Question 10 on networks from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper.

Q10. The average power delivered to an impedance  by a current  is

(A) 44.2W

(B) 50W

(C) 62.5W

(D) 125W

Solution

The average power delivered to the load is integral of the instantaneous power over some period , i.e.

Given that , the equation reduces to,

.

In our example with 

Plugging in,

.

Based on the above, the right choice is (B) 50W

References

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

[2] http://en.wikipedia.org/wiki/Maximum_power_transfer_theorem

[3] http://en.wikipedia.org/wiki/AC_power

Related posts:

GATE-2012 ECE Q25 (math) GATE-2012 ECE Q12 (math) GATE-2012 ECE Q7 (digital) GATE-2012 ECE Q24 (math)

Question 18 on signals from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper.

Q18. If , then the region of convergence (ROC) of its Z transform  in the z-plane will be

(A) 

(B)

(C)

(D)

Solution

As discussed in Chapter 3 of Discrete-Time Signal Processing, 2nd Edition, Alan V. Oppenheim, John R. Buck, Ronald W. Schafer (Buy from Amazon.com, Buy from Flipkart.com), the z transform of a sequence  is defined as

.

For any given sequence, the values of z for which the z transform converges is called the Region Of Convergence (ROC) i.e.

In our case,

.

Finding the z transform of ,

The first term in the series converges for and the second term will be converging for .

Now, finding the z transform of ,

The term in the series converges for .

Combining both the terms,  the Region Of Convergence is the region of overlap of the two terms, i.e.  and .

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

References

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

[2] Discrete-Time Signal Processing, 2nd Edition, Alan V. Oppenheim, John R. Buck, Ronald W. Schafer (Buy from Amazon.com, Buy from Flipkart.com)

Related posts:

Books GATE-2012 ECE Q11 (signals) GATE-2012 ECE Q34 (signals) GATE-2012 ECE Q3 (communication)

Being not too happy with the speed of the shared hosting, decided to move the blog to an Amazon Elastic Compute Cloud (Amazon EC2) instance.  Given this is a baby step, picked up a micro instance running an Ubuntu server and installed Apache web server, MySQL, PHP . After doing a bit of tweaking with this new instance, imported the SQL database and other files from the shared hosting and pointed the A name record to the new IP address. This switch happened over this weekend.

One particular issue which I faced was frequent crashing of MySQL due to memory limitations. Followed few online instructions to improve the situation and the current configuration seems to be holding up (but this is a cause of worry – need to figure the right solution).

Anyhow, hope you like the decreased page load time!

Some helpful links from the web:

a) How to install WordPress on Amazone EC2

b) Move WordPress site from shared hosting to Amazon EC2

c) DIY: Enable CGI on your Apache server

d) Import MySQL Dumpfile, SQL Datafile Into My Database

e) Making WordPress Stable on EC2-Micro

f) how to enable mod_rewrite in apache2.2 (debian/ubuntu)

Related posts:

Migration to Who’s Who theme OT: Migration to a Deep Blue template LaTex on blogger not working? Migration to new template (skin)

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

Q28. A transmission line with a characteristic impedance of 100 is used to match a 50 section to a 200 section. If the matching is to be done both at 429MHz and 1GHz, the length of the transmission line can be approximately

(A) 82.5cm

(B) 1.05m

(C) 1.58m

(D) 1.75m

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, Buy from Flipkart.com)  is used as reference.

Consider a transmission line of very small length  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, Buy from Flipkart.com)

is the resistance per unit length ,

 is the inductance per unit length ,

 is the conductance per unit length ,

 is the capacitance per unit length.

Applying Kirchoff’s voltage law,

Applying Kirchoff’s current law,

Dividing the above equations by  and taking the limit ,

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

Substituting,

,

,

where

.

The solution to the above equations are,

.

The current on the line can be alternately expressed as,

,

where the characteristic impedance of the line is defined as,

.

The wavelength on the line is,

.

Loss less transmission line case

For a lossless transmission line, we can set .

Then the propagation constant reduces to, the characteristic impedance is  and the voltage and current on the line can be written as,

Terminated lossless transmission line

Consider a transmission line terminated with load impedance  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, Buy from Flipkart.com)

At the load , the relation between the total voltage and current is related to the load impedance 

.

Alternatively,

.

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

.

For no reflection to happen, i.e , the load impedance should be equal to the characteristic impedance of the transmission line. The above equation captures the impedance seen at the load . 

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

When looking from a point  from the load, the input impedance seen is,

.

Substituting for , 

.

Special case when  (and it’s odd multiples)

For the case when  the input impedance seen is,

.

This result can be used to for impedance matching.

