## Thursday, September 16, 2021

### Does Source Impedance affect SWR?

[This post is a response to the article, "SWR Dependence on Output Impedance," by Wright, M., in the September-October issue of QEX magazine]

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:

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 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:

Note that, because it takes time for the signal to travel from the source end of a transmission line to the load end (and vice-versa), the phase of the signal will change as it moves from source to load or from load to source.

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.

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:

This point on the Vfs waveform, as it travels from the source end of the transmission line to the load end and back again, 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 only a function of the two reflection coefficients (Γs, Γld), and the one-way transmission line delay and loss (e-γl).

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.  Also, as long as 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 sum of Vfs (at that point in time) plus the infinite sum 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.

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:

We can further reduce the equations for Vf and Vr:

Now, 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:

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):

Therefore, SWR for a lossy transmission line is:

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

Note that 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 in neither SWR equation 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).

Great, 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, is 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 two circuits, one with the source impedance matched to the transmission line and one with the source impedance mismatched.  These two simulations give values for Vf, Vr, and SWR that match the values calculated using the equations derived above.

And SWR is the same in either case.

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.

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.

## Friday, May 14, 2021

### LC Network Reflection and Transmission Coefficients

This blog post is inspired by a previous blog post in which I looked at the reflections and re-reflections on a transmission line given an LC tuner circuit placed at the transmitter.

In other words, the combined LC tuner and transmitter were considered to be a single "lumped-element" network.

I was wondering how these reflections would change if the LC tuner was moved further away from the transmitter and connected to it with a second transmission line (rather than being connected directly to the transmitter at the transmitter's output).  And I realized, although I knew how to calculate the Reflection Coefficients at the LC network's two ports, I did not know how to calculate the Transmission Coefficients through the network (either from the input port to the output port, or from the output port to the input port).

A quick googling of the internet did not reveal anything, so I did some pencil and paper noodling and came up with a way to calculate these coefficients for the general case of  a lossless, two-port network consisting of any combination of lumped-element inductors, capacitors, and transformers.

And thus this post.

First, a review the Reflection and Transmission Coefficients when two transmission lines of different characteristic impedances are connected in series:

In the system above energy must be conserved.  If the system is lossless (ideal transmission lines with no loss), then the power of the incident wave will equal the sum of the powers of the forward and the reflected waves:

P(incident) = P(forward) + P(reflected)

Therefore, for the diagram, above:

(|V(incident)|^2 ) / Za = (|V(forward)|^2) / Zb + (|V(reflected)|^2 ) / Za

But suppose I insert a lumped-element, lossless, two-port network between the two transmission lines?  How do I calculate the new reflection and transmission coefficients if this network is now in-line?

The Reflection Coefficient should be obvious -- it is just the normal equation for a Reflection Coefficient but using the network's Zin (with the network terminated with Zb) in place of the second transmission line's Zo.

The Transmission Coefficient is not so obvious.  But considering that the incident wave sees the network's Zin at the network's input, I make the assumption that the transmitted wave is first transformed using the standard "Transmission Coefficient" equation using the network's Zin.

But more is required than just this calculation, -- a quick check of the results will show that energy (calculated as power) is not conserved.

Another factor is needed, and that is the voltage attenuation (or gain) from the network's input port to its output port when the output port is terminated in the characteristic impedance of the second transmission line.

Including this second factor will result in voltage values in which energy is conserved.

The formulas for calculating the Reflection and Transmission Coefficients for general 2-port, lossless, lumped-element networks are shown in the figure, below:

Next, an example applying these equations...

1.  Example, LC Network:

Let's insert a lossless LC network between the two transmission lines.  We can calculate the resultant Reflection and Transmission Coefficients as shown in the figure, below:

2.  Results:

I will assign L = 1.378 uH and C = 137.8 pF.  At 10 MHz their impedances are:

Zl =  + j86.58 ohms

Zc = -j115.6 ohms

And I will define Za = 50 ohms and Zb = 25 ohms.

