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GATE-2012 ECE Q28 (electromagnetics)

Posted By Krishna Sankar On February 20, 2013 @ 6:50 am In GATE | No Comments

Question 28 on electromagnetics from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper.

## Solution

To answer this question, let us first understand the propagation in a transmission line,  termination and the concept of impedance matching. The section 2.1 in Microwave Engineering, David M Pozar (buy from Amazon.com [1], Buy from Flipkart.com [2] is used as reference.

Consider a transmission line of very small length $\Delta z$ having the parameters as show in figure below.

Figure : Transmission line model  (Reference Figure 2.1 in Microwave Engineering, David M Pozar (buy from Amazon.com [1]Buy from Flipkart.com [2])

$R$ is the resistance per unit length $\Omega/m$,

$L$ is the inductance per unit length $H/m$,

$G$ is the conductance per unit length $S/m$,

$C$ is the capacitance per unit length$F/m$.

Applying Kirchoff’s voltage law,

$v(z,t)-R\Delta zi(z,t)-L\Delta z\frac{\partial i(z,t)}{\partial t}-v(z+\Delta z ,t)=0$

Applying Kirchoff’s current law,

$i(z,t)-G\Delta zv(z+\Delta z,t)-C\Delta z\frac{\partial v(z+\Delta z,t)}{\partial t}-i(z+\Delta z ,t)=0$

Dividing the above equations by $\Delta z$ and taking the limit $\Delta z \rightarrow 0$,

$\frac{\partial v(z,t)}{\partial z}=-Ri(z,t) - L\frac{\partial i(z,t)}{\partial t}$

$\frac{\partial i(z,t)}{\partial z}=-Gv(z,t) - C\frac{\partial v(z,t)}{\partial t}$

If we assume that the inputs are sinusoidal, then the above equation can be re-written as

$\frac{dV(z)}{dz}=-(R+jwL)I(z)$

$\frac{dI(z)}{dz}=-(G+jwC)V(z)$

Substituting,

$\frac{d^2V(z)}{dz}-\gamma^2V(z)=0$,

$\frac{d^2I(z)}{dz}-\gamma^2I(z)=0$,

where

$\gamma=\alpha+j\beta = \sqrt{(R+jwL)(G+jwC)}$.

The solution to the above equations are,

$V(z) = V_0^+e^{-\gamma z}+V_0^-e^{\gamma z}$

$I(z) = I_0^+e^{-\gamma z}+I_0^-e^{\gamma z}$.

The current on the line can be alternately expressed as,

$I(z) =\frac{\gamma}{R+jwL}$$V_0^+e^{-\gamma z}-V_0^-e^{\gamma z}$$=\frac{1}{Z_0}$$V_0^+e^{-\gamma z}-V_0^-e^{\gamma z}$$$,

where the characteristic impedance of the line is defined as,

$Z_0=\frac{R+jwL}{\gamma}=\sqrt{\frac{R+jwL}{G+jwC}}$.

The wavelength on the line is,

$\lambda=\frac{2\pi}{\beta}$.

Loss less transmission line case

For a lossless transmission line, we can set $R=G=0$.

Then the propagation constant reduces to$\begin{array}\alpha=0,&\beta = \omega\sqrt{LC}\end{array}$, the characteristic impedance is $Z_0=\sqrt{\frac{L}{C}}$ and the voltage and current on the line can be written as,

$V(z) = V_0^+e^{-j\beta z}+V_0^-e^{j\beta z}$

$I(z) =\frac{1}{Z_0}$$V_0^+e^{-j\beta z}-V_0^-e^{j\beta z}$$$

## Terminated lossless transmission line

Consider a transmission line terminated with load impedance $Z_L$ as shown in figure below.

Figure: Transmission line with load impedance (Reference Figure 2.4 in Microwave Engineering, David M Pozar (buy from Amazon.com [1]Buy from Flipkart.com [2])

At the load $z=0$, the relation between the total voltage and current is related to the load impedance $Z_L$

$Z_L=\frac{V(0)}{I(0)}=\frac{V_0^{+}+V_0^{-}}{V_0^{+}-V_0^-}Z_0$.

Alternatively,

$V_0^{-}=\frac{Z_L-Z_0}{Z_L+Z_0}V_0^{+}$.

The reflection coefficient $Z_L$ is defined as the amplitude of the reflected voltage to the incident voltage,

$\Gamma=\frac{V_0^{-}}{V_0^{+}}=\frac{Z_L-Z_0}{Z_L+Z_0}$.

For no reflection to happen, i.e $\Gamma=0$, the load impedance $Z_L$ should be equal to the characteristic impedance $Z_0$ of the transmission line. The above equation captures the impedance seen at the load $z=0$

The voltage and current on the line can be represented using $\Gamma$ as,

$V(z) = V_0^+$$e^{-j\beta z}+\Gamma e^{j\beta z}$$$

$I(z) =\frac{V_0^+}{Z_0}$$e^{-j\beta z}-\Gamma e^{j\beta z}$$$

When looking from a point $z=-l$ from the load, the input impedance seen is,

$\begin{array}{llllll}Z_{in}&=&\frac{V(-l)}{I(-l)}&=&\frac{V_0^{+}$$e^{j\beta l} + \Gamma e^{-j\beta l}$$}{\frac{V_0^{+}}{Z_0}$$e^{j\beta l} - \Gamma e^{-j\beta l}$$}\end{array}$.

