line with positive slope (figure 3.6c), the frequency is found. When the frequency is increased further, this line can be achieved again for integer multiples of 360◦ . It should be noted that it is more accurate to find the frequencies at which the phase shift is an integer multiple of 180◦ because the resulting Lissajous figure is a straight line. This is in contrast to finding frequencies where the phase shift is n · 180◦ + 90◦ when the resulting figure is a circle. The observed at which the radfrequency = 3.46 · 10−9 ω rad , 360◦ phase shift is achieved was 38.1 kHz, suggesting β = 26.2 · 10−6 line m which is close to the theoretical value. Lissajous figure

Lissajous figure

0.15

0.1

0.05

0.05

0.05

Output e(t) [V]

0.1

0

0

0

-0.05

-0.05

-0.05

-0.1

-0.1

-0.1

-0.15 -0.15

-0.1

-0.05

0

0.05

0.1

0.15

Input e(t) [V]

(a) f = 10.0 kHz, phase shift is 90◦ . The same pattern is observed at f = 30.0 kHz when the phase shift is 270◦ .

-0.15 -0.15

-0.1

-0.05

0

Lissajous figure

0.15

0.1

Output e(t) [V]

Output e(t) [V]

0.15

0.05

0.1

0.15

Input e(t) [V]

(b) f = 19.2 kHz, phase shift is 180◦ . The output is equal to the inverted input. The Lissajous figure is straight line.

-0.15 -0.15

-0.1

-0.05

0

0.05

0.1

0.15

Input e(t) [V]

(c) f = 38.1 kHz, phase shift is 360◦ . The output is equal to the input. The Lissajous figure is straight line.

Figure 3.6: Lissajous figures showing the relation between the transmission line input and output at various frequencies.


3.2.2

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Deduction of Z0

√ The value of Z0 may be determined using the equation Z0 = Zsc Zoc where Zsc is the sending end impedance with a short circuit termination, and Zoc with an open circuit termination, with frequency chosen so that resonance does not occur. We chose frequency at which the length of the transmission line is 38 λ, which is f = 14.3 kHz. The sending end impedance was found by measuring the input voltage and current, then Zs = EIss . The input current can be measured by measuring the voltage across the R2 resistor, then Is = VRR22 . The following values were measured:

Es [mV] Is [µA] Zs [Ω]

Open circuit 132 489 270

Short circuit 140 465 301

Therefore, the characteristic impedance is measured to be Z0 = 285 Ω, which is reasonably close to the value of Z0 = 324 Ω calculated in Section 3. 3.2.3

Standing waves

The frequency was adjusted so that the simulated line length is l = 34 λ, which corresponds to frequency f = 28.2 kHz. Various termination loads were then applied, and peak-to-peak voltages were measured at various points on the transmission line. The measured values are attached in Appendix B. 3.2.3.1 Short circuit termination When the termination is a short circuit, the reflection coefficient is kr = −1. Therefore the voltage minimum is at the receiving end, and is equal λ. Therefore to 0 Vpk-pk . The minima repeat every n2 λ from the load, with maxima at 2n+1 4 the input is located at one of these maxima. Figure 3.7 shows the standing wave pattern for +j1.0

Standing wave pattern

Impedance +j2.0 Receiving end Sending end

+j0.5

0.25 Datapoints Simulation

0.1

5.0

2.0

0.0

1.0

0.15

+j5.0

0.5

+j0.2

0.2

epk−pk [V]

0.2

-j0.2

∞

-j5.0

0.05 -j0.5

-j2.0 -j1.0

0 0

1000

2000

3000

4000

5000

6000

7000

8000

x [m]

(a) Measured peak-to-peak voltages at various points along the transmission line, compared with the simulation.

(b) The lattice diagram using 80 simulated lumped LC sections.

(c) Smith chart showing the sending and receiving impedance, as well as all other apparent impedances along the line.

