Difference between revisions of "RF Tutorial Lesson 2: Transient Analysis of a Simple Transmission Line Circuit"

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{{projectinfo|Tutorial| Transient Analysis of a Simple Transmission Line Circuit |RF176.png|In this project, the basic concepts of RF.Spice A/D are demonstrated, and a simple voltage divider is modeled and examined.|
+
{{projectinfo|Tutorial| Transient Analysis of a Simple Transmission Line Circuit |RFTUT2 6.png|In this project, you will perform transient simulation and Fourier analysis of simple transmission line circuits.|
  
−
*[[CubeCAD]]
+
*Transmission Line
−
*Visualization
+
*Generic T-Line
−
*[[EM.Tempo#Lumped Sources | Lumped Sources]]
+
*Transient Test
−
*[[EM.Tempo#Scattering Parameters and Port Characteristics | S-Parameters]] 
+
*Sinusoidal Waveform
−
*[[EM.Tempo#Far Field Calculations in FDTD | Far Fields]]
+
*Pulse Waveform
−
*[[Advanced Meshing in EM.Tempo]]
+
*Characteristic Impedance
−
|All versions|{{download|http://www.emagtech.com/content/project-file-download-repository|EM.Tempo Lesson 1|[[EM.Cube]] 14.8}} }}
+
*Reflection Coefficient
 +
*Fourier Analysis
 +
*Fundamental Frequency
 +
*Harmonic
 +
|All versions|{{download|http://www.emagtech.com/downloads/ProjectRepo/RFLesson2.zip RF Lesson 2}} }}
  
−
=== What You Will Learn ===
+
== 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.  
 
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.  
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<tr>
 
<tr>
 
<td>
 
<td>
−
[[File:RF160.png|thumb|500px|The basic transmission line circuit with three voltage probe markers.]]
+
[[File:RF160.png|thumb|left|550px|The basic transmission line circuit with three voltage probe markers.]]
 
</td>
 
</td>
 
</tr>
 
</tr>
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== Transient Simulation of a Simple Transmission Line Circuit ==
 
== 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.   
+
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.   
  
 
{| border="0"
 
{| border="0"
Line 104: Line 108:
 
</table>
 
</table>
  
−
[[File:RFTUT2 5.png|thumb|500px|Changing the waveform in the property dialog of the voltage source.]]
+
<table>
−
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.           
+
<tr>
 +
<td>
 +
[[File:RFTUT2 5.png|thumb|left|640px|Changing the waveform in the property dialog of the voltage source.]]
 +
</td>
 +
</tr>
 +
</table>
 +
 
 +
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 waveform parameters as specified below:
 +
 
 +
{| border="0"
 +
|-
 +
| valign="top"|
 +
|-
 +
{| class="wikitable"
 +
|-
 +
! scope="row"| Initial Voltage
 +
| 0
 +
|-
 +
! scope="row"| Peak Voltage
 +
| 1
 +
|-
 +
! scope="row"| Delay Time
 +
| 0
 +
|-
 +
! scope="row"| Rise Time
 +
| 1p
 +
|-
 +
! scope="row"| Fall Time
 +
| 1p
 +
|-
 +
! scope="row"| Pulse Width
 +
| 50p
 +
|-
 +
! scope="row"| Pulse Period
 +
| 500p
 +
|}
 +
 
 +
The duty cycle of the pulse train waveform is therefore 10%. Run a new transient test of your circuit with the same test parameters as before and compare the results to the previous case of a sinusoidal waveform. Here, too, there is a 200ps delay between the input and 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, v(in) is simply is half-scaled replica of the source voltage.           
  
 
<table>
 
<table>
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== Investigating the Effect of Load Mismatch ==
 
== Investigating the Effect of Load Mismatch ==
  
−
Now change the load resistor to RL = 100&Omega;. 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:  
+
Now change the load resistor to RL = 100&Omega;. Run the Transient Test with the same settings as before for both cases of sinusoidal and pulse waveforms:
 +
 
 +
{| border="0"
 +
|-
 +
| valign="top"|
 +
|-
 +
{| class="wikitable"
 +
|-
 +
! scope="row"| Start Time
 +
| 0
 +
|-
 +
! scope="row"| Stop Time
 +
| 3n
 +
|-
 +
! scope="row"| Linearize Step
 +
| 1p
 +
|-
 +
! scope="row"| Step Ceiling
 +
| 1p
 +
|-
 +
! scope="row"| Preset Graph Plots
 +
| v(source), v(in), v(out)
 +
|}
 +
 