Substituting for $\Gamma$

$\begin{array}{llllll}Z_{in}&=&Z_0\frac{$$Z_L+Z_0$$e^{j\beta l} +(Z_L-Z_0) e^{-j\beta l}}{$$Z_L+Z_0$$e^{j\beta l} -$$Z_L-Z_0$$ e^{-j\beta l}}\\&=&Z_0\frac{Z_L+jZ_0\tan \beta l}{Z_0+jZ_L\tan \beta l}\end{array}$.

Special case when ${l =\lambda/4}$ (and it’s odd multiples)

For the case when ${l =$$2n+1$$\frac{\lambda}{4}}$ the input impedance seen is,

$\begin{array}{llll}Z_{in}&=&Z_0\frac{Z_L+jZ_0\tan \beta l}{Z_0+jZ_L\tan \beta l}\\&=&Z_0\frac{Z_L+jZ_0\tan$$\frac{2\pi}{\lambda}\frac{(2n+1)\lambda}{4}$$}{Z_0+jZ_L\tan$$\frac{2\pi}{\lambda}\frac{(2n+1)\lambda}{4}$$}\\&=&Z_0\frac{Z_0}{Z_L}=\frac{Z_0^2}{Z_L}\end{array}$.

This result can be used to for impedance matching.

## Quarter wave transformer

Consider a circuit with load $R_L$ and a line with characteristic impedance $Z_0$ connected by a transmission line of characteristic impedance $Z_1$ with length $\lambda/4$.

Figure: Quarter Wave Matching transformer (Reference Figure 2.16 in Microwave Engineering, David M Pozar (buy from Amazon.com [1]Buy from Flipkart.com [2])

The input impedance seen is,

$\begin{array}{llll}Z_{in}&=&Z_1\frac{R_L+jZ_1\tan \beta l}{Z_1+jR_L\tan \beta l} \\&=&Z_0\frac{Z_L+jZ_0\tan$$\frac{2\pi}{\lambda}\frac{\lambda}{4}$$}{Z_0+jZ_L\tan$$\frac{2\pi}{\lambda}\frac{\lambda}{4}$$}& \mbox{ } & $$\beta l =\frac{2\pi}{\lambda}\frac{\lambda}{4}\right \pi/2$$ \\&=&Z_1\frac{Z_1}{R_L}=\frac{Z_1^2}{R_L}\end{array}$.

So if we choose $Z_1 = \sqrt{Z_0R_L}$, then the input impedance seen is $Z_{in} = Z_0$ which is the condition required for having no reflection i.e. $\Gamma=0$.

One important aspect to note here is that $\Gamma=0$ is not guaranteed for all frequencies, but rather only for certain frequencies. The frequency dependence can be found by finding the frequencies for which $\beta l=$$2n+1$$\frac{\pi}{2}$ .

Replacing $l=\frac{\lambda_0}{4}$ where $\lambda_0$ is the wave length corresponding to frequency $f_0$,

$\beta l = $$\frac{2\pi}{\lambda}$$$$\frac{\lambda_0}{4}$$=$$\frac{2\pi f}{v_p}$$$$\frac{v_p}{4f_0}$$=\frac{\pi f}{2 f_0}$.

It can be seen that only for $f=$$2n+1$$f_0$ , the term $\beta l=$$2n+1$$\frac{\pi}{2}$ resulting in reflection coefficient $\Gamma=0$ only for those frequencies.

## Solving the GATE question

Applying all this to the problem at hand, we have $Z_0=50$$R_L=200$  and $R_1=100$.

Given that $R_1=\sqrt{Z_0 R_L} = \sqrt{50 * 200 }=100$, we know that a quarter wave transformer is used to achieve impedance matching.

Now we also know that we need to match for two frequencies $f_{1}=429\mbox{ MHz}$ and $f_{2}=1\mbox{ GHz}$.

The wavelength for each frequencies are,

$\lambda_{1}=\frac{3e^8}{429e^6}*100\simeq 70\mbox{ cm}$

$\lambda_{2}=\frac{3e^8}{1e^9}*100\simeq 30\mbox{ cm}$

The least common multiple of these two wavelength is, $\lambda_{lcm}=\mbox{lcm}$$70,30$$=210\mbox{ cm}$ and the corresponding quarter wave length is $\frac{\lambda_{lcm}}{4}=52.5\mbox{ cm}$.

Given than $\frac{\lambda_{lcm}}{4}=52.5\mbox{ cm}$ is not listed in the options, we can go for the next higher odd multiple i.e. $52.5*3=157.5 \simeq 1.58\mbox{ m}$

Based on the above, the right choice is (C) 1.58m

## References

[1] GATE Examination Question Papers [Previous Years] from Indian Institute of Technology, Madras http://gate.iitm.ac.in/gateqps/2012/ec.pdf [3]

[2] Microwave Engineering, David M Pozar (buy from Amazon.com [1]Buy from Flipkart.com [2]

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