Using the equations in the LC network figure, above, let's do the following calculations:

Calculating Γ11:

To calculate Γ11 (the Reflection Coefficient looking into the network's input,) we first need to calculate Zin of the LC Network.  Using the equation in the figure, above, the result is:

Zin =  23.88 + j81.41 ohms (using MATLAB).

Therefore, using the equation for Γ11 and substituting in the values for Zin and Za, the result is:

Γ11 = 0.389 + j0.674.

Calculating T21:

To calculate T21 I first calculate T1 using Zin and Za.  The result is:

T1 = 1.389 + j0.674

Then I calculate the voltage gain, AV21, with the network terminated with Zb.  The result is::

AV21 =  0.0207 - j0.2872

Next, let's calculate voltages...

Set Vincident = 1 volt and Calculate Vf and Vr:

Vr =   Γ11 * Vincident =  0.389 + j0.674 volts

Vf = Γ11 * AV21 * Vincident =    0.222 - j0.385 volts

And note that the magnitudes of these voltages are:

|Vincident| = 1 volt

|Vr| = 0.778 volts

|Vf| = 0.445 volts

Check if Energy is Conserved:

Let's verify that power, and thus energy, is conserved:

Pincident = (1^2) / 50 = 20 mW.

Pr = (|Vr|^2) / 50 = 12.1 mW

Pf =  (|Vf|^2) / 25 = 7.9 mW

Pf + Pr = 7.9 + 12.1 mW = 20 mW = Pincident.

Energy is conserved!

To verify my results, I'll use the following  Simulink model:

The Simulation waveforms are below (note that the simulation voltages are spec'd as peak values, not RMS):

First, note that Vincident = 1 V (peak).  The simulated Vf equals 0.444 volts (peak), which is essentially identical to the value calculated, above.

Below, Vr equals 0.777 volts peak, which is essentially the same as the result calculated, above.

So simulated results match calculated results!

(Note that these are the steady-state values.  You can see that there is a short initial transient in both Vr and Vf when the sine-wave source is first gated on.  The length and amplitude of these transients are related to the network type (e.g. L-network, PI, T, etc.), the component values, and impedances seen by each network port.)

Other Notes:

1.  The technique I present, above, has not been rigorously proven mathematically, so take it with a grain of salt.  Never the less, in the few cases where I have applied it, the calculated results match the simulated results.

2.  For Reflection and Transmission Coefficients from the network's Port 2 to its Port 1 (in other words, in the opposite direction), use the same technique.  Note that the impedance (Zin) looking into the network's output (Port 2) equals Zc || (Zl + Za), where Za is the impedance of the transmission line connected to Port 1, and AV12 (the voltage gain from Port 2 to Port 1) = Za / (Za + Zl).

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.

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.

## Monday, May 10, 2021

### Antenna Tuners: Lumped-Element Tuner & Transmission Line Wave Interactions

Summary:  This blog post examines the operation of a typical "lumped-element" antenna tuner (for example, an L-network tuner) and its effect on forward and reflected wave interactions on a transmission line.

I also analyze at the effect of source impedance (i.e. a transmitter's impedance looking into its output port) on the transmission line waves and conclude that they are only affected by the source impedance during transient conditions, but not in steady-state.  For an impedance matching network that has been "tuned" to provide a match, for example, to transform a load to 50 ohms, the source impedance has no effect on the matching-network's input impedance in steady-state.

I usually don't care what the impedance of an antenna tuner is looking backwards into its output port (towards the transmitter).  I adjust the tuner so that its input impedance looks like 50 ohms, and with that accomplished, I'm a happy camper.

But recently I was wondering if there was a way to calculate this impedance (assuming a load attached to the tuner's output port via a transmission line) and how this impedance would affect the forward and reflected voltages on the transmission line.  And how the transmitter's output impedance (i.e. the system's source impedance) might affect the value of the tuner's output impedance.