Figure 3.7: Short circuit termination places the voltage minimum at the receiving end, and will be equal to 0 Vpk-pk . Therefore the sending end is at the voltage maximum. the short circuit termination. The voltage standing wave ratio was found to be VSWR = |Emax | = 0.216 = 18, so a circle was drawn on the Smith Chart with radius corresponding to that |Emin | 0.012


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VSWR. The minimum was located at the receiving end, so the load was placed on the smaller intersection of the circle and the real axis of the Smith Chart. The length of 34 λ corresponds to 3 · 180◦ = 540◦ rotation on the Smith Chart, meaning that the impedance at the sending end is located 180◦ away from the impedance at the receiving end. Therefore, the sending end impedance is located at the larger intersection of the circle and the real axis. This yields the sending end impedance of Zs = 18Z0 = 5.83 kΩ. This value is comparable to the measured mV = 1.95 kΩ. value Zs = 194 99.2 µA 3.2.3.2 21 Z0 termination When the termination is one half of the line’s characteristic impedance, the reflection coefficient is kr = − 31 . The voltage minimum is located at the λ. Therereceiving end, and the minima repeat every n2 λ from the load, with maxima at 2n+1 4 fore the input is located at one of these maxima. Figure 3.8 shows the standing wave pattern +j1.0

Standing wave pattern Datapoints Simulation

0.14

+j5.0

5.0

2.0

1.0

0.0

0.5

0.1

0.2

+j0.2

0.12

epk−pk [V]


+j0.5

0.16

∞

0.08 0.06

-j0.2

-j5.0

0.04 -j0.5

0.02

-j2.0 -j1.0

0 0

1000

2000

3000

4000

5000

6000

7000

8000

x [m]




Figure 3.8: Termination with a value lower than the line’s characteristic impedance places the voltage minimum at the receiving end. Therefore the sending end is at the voltage maximum. for the termination ZL = 21 Z0 . The voltage standing wave ratio was found to be VSWR = |Emax | 0.142 = 0.072 = 1.97, so a circle was drawn on the Smith Chart with radius corresponding |Emin | to that VSWR. The minimum was located at the receiving end, so the load was placed on the smaller intersection of the circle and the real axis of the Smith Chart. The length of 43 λ corresponds to 3 · 180◦ = 540◦ rotation on the Smith Chart, meaning that the impedance at the sending end is located 180◦ away from the impedance at the receiving end. Therefore, the sending end impedance was located at the larger intersection of the circle and the real axis. This yields the sending end impedance of Zs = 2Z0 = 648 Ω. This value is comparable to the mV measured value Zs = 188 = 879 Ω. 213 µA 3.2.3.3 Z0 (matched) termination When the termination is matched to the line’s characteristic impedance, the reflection coefficient is kr = 0. Therefore the same voltage should be measured across the whole line, and no standing waves should occur. Therefore, there are no minima and maxima. Figure 3.9 shows that there is no standing wave when the termination is max | matched (ZL = Z0 ). The voltage standing wave ratio was found to be VSWR = |E = 0.120 = |Emin | 0.096 1.25, so a circle was drawn on the Smith Chart with radius corresponding to that VSWR. The maximum was located at the 10th segment, which corresponds to 81 λ away from the load, or 90◦ clockwise rotation from the open circuit. Therefore, a line was drawn at 90◦ , and the load was placed on the intersection of the circle and the line. The length of 34 λ corresponds to


12

+j1.0



+j0.5

0.14

0.12 +j5.0

+j0.2

5.0

2.0

1.0

0.5

0.0

0.08

0.2

epk−pk [V]

0.1

∞

0.06 -j0.2

-j5.0

0.04

0.02

-j0.5

Datapoints Simulation

-j2.0 -j1.0

0 0

1000

2000

3000

4000

5000

6000

7000

8000

x [m]




Figure 3.9: When the termination impedance is matched, no reflection occurs and therefore there is no standing wave. 3 · 180◦ = 540◦ rotation on the Smith Chart, meaning that the impedance at the sending end is located 180◦ away from the impedance at the receiving end. This yields the sending end impedance of Zs = (0.99 + 0.21j)Z0 = (321 + j68) Ω. This value is comparable in magnitude mV to the measured value Zs = 184 = 574 Ω. Ideally, the sending end input impedance should 321 µA be equal to line’s characteristic impedance. 3.2.3.4 Termination with a mystery load The voltage standing wave ratio was found 0.15 max | = 0.074 = 2.07, so a circle was drawn on the Smith Chart with radius to be VSWR = |E |Emin | corresponding to that VSWR. The minimum was located at the 10th segment, which corresponds to 81 λ away from the load, or 90◦ clockwise rotation from the short circuit. Therefore, a line was drawn at 90◦ , and the load was placed on the intersection of the circle and the line. This impedance corresponds to load ZL = (0.775 − 0.625j)Z0 = (251 − 202j) Ω, which is equivalent to a 251 Ω resistor in series with a 27.9 nF capacitor. Because the load is complex, the minima and maxima are not located exactly on the sending or receiving end. Figure 3.10 shows the +j1.0