Quarter wave transformer

Consider a circuit with load  and a line with characteristic impedance connected by a transmission line of characteristic impedance with length .

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

The input impedance seen is,

.

So if we choose , then the input impedance seen is  which is the condition required for having no reflection i.e. .

One important aspect to note here is that  is not guaranteed for all frequencies, but rather only for certain frequencies. The frequency dependence can be found by finding the frequencies for which  .

Replacing  where is the wave length corresponding to frequency ,

.

It can be seen that only for  , the term resulting in reflection coefficient only for those frequencies.

Solving the GATE question

Applying all this to the problem at hand, we have ,   and .

Given that , 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  and .

The wavelength for each frequencies are,

The least common multiple of these two wavelength is,  and the corresponding quarter wave length is .

Given than  is not listed in the options, we can go for the next higher odd multiple i.e. 

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

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

Related posts:

GATE-2012 ECE Q16 (electromagnetics) GATE-2012 ECE Q47 (math) GATE-2012 ECE Q13 (circuits) GATE-2012 ECE Q52 (electromagnetics)

The post on IQ imbalance in transmitter, briefly discussed the effect of amplitude and phase imbalance and also showed that IQ imbalance results in spectrum at the image frequency. In this article, we will quantify the power of the image with respect to the desired tone (also known as IMage Rejection Ratio IMRR) for different values of gain and phase imbalance.

System Model

Consider an IQ modulator having gain of  and on each arm and phase imbalance of  as shown in figure below.

Figure : IQ modulator with gain and phase imbalance

The output signal is,

.

Considering an ideal IQ demodulator multiplying the received signal  with and respectively,

.

.

Ignoring the common term  and writing the base band equivalent form,

.

This is the model for transmit IQ imbalance. 

Image Rejection Ratio (IMRR) with transmit IQ imbalance

By sending a complex sinusoidal , and by taking ratio of the power of the signal at the image frequency   and desired frequency  , the image rejection ratio can be computed.

Let and correspondingly,  and .

Finding the  component

To find the  component, multiply the received signal  with  and integrate over period .

The power of the  component is,

Finding the  component

To find the  component, multiply the received signal  with  and integrate over period .

The Image Rejection Ratio (IMRR) is

.

Substituting  and  with variable  and , the equation simplifies to,

.

A useful approximation to IMRR

When there is no phase imbalance i.e , the equation reduces to,

.

When there is no gain imbalance i.e , the equation reduces to,

.

As these two are independent, they can be added to give an approximate value of Image Rejection Ratio.

Summarizing, the Image Rejection Ratio for a given value of gain imbalance  and phase imbalance is,

.

Simulation Results

Simple Matlab/Octave code plotting the simulated and theoretical values of Image Rejection for different values of gain and phase imbalance.

clear; close all

N  = 64;
fm = 2;
gammadB_v = [-3:.1:3];
phiDeg_v  = [-6:.2:6];
[tt gammadB_zeroIdx ] = min(abs((gammadB_v-0)));
[tt phiDeg_zeroIdx  ] = min(abs((phiDeg_v-0)));

for (ii = 1:length(gammadB_v))
   for (jj = 1:length(phiDeg_v))
      gammadB    = gammadB_v(ii); 
      phiDeg     = phiDeg_v(jj);
      gammaLin   = 10^(gammadB/20); 
      phiRad     = phiDeg*pi/180;
      epsilonLin = gammaLin -1 ;

      % transmitted signal
      xt        = exp(j*2*pi*fm*[0:N-1]/N);
      % received signal with IQ imbalance
      xht_re    = gammaLin*cos(phiRad/2)*real(xt) + sin(phiRad/2)*imag(xt);
      xht_im    = gammaLin*sin(phiRad/2)*real(xt) + cos(phiRad/2)*imag(xt);
      xht       = xht_re + j*xht_im; 

      % taking ifft() to find the +fm and -fm components
      yF        = fft(xht,N);
      y_pfm     = yF(fm+1);
      y_nfm     = yF(N-fm+1);

      est_imrr_lin    = (abs(y_nfm)./abs(y_pfm))^2;
      theory_imrr_lin = (gammaLin^2 + 1  - 2*gammaLin*cos(phiRad))./(gammaLin^2 + 1  + 2*gammaLin*cos(phiRad));
      approx_imrr_lin = (epsilonLin^2 + phiRad^2)/4;

      est_imrr_dB(ii,jj)    = 10*log10(est_imrr_lin);
      theory_imrr_dB(ii,jj) = 10*log10(theory_imrr_lin);
      approx_imrr_dB(ii,jj) = 10*log10(approx_imrr_lin);
   end
end