 +
The results of the two cases are shown in the figures below:  
  
 
<table>
 
<table>
 
<tr>
 
<tr>
 
<td>
 
<td>
−
[[File:RF166.png|thumb|left|720px|The graph of the source, input and output voltages for a sinusoidal waveform with f<sub>0</sub> = 2GHz when RL = 100&Omega;.]]
+
[[File:RFTUT2 7.png|thumb|left|720px|The graph of the source, input and output voltages for a sinusoidal waveform with f<sub>0</sub> = 2GHz when RL = 100&Omega;.]]
 
</td>
 
</td>
 
</tr>
 
</tr>
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</table>
 
</table>
  
−
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 &Gamma;<sub>L</sub> = (Z<sub>L</sub> - Z<sub>0</sub>) / (Z<sub>L</sub>+ Z<sub>0</sub>) = 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.       
+
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:
 +
 
 +
<math> \Gamma_L = \frac{ Z_L - Z_0 }{ Z_L + Z_0 } = \frac{ 100 - 50 }{ 100 + 50 } = \frac{1}{3} </math>
 +
 
 +
In the case of the pulse waveform, the impedance mismatch introduces a reflected pulse that reaches node "IN" after a delay of 400ps, which is the round-trip time from the input point to the load and back. The amplitude of the smaller reflected pulse (166.7mv) is 1/3 the amplitude of the larger incident pulse (500mV) as you would expect from the value of the reflection coefficient.       
 +
 
 +
==Analyzing a Quarter-Wave Impedance Transformer ==
 +
 
 +
In RF Tutorial lesson 1, you designed a quarter-wave impedance transform to match an arbitrary resistive load to a 50&Omega; source. Set the length of the T-Line segment to L = &lambda;<sub>0</sub>/4 = 37.5mm for an operating frequency of 2GHz. Also, set the characteristic impedance of the T-line to Z<sub>0</sub> = &radic;(100.50) = 70.71&Omega;. Run a new transient test with the same settings as before for both cases of sinusoidal and pulse waveforms:
 +
 
 +
{| border="0"
 +
|-
 +
| valign="top"|
 +
|-
 +
{| class="wikitable"
 +
|-
 +
! scope="row"| Start Time
 +
| 0
 +
|-
 +
! scope="row"| Stop Time
 +
| 3n
 +
|-
 +
! scope="row"| Linearize Step
 +
| 1p
 +
|-
 +
! scope="row"| Step Ceiling
 +
| 1p
 +
|-
 +
! scope="row"| Preset Graph Plots
 +
| v(source), v(in), v(out)
 +
|}
  
 
<table>
 
<table>
 
<tr>
 
<tr>
 
<td>
 
<td>
−
[[File:RF165.png|thumb|left|720px|A close-up of the graph of the input voltage for a pulse waveform with a period of T = 500ps when RL = 100&Omega;.]]
+
[[File:RF167.png|thumb|left|550px|The quarter-wave impedance transformer circuit designed for 2GHz operation.]]
 
</td>
 
</td>
 
</tr>
 
</tr>
 
</table>
 
</table>
  
−
[[File:RF167.png|thumb|450px|The Quarter-Wave Impedance Transformer circuit.]]
+
The results are shown and compared in the figures below:
−
 
+
−
==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 50&Omega; source or a 50&Omega; transmission line. Set the length of the T-Line segment to  L = &lambda;<sub>0</sub>/2 = 37.5mm. Also, set the characteristic impedance of the T-line to Z<sub>0</sub> = &radic;(100.50) = 0.71&Omega;. 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.
+
  
 
<table>
 
<table>
 
<tr>
 
<tr>
 
<td>
 
<td>
−
[[File:RF168.png|thumb|left|900px|The graph of the source, input and output voltages for a sinusoidal waveform with f<sub>0</sub> = 2GHz when the T-Line segment acts as a quarter-wave transformer.]]</td>
+
[[File:RFTUT2 8.png|thumb|left|720px|The graph of the source, input and output voltages for a sinusoidal waveform with f<sub>0</sub> = 2GHz when the T-Line segment acts as a quarter-wave transformer.]]</td>
 
</tr>
 
</tr>
 
<tr>
 
<tr>
 
<td>
 
<td>
−
[[File:RF169.png|thumb|left|900px|The graph of the source, input and output voltages for a pulse waveform with a period of T = 500ps when the T-Line segment acts as a quarter-wave transformer.]]
+
[[File:RFTUT2 9.png|thumb|left|720px|The graph of the source, input and output voltages for a pulse waveform with a period of T = 500ps when the T-Line segment acts as a quarter-wave transformer.]]
 