And thus this post.

This post is a continuation of two earlier posts exploring transient and steady-state responses in Antenna Tuners (i.e. impedance matching devices) attached to transmission lines.

The first post calculates the transient and steady-state responses of an Antenna Tuner that, for ease of analysis, consists of a single wide-band transformer.

The transformer, being a wide-band device, allows one to easily calculate the system's impulse response, from which the transient and steady-state responses are calculated using the convolution integral.

The second post is a continuation of the first post in which a wide-band matching device is used.  In this case, though, the wide-band matching device is a quarter-wave transformer.  The quarter-wave transformer, in this sense, is a wide-band device in that its frequency response is wide band (there is no attenuation if the transmission line is assumed to be ideal), although it only matches impedances over specific narrow bands of frequency.

Because of its wide-band frequency response, the quarter-wave transformer's impulse response (in a simple system) is easy to determine by hand and transient and steady-state responses for a sinusoidal stimulus easily calculated.

This new post examines the startup and steady-state wave interactions with the more general case of a "Lumped-element" Antenna Tuning network.  The tuner might be an L-network, a T-network, a PI-network, or some other topology -- the approach described below should work for any two-port lumped-element impedance matching topology.

Lumped-element networks such as these are band-limited networks, and an accurate impulse response is significantly more painful to calculate (because an impulse is no longer an impulse after passing through such a network).

And because of this frequency-dependency, a true time-domain transient response (that includes, for example network "ringing") is difficult to calculate by hand.  But a "limited" transient response that simply shows how reflections and re-reflections build over time on a transmission line is easier to calculate.  This is the path this post will.

The goal of these posts is to demonstrate the underlying wave mechanics that occur on the transmission lines and at the impedance matching devices (rather than a deep-dive into a rigorous mathematical derivation), and some assumptions are made to keep the math manageable.

These assumptions are:

1. Transmission lines are assumed to be lossless.
2. Transmission line delays (and thus their lengths) are assumed to delay a sine-wave by multiples of 90 degrees or 180 degrees, depending upon the application.  E.g. the quarter-wave transformer has a length that adds a phase-shift of 90 degrees when comparing the phase of the signal at its input versus the phase of the signal at its output.
3. Antenna Tuning networks are assumed to consist of ideal components.  E.g. wide-band lossless transformers, or lossless inductors and capacitors.  No resistors.
4. Transmitters are assumed to be ideal, consisting of a voltage source with a source impedance Zs.
5. Loads are assumed to be resistive, with no reactive component.
6. Directional Couplers (i.e. SWR meters) are assumed to be lumped-element circuits consisting of ideal components (e.g. ideal lossless transformers and resistors without parasitics).
7. Unless otherwise specified, Directional Couplers are assumed to be referenced to 50 ohms.  That is, the voltage from their "Reflected Voltage" port will be zero when the impedance connected to their Output port is 50 + j0 ohms.

Let's look at a basic station setup consisting of a Transmitter, an SWR Meter, an Antenna Tuner, and a long Transmission Line connecting the Tuner's output to a load (antenna).

If the interconnects between the transmitter, SWR meter, and antenna tuner are short (my rule-of thumb is that lengths are less than 1/20 of the wavelength of the highest frequency), then these system elements and their interconnects can be treated as a lumped-element system, and lumped-element circuit analysis techniques can be used to calculate the voltages and currents at the various nodes of this part of the system.

On the other hand, the transmission line portion of the system is best analyzed using the familiar concepts of reflection and transmission coefficients of transmission-line systems.

Goals of this Post:

One goal of this post is to show a method of calculating, and then calculate, the following voltages:

1. The Forward and Reflected voltages on the transmission line and how they change from startup to steady-state.
2. The Forward and Reflected voltages that would be measured by the lumped-element Directional Coupler (i.e. SWR meter) inserted between the Transmitter and the Antenna Tuner, and how these voltages change from startup to steady-state.