Standing wave pattern Datapoints Simulation

0.14

5.0

2.0

1.0

0.0

+j5.0

0.5

0.1

0.2

+j0.2

0.12

epk−pk [V]


+j0.5

0.16

∞

0.08 0.06

-j0.2

-j5.0

0.04 -j0.5

0.02

-j2.0 -j1.0

0 0

1000

2000

3000

4000

5000

6000

7000

8000

x [m]




Figure 3.10: The mystery load causes the minimum to shift away from the load, so it must contain a reactive component. It also doesn’t cause an infinite VSWR, so the real component must be non-zero.


13

standing wave pattern for the termination with the mystery load, which was determined to be ZL = (251 − 202j) Ω (calculation above). The length of 34 λ corresponds to 3 · 180◦ = 540◦ rotation on the Smith Chart, meaning that the impedance at the sending end is located 180◦ away from the impedance at the receiving end. This yields the sending end impedance of Zs = 2Z0 = (251 − 202j) Ω. This value is comparable in magnitude to the measured value mV = 307 Ω. Zs = 176 573 µA 3.2.3.5 Capacitive termination −jZ0 At frequency 28.2 kHz the capacitor with impedance −j324 Ω has value 17.4 nF. The closest value available in the lab was 18.1 nF, which has an impedance of −j311 Ω. Figure 3.11 shows the standing wave pattern for the termination max | ZL = −jZ0 . The voltage standing wave ratio was found to be VSWR = |E = 0.23 = 7.66, so |Emin | 0.03 +j1.0



+j0.5

0.25 Datapoints Simulation

+j5.0

0.1

5.0

2.0

0.0

1.0

0.15

0.5

+j0.2

0.2

epk−pk [V]

0.2

-j0.2

∞

-j5.0

0.05 -j0.5

-j2.0 -j1.0

0 0

1000

2000

3000

4000

5000

6000

7000

8000

x [m]




Figure 3.11: Purely reactive load impedance causes the minimum to shift away from the load. As there is no real load in which the power could dissipate, the whole signal is reflected and VSWR approaches infinity. a circle was drawn on the Smith Chart with radius corresponding to that VSWR. The minimum was located at the 10th segment, which corresponds to 81 λ away from the load, or 90◦ clockwise rotation from the short circuit. Therefore, a line was drawn at 90◦ , and the load was placed on the intersection of the circle and the line. The length of 43 λ corresponds to 3 · 180◦ = 540◦ rotation on the Smith Chart, meaning that the impedance at the sending end is located 180◦ away from the impedance at the receiving end. This yields the sending end impedance of Zs = (0.25 + j0.95)Z0 = (81 + j308) Ω. This value is comparable in magnitude to the measured mV value Zs = 184 = 415 Ω. 442 µA

3.3

Quarter-wavelength Lines

When the length of a transmission line is

2n+1 λ, the input impedance of the line can be expressed 4 Z02 . Two special cases are considered in this lab, where ZL

in terms of the load impedance as Zs = the load impedance ZL is an open circuit (ZL → ∞), and a short circuit (ZL = 0). 3.3.1

Resonance

The resonance occurs when the input impedance of the transmission line is zero for an open circuit termination. This state can be found by tuning the frequency until the input voltage is


14

zero. We found this frequency to be f = 9.92 kHz, at which the input impedance for an open 24 mV circuit termination was Zs = 64 = 0.37 Ω. This impedance is very low and can be regarded mA as a short circuit. It is not exactly zero because the line is not perfectly loss-less (α 6= 0), and because a lumped approximation to a real transmission line was used. 3.3.2

Anti-resonance

The anti-resonance occurs when the input impedance of the transmission line is infinite for a short circuit termination. The frequency is the same as for the resonance. The input impedance mV = 8521 Ω. This impedance is very large compared to the characwas found to be Zs = 196 23 mA teristic impedance of the line, and as such can be regarded as approximately open circuit. 3.3.3

Deduction of α

The attenuation constant α may be deduced from the previous quarter-wavelength line measurements. For an open-circuit stub of length l = 41 λ, the following equation is true: Z0 tanh(γl) γl e + e−γl = Z0 γl e − e−γl

