**Points of Discussion: Microwave Resonators**

**Introduction to Microwave Resonators****Series Resonator Circuit****Parallel Resonator Circuit****Transmission Line Resonators****Solved Mathematical Example of Microwave Resonators**

**Introduction to Microwave Resonators**

Microwave resonators are one of the crucial elements in microwave communication circuit. They can create, filter out, and select frequencies in various applications, including oscillators, filters, frequency meters, and tuned oscillators.

Operations of microwave-resonators are very much like the resonators used in network-theory. We will discuss the series and parallel RLC resonant circuits at first. Then, we will find out various applications of resonators at microwave frequencies.

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**Series Resonator Circuit**

A series resonator circuit is made by arranging a resistor, an inductor, and a capacitor in series connection with a voltage source. The circuit diagram of a series RLC is given below. It is one of the type of microwave resonators.

The input impedance of the circuit is given as **Z _{in} = R + jωL − j /ωC**

The complex power from the resonator is given by P_{in}.

**P _{in} = ½ VI* = ½ Z_{in} | I|2 = ½ Z_{in} | (V/Z_{in}) |^{2}**

**Or, P _{in} = ½ |I|^{2} (R + jωL − j /ωC)**

The power by the resistor is: **P _{loss} = ½ |I|^{2} R**

The average magnetic energy stored by the inductor L is:

**W _{e} = ¼ |V_{c}|^{2} C = ¼ |I|^{2} (1/ω^{2}C)**

Here, V_{c} is the voltage across the capacitor.

Now, complex power can be written as follow.

**P _{in} = P_{loss} + 2 jω (W_{m} − W_{e})**

Also, the input impedance can be written as: **Z _{in} = 2P_{in}/ |I|^{2}**

**Or, Z _{in} = [P_{loss} + 2 jω (W_{m} − W_{e})] / [½ |I|^{2}]**

In a circuit, resonance occurs when the stored average magnetic field and the electric charges are equal. That means, W_{m} = W_{e}. The input impedance at resonance is: **Z _{in} = P_{loss} / [½ |I|^{2}] = R.**

R is a pure real value.

At W_{m} = W_{e}, the resonance frequency ω_{0} can be written as **ω _{ 0} = 1/ √(LC)**

Another critical parameter of the resonant circuit is the Q factor or quality factor. It is defined as the ratio of the average energy stored to the energy loss per second. Mathematically,

**Q = ω * Average energy change**

**Or Q = ω *(W _{m} + W_{e}) / P_{loss}**

Q is a parameter which gives us the loss. Higher Q value implies the lower loss of the circuit. Losses in a resonator may occur due to loss in conductors, dielectric loss, or radiation loss. An externally connected network may also introduce losses to the circuit. Each of the losses contributes to the lowering of the Q factor.

**The Resonator’s Q is known as Unloaded q. It is given by Q _{0}.**

The unloaded Q or Q_{0} can be calculated from the previous equations of Q factor and Power loss.

**Q _{0} = ω_{ 0} 2W_{m} / P_{loss} = w_{0}L / R = 1/ w_{0}Rc**

From the above expression, we can say that the Q decreases with the increase of R.

We will now study the behaviour of the input impedance of the resonator circuit when it is near its resonance frequency. Let w = w0 + Δω, here Δω represents a minimal amount. Now, the input impedance can be written as:

**Z _{in} = R + jωL (1 − 1/ω^{2}LC)**

**Or Z _{in} = R + jωL ((ω^{2} – ω_{0}^{2})/ω^{2})**

**Now, ω ^{2}_{0} = 1/LC and ω^{2} − ω^{2}_{0} = (ω − ω_{0}) (ω + ω_{0}) = Δω (2ω − Δω)2ω Δω**

**Z _{in} ~ R + j^{2}L Δω**

**Z _{in} ~ R + j^{2}RQ_{0}L Δω / ω_{0}**

Now, the calculation for half-power fractional bandwidth of the resonator. Now, if the frequency becomes |Z_{in}|^{ 2} = 2R^{2}, the resonance receives 50% of the total delivered power.

One more condition is such that when the Band Width value is in fraction, the value of **Δω/ω _{0}** becomes half of the Band Width.

**|R + jRQ _{0}(BW)| ^{2} = 2R^{2},**

**or BW = 1/Q _{0}**

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**Parallel Resonant Circuit**

A parallel resonator circuit is made by arranging a resistor, an inductor, and a capacitor in parallel with a voltage source. The circuit diagram of a parallel RLC is given below. It is one of the type of microwave resonators.

Z_{in} gives the input impedance of the circuit.

**Z _{in} = [1/R + 1/jωL + jωC] ^{−1}**

The complex power delivered from resonator is given as P_{in}.

**P _{loss} = ½ VI* = ½ Z_{in} | I|^{2} = ½ Z_{in} | V|^{2} / Z_{in}***

**Or P _{in} = ½ |V|^{2} (1/R + j/wL – jωC)**

The power from resistor R is P_{loss}.