Note that in a lumped-element system there actually aren't forward and reflected waves.  But the SWR meter, itself being a lumped-element circuit, does not know that they do not exist -- it is measuring the voltages and currents at a single point (node) and from these deriving what the forward and reflected voltages would be if it were connected in a 50 ohm transmission line.  (More on this topic, below).

Vf and Vr (forward and reflected) voltages derived with the equations I will present will be verified with simulated results generated with Simulink.

Another goal is to determine, when the Antenna Tuner is tuned for a 1:1 SWR at its input (i.e. the impedance looking into the antenna tuner's input port is 50 ohms), the value of the impedance looking back into the OUT port of the antenna tuner (towards the transmitter).  This is the impedance that reflections from the load will encounter as they travel backwards on the transmission line towards the tuner, and this is the impedance that then re-reflects these reflections back to the load.

In other words, when the system is in steady-state with the Antenna Tuner tuned so that its input impedance is 50 + j0 ohms...

Foundation Concepts:

The following concepts form the foundation for my equations and calculations later in this post.

1.  On a Transmission Line, the voltage at any point on that line is the sum of the Forward Voltage and the Reflected Voltage:

V(at point) = Vf(at point) + Vr(at point)

2.  On a Transmission Line, when a traveling wave arrives at an impedance discontinuity, this "incident" wave splits into two parts.  One part is reflected back (Vr), and the other part continues forward (Vf) through the discontinuity.  The amplitudes of Vr and Vf with respect to the incident wave's voltage are such that energy is conserved.

The amplitudes of the signals traveling in the two directions can be determined using the calculated values of the Reflection Coefficient and the Transmission Coefficient at the discontinuity

We can use the same V(transmitted) equation for a transmission line with a resistive load.  After all, a resistor simply looks like an infinitely long transmission line whose characteristic impedance equals the resistor's resistance:

3.  On a Transmission Line, delay can be expressed as a complex number:

On an ideal lossless transmission line with a sinusoidal waveform, we can mathematically treat delay as e-jφ, where φ is the phase shift introduced by the transmission line's delay (note: e-jφ = cos(φ) - j*sin(φ)).

For example, if a transmission line is one wavelength long, a sine-wave signal's phase will shift by 360 degrees (2 pi radians).  And so in this case, the delay factor would be e-j2π =  1 + j0.

If the transmission line were 1/4  λ long, the delay factor would be e-jπ/2 =  0 + j1.

4.  The Principle of Superposition:

The "Principle of Superposition" is a fundamental principle of circuit analysis which states that, in a circuit with multiple voltage and current sources, we can calculate the voltage and current at any node in the system by first calculating the individual contributions of each voltage or current source to the voltage (or current) at that node (by first turning all other sources off -- replacing voltage sources with shorts and current sources with opens), and then summing the individual contributions together to get the final value.

This same principle can be applied to a transmission line upon which multiple reflections might exist.  For example, on a mismatched transmission line, the total reflected voltage on the line can be represented by the sum of an infinite series of individual reflected voltages.  (More on this in a bit).

5.  An ideal lumped-element Directional Coupler can be represented by the following circuit:

The circuit above is the Directional Coupler I will use in my Simulink simulations.

Lumped-element directional couplers measure current and voltage at a single point.  As such, they have no knowledge of the direction of wave travel or even the existence of waves.  But they will generate two voltages that would represent Vfwd and Vref on a transmission line if the directional coupler were installed in a transmission line whose characteristic impedance Zo equals the directional coupler's "Rref".

Another way to look at this circuit's operation is to consider it to be a lumped-element bridge circuit, and the voltage Vref will equal zero when Zload (connected to the directional coupler's Vout port) equals 50 + j0 ohms (assuming Rref = 50 ohms).

Therefore, if this directional coupler is attached to the input port of an antenna tuner and the tuner adjusted until Vref equals 0, then the impedance looking into the antenna tuner's input will be 50 + j0 ohms.