Zs =

(3.6) (3.7)

π

Because eγl = eαl ej 2 = jeαl ∴ Zs = Z0

jeαl + je−αl jeαl − je−αl

(3.8)

For a low-loss line where αl 1, first order Taylor approximation can be made (eαl ≈ 1 + αl, e−αl ≈ 1 − αl): ∴ Zs ≈ Z0 αl

(3.9)

Substituting the input impedance Np obtained in Section 3.3.1, the propagation constant was −7 . This is a very small number, and as such the assumption determined to be α = 1.5 · 10 m of the lossless line is a valid approximation. 3.3.4

Quarter-wavelength transformer

A quarter-wavelength transformer is a technique used to match transmission lines and loads of Z2 different impedances to minimise reflections. In general, the input impedance is Zs = ZL0 . A load impedance ZL = 2Z0 = 648 Ω was applied, and the sending end impedance was found to Z02 mV be Zs = 640 = 156 Ω. This is very close to the expected Z = = Z20 = 162 Ω. Therefore, s 410 mA 2Z0 if the load would be connected to a line with characteristic impedance of Z0 = 156 Ω through this transformer, the line would be matched.

3.4

Limitations of the lumped model

The use of lumped LC section model of a transmission line is a reasonable model at low frequencies, where the voltage and current waveforms vary little from section to section. As the frequency is increased, the approximation breaks down and the system begins to act as a


15

low-pass filter. A very definite breakdown occurred at 240 kHz, which is approximately 6 wavelengths, where each section corresponds to a half wavelength. Increasing the number of sections and decreasing the capacitance and inductance of each section proportionally produces a better approximation. As the number of sections approaches infinity, the behaviour approaches that of a real transmission line. 10 4

10 4 ZL ZL ZL ZL ZL

=0 = 12 Z0 = Z0 = 2Z0 =∞

ZL ZL ZL ZL ZL

Zs (f) [Ω]

10 3

Zs (f) [Ω]

10 3

=0 = 12 Z0 = Z0 = 2Z0 =∞

10 2

10 1

10 2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

f [Hz]

1.8

2 ×10

5

(a) 12 segment lumped transmission line

10 1

0

0.2

0.4

0.6

0.8

1

f [Hz]

1.2

1.4

1.6

1.8

2 ×10 5

(b) 500 segment lumped transmission line

Figure 3.12: Plots showing input impedances of the simulated transmission lines as function of frequency. It can be seen that the 12 segment line (figure 3.12a) serves as a good approximation at low frequencies, but introduces distortion at higher frequencies, and stops working completely when the number of wavelengths per line reaches 12. Lattice diagram 0

×10 -4 0.08

t [s]

0.2 0.4

0.06

0.6

0.04

0.8

0.02

1

0

1.2

-0.02

1.4

-0.04

1.6

-0.06

1.8

-0.08

2 0

2

4

6

8

10

12

δx

(a) 12 segment lumped transmission line

(b) 120 segment lumped transmission line

Figure 3.13: Plots showing lattice diagrams of the simulated transmission lines running at 225 kHz, which corresponds to 6 wavelengths per line. It can be seen that in the 12 segment line (figure 3.13a) the signal doesn’t even propagate to the second section, whereas in the 120 segment line (figure 3.13b) the signal propagates to the output.

4

Discussion

The lumped model approximation of an ideal transmission line shows all the main characteristics of the transmission line, but breaks down for larger frequencies (smaller wavelengths) where the


16

voltage differences between adjacent lumps become significant. This lumped model could also be used to accurately approximate the overhead transmission lines, where each tower/pole is represented by a discrete capacitor (and a conductor in lossy lines), and each line by a discrete inductor (and a resistor). The model is useful in the laboratory as it allows for fast and easy experimentation with transmission lines without having to handle kilometres of cabling.


