Does Source Impedance affect SWR?

[This post is a response to the article, "SWR Dependence on Output Impedance," by Wright, M., W6PAP, in the September-October, 2021, issue of QEX magazine.  Although that article focused on source impedance affecting the measurement of SWR (a conclusion I do not agree with), I thought I should verify mathematically for myself that a transmission line's SWR is truly independent of source impedance.]

Does a source-impedance mismatch at the input of a transmission line affect the transmission line's SWR?

The answer is no, it does not.  Let's examine this conclusion both mathematically and with Simulink simulations.

Mathematical Derivation of SWR with Mismatched Load and Source Impedances:

Consider a transmission line of characteristic impedance Zo and a length of l.  Let the source impedance Zs and the load impedance Zld be mismatched from Zo, with reflection coefficients Γs and Γld, respectively: 

The Reflection Coefficients are:

Voltage Vfs is the voltage entering the transmission line from the source Vs.  Imagine that Vs (a sine-wave source) has been on long enough to allow any transient conditions to have settled.  In other words, the transmission line voltages are in steady-state.

Now let's consider a point on the Vfs waveform just as it enters the transmission line.  Even though there may be other voltages that represent reflections or re-reflections on the transmission line, from the point of view of this single point on Vfs, it only sees the transmission line's impedance Zo (per the Principle of Superposition).

Therefore, one can calculate Vfs using the equation of a simple voltage divider:

Vfs = Vs*(Zo/(Zo+Zs))

This point on Vfs will travel down the transmission line until it reaches the mismatch at the load end (at which point it will be reflected back to the source (with a changed value)).  

Before I go any further with reflections, though, let me quickly introduce transmission-line Propagation Constants.  These will be useful for calculating the values of the reflections.

Propagation Constants  γ, α, and β:

If one simultaneously compares (at, say, time t = t₀ ) the phase of the steady-state signal at the source-end of the transmission line to the phase of the signal further down the line, there will be a phase difference between the two, due to the time it took the earlier Vfs to travel along the transmission line.

Similarly, if the transmission line is lossy, the amplitude of the voltage will decrease (due to loss) as it travels along the transmission line, compared to its original value at the source end.

We can represent this attenuation and phase-shift with the term e^-γl.  This term is the one-way attenuation and phase-shift on a transmission line of length l.  (Round-trip attenuation and phase-shift is simply e^-2γl).

Note that γ = α + j*β, where α represents loss in nepers per unit-length, and β represents phase-shift in radians per unit-length.

Therefore, e^-γl =  e^-(α+jβl)  = e^-αl  * e^-j*βl.  If there is loss (α not equal to 0), then e^-αl  will be a positive real number whose value is less than 1.  And e^-j*βl is simply a phasor of magnitude 1 and angle of -βl radians.

Note that there is a negative sign in front of βl because the signal at the further point represents an earlier version (in time) of Vf, given it being measured simultaneously with Vf at the near end.  That is, the phase of Vf at the further point would be negative with respect to the phase at the near end.

The figure below demonstrates their meaning:

(You can find more on these propagation constants here: https://www.microwaves101.com/encyclopedias/propagation-constant)

Reflections on the Transmission Line:

Because of the mismatched source and load impedances, this point on the Vfs waveform, when it arrives at the load end of the transmission line, will be reflected back towards the source by the load mismatch.  When it arrives at the source end of the line, it will be reflected back to towards the load by the source mismatch.  

The result -- this point will bounce back and forth between the two mismatched ends ad infinitum.

The figure below shows the change in Vfs from the time when it enters the transmission line to when arrives at one end of the line and is reflected back towards the other end, etc.  You can see that each new amplitude and phase is a function of only three factors: the two reflection coefficients (Γs, Γld), and the one-way transmission line loss (e^-αl).

(Note that transmission line loss and delay is always referenced with respect to the Vfs that is currently entering the transmission line at the source).

In steady-state there will be large (approaching infinite) number of past reflections on the transmission line (because both ends are mismatched), and each reflection represents a version of the Vfs waveform that had entered the transmission line at some time earlier than the current Vfs entering the line.

If we were to take a "snapshot" in time of the state of the current Vfs entering the line and all of its reflections (from previous versions of itself) on the line, we would see that each reflection has a phase offset associated with it -- the phase offset represents the amount of time each reflection has "been on the line", compared to the current Vfs entering the line).  This phase offset, -βl, is incorporated into the exponential function for line loss (as a phasor), and the exponential function becomes a combination of line loss and line delay, represented as e^-γl , where γ = α + j*β.