Note that for large transformer turns-ratios, the voltage drop across the current-sense transformer will essentially be 0, and Vin will essentially equal Vout (which we want).  We can derive the following equations for Vfwd and Vref based upon the voltage at Vout and the load impedance attached to it.  (Note that the turns ratio 'n' results in the attenuation of Vfwd and Vref and so my "ideal" directional coupler adds multiplication factors that remove this attenuation effect):

(More on the derivation of these equations, here: http://k6jca.blogspot.com/2015/01/notes-on-directional-couplers-for-hf.html )

If we know Vfwd and Vref, we can derive the Zload connected to the directional coupler's output port:

Zload = Rref * (Vfwd + Vref) / (Vfwd - Vref)

The drawing below illustrates these equations:

6.  Any waveform can be represented by an infinite series of impulses, and we can examine the operation of a system by taking these impulses, one at a time, and investigating their effect on the system's voltages, and then summing their contributions:

With these definitions out of the way, let's apply these principles to an example...

Example with a Source Impedance Equal to 50 Ohms and a 200 Ohm Load:

Let's take an example with a 200 ohm load at the end of a 50-ohm transmission line that is one wavelength long (taking its velocity factor into account).

Defining the Circuit Model:

I will represent the transmitter as a voltage source with a series source impedance Zs.  In this example I will set Zs to 50 + j0 ohms.

The transmission line's characteristic impedance will be defined to be 50 ohms, and the line lossless.  Because the transmission line is one wavelength long, the load impedance presented by the input of the line to the output port of the Antenna Tuner is also, conveniently, 200 ohms.

The Transmitter, SWR Meter, and Antenna Tuner are assumed to be connected together with very short interconnects, so I will combine these three elements together and treat them as a single lumped-element network.

I will represent the antenna tuner with an L-network that will transform the 200 + j0 ohms at its output port to 50 + j0 ohms looking into its input port.

In the illustration, below, I've removed the directional coupler (in this example considered to be ideal) because it has no effect on system voltages and currents (but I will present equations that allow the calculation of the Vf and Vr it would generate, given the voltage at a node into which it would be inserted and the load impedance it would see at that node.)

Within the lumped-element network there are two nodes whose voltages I will use for calculations.  One is node 'A' (at the L-network's input), and the other is node 'b' (at the L-network's output, which is also the transmission line's input).  A directional coupler could be inserted into the circuit at either of these two nodes to determine either the forward and reflected voltages at the input end of the transmission line (node B) or the "equivalent" forward and reflected voltages at node A.

The Transmission Line's Reflection and Transmission Coefficients are shown, below.

The illustration below shows how the Reflection Coefficient looking into the Output port of the L-network was calculated:

SimSmith can be used as a quick check of this value:

Calculating  the Model's Voltages and Wave Reflections:

1.  Startup:

Let's say that at time t = 0 a sine-wave of amplitude 2 volts is started.  From a lumped-element circuit analysis perspective, the voltages in the lumped-element network can be calculated.

First, note that the output of the L-network is attached to the transmission line.  At startup the impedance that the L-network sees at this output (labeled below as "Z_TLin") is Zo, because there are not yet any reflections on the transmission line.

Given this load impedance, the voltages at the two nodes ('a' and 'b') can be calculated, and from these we can calculated the "equivalent" Vfwd and Vref voltages that ideal "Tandem-match" lumped-element directional couplers would generate if placed at either node 'a' or node 'b' (the latter being the Transmission line's input).

(Note:  after startup, as reflections on the transmission line arrive back at the L-network's output, the impedance seen at the transmission line's input by the L-network will change.  The equation to calculate this new impedance is shown at the bottom of the illustration, above).

2.  Reflections on the Transmission Line:

For a sine-wave entering a transmission line, we can take a point on that sine-wave and follow it as it travels down the line to the load and then reflects back towards the source.