17

Appendices A

MATLAB code for simulations

To produce the lattice diagrams, Lissajous figures, and input/output waveforms, the following MATLAB code was developed which employs the Euler’s method to numerically solve the differential equations corresponding to the state space of the circuit. Listing A.1: Main simulator procedure 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61

% - - - - Config n =12; dt =1 e -10; step =2000; t =0: dt :4 e -4; fc =37573.8*3/4; % fc =1 e3 ; u =0.278* sin (2* pi * fc * t ) ; % u =( sign ( u ) +1) /2; Rg =18; Ls =12/ n *0.713 e -3; Cs =12/ n *6.79 e -9; Rs =12/ n *2* pi *100* Ls *0.055; Gs =12/ n *1.0 e -12; Z0 = real ( sqrt (( Rs +1 j *2* pi * fc * Ls ) /( Gs +1 j *2* pi * fc * Cs ) ) ) ; R1 =117.9; R2 =262*1; Zl = Z0 *1 e0 ; % - - - - Bode plot of line f =10.^(0:0.0005:6) ; Zls =[0 0.5* Z0 Z0 2* Z0 1 e12 ]; Z = zeros ( length ( Zls ) , length ( f ) ) ; for i =1: length ( f ) for j =1: length ( Zls ) Z (j , i ) = buildZ ( Rs , Cs , Ls ,1 e -9 , Zls ( j ) ,2* pi * f ( i ) ,n ) ; end end figure ; subplot (2 ,1 ,1) ; plot (f , abs ( Z ) ) ; % set ( gca , ’ xscale ’ , ’ log ’) ; set ( gca , ’ yscale ’ , ’ log ’) ; grid on ; subplot (2 ,1 ,2) ; plot (f , angle ( Z ) ) ; % set ( gca , ’ xscale ’ , ’ log ’) ; grid on ; % figure ; plot (f ,[ real ( Z (1 ,:) ) ; imag ( Z (1 ,:) ) ]) ; % set ( gca , ’ yscale ’ , ’ log ’) ; % - - - - Magic below et =1* R1 /( R1 + Rg ) ; Rt = R2 + Rg + R2 * Rg / R1 ; A = buildA ( Rs , Cs , Ls , Gs , Zl , Rt , n ) ; B =[ et / Rt / Cs ; zeros (2* n -1 ,1) ]; x = zeros (2* n ,1) ; xres = zeros ( length ( x ) , length ( t (1: step : end ) ) ) ; j =1; for i =1: length ( t ) xdot = A * x + B * u ( i ) ; x =( x + xdot * dt ) ; if ( mod (i , step ) ==1) xres (: , j ) = x ; j = j +1; end end % - - - - End of magic . P l o t t i n g


62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102

18

figure ; subplot (3 ,2 ,1) ; plot ( t (1: step : end ) ,[ xres (1 ,:) ; xres (2* n ,:) * Zl ]) ; grid on ; title ( ’ Voltage ’ , ’ interpreter ’ , ’ latex ’) ; xlabel ( ’ $t$ [ s ] ’ , ’ interpreter ’ , ’ latex ’) ; ylabel ( ’ $e ( t ) $ [ V ] ’ , ’ interpreter ’ , ’ latex ’) ; legend ({ ’ Input ’ , ’ Output ’} , ’ interpreter ’ , ’ latex ’) ; subplot (3 ,2 ,2) ; plot ( t (1: step : end ) ,[ xres ( n +1 ,:) ; xres (2* n ,:) ]) ; grid on ; title ( ’ Current ’ , ’ interpreter ’ , ’ latex ’) ; xlabel ( ’ $t$ [ s ] ’ , ’ interpreter ’ , ’ latex ’) ; ylabel ( ’ $i ( t ) $ [ A ] ’ , ’ interpreter ’ , ’ latex ’) ; legend ({ ’ Input ’ , ’ Output ’} , ’ interpreter ’ , ’ latex ’) ; subplot (3 ,2 ,3) ; plot ( xres (1 , round ( length ( xres (1 ,:) ) /2) : end ) , xres (2* n , round ( length ( xres (1 ,:) ) /2) : end ) * Zl ) ; title ( ’ Lissajous figure ’ , ’ interpreter ’ , ’ latex ’) ; xlabel ( ’ Input $e ( t ) $ [ V ] ’ , ’ interpreter ’ , ’ latex ’) ; ylabel ( ’ Output $e ( t ) $ [ V ] ’ , ’ interpreter ’ , ’ latex ’) ; grid on ; subplot (3 ,2 ,4) ; plot ( xres ( n +1 , round ( length ( xres (1 ,:) ) /2) : end ) , xres (2* n , round ( length ( xres (1 ,:) ) /2) : end ) ) ; title ( ’ Lissajous figure ’ , ’ interpreter ’ , ’ latex ’) ; xlabel ( ’ Input $i ( t ) $ [ A ] ’ , ’ interpreter ’ , ’ latex ’) ; ylabel ( ’ Output $i ( t ) $ [ A ] ’ , ’ interpreter ’ , ’ latex ’) ; grid on ; subplot (3 ,2 ,5) ; surf ( t (1: step : end ) ,0:n ,[ xres (1: n ,:) ; xres (n ,:) ] , ’ linestyle ’ , ’ none ’) ; xlabel ( ’ $t$ [ s ] ’ , ’ interpreter ’ , ’ latex ’) ; ylabel ( ’$ \ delta x$ ’ , ’ interpreter ’ , ’ latex ’) ; zlabel ( ’ $e (t , x ) $ [ V ] ’ , ’ interpreter ’ , ’ latex ’) ; title ( ’ Lattice diagram ’ , ’ interpreter ’ , ’ latex ’) ; axis ( ’ tight ’) ; view ([0 90]) ; subplot (3 ,2 ,6) ; surf ( t (1: step : end ) ,0:n ,[ xres ( n +1:2* n ,:) ; xres (2* n ,:) ] , ’ linestyle ’ , ’ none ’) ; xlabel ( ’ $t$ [ s ] ’ , ’ interpreter ’ , ’ latex ’) ; ylabel ( ’$ \ delta x$ ’ , ’ interpreter ’ , ’ latex ’) ; zlabel ( ’ $i (t , x ) $ [ V ] ’ , ’ interpreter ’ , ’ latex ’) ; title ( ’ Lattice diagram ’ , ’ interpreter ’ , ’ latex ’) ; axis ( ’ tight ’) ; view ([0 90]) ;