(Note that phase = βl (i.e. radians per unit-length times length), and phase also equals ωt_d, where ω is the frequency (in radians per second) of Vfs and t_d is time-delay (in seconds).  Therefore, because ωt_d  = βl, the transmission line's time delay from end to end is:  t_d = βl/ω.)

The Lattice Diagram, below, shows a continuation of these reflections:

Here's a zoomed-in view of the reflections:

(The voltage values of the reflections in the lattice diagram, above, are assumed to be measured at the source end of the transmission line (i.e. at the SWR meter's position in the earlier figure).  Also, if there is some loss in the system (either in Zs, Zld, or within the transmission line, itself),  these reflections will decay towards 0 over time.)

So, in steady state, the total voltage traveling from the source-end of the transmission line to the load is the voltage Vfs (at a point in time, say, t = t₀ ) summed with the infinite series of all of the re-reflections that have just been reflected back towards the load by the source mismatch at that same point in time (i.e. at time t = t₀ ) .

The sum of these voltages is Vf, which represents the total voltage of the wave traveling from source to load.  

This sum can be expressed as the following infinite series:

Similarly, the steady-state total voltage flowing from the load back to the source, Vr, can be calculated by summing all of the reflections from the load at that same point in time (i.e. t = t₀ ):

We can further reduce the equations for Vf and Vr:

Important Note!  If we were to calculate Γ as being Vr/Vf, using the two equations above, we would see that Γ is independent of Γs. In other words, Gamma is independent of Source Impedance.

Let's check if SWR is also independent of Γs (it should be).  Given the following equation for SWR (from Wikipedia):

Let's substitute our equations for Vf and Vr into the above equation.  I'll use "D" to represent the denominator of Vf and Vr:

Let's start simplifying this equation:

Already we can see that SWR is not a function of Γs.

Next, let's look at the exponential part of the equation:

Note the delay term.  We can remove this from the equation because its magnitude is 1 (and in the SWR equation we only care about magnitudes):

SWR on a Lossy Transmission Line:

Therefore, SWR for a lossy transmission line, measured at the source-end of the line, is:

where α is line loss in nepers/unit-length, and l is line length.

The SWR of a lossy transmission line will vary along its length.  Thus, the general form of SWR for a lossy transmission line, as a function of position along that line, is:

SWR on a Lossless Transmission Line:

If the line is lossless:

Thus, if the line is lossless (or essentially lossless), we can remove the exponential term representing line loss, resulting in the following SWR equation for lossless lines.  Note that this equation is independent of position along the line:

Most importantly, in none of these SWR equations does the SWR depend upon the source's Reflection Coefficient.  Therefore, SWR is independent of source impedance. 

Regarding the infinite geometric series constraint earlier in this post:

This condition is met if the line is lossy (and the magnitude of each reflection coefficient is less than or equal to 1), or, if the line is lossless, at least one of the reflection coefficients has a magnitude less than 1 (which is pretty much always, in real life). 

Simulating SWR Independence from Source Impedance:

To simulate SWR independence from source impedance, I'm going to tweak a Simulink model found in this blog post of mine:  http://k6jca.blogspot.com/2021/05/antenna-tuners-lumped-element-tuner.html

This simulation model consists of a voltage source, a series source impedance, a lumped-element directional coupler (tandem-match topology) referenced to 50 ohms, a lossless, two-wavelength long 50 ohm transmission line, and a load impedance.  Below is the simulation model with the source impedance set to 50 ohms and a mismatched load impedance:

This model performs a time-domain simulation in which the source is a 10 MHz gated sine-wave whose amplitude is 1.09 volts (both the gating and the voltage are hold-overs from my previous modeling).

The directional coupler is modeled using a "tandem match" topology, implemented with ideal transformers having a 1:N turns ratio (N in this case being 50), with the outputs amplified by this same amount, N, to compensate for the coupler's "coupling factor."

Thus, the directional coupler's Vfwd and Vref outputs are scaled to be a 1:1 representation of the transmission line's actual Vfwd and Vref voltages.

Here is the model of my directional coupler:

The Simulink model lets me plot the time-domain values of Vfwd and Vref as "measured" by the directional coupler at the transmission line's input.

From the plots of these two waveforms I can manually find their peak values (using cursors to measure the waveforms when they are in steady state, after transient behavior has died down), and then calculate SWR.

To demonstrate, below are two simulations, each with a different source impedance but the same mismatched load impedance.  A third simulation will verify the phase-shift predicted by the math.

