RF Tutorial Lesson 2: Transient Analysis of a Simple Transmission Line Circuit
Contents
What You Will Learn
In this tutorial you will explore the transient response of transmission line circuits with various load configurations. You will also perform a Fourier analysis of non-sinusoidal signals.
Putting the Circuit Together
You will use the simple transmission line circuit of RF Tutorial Lesson 1 to start this project. However, instead of an AC voltage source, you will use a regular voltage source. The following is a list of parts needed for this part of the tutorial lesson:
| Part Name | Part Type | Part Value |
|---|---|---|
| VS | Voltage Source | Waveform: TBD |
| XTL1 | Generic T-Line | Defaults: Z0 = 50, eeff = 1, len = 10 |
| RS | Resistor | 50 |
| RL | Resistor | 50 |
| SOURCE | Voltage Probe Marker | N/A |
| IN | Voltage Probe Marker | N/A |
| OUT | Voltage Probe Marker | N/A |
Set the length of the T-Line segment to 60mm. Set the waveform of the voltage source to "Sinusoidal" with a frequency of 2GHz and a peak amplitude of 1V. The free-space wavelength at this frequency is λ0 = 150mm. Therefore, the length of your T-Line segment is 0.4λ0. Place three voltage probe markers (keyboard shortcut: Alt+L) called "SOURCE", "IN" and "OUT" as shown in the figure.
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Transient Simulation of a Simple Transmission Line Circuit
Run a Transient Test of your circuit with the specified parameters below. Note that your Plot Edit List must already contain v(SOURCE), v(IN) and v(OUT). The node voltages of all voltage probe markers are automatically added to the Plot Edit List.
| Start Time | 0 |
|---|---|
| Stop Time | 3n |
| Linearize Step | 1p |
| Step Ceiling | 1p |
| Preset Graph Plots | v(source), v(in), v(out) |
Run the simulation and observe the time-domain voltages. The period of your sinusoidal waveform is T = 1/f0 = 1 / 2GHz = 500ps. As you would expect, the T-Line segment is matched to both the source and the load. There are no reflections off either the input or the output of the transmission line. You can see that the output voltage has a delay of 200ps with respect to the input voltage, which is consistent with the calculation given below:
[math]\tau = \frac{L}{c} = \frac{60 mm}{3\times 10^8 m/s} = 200ps [/math]
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Next, you will try out a rectangular pulse waveform as your voltage source. Open the property dialog of VS and change the waveform type to "Pulse". Set the period of the pulse train to 500ps and set the pulse width to 100ps. This means a duty cycle of 20%. Set the rise time and fall time of the pulse both to 1ps. Set the initial and peak voltages of the pulse to 0 and 1V, respectively. Run a new transient test of your circuit and compare the results to the previous case of a sinusoidal waveform. Here, too, there is a 200ps delay between the input ad output voltages. Due to the perfect impedance match at both the input and output, v(in) and v(out) have equal amplitudes of 0.5V. Moreover, the v(in) is simply is half-scaled replica of the source voltage.
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Investigating the Effect of Load Mismatch
Now change the load resistor to RL = 100Ω. Run the Transient Test with the same settings as before for both cases of sinusoidal and pulse waveforms. The results of the two case are shown in the figures below:
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In the top graph, the "Show Maxima" feature of the graph window has been enabled. This can be done from the first tab of the "Edit Plots" Panel. As you can see from these figures, the effect of the load mismatch in the sinusoidal signal case is the difference in the amplitudes of the input voltage and load voltage. This is due to the nonzero load reflection coefficient ΓL = (ZL - Z0) / (ZL+ Z0) = 1/3. In the case of the pulse waveform, the impedance mismatch introduces a dispersion of the waveform as can be clearly seen in the shape of v(in). Besides the additional reflected pulse in each period, you can also see a sizable overshoot in the following close-up of the last graph. Note that the second smaller pulse has a delay of 400ps, which is the round-trip time from the input point to the load and back. Moreover, the amplitude of the smaller reflected pulse is 1/3 the amplitude of the larger incident pulse as you would expect from the value of the reflection coefficient.
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Analyzing a Quarter-Wave Impedance Transformer and Effect of Source Mismatch
In Tutorial lesson 1 you designed a quarter-wave impedance transform to match an arbitrary resistive load to a 50Ω source or a 50Ω transmission line. Set the length of the T-Line segment to L = λ0/2 = 37.5mm. Also, set the characteristic impedance of the T-line to Z0 = √(100.50) = 0.71Ω. Run a new Transient Test with the same settings as before for both cases of sinusoidal and pulse waveforms. The results are shown and compared in the figure below.
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The input impedance of the quarter-wave T-Line segment (at 2GHz) is given by:
[math] Z_{in} = \frac{Z_{0c}^2}{Z_L} = \frac{70.71^2}{100} = 50\Omega [/math]
As you can see from the above figure for the "pure harmonic" sinusoidal excitation, the source voltage is equally split between the source resistor RS and the input port of the T-Line. However, the situation is slightly different in the case of pulse waveform. The dispersive effects of the transmission line are in full display in this case. In other words, the matching condition is satisfied only at 2GHz and not at its harmonics present in the pulse waveform. Note that both load and source reflection coefficients are nonzero for this circuit:
[math] \Gamma_L = \frac{ Z_L - Z_{0c} }{ Z_L + Z_{0c} } = \frac{ 100 - 70.71 }{ 100 + 70.71 } = 0.172 [/math]
[math] \Gamma_S = \frac{ Z_S - Z_{0c} }{ Z_S + Z_{0c} } = \frac{ 50 - 70.71 }{ 50 + 70.71 } = -0.172 [/math]
To better understand this point, run the Transient Test for the pulse waveform two more times, once with the output plot set to v(source) and next with the output plot set to v(in). However, this time you will also enable the "Fourier Analysis" feature of Transient Test. This can be done from the Transient Test Panel. Check the "Apply Fourier" checkbox and click the "Fourier Setup" button to open the Fourier Transform Settings dialog. Set the Fundamental Frequency to 2GHz and set the reference output node to "0" for the ground. Then, set the positive output node to "1" for the source voltage in the first run and then to "2" for the input voltage in the second run. The spectral contents of the source and input voltages are shown in the figures below. You can clearly see that both signal have significant DC contents as well as sizable higher harmonics.
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Effect of Inductive and Capacitive Loads
In Tutorial Lesson 3 you used open and short stubs for tuning and matching of loads to transmission lines and sources. You saw that a given frequency, an open or short stub may behave as an inductive or capacitive load depending on its length. The impedance of an open or short stub is given by:
[math] Z_{open} = -jZ_0 cot(\beta L) [/math]
and
[math] Z_{short} = jZ_0 tan(\beta L) [/math]
In this part of the tutorial lesson you will add very small open and short stubs with len = 5mm and Z0 = 70.71Ω to the 100Ω load of the previous circuit. At f0 = 2GHz, you have tan(βL) = tan(2π L/λg) = 0.213 and cot(βL) = 4.70.
First, you will add a series short stub between the output of the T-Line segment and the resistive load RL as shown in the above figures. The small short stub introduces a series reactance of -j15.03Ω to be added to the 100Ω resistive load. Next, you will add a shunt open stub between the output of the T-Line segment and the resistive load RL also shown in the above figures. The small open stub introduces a shunt susceptance of j0.003S to be added to the 0.01S conductance of the resistive load. Run new Transient Tests with these additional series short and shunt open stubs over the time interval [0 - 3ns]. The time-domain voltages have been plotted in the figures below. The dispersive effects gravely affect the input and output voltages.
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