If the impedance at the source end of the line is also mismatched from Zo, this point will re-reflect and head back towards the load.

These reflections and re-reflections of this point on the sine-wave will continue indefinitely over time, with each reflection becoming smaller (assuming a source or load impedance with some resistive element, because I've defined the transmission line to be lossless).  And the amplitude of each new reflection can be calculated using a Lattice Diagram.

So, at any time 't', the forward voltage or the reflected voltage on a transmission line can be calculated by summing all of the contributions either from all of the reflections traveling in the forward direction (plus the voltage currently being sent by the source), or from all of the reflections traveling in the reverse direction (from the load).

At startup, given the initial Vfwd on the transmission line (calculated with the lumped-element equations above),  the resulting amplitude and phase of the reflections and re-reflections can be calculated using the equations shown in the Lattice Diagram, below.

Here's the Lattice Diagram by itself:

Note that equations in the lattice diagram are easily determined and are based solely upon the calculated Reflection coefficients at either end of the Transmission line and also the Transmission coefficient at the load.

Note that the e-jφ factor is simply the phase-shift of the signal due to the transmission line's delay.  In this example in which the transmission line is one wavelength long, the delay is 360 degrees, and so the delay factor e-j2π equals 1 + j0.  In other words, it has no effect

Note that the reflections described by the lattice diagram continue forever.  Thus, when summed, they are infinite series.

Never the less, these infinite series converge to values (see http://k6jca.blogspot.com/2021/02/antenna-tuners-transient-and-steady.html for the math describing how to determine the equations to calculate the values to which they converge).

For the Lattice Diagram, above, the voltages converge to the following values:

Regarding Vf2 in the equation above, it represents the sum of all re-reflections (reflected from the Output port of the impedance matching network) traveling towards the load.  Thus it excludes the very first forward signal (called Vf1) which represents the current signal from the generator, passing through the matching network, to the transmission line.

3:  Procedure to Perform Calculations

I calculate the various voltages (nodal and transmission-line) in time increments of Td, where Td is the period of the 10 MHz signal.  The Transmission Line in this example is ideal (thus the velocity factor equals 1), and so Td is also the amount of time it takes a "point" on the sine-wave to travel from one end of the one-wavelength long transmission line to the other end (and so a round-trip on the transmission line is 2*Td).

Below is the procedure I used to calculate the system's voltages over time.

1. Calculate the reflection and transmission coefficients Γ1, Γ2,, and Tld.
2. At startup (time 't' = 0) calculate impedances Za and Zb, and voltages Va and Vb, using the lumped-element equations and impedances, with Z_TLin equal to 50 ohms.
3. Given Va, Vb, and Z_TLin (= 50 ohms at startup), calculate the "equivalent" forward and reflected voltages that lumped-element Directional Couplers would generate:  Vf_LCin, Vr_LCin, and Vf_TLin using the lumped-element equations (Vr_TLin will be 0 as there are no  reflections from the load yet).
4. Set Vf1 equal to Vf_TLin.  Vf1 represents the forward voltage on the transmission line generated at that moment by the source.
5. Create three values for keeping track of running sums of Transmission Line Forward and Reflected voltages and the Load voltage:  Vf, Vr, and Vload. For time t = 0, Set Vf(t) equal to Vf1 and set Vr(t) and Vload(t) to 0 (there are no reflections yet, and no signal has yet arrived at the load)
6. Increment time t by Td (t = t + Td).  Calculate Vld(t) using the Lattice diagram equation for this time increment.
8. Increment time t by Td (t = t + Td).  Calculate Vr_TLin(t) and Vf_TLin(t) using Lattice diagram equations for this time increment.  Vr_TLin represents the reflection from the load when it arrives back at the transmission line input (i.e. the line's "source" end) and Vf_TLin represents the re-reflection of the Vr_TLin signal off of the impedance discontinuity at the transmission line's input as it starts its return journey on the line back towards the load.
9. Add these values to the Vf, Vr, and Vload  running sums:  Vf(t) = Vf (t-1) + Vf_TLin(t); Vr(t) = Vr(t-1) + Vr_TLin(t).
10. Calculate the new Z_TLin using Vf, Vr, and Rref (see lumped-element equations).
11. Using this new Z_TLin, calculate impedances Za, Zb, and voltages Va, Vb, Vf_LCin(t), and Vr_LCin(t) using lumped-element equations.
12. Go to Step 6 and repeat.
(Note:  tracking running sums is not required for the voltages Vf_LCin and Vr_LCin because their values are based upon Z_TLin, which itself is calculated from running sums of the Transmission line Vf and Vr voltages).