</td>
 
</td>
 
</tr>
 
</tr>
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<math> Z_{in} = \frac{Z_{0c}^2}{Z_L} = \frac{70.71^2}{100} = 50\Omega </math>
 
<math> Z_{in} = \frac{Z_{0c}^2}{Z_L} = \frac{70.71^2}{100} = 50\Omega </math>
  
−
[[File:RF170.png|thumb|450px|The Fourier Transform Settings dialog.]]
+
As you can see from the above figure, in the case of 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:
−
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_L = \frac{ Z_L - Z_{0c} }{ Z_L + Z_{0c} } = \frac{ 100 - 70.71 }{ 100 + 70.71 } = 0.172 </math>
Line 171: Line 261:
 
<math> \Gamma_S = \frac{ Z_S - Z_{0c} }{ Z_S + Z_{0c} } = \frac{ 50 - 70.71 }{ 50 + 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.       
+
<table>
 +
<tr>
 +
<td>
 +
[[File:RF170.png|thumb|left|420px|The Fourier Transform Settings dialog.]]
 +
</td>
 +
</tr>
 +
</table>
 +
 
 +
To better understand this point, you can use RF.Spice's Fourier analysis feature, which is part of the transient test. You will run the "Transient Test" for the pulse waveform two more times with the "Fourier Analysis" enabled in 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. At the end of the simulation, additional bar chart graphs are added to the Data Manager window. The spectral contents of the source and input voltages are shown in the figures below. You can clearly see that both signals have significant DC contents as well as sizable higher harmonics.       
  
 
<table>
 
<table>
 
<tr>
 
<tr>
 
<td>
 
<td>
−
[[File:RF171.png|thumb|left|900px|The spectral contents of the source voltage v(source) with a pulse waveform.]]
+
[[File:RFTUT2 10.png|thumb|left|720px|The spectral contents of the source voltage v(source) with a pulse waveform.]]
 
</td>
 
</td>
 
</tr>
 
</tr>
 
<tr>
 
<tr>
 
<td>
 
<td>
−
[[File:RF172.png|thumb|left|900px|The spectral contents of the input voltage v(in) with a pulse waveform.]]
+
[[File:RFTUT2 11.png|thumb|left|720px|The spectral contents of the input voltage v(in) with a pulse waveform.]]
 
</td>
 
</td>
 
</tr>
 
</tr>
 
</table>
 
</table>
  
−
[[File:RF173.png|thumb|450px|The Quarter-Wave Impedance Transformer circuit with a series short stub before the load.]]
+
== Investigating the Effect of a Capacitive Load ==
−
[[File:RF175.png|thumb|450px|The Quarter-Wave Impedance Transformer circuit with a shunt open stub before the load.]]
+
−
== 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|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:
+
In the last part of this tutorial lesson, you add a 0.1pF shunt capacitor called "CL" to the load and see its effect in the case of pulse train signal. The modified circuit is shown in the opposite figure. The impedance of the shunt capacitor at the <I>n</I>th harmonic of the source's 2GHz fundamental frequency is given by:  
  
−
<math> Z_{open} = -jZ_0 cot(\beta L) </math>
+
<math> Z_{Cap} = -\frac{j}{\omega C_L} = -\frac{j}{2\pi n (2\times 10^9)(0.1\times 10^{-12})} \approx -j(796/n) \ \Omega </math>
  
−
and
+
It can be seen that at the fundamental frequency, the capacitor has very negligible effect, but at higher harmonics, its impedance becomes comparable to the 100&Omega; resistive loads. 
  