Simulation 1:

First, let's match the source impedance to the transmission line's Zo (i.e. 50 ohms).  Note that the load consists of a 75 ohm resistor in parallel with a 3.049 uH inductor, so it is mismatched from the transmission line's Zo of 50 ohms (i.e. Zld = 65.033 + j25.46 ohms).

Using the model from above:

Here are the simulation's time-domain waveforms:

Zooming in and measuring peak voltage values:

Note that the SWR is 1.67 with the matched source impedance.

As a check, let's solve the equations presented earlier for Vf, Vr, and SWR, given Vs = 1.09, Zs = 50, and Zld = 65.033 + j25.46 ohms (i.e. Γld of 0.1713 + j0.1834):

• Vfs = Vs*(Zo/(Zo+Zs) = 0.545 volts.
• Γs = (Zs-Zo)/(Zs+Zo) = 0.
• Γld = (Zld-Zo)/(Zld+Zo) = 0.1713 + j0.1834.
• For the exponential factor e^-2γl, let α = 0 (no loss) and βl = 4pi radians (because the transmission line is 2 wavelengths long, the one-way phase delay is 2λ*2pi/λ = 4pi radians).  So e^-2γl =  e^-j2βl = e^-j2*(4pi)  = 1.
• Vf = (Vfs)/(1-Γs*Γld*e^-2γl) = 0.545 volts.
• |Vf| = 0.545 volts (Check).
• Vr = (Vfs*Γld)/(1-Γs*Γld*e^-2γl)  = 0.0933 + j0.1 volts.
• |Vr| = 0.1368 volts. (Check).
• SWR = (0.545+0.1368)/(0.545-0.1368) = 1.67 (Check).
• Time  delay from Vf(peak) to Vr(peak) = (angle(Vf)-angle(Vr))/(2pi)/10MHz = -13.04 ns (Check, given that simulation's cursor resolution is 0.238 ns, and so the measurement of -13.3 ns is within an increment of the cursor's resolution).

So, the simulated results match the mathematically-derived results!

Simulation 2:

Now let's change the source impedance so that it is 17.928 + j31.423 ohms (i.e. no longer matched to 50 ohms), while keeping the load the same mismatched value that was used above. 

Here's the model:

 And here are the time-domain plots of Vfwd and Vref:

Zooming in...

Again, the SWR is unchanged at 1.67, even though the source impedance is now mismatched from the transmission line's Zo.

Again, let's calculate Vf, Vr, and SWR using the mathematical formulas presented earlier in this post and compare them to the simulated values, above:

• Vfs = Vs*(Zo/(Zo+Zs) = 1.09*(50/(50+17.928+j31.423) = 0.6609 - j0.3057 volts.
• Γs = (Zs-Zo)/(Zs+Zo) = -0.2126 + j0.561.
• Γld = (Zld-Zo)/(Zld+Zo) = 0.1713 + j0.1834.
• For the exponential factor e^-2γl, let α = 0 (no loss) and βl = 4pi radians (because the transmission line is 2 wavelengths long, the one-way phase delay is 2λ*2pi/λ = 4pi radians).  So e^-2γl =  e^-j2βl = e^-j2*(4pi)  = 1.
• Vf = (Vfs)/(1-Γs*Γld*e^-2γl) = 0.5920 - j0.2387 volts.
• |Vf| = 0.6383 volts (Check).
• Vr = (Vfs*Γld)/(1-Γs*Γld*e^-2γl) = 0.1452 + j0.0677 volts.
• |Vr| = 0.1602 volts. (Check).
• SWR = (0.6383+0.1602)/(0.6383-0.1602) = 1.67 (Check).
• Time  delay from Vf(peak) to Vr(peak) = (angle(Vf)-angle(Vr))/(2pi)/10MHz =  -13.04 ns (Check, given that simulation's cursor resolution is 0.238 ns, and so the cursor measurement of -12.6 ns is within an increment or two of the cursor's resolution).

Again, the simulated results match the mathematically-derived results!

Simulation 3:

In the two simulations, above, the transmission line was two wavelengths long.  This means that the one-way end-to-end phase shift was 4pi, and therefore the quantity e^-γl, given a lossless line, equals 

e^-j(4pi) which equals 1.

Let's change Simulation 2's transmission line length so that e^-γl does not equal 1 and verify that the mathematical model still correctly predicts Vf, Vr, and SWR.  Note that I am not changing Simulation 2's source and load impedances.

