There are all sorts of resonances around us, in the world, in our culture, and in our technology. A tidal resonance causes the 55 foot tides in the Bay of Fundy. Mechanical and acoustical resonances and their control are at the center of practically every musical instrument that ever existed. Even our voices and speech are based on controlling the resonances in our throat and mouth. Technology is also a heavy user of resonance. All clocks, radios, televisions, and gps navigating systems use electronic resonators at their very core. Doctors use magnetic resonance imaging or MRI to sense the resonances in atomic nuclei to map the insides of their patients. In spite of the great diversity of resonators, they all share many common properties. In this blog, we will delve into their various aspects. It is hoped that this will serve both the students and professionals who would like to understand more about resonators. I hope all will enjoy the animations.

Origins of Newton's laws of motion

History of mechanical clocks with animations
Understanding a mechanical clock with animations
includes pendulum, balance wheel, and quartz clocks

## Friday, June 28, 2013

### 3.16e Steady state response of a transmission line resonator via a Smith Chart

 All postings by author previous: 3.16d Transmission line resonators up: Contents next: 3.16f Transient response

3.16e Steady state response of a transmission line resonator via a Smith Chart

Keywords: Smith chart, resonator, input impedance, matching, Q0

Summary: The Smith chart is one of the most important tools for designing and analyzing the transmission line circuits that are used throughout modern communication hardware. In this posting we use this tool to analyze and better understand a transmission line resonator.

Topics covered in this posting

• Analysis of a transmission line resonator with a Smith chart.
• Qualitative explanation of how an inductor can accomplish matching the transmission line to the resonator.

Contents of this posting

1. Analysis of the circuit with a Smith chart

While we could solve this transmission line problem with equations, they would be somewhat repetitious of previous postings. A nice alternative is the Smith chart which allows graphical solutions to transmission line problems. Once a person is familiar with Smith charts they allow him or her to "see" the solution. The text in this posting may be a little hard to follow for a person new to Smith charts, perhaps requiring some independent studying of Smith charts.

 ⇑ Fig. 4. Smith Chart "calculation" of the transmission line circuit of Fig. 2 of posting 3.16d.

At the right we show a Smith chart marked up for the combination of frequency and LC values that will allow the resonator to completely absorb the waves traveling down the transmission line for the setup shown in Fig. 2 of posting 3.16d. For the sake of comparison, we will assume component values equal to that used in Fig. 2 of posting 3.16c: R0 = 5Ω (of the left transmission line), R = 100Ω and ω0 = 2.5rad/sec.

Instead of the characteristic impedance of the resonator part of the transmission line set at R0 as shown in Fig. 2 of the previous posting, we have set Z0 = 7.85Ω for the resonator as Eqn. 37 of the previous posting implies to keep the intrinsic Q0 as before, i.e. so that Q0 = 20.

The approach on the Smith chart (and with most other transmission line solutions) is to start at the load end (at RL) and work our way towards the wave source. We normalize all resistance and reactive values to the characteristic impedance of the transmission line Z0 = 7.85Ω , i.e. divide all by Z0. So the normalized load resistance is RL/Z0 = 100Ω/7.85Ω = 12.75 .

We plot this point on the Smith chart as point 1 at the far right mid line (real value is 12.75, imaginary value is 0, i.e. 0 reactance). We then move towards the left in the transmission line model, to the left end of the resonator, towards the generator. On the Smith chart this means moving around clockwise in a circle, centered on the middle of the chart. A distance of λ0/2 (half a wavelength) would take us all the way around the chart, back to point 1, which is also labeled as point 2.

While point 2 is correct for the system operating at a frequency of ω0 = 2.5 radians/sec, the resonant frequency of the resonator part in isolation, for a perfect match (that uses the wave energy most efficiently) we would like to move beyond point 2 to point 3, for reasons we will explain in a moment.

We (and the system) can accomplish a perfect match by increasing the drive frequency of the source to a frequency ω a little greater than ω0 which will shorten the wavelength and allow more "wavelengths" to fit in the resonator's length  . If this is done just right, we will be at point 3 on the chart when we are at the entrance to the resonator, at the right end of the coupling inductor LC .

We can figure out the new frequency by looking at the purple radial line that ends at point 5. Point 5 is labeled 0.305 wavelengths from the 0 mark at the left mid line. We started at point 1 which means we need to add another 0.25 wavelengths, giving us a total of 0.25 + 0.305 = 0.555 wavelengths to go from point 1 all the way around to point 3. So the ratio of the resonator's length in new wavelengths to its length in old wavelengths equals 0.555/0.5 = 1.11, i.e. 11% more wavelengths, which means the new wavelength is 11% shorter than the old and since frequency is inversely proportional to wavelength, this means 11% higher frequency. So we expect the new frequency to be 1.11×2.5 = 2.78 radians/sec.

Point 3 is specially chosen so as to lie on the loci of all points with a normalized real impedance of 0.636  or an actual impedance of 7.85Ω×0.636 = 5Ω's. That means that at this point, the resonator presents an impedance with a real part equal to the characteristic impedance of the waveguide at the left (the driving waveguide). The imaginary part of this impedance is negative (in the lower half of the chart) and can be read off the chart. We see from the chart that the normalized impedance Zin = 0.636 − 2.65 j  where j = √−1 . This means that if we add a positive normalized reactance of 2.65 in series with the resonator at this point, we can cancel out the −2.65 capacitative reactance and have a perfect match condition. This we do with the proper choice of coupling inductor, chosen so that ωLC/Z0 = 2.65 , i.e. LC = 2.65Z0/ω = 2.65R0/ω = 2.65×7.85/2.78 = 7.5 H. This agrees with the 8H inductance shown in Fig. 4 of posting 3.16c within the accuracy we might expect. The predicted peak frequency of 2.78radians/sec from the previous paragraph also agrees with the location of the peak in Fig. 2 of that same posting.

2. How does an inductor really accomplish matching?

The Smith chart provides one way to understand the dilemma of how can an inductor be used to match a transmission line to a resonator. Normally we think of a resonator as behaving as a pure resistance when driven at its resonant frequency and because of that we would expect to need a transformer to match the transmission line's characteristic impedance to this pure resistance. The reader needs to understand that ideally a transformer can "transform" a pure resistance from one value to another... allowing a load resistor to appear to have its resistance value changed from the point of view of a driving transmission line. Our Smith chart exercise shows us that we can exploit the tremendous variation in impedance that a resonator displays on either side of resonance to allow an inductor (or capacitor on the other side of resonance) to match the transmission line to the resonator. The inductor participates in the resonance and changes the resonant frequency, shifting the response curve as shown in Fig. 2 of posting 3.16c to the new frequency. This shift in frequency becomes progressively less as one uses higher Q0 resonators (larger values of RL ) which have sharper resonance curves. Experimentally, when dealing with a high Q0 resonator, one need not be aware of the exact method of matching, only that many coupling methods will shift the resonant frequency a little, probably on the order of one bandwidth of the resonator: df½ = f0 / Q0 .

 All postings by author previous: 3.16d Transmission line resonators up: Contents next: 3.16f Transient response