 +
<table>
 +
<tr>
 +
<td>
 +
[[File:RFTUT2_12.png|thumb|left|550px|The quarter-wave impedance transformer circuit with a parallel capacitor at the load.]]
 +
</td>
 +
</tr>
 +
</table>
  
−
<math> Z_{short} = jZ_0 tan(\beta L) </math>
+
Run a transient test of your modified circuit with the same settings as before only for the case of pulse waveform:
  
−
In this part of the tutorial lesson you will add very small open and short stubs with len = 5mm and Z<sub>0</sub> = 70.71&Omega; to the 100&Omega; load of the previous circuit. At f<sub>0</sub> = 2GHz, you have tan(&beta;L) = tan(2&pi; L/&lambda;<sub>g</sub>) = 0.213 and cot(&beta;L) = 4.70.
+
{| border="0"
 +
|-
 +
| valign="top"|
 +
|-
 +
{| class="wikitable"
 +
|-
 +
! scope="row"| Start Time
 +
| 0
 +
|-
 +
! scope="row"| Stop Time
 +
| 3n
 +
|-
 +
! scope="row"| Linearize Step
 +
| 1p
 +
|-
 +
! scope="row"| Step Ceiling
 +
| 1p
 +
|-
 +
! scope="row"| Preset Graph Plots
 +
| v(source), v(in), v(out)
 +
|}
  
−
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 -j15.03&Omega; to be added to the 100&Omega; 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.       
+
Enable the "Fourier Analysis" for Node 2 with a fundamental frequency of 2GHz. The figures below show the simulation results. You can see from the top figure that the output voltage has been slightly deformed with a rounded rising edge. Also, the reflected pulse now has significant undershoot and overshoot at its rising and falling edges, respectively. Next, compare the spectral contents of v(in) for the two cases without and with the shunt capacitor. The contents at DC, fundamental frequency and first few harmonics are almost the same, but the results start to differ at higher harmonics as we justified earlier.       
  
 
<table>
 
<table>
 
<tr>
 
<tr>
 
<td>
 
<td>
−
[[File:RF174.png|thumb|left|900px|The graph of the source, input and output voltages in the quarter-wave transformer circuit with a series short stub before the load.]]
+
[[File:RFTUT2_13.png|thumb|left|720px|The graph of the source, input and output voltages in the quarter-wave transformer circuit with a series short stub before the load.]]
 
</td>
 
</td>
 
</tr>
 
</tr>
 
<tr>
 
<tr>
 
<td>
 
<td>
−
[[File:RF176.png|thumb|left|900px|The graph of the source, input and output voltages in the quarter-wave transformer circuit with a shunt open stub before the load.]]
+
[[File:RFTUT2_14.png|thumb|left|720px|The graph of the source, input and output voltages in the quarter-wave transformer circuit with a shunt open stub before the load.]]
 
</td>
 
</td>
 
</tr>
 
</tr>
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<p>&nbsp;</p>
 
<p>&nbsp;</p>
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[[Image:Back_icon.png|40px]] '''[[RF.Spice_A/D#RF.Spice_A.2FD_Tutorial | Back to RF.Spice A/D Tutorial Gateway]]'''
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[[Image:Back_icon.png|40px]] '''[[RF.Spice_A/D#RF.Spice_A.2FD_Tutorials | Back to RF.Spice A/D Tutorial Gateway]]'''

Latest revision as of 23:44, 8 November 2016

Tutorial Project: Transient Analysis of a Simple Transmission Line Circuit
RFTUT2 6.png

Objective: In this project, you will perform transient simulation and Fourier analysis of simple transmission line circuits.

Concepts/Features:

  • Transmission Line
  • Generic T-Line
  • Transient Test
  • Sinusoidal Waveform
  • Pulse Waveform
  • Characteristic Impedance
  • Reflection Coefficient
  • Fourier Analysis
  • Fundamental Frequency
  • Harmonic

Minimum Version Required: All versions

'Download2x.png Download Link: RF Lesson 2

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.

The basic transmission line circuit with three voltage probe markers.

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]

The graph of the source, input and output voltages for a sinusoidal waveform with f0 = 2GHz when RL = 50Ω.
Changing the waveform in the property dialog of the voltage source.

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 waveform parameters as specified below:

Initial Voltage 0
Peak Voltage 1
Delay Time 0
Rise Time 1p
Fall Time 1p
Pulse Width 50p
Pulse Period 500p

The duty cycle of the pulse train waveform is therefore 10%. Run a new transient test of your circuit with the same test parameters as before and compare the results to the previous case of a sinusoidal waveform. Here, too, there is a 200ps delay between the input and 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, v(in) is simply is half-scaled replica of the source voltage.