I'll make the length equal to 0.3λ.  This will represent an end-to-end phase shift of 0.6pi radians (i.e. 0.3λ*2pi radians/λ).

Here's the model:

And here is the model's time-domain response:

And the time-domain response, zoomed in:

Note that the simulated SWR is still 1.67.

Let's now use the mathematical equations derived above to run some calculations. I will check them by comparing the calculated values to the values in the zoomed-in time-domain plots, above.

First, recall that Zs =  17.928 + j31.423 ohms, Zld = 65.033 + j25.46 ohms, Vs = 1.09 volts, Zo = 50 ohms, and and βl = 0.6pi.

• Vfs = Vs*(Zo/(Zo+Zs) = 1.09*(50/(50+17.928+j31.423) = 0.6609 - j0.3057 volts.
• Γs = (Zs-Zo)/(Zs+Zo) = -0.2126 + j0.561.
• Γld = (Zld-Zo)/(Zld+Zo) = 0.1713 + j0.1834.
• For the exponential factor e^-2γl, let α = 0 (no loss) and βl = 0.6pi radians (because the transmission line is 0.3 wavelengths long, the one-way phase delay is 0.3λ*2pi/λ = 0.6pi radians).  So e^-2γl =  e^-j2βl = e^-j2*(0.6pi)  = -0.809 + j0.5878.
• Vf = (Vfs)/(1-Γs*Γld*e^-2γl) = 0.6588 - j0.4236 volts.
• |Vf| = 0.7832 volts (Check).
• Vr = (Vfs*Γld)/(1-Γs*Γld*e^-2γl) = -0.1825 + j0.0729 volts.
• |Vr| = 0.1966 volts. (Check).
• SWR = (0.7832+0.1966)/(0.7832-0.1966) = 1.67 (Check).
• Time  delay from Vf(peak) to Vr(peak) = (angle(Vf)-angle(Vr))/(2pi)/10MHz =  46.956 ns (Check).

Again, SWR is unchanged at 1.67, and the magnitudes of Vf and Vr, as well as the time delay between their peaks, are confirmed by the Simulink simulation.

Conclusion:

I have mathematically derived an equation for SWR that shows that SWR is independent of source impedance.  And I have simulated three circuits, one with the source impedance matched to the transmission line, one with the source impedance mismatched, and one with the source impedance still mismatched but with a different transmission line length.  These three simulations give values for Vf, Vr, and SWR that match the values calculated using the equations derived above.  

And SWR is the same for all.

Other Transmission-Line Posts:

http://k6jca.blogspot.com/2021/02/antenna-tuners-transient-and-steady.html.  This post analyzes the transient and steady-state response of a simple impedance matching system consisting of a wide-band transformer.  I calculate the system's impulse response and find the time-domain response by convolving this impulse-response with a stimulus signal.

http://k6jca.blogspot.com/2021/02/the-quarter-wave-transformer-transient.html.   This post analyzes the transient and steady-state response of a Quarter-Wave Transformer impedance matching device.  I calculate the system's impulse response and find the time-domain response by convolving this impulse-response with a stimulus signal.

http://k6jca.blogspot.com/2021/03/useful-swr-voltage-and-power-equations.html.  This post lists (in an easily accessible location that I can find!) some equations that I find useful

http://k6jca.blogspot.com/2021/05/antenna-tuners-lumped-element-tuner.html.  This post analyzes the transient and steady-state reflections of a lumped-element tuner (i.e. the common antenna tuner).  I describe a method for making these calculations, and I note that the tuner's match is independent of the source impedance.

http://k6jca.blogspot.com/2021/05/lc-network-reflection-and-transmission.html.  This post describes how to calculate the "Transmission Coefficient" through a lumped-element network (and also its Reflection Coefficient) if it were inserted into a transmission line.  

http://k6jca.blogspot.com/2021/09/does-source-impedance-affect-swr.html.  This post shows mathematically that source impedance does not affect a transmission line's SWR.  This conclusion is then demonstrated with Simulink simulations.

https://k6jca.blogspot.com/2021/10/revisiting-maxwells-tutorial-concerning.html  This posts revisits Walt Maxwell's 2004  QEX rebuttal of Steven Best's 2001 3-part series on Transmission Line Wave Mechanics.  In this post I show simulation results which support Best's conclusions.

Standard Caveat:

I might have made a mistake in my designs, equations, schematics, models, etc. If anything looks confusing or wrong to you, please feel free to comment below or send me an email.

Also, I will note:

This design and any associated information is distributed in the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.