**P _{loss} = ½ |V|^{2} / R**

Now, the Capacitor also stores the energy, it is given by –

**W _{e} = ¼ |V|^{2}C**

The inductor also stores the magnetic energy, it is given by –

**W _{m} = ¼ |I_{L}|^{2} L = ¼ |V|^{2} (1/ ω^{2}L)**

IL is the current through the inductor. Now, the complex power can be written as: **P _{in} = P_{loss} + + 2 jω (W_{m} − W_{e})**

The input impedance can also be written as: **Z _{in} = 2P_{in}/ |I|^{2} = (P_{loss} + 2 jω (W_{m} − W_{e}))/ ½ |I|^{2}**

In the series circuit, the resonance occurs at W_{m} = W_{e}. Then the input impedance at resonance is Z_{in} = P_{loss} / ½ |I|^{2} = R

And the resonant frequency at W_{m} = W_{e} can be written as **w _{0} = 1 / √ (LC)**

It is same as the value of series resistance. Resonance for the parallel RLC circuit is known as an antiresonance.

The concept of unloaded Q, as discussed early, is also applicable here. The unloaded Q for the parallel RLC circuit is represented as **Q _{0} = ω_{0}^{2}W_{m}/ P_{loss}.**

**Or Q _{0} = R /ω_{0}L = ω_{0}RC**

Now, at antiresonance, “W_{e} = W_{m}”, and the value of the Q factor decreases with the decrease in R’s value.

Again, for input impedance near resonance frequency consider ω = ω_{0} +Δω. Here, Δω is assumed as a small value. The input impedance is again rewritten as Z_{in}.

**Z _{in} = [ 1/R + (1 – Δω / ω_{0}) / jω_{0}L + jω_{0}C + jΔωC]^{ -1}**

**Or Z _{in} = [ 1/R + j Δω / ω2L + jΔωC] ^{– 1}**

**Or Z _{in} = [ 1/R + 2jΔωC]^{-1}**

**Or Z _{in} = R / (1 + 2jQ_{0}Δω/ω_{0})**

Since **ω ^{2} = 1/LC** and R = infinite.

**Z _{in }= 1 / (j^{2}C (ω – ω_{0}))**

The half-power bandwidth edges occur at frequencies (Δω / ω_{0} = BW/2) such that, **|Z _{in}|^{2} = R^{2}/ 2**

**Band Width = 1 / Q _{0}.**

**Transmission Line Resonators**

Almost always, the perfect lumped components can not deal in the range of microwave frequencies. That is why distributed elements are used at microwave frequency ranges. Let us discuss various parts of transmission lines. We will also take into consideration of the loss of transmission lines as we have to calculate the resonators’ Q value.

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**Short circuited λ/2 Line**

Let us take a transmission line which suffers loss and also it is short-circuited at one of its terminal.

Let us assume the transmission line has a characteristic impedance of Z_{0}, the propagation constant of β and attenuation constant is α.

We know that, at resonance, the resonant frequency is ω = ω_{0}. The length of the line ‘l’ is λ/2.

The input impedance can be written as **Z _{in} = Z_{0} tanh (α + jβ)l**

Simplifying the tangential hyperbolic function, we get Z_{in}.

**Z _{in} = Z_{0} (tanh αl + j tan βl) / (1 + j tan βl tanh αl).**

For a lossless line, we know that **Z _{in} = jZ_{0} tan βl if α = 0.**

As discussed earlier, we will consider the loss. That I why, we will take,

**αl << 1 **and **tanh αl = αl**.

For a TEM line,

**βl = ωl/ v _{p} = ω_{0}l/ v_{p} + Δωl/ v_{p}**

v_{p} is an important parameter which represents the transmission line’s phase velocity. ** L = λ/2 = πv_{p}/ω_{0} for ω = ω_{0}**, we can write,

**βl = π + Δωπ/ ω _{0}**

Then, **tan βl = tan (π + ωπ/ ω _{0}) = tan (ωπ/ ω_{0}) = ωπ/ ω_{0}**

Finally, **Z _{in} = R + 2 jLω**

At last, the value of resistance comes as: **R = Z _{0}αl**

The value of inductance comes as : **L = Z _{0}π/ 2ω_{0}**

And, the value of Capacitance comes as : **C = 1/ ω ^{2}_{0}L**

The unloaded Q of this resonator is, **Q _{0} = ω_{0}L/ R = π/ 2αl = β/ 2α**

**Solved mathematical Example of Microwave Resonators**

**1. A λ/2 resonator is made up of copper coaxial line. Its inner radius is 1 mm, and the outer radius is 4mm. The value of the resonant frequency is given as 5 GHz. Comment on the calculated Q value of two coaxial line among which one is filled with air another filled with Teflon. **

**Solution:**

a = 0.001, b = 0.004, η = 377 ohm

We know that the conductivity of the copper is 5.81 x 107 S/m.

Thus, the surface resistivity at 5GHz = Rs.

Rs = root (ωµ0/ 2σ)

Or Rs = 1.84 x 10-2 ohm

Air filled attenuation,

αc = Rs /2η ln b/a {1/ a + 1/ b}

Or αc = 0.22 Np/m.

For Teflon,

Epr = 2.08 and tan δ = 0.0004

αc = 0.032 Np/m.

There is no dielectric los due to air filled, But for Teflon-filled,

αd = k0 √epr/2 * tan δ

αd = 0.030 Np/m

So, Q_{air} = 104.7 / 2 * 0.022 = 2380

Q_{tefflon} = 104.7 * root(2.008) / 2 * 0.062 = 1218