The graph of the source, input and output voltages for a pulse waveform with a period of T = 500ps when RL = 50Ω.

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:

Start Time 0
Stop Time 3n
Linearize Step 1p
Step Ceiling 1p
Preset Graph Plots v(source), v(in), v(out)

The results of the two cases are shown in the figures below:

The graph of the source, input and output voltages for a sinusoidal waveform with f0 = 2GHz when RL = 100Ω.
The graph of the source, input and output voltages for a pulse waveform with a period of T = 500ps when RL = 100Ω.

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:

[math] \Gamma_L = \frac{ Z_L - Z_0 }{ Z_L + Z_0 } = \frac{ 100 - 50 }{ 100 + 50 } = \frac{1}{3} [/math]

In the case of the pulse waveform, the impedance mismatch introduces a reflected pulse that reaches node "IN" after a delay of 400ps, which is the round-trip time from the input point to the load and back. The amplitude of the smaller reflected pulse (166.7mv) is 1/3 the amplitude of the larger incident pulse (500mV) as you would expect from the value of the reflection coefficient.

Analyzing a Quarter-Wave Impedance Transformer

In RF Tutorial lesson 1, you designed a quarter-wave impedance transform to match an arbitrary resistive load to a 50Ω source. Set the length of the T-Line segment to L = λ0/4 = 37.5mm for an operating frequency of 2GHz. Also, set the characteristic impedance of the T-line to Z0 = √(100.50) = 70.71Ω. Run a new transient test with the same settings as before for both cases of sinusoidal and pulse waveforms:

Start Time 0
Stop Time 3n
Linearize Step 1p
Step Ceiling 1p
Preset Graph Plots v(source), v(in), v(out)
The quarter-wave impedance transformer circuit designed for 2GHz operation.

The results are shown and compared in the figures below:

The graph of the source, input and output voltages for a sinusoidal waveform with f0 = 2GHz when the T-Line segment acts as a quarter-wave transformer.
The graph of the source, input and output voltages for a pulse waveform with a period of T = 500ps when the T-Line segment acts as a quarter-wave transformer.

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, in the case of 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]

The Fourier Transform Settings dialog.

To better understand this point, you can use RF.Spice's Fourier analysis feature, which is part of the transient test. You will run the "Transient Test" for the pulse waveform two more times with the "Fourier Analysis" enabled in 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. At the end of the simulation, additional bar chart graphs are added to the Data Manager window. The spectral contents of the source and input voltages are shown in the figures below. You can clearly see that both signals have significant DC contents as well as sizable higher harmonics.

The spectral contents of the source voltage v(source) with a pulse waveform.
The spectral contents of the input voltage v(in) with a pulse waveform.

Investigating the Effect of a Capacitive Load

In the last part of this tutorial lesson, you add a 0.1pF shunt capacitor called "CL" to the load and see its effect in the case of pulse train signal. The modified circuit is shown in the opposite figure. The impedance of the shunt capacitor at the nth harmonic of the source's 2GHz fundamental frequency is given by:

[math] Z_{Cap} = -\frac{j}{\omega C_L} = -\frac{j}{2\pi n (2\times 10^9)(0.1\times 10^{-12})} \approx -j(796/n) \ \Omega [/math]

It can be seen that at the fundamental frequency, the capacitor has very negligible effect, but at higher harmonics, its impedance becomes comparable to the 100Ω resistive loads.

The quarter-wave impedance transformer circuit with a parallel capacitor at the load.

Run a transient test of your modified circuit with the same settings as before only for the case of pulse waveform:

Start Time 0
Stop Time 3n
Linearize Step 1p
Step Ceiling 1p
Preset Graph Plots v(source), v(in), v(out)

Enable the "Fourier Analysis" for Node 2 with a fundamental frequency of 2GHz. The figures below show the simulation results. You can see from the top figure that the output voltage has been slightly deformed with a rounded rising edge. Also, the reflected pulse now has significant undershoot and overshoot at its rising and falling edges, respectively. Next, compare the spectral contents of v(in) for the two cases without and with the shunt capacitor. The contents at DC, fundamental frequency and first few harmonics are almost the same, but the results start to differ at higher harmonics as we justified earlier.

The graph of the source, input and output voltages in the quarter-wave transformer circuit with a series short stub before the load.
The graph of the source, input and output voltages in the quarter-wave transformer circuit with a shunt open stub before the load.

 

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