This post is the second post of two analyzing transient and steady-state reflections in systems with transmission lines and impedance matching devices.

This analysis is done using the system's Impulse Response and convolving that with a Stimulus that is bounded in time (that is, the stimulus has a defined start time and a defined stop time).

This method of using the Convolution Integral allows us to examine both the system's transient response (when the stimulus switches from absent to present or from present to absent) and its steady-state response. For more background on this technique, please refer to my first post in this series: Antenna Tuners: Transient and Steady-state Analysis.

__The Quarter-wave Transformer:__The quarter-wave impedance transformer is a two-port device consisting of a quarter wavelength of transmission line whose characteristic impedance is the square-root of the product of the impedances attached at its two ports.

__Calculating the System's Impulse Response:__

To analyze the transient-state of the reflections within the matching system using the Convolution Integral, I first need to calculate the system's impulse response (i.e. its response to a single impulse).

To simplify the analysis, I'll assume that the transmission lines are lossless and behave like perfect delay lines. Also, I will assume that the Source is matched to Transmission Line 1.

Therefore, an impulse traveling on one transmission line, when it arrives at the second transmission line of a different impedance, will be divided into two parts at the impedance discontinuity. One part is reflected backwards in the opposite direction of travel and the other part continues forward. Their amplitudes as they travel along the transmission lines will only be a function of the various Reflection and Transmission Coefficients within the antenna system and not due to transmission line loss.

The Reflection and Transmission Coefficients that a single impulse sees as
it travels through the antenna system are solely the function of the
impedances *immediately at either side* of an impedance discontinuity,
and they are calculated using the standard formulas:

R = (Zb - Za)/(Zb + Za)

T = 2Zb/(Zb + Za)

where Zb is the impedance that the impulse is traveling *to*, and Za the impedance it is traveling *from.*

The drawing below shows the calculated Reflection and Transmission Coefficients at each impedance discontinuity, assuming the load is 200 ohms, the characteristic impedance of Transmission Line 1 is 50 ohms, and the characteristic impedance of the Quarter-wave Transformer (Transmission Line 2) is 100 ohms:

If I shoot a single impulse from the source into Transmission Line 1, it will move from left to right until it reaches the discontinuity at Transmission Line 2. At that point, a portion of the impulse will continue onwards, towards the load, while a second portion is reflected back to the source.

(Note that I've defined the source to be matched to Transmission Line 1, so there is no re-reflection back towards the load when this part of the impulse arrives back at the source).

The first portion, when it reaches the load, will again see an impedance discontinuity and it will again be divided into two parts. One part is dissipated by the load's resistance, and the other part is reflected back towards Transmission Line 1.

When this reflection from the load reaches Transmission Line 1, it is again divided into to parts. One part reflects back to the load, and one part passes through the impedance discontinuity and continues back to the source.

And thus it goes -- being lossless, there will be an infinite number of reflections and re-reflections on Transmission Line 2.

Below is a diagram showing the amplitudes of the impulse (in terms of Reflection and Transmission coefficients) as it is divided into two parts whenever it reaches an impedance discontinuity:

For example, let Vi(t) be a single impulse in time. Let's define its arrival (from the source) at the impedance discontinuity between Transmission Line 1 and Transmission Line 2 to be at time t = 0.

From the lattice diagram above I can calculate impulse responses for Vr1, Vf2, Vr2, and Vld.

__Convolving the System's Impulse Response with a Sinusoidal Stimulus:__Given a sine-wave signal Vi(t) of amplitude equal to 1, the steady-solutions for these equations are shown at the bottom of the lattice diagram, below:

_{2}" to find the equation that calculates the required characteristic impedance of Transmission Line 2, given the load impedance and the characteristic impedance of Transmission Line 1, that will result in an "Apparent Γ

_{2}" equal to 0. See: Multiple Reflections and the Quarter-wave Transformer.)

__Convolution Results:__

I've used MATLAB to first calculate the impulse responses of Vr1, Vf1, Vr2, and Vld, and then to convolve these impulse responses with the sinusoidal stimulus. I've plotted the results, below.

Below is the stimulus, Vi(t). It is present for 12 cycles and otherwise off:

Below are the MATLAB plots of Vr1's impulse response and resulting output when convolved with the stimulus. You can see the non-zero transient response at the start and finish of the sine-wave. And in steady state Vr1 goes to 0.

For convenience, I've plotted the stimulus and the Vr1 output together, below:

Below are Vf2's impulse response and resulting output when it is convolved with the stimulus.

Below are Vr2's impulse response and resulting output when it is convolved with the stimulus.

And finally, below are Vld's impulse response and resulting output when it is convolved with the stimulus.

Again, for convenience, I've plotted the stimulus and the Vld output together, below:

You can see the quarter-cycle delay between the stimulus and Vld in the plot, above.

__The Re-reflection of Vr2:__

In steady-state it is clear that some portion of the Vr2 voltage continues through the discontinuity towards the source. This is why the wave heading back to the source (Vr1) goes to 0 in steady-state (when load is matched to Transmission Line 1) -- this transmitted part of Vr2 cancels the part of Vi that is reflected by the transformer's input discontinuity.

But some part of Vr2 is reflected back to the load. How much? And (assuming that the quarter-wave line is matching load to Transmission Line 1) does Γ_{3} change in steady-state (in a fashion similar to how Γ_{2} changes to to an "apparent" Γ_{2} of 0)?

We can calculate the steady-state Γ_{3} using the same
steady-state equations for Vf2 and Vr2 that were derived earlier.

First, let's call this reflection of Vr2 off of the quarter-wave transformer's "input" impedance discontinuity "Vr2Re_Reflected". This is the portion of Vr2 that is re-reflected back to the load by the impedance discontinuity (the other part continues through the discontinuity to the source)

This re-reflected voltage sums with the forward voltage from the source that is passing through the input discontinuity towards the load. In other words, Vf2 = Vi(0°)Tf + Vr2Re_Reflected.

Therefore Vr2Re_Reflected = Vf2 - Vi(0°)Tf , and the "apparent" Γ_{3} should therefore be the ratio of Vr2Re_Reflected to Vr2.

Using these relationships and recognizing that Vr2 is 180 degrees out of phase with Vi when Vr2 arrives back at the input to the quarter-wave transformer, we can derive the "apparent" Γ_{3} as follows:

Therefore, the steady-state "apparent" Γ

_{3}is the same as the original Γ

_{3}calculated for a single impulse on the line.

__Conclusions:__

1. As with any other linear system, the Convolution Integral allows us to calculate the time-domain transient and steady-state responses of the system to a stimulus.

2. Given a sine-wave stimulus that is bounded in time, there are
reflections back to the source during the system's transient states.
These reflections decay to 0 in the steady-state, resulting in an "apparent" Γ_{2 }of 0.

3. In the steady-state, Vf2 is a combination of the forward signal
arriving from the source (scaled by the Transmission factor of the impedance discontinuity between the two transmission lines) and the *re-reflection* by Γ_{3}, at the impedance discontinuity between the two transmission lines, of the backwards traveling Vr2
wave (from the load). The amplitude of this re-reflection equals Vr2 * |Γ_{3}|.

4. Assuming that the quarter-wave transformer is matching the load to Transmission Line 1's characteristic impedance, in steady-state Γ_{3} remains unchanged from the value
calculated for an impulse response, and Γ_{3} is independent of the source
impedance. In other words, because there are no signals sent back *to* the
source in steady-state, it does not matter what the source impedance is
because there is no signal to be reflected back by a possible source mismatch.

__Standard Caveat:__

Also, I will note:

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