GATEECE2012(1)

GATE 2012 ECE paper fully solved.Step by step, detailed solutions are given.Corrections and suggestions are appreciated. 

For reference material, quote the question and send a mail to arunsblog4ece@gmail.com.

1)  The current ib through the base of a silicon npn transistor is 1+0.1 cos(10000πt) mA. At 300 K, the r_π in the small signal model of the transistor is

                                                                  

 (A) 250 Ω       (B) 27.5 Ω      (C) 25 Ω          (D) 22.5 Ω

Solution

r_π =   β/g_m

g_m =  I_c/V_T

substituting second equation in first              

r_π            = β / (I_c/V_T)

         

                       = β V_T/ I_c

                          

                          = (β/I_c )V_T

                         

                          = V_T / I_b                           (because I_c = βI_b, So 1/I_b = β/I_c)

                    

                          = 25mV/1mA                 ( I_c=1mA)

                          

                       = 25 Ω

2) In a baseband communications link, frequencies upto 3500 Hz are used for signalling. Using a raised cosine pulse with 75% excess bandwidth and for no inter-symbol interference, the maximum possible signalling rate in symbols per second is

(A) 1750          (B) 2625         (C) 4000          (D) 5250

Solution

Raised cosine filter is an example of Pulse shaping filter. Pulse shaping is the process of changing the waveform of transmitted pulses. Its purpose is to make the transmitted signal better suited to its purpose or the communication channel, typically by limiting the effective bandwidth of the transmission. By filtering the transmitted pulses this way, the InterSymbol Interference caused by the channel can be kept in control.

The roll-off factor, β, is a measure of the excess bandwidth of the filter, i.e. ratio of the excess bandwidth occupied to the Nyquist bandwidth B

Bandwidth       B = 3500 Hz

ie;                  β =  Δf/B = 0.75 (given)

Also Bandwidth B is defined by

B = R_S(β+1)/2

Where R_S is the symbol rate 

                   

              3500 = R_S(0.75+1)/2

R_S =     7000/1.75= 4000

 3)  The i-v characteristics of the diode in the circuit given below are

The current in the circuit is

(A) 10 mA       (B) 9.3 mA      (C) 6.67 mA    (D) 6.2 mA

Solution

Applying KVL in the loop

10 - 1000i - v = 0

10 - 1000*[(v - 0.7) / 500] - v = 0

10 - 2(v - 0.7) - v = 0

10 - 2v + 1.4 – v = 0

v = 11.4 / 3 = 3.8

Now,    i = (v - 0.7) / 500               (given)

              = (3.8 - 0.7) / 500

              = 6.2 mA

4)  The average power delivered to an impedance (4 - j3)Ω  by a current 5cos(100p t +100) A is

 

       (A) 44.2 W      (B) 50 W          (C) 62.5 W       (D) 125 W

   Solution

I = 5cos(100p t +100) A

I_rms= I _peak/√2

      = 5 /√2

Z = (4 - j3) W

Average power= (I_rms)² *R               (R is the real part of Z)

    =(25/2)*4

    = 50W

 

5)  The diodes and capacitors in the circuit shown are ideal. The voltage v(t) across the diode D1 is

(A) cos(wt) -1       (B) sin(w t)           (C) 1-cos(wt)            (D) 1-sin(wt)

Solution

The C1 and D1 acts as a negative clamper.

During positive cycle of coswt, D1 is forward biased capacitor left plate charges to +1V. So on right plate of C1 voltage will be -1V.This is the DC shift value of clamper. During negative cycle diode is reverse biased and capacitor retains DC value.

The output of clamper section will be dc value + ac value riding over it i.e.;    -1+coswt or (coswt-1).

C2, D2 section is a filter which allows only negative cycles to pass through. But our waveform   (coswt-1) contains negative values only and it is passed through filter as such.

Output is (coswt-1)

6)  The impedance looking into nodes 1 and 2 in the given circuit is

 

(A) 50 ohm   (B) 100 ohm   (C) 5 K   (D) 10.1 K 

 Solution

The circuit given is in CC configuration (no collector resistance).

β = 99 (given)

We have to find impedance in emitter side. There is already one 100ohm at emitter and there is a 9k and 1k in the base side.

Base resistance 9K + 1K = 10K because they are in series

But this 10K and emitter 100ohm are not in series because former is in base terminal where I_b is the current and latter is in emitter side where Ie is the current.

So we have to move this 10K to emitter side first. For that we have to divide it by β+1(using impedance reflection, and I_b = Ie/ (β+1))

Moving 10K to emitter  = 10K /β+1  = 10K / 100 = 100ohm

Now this 100ohm is in parallel with emitter’s 100ohm,

So equivalent resistance at emitter(between terminals 1 and 2)                 

=100||100 = 50 ohm

7) Consider the given circuit.

 

In this circuit, the race around

 

(A) does not occur                                        (B) occurs when CLK = 0

(C) occurs when CLK = 1 and A = B = 1         (D) occurs when CLK = 1 and A = B = 0

Solution

(A)

Race around occurs in JK flip flop.

The given flip flop is SR.

8) The output Y of a 2-bit comparator is logic 1 whenever the 2-bit input A is greater than the 2-bit

input B. The number of combinations for which the output is logic 1, is

(A) 4                (B) 6               (C) 8                (D) 10

Solution

Two bit numbers A and B

A             B             A>B

00           00              0

01           01              1           (when A=01, B=00)

10           10              2           (when A=10, B=00, 01)

11           11              3           (when A=11, B=00, 01, 10)

A>B can occur 0+1+2+3 = 6 times

9)  In the circuit shown

            _ _   _

(A) Y = A B +C                                                 (B) Y = (A+ B)C

              _   _  _

(C) Y = (A + B)C                                               (D) Y = AB +C

Solution

CMOS circuit given is

     ______

Y= A+B.C

     ____   _

Y= A+B + C                          (Using De Morgan’s theorem)

     _   _     _

Y= A . B + C                         (Using De Morgan’s theorem)

10)  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 e and decreases that of the second by e . After encoding, the entropy of the source

(A) increases                            (B) remains the same

(C) increases only if N = 2         (D) decreases

Solution

Maximum Entropy is obtained when all symbols have same probability.

Here we are changing two equiprobable symbols to distinct ones. So always entropy will decreases compared to previous case.