The above procedure relies on the function buildA which produces the system matrix A. Below is the code that produces such matrix corresponding to the circuit with an arbitrary number of lumped sections. Listing A.2: Procedure to generate the state matrix 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

function A = buildA ( Rs , Cs , Ls , Gs , Zl , Rt , n ) A = zeros (2* n ) ; A (1 ,1) = -1/ Rt / Cs - Gs / Cs ; A (1 , n +1) = -1/ Cs ; for i =2: n A (i , i ) = - Gs / Cs ; A (i , n +i -1) =1/ Cs ; A (i , n + i ) = -1/ Cs ; end for i =1: n -1 A ( n +i , i ) =1/ Ls ; A ( n +i , i +1) = -1/ Ls ; A ( n +i , n + i ) = - Rs / Ls ; end A (2* n , n ) =1/ Ls ; A (2* n ,2* n ) = -( Zl + Rs ) / Ls ; end

It is also possible to determine the input impedance of the simulated transmission line by simple circuit analysis. This can be done for an arbitrary number of sections by the procedure below, and can be plotted as a function of frequency and load impedance.


19

Listing A.3: Procedure to calculate the input impedance 1 2 3 4 5 6 7 8 9 10 11

function Zl = buildZ ( Rs , Cs , Ls , Gs , Zl , omega , n ) ZL =1 j * omega * Ls + Rs ; ZC = par (1/(1 j * omega * Cs ) ,1/ Gs ) ; for i =1: n Zl = par ( ZC , ZL + Zl ) ; end end function Z = par ( Z1 , Z2 ) Z = Z1 * Z2 /( Z1 + Z2 ) ; end

B

Raw datapoints

In Section 3.2.3 we measured the peak-to-peak voltages at various points of the transmission line with different terminations. Table below lists these voltages.

v1 v2 v3 v4 v5 v6 v7 v8 v9 v10 v11 v12 vo

sc 0.208 0.182 0.15 0.086 0.012 0.0856 0.151 0.198 0.216 0.186 0.148 0.084 0.015

1 Z 2 0

0.14 0.14 0.115 0.092 0.072 0.083 0.114 0.138 0.142 0.135 0.116 0.092 0.072

Z0 0.111 0.118 0.12 0.115 0.105 0.096 0.096 0.1 0.108 0.115 0.12 0.116 0.109

? 0.116 0.09 0.079 0.095 0.125 0.145 0.15 0.145 0.118 0.092 0.074 0.094 0.124

jZ0 (18 nF) 0.15 0.071 0.03 0.033 0.176 0.22 0.23 0.204 0.15 0.076 0.03 0.11 0.178