4.  Equation Results:

Below are the results from the procedure described above (calculated using MATLAB and tabulated in an Excel spreadsheet):

As a check, let's take a look at the steady-state results predicted by the following   Lattice Diagram equations:

The results of these equations are:

Vr = 0.8*0.6/(1 - 0.6*0.6) = 0.75

Vld = 0.8*1.6/(1 - 0.6*0.6) = 2.0

Vf2 = 0.8*0.6*0.6/(1 - 0.6*0.6) = 0.45

Vf = Vf1 + Vf2 = 0.8 + 0.45 = 1.25

These results match the steady-state values shown in the Excel table, above.

The results derived with the above procedure can be verified with a Simulink Simulation.

Here is the Simulink model that I used for verification:

(Note that the model for the Directional Couplers has been shown earlier in this post).

The input is a gated sine-wave.  This lets me determine Vf1 and Vf2 from the simulation results.  Vf1 is the forward voltage on the Transmission line at the start of the gated sine-wave.  And Vf2 is the forward voltage on the Transmission line immediately after the gated-sine has been turned off (in other words, there is no longer a Vf1, but the earlier re-reflections are still traveling from the tuner-end of the transmission line towards the load).

You can see below that the amplitude of Vf1 is 0.8 volts and Vf2 is 0.45 volts.  These sum to the steady-state value Vf value of 1.25.

(Note:  the magnitude of Vf, |Vf| only equals |Vf1| + |Vf2| if the reflection coefficients Γ1 and Γ1 at the two ends of the transmission line are real, without imaginary terms, and if the transmission line is a multiple of a half-wavelength in length.  Otherwise, phase-shift is introduced between the two ends of the line and it is very likely that Vf2 will not be in phase with Vf1.  In this case, vector addition still works (i.e. Vf = Vf1 + Vf2), but the sum of the magnitudes of Vf1 and Vf2 no longer equals the magnitude of Vf.)

And below we can see that the directional coupler's "equivalent Vref" at the L-network's input decays to 0 in steady-state, indicating that the L-network's input impedance looks like 50 ohms in steady-state.

The resulting Simulink wave-forms and their amplitudes match my calculated results.

6.  Summary of Forward and Reflected Voltages:

The diagram below illustrates the steady-state Forward and Reflected voltages and power when Zs equals 50 ohms:

Example with a Source Impedance Not Equal to 50 Ohms:

Let's set Zs to 5 ohms.  Here's the new circuit:

The Reflection and Transmission Coefficients at the load remain unchanged.  But the Reflection Coefficient looking into the output of the L-network (towards the source) has changed from 0.6 to the complex number 0.8995 + j0.2621.

Note that this reflection coefficient can be found two ways:

1.  Circuit analysis method (in which Vs is replaced by a short):

2.  SimSmith method:

Calculated Results:

The following results were calculated using the procedure described above (and MATLAB).

Note the steady-state magnitudes of Vf, Vr, and Vload are the same irrespective of Zs.

Examining the calculated values in their complex form, I can derive the following diagram showing the steady-state voltages and power on the transmission line.