figure
plot(gammadB_v,theory_imrr_dB(:,phiDeg_zeroIdx),'bs-'); hold on
plot(gammadB_v,est_imrr_dB(:,phiDeg_zeroIdx),'md-');
plot(gammadB_v,approx_imrr_dB(:,phiDeg_zeroIdx),'gx-');
xlabel('gain imbalance, dB'); ylabel('image rejection, dB'); grid on;
legend('theory','estimated','approx')
title('Image Rejection Ratio with gain imbalance alone');
axis([-3 3 -50 -10]);

figure
plot(phiDeg_v,theory_imrr_dB(gammadB_zeroIdx,:),'bs-'); hold on
plot(phiDeg_v,est_imrr_dB(gammadB_zeroIdx,:),'md-');
plot(phiDeg_v,approx_imrr_dB(gammadB_zeroIdx,:),'gx-');
xlabel('phase imbalance, degree'); ylabel('image rejection, dB'); grid on;
legend('theory','estimated','approx')
title('Image Rejection Ratio with phase imbalance alone');
axis([-6 6 -50 -20]);

Figure : Image Rejection Ratio (IMRR) with gain imbalance alone

Figure : Image Rejection Ratio (IMRR) with phase imbalance alone

Observations

1) The approximate expression holds good for reasonable values of gain and phase imbalance.

2) As a rule of thumb, the following numbers are useful :

   - For 1 degree of phase imbalance, the Image Rejection Ratio (IMRR) is around -41dB

   - For 1dB of gain imbalance, the Image Rejection Ratio (IMRR) is around -25dB

References

[1] Cavers, J.K.; Liao, M.W.; , “Adaptive compensation for imbalance and offset losses in direct conversion transceivers,” Vehicular Technology, IEEE Transactions on , vol.42, no.4, pp.581-588, Nov 1993 doi: 10.1109/25.260752

[2] Table of trignometric identities http://www.sosmath.com/trig/Trig5/trig5/trig5.html

Related posts:

IQ imbalance in transmitter First order digital PLL for tracking constant phase offset Using CORDIC for phase and magnitude computation Equal Gain Combining (EGC)

Question 15 on communication from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper.

Q15. A source alphabet consists of N symbols with the probability of the first two symbols being the same. A source encoder increases the probability of the first symbol by a small amount  and decreases that of the second by . After encoding, the entropy of the source

(A) increases

(B) remains the same

(C) increases only if N=2

(D) decreases

Solution

Entropy of a random variable  is defined as ,

 .

Refer Chapter 2 in Elements of Information Theory, Thomas M. Cover, Joy A. Thomas (Buy from Amazon.com, Buy from Flipkart.com)

Let us consider a simple case where  can take two values 1 and 0 with probability  and respectively, i.e.

.

The entropy of  is,

.

The plot of the entropy versus the probability is shown in the figure below.

clear all; close
p = [0:.001:1];
hx = -p.*log2(p) - (1-p).*log2(1-p);
plot(p,hx);
xlabel('probability, p'); ylabel('H(X)');
title('entropy versus probability, p'); 
axis([0 1 0 1]);grid on;

Figure : Entropy versus probability for binary symmetric source

It can be see that the entropy (also termed as uncertainty) is maximum when  and for other values of , the entropy is lower. The entropy becomes 0 when  i.e. when the value of  becomes deterministic. If we extend this to a source with more than two symbols,  when probability of one of the symbols  becomes more higher than the other, the uncertainty decreases and hence entropy also decreases.

Based on the above, the right choice is (D) decreases

References

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

[2] Elements of Information Theory, Thomas M. Cover, Joy A. Thomas (Buy from Amazon.com, Buy from Flipkart.com)

Related posts:

GATE-2012 ECE Q38 (communication) GATE-2012 ECE Q3 (communication) GATE-2012 ECE Q39 (communication) GATE-2012 ECE Q26 (electronic devices)

Question 7 on digital from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper.

Q7. The output Y of a 2-bit comparator is logic 1 whenever the 2 bit input A is greater than 2 bit input B. The number of combinations for which output is logic 1 is

(A) 4

(B) 6

(C) 8

(D) 10

Solution

Let’s write all the 16 possible combinations of A and B and count the number of times A is greater than B. Can easily see that the count is 16.

Alternately, pick each value of A and count the number of instances when A is greater than B. The count is 0 + 1 + 2 + 3 for A = 0,1,2,3 respectively.

Based on the above, the right choice is (B) 6

References

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

Related posts:

Update : Correction to solution of GATE-2012 ECE Q38 GATE-2012 ECE Q6 (digital) GATE-2012 ECE Q12 (math) GATE-2012 ECE Q25 (math)