Note that Vf1 is now a complex number (compared to the value calculated when Zs was 50 ohms).  Never the less, I can still calculate Vf2 and, with Vr, calculate the Reflection Coefficient looking into the LC network's Output port.  Specifically, I made this calculation as follows:
• Vf1 (calculated at time t = 0) = 0.1166 - j0.5911
• Vf (at time t = 28*Td, which is as far as I took my calculations) = 0.6199 - j1.0724.
• Subtracting these two quantities, Vf2 = Vf - Vf1 = 0.5033 - j0.4814
• Vr (at time t = 28*Td) = 0.3721 - j0.6435
I can derive Γ1 by dividing Vf2 by Vr:  Γ1 = Vf2/Vr = 0.8995 + j0.2621, which is the same value calculated using circuit analysis techniques.

(Note that because Γ1 now has an imaginary part, the magnitudes of Vf1 and Vf2 no longer sum to be the magnitude of Vf, as it did in the Zs = 50 ohm example, above.  But Vf still equals Vf1 + Vf2, but one needs to use vector arithmetic for the calculations.)

Simulated Results:

The circuit was simulated with Simulink:

I adjusted the source amplitude from 2 to 1.09 to set the steady-state Load voltage very close to 2.0 volts peak -- this is the same Vload level for the 50 ohm Zs simulations).

Note that I've increased the length of the transmission line from 1 lambda to 2 lambda.  This gives the signal a bit more time (i.e. 4 cycles) to settle down before the next reflection arrives (it takes a bit longer for the signal to settle when Zs is 5 ohms).

The Simulation results are below.  Note that an SWR meter at the input to the LC network would still indicate an SWR of 1:1 after a dozen or so cycles following startup:

The Transmission Line voltages Vf  and Vr, and the output voltage Vload, have the same steady-state values as they had in the Zs = 50 ohm simulation.  But the transient voltage levels at startup and shut-down differ from the 50 ohm simulation.  However, they are consistent with the calculations tabulated for Zs = 5 ohms, above.  So we can say that the simulation validates the calculations performed using the lumped-element and transmission line models.

From these plots we can still get an idea of the magnitude of reflection coefficient looking into the L-network's output port (we cannot get the angle because these plots don't show us the phase relationship between Vr and Vf2).  Note that |Vf2| is about 0.69 volts (this is measured on the "back porch" of Vf_TLin, just after the source turns off (and waiting a cycle or two for the level to stabilize), while |Vr| (in steady-state) is about 0.744 volts.

1| equals |Vf2|/|Vr| = 0.69/0.74 = 0.93, which is quite close to the magnitude of the calculated Γ1 (= 0.9369)

Conclusions:

1.  The Reflection Coefficient looking into the output of the Impedance Matching network (Antenna Tuner) is a function of the source impedance and, for a source modeled as a voltage with a series source impedance, can be calculated using basic circuit analysis techniques.

2.  This Reflection Coefficient may or may not be the Complex Conjugate of the load impedance connected to the Output port of the Impedance Matching network.  If the source impedance is 50 ohms (and there is no other loss in the system), then it is the complex conjugate and thus there is a Conjugate Match (this can be quickly demonstrated with SimSmith).  But for all other possible source impedances, there is no Conjugate Match.

3.  Irrespective of presence or absence of a Conjugate Match, an SWR meter at the input of the Impedance Matching network will show an SWR of 1:1 when the Impedance Matching network has been tuned so that the impedance looking into the Impedance Matching network's input port is 50 + j0 ohms.  The transmitter's source resistance has no effect on this tuning.

Notes:

If you'd like to play with the Simulink models, you can find them here:   https://github.com/k6jca/Antenna-Tuner-Simulink-Models

They were created with Simulink R2020a (version 10.1), so if you have an earlier version of Simulink, you might not be able to run them.

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.

Standard Caveat:

As always, I might have made a mistake in my equations, assumptions, drawings, or interpretations.  If you see anything you believe to be in error or if anything is confusing, please feel free to contact me or comment below.

And so I should add -- this 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.