GATE2012ECE(4)

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

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18)  The signal m(t) as shown is applied both to a phase modulator (with  k_p  as the phase constant) and a frequency modulator (with  k_f  as the frequency 
 constant) having the same carrier frequency.

The ratio k_p / k_f  (in rad/Hz) for the same maximum phase deviation is

(A) 8π         (B) 4π        (C) 2π        (D) π

Solution

                                       

$$\text{Phase Modulation equation is , } s(t)=Acos(2πf_{c}t+k_{p}m(t))$$

                                           

  $$\text{Phase Deviation of PM }= k_{p}m(t)$$

                                                

$$\text{Frequency Modulation equation is, }s(t)=Acos(2πf_{c}t+2πk_{f}\int_{0}^{t}m(t)dt)$$

  $$\text{Phase Deviation of FM }= 2πk_{f}\int_{0}^{t}m(t)dt)$$

Max. phase deviation occurs at m(t) maximum value

Max. Phase Deviation of PM } =  2* k_p

Integral of square wave is triangular wave, with peak value in this case as +4 

Max. Phase Deviation of FM =  4*2π k_f

Now given condition is, Max. Phase Deviation of PM=Max. Phase Deviation of FM

                                           2 k_p = 8π k_f

     

                                         k_p / k_f = 8π/2 = 4π

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19)  In the sum of products function f (X, Y, Z) = ∑(2, 3, 4, 5) , the prime implicants are

      _       _                                            _     _ _    _

(A) X Y, X Y                                   (B) X Y, X Y Z, X Y Z
     _    _   _          _                             _ _        _        _ _     _
(C) X Y Z, X Y Z, X Y                       (D) X Y Z, X Y Z, X Y Z, X Y Z

Solution 

X    Y    Z   minterm
0    0    0       0
0    0    1       1

0    1    0       2

0    1    1       3

1    0    0       4

1    0    1       5
1    1    0       6
1    1    1       7

          
f(X,Y,Z) = (X’ Y Z’ + X’ Y Z + X Y’ Z’ + X Y’ Z)

f(X,Y,Z) = (X’ Y Z’ + X’ Y Z) + (X Y’ Z’ + X Y’ Z)

f(X,Y,Z) = X’ Y (Z’+Z) + X Y’ (Z’+Z)

f(X,Y,Z) = X’ Y + X Y’    
These cannot be further reduced and hence are prime implicants

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20)  The source of a silicon (ni = 10¹⁰  per cm³ ) n-channel MOS transistor has an area of 1 sq μm and a depth of 1 μm. If the dopant density in the source is 10¹⁹ /cm³ , the number of holes in the source region with the above volume is approximately

(A) 10⁷                                                (B) 100                        (C) 10             (D) 0

Solution 

n_i² = np

n_i² = N_{D *} p   (For an n-type semiconductor with N_D donor dopant atoms)

p = n_i² / N_D

     

    = (10¹⁰ per cm³)²  / (10¹⁹ per cm³)

   

    = 10 per cm³

Total holes,  p = 10 per cm³ * volume

                        = 10 * Area * depth

                        = 10 * (1*10^-4)² * (1*10^-4)       (Now all units are in cm³)

                         

                        = 10^-11

                           = 0 (Approx.)

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21)  Let y[n] denote the convolution of h[n] and g[n], where h[n] = (1/2)^n u[n] and g[n] is a causal
sequence. If y[0] = 1 and y[1] = 1/2, then g[1] equals

(A) 0             (B) 1/2            (C) 1          (D) 3/2

Solution

h[n] = (1/2)ⁿ u[n] is drawn below

 

g[n] is causal, therefore it is right sided with no values at negative time

It might be like this a, b, c are unknowns, they can take any values .

Now convolution is  y[n] = ∑_{-∞ to ∞} h[k] g[n-k]

Or in simple words we have to time reverse one of the signal, multiply overlap of signals and then sum it.

Here assume we are time reversing g[n]

For n=0

                                 h[k]

 

                                           

                                g[0-k]

y[0] = ∑_{-∞ to ∞} h[k] g[-k] = 1

y[0] = 1 * a = 1

or a = 1  => g[0] = 1

For n=1 

                                  h[k]

                                           

                                  g[1-k]

y[1] = ∑_{-∞ to ∞} h[k] g[1-k] = ½     g[n ] is shifted one instant towards right

y[1] = (a * ½) + (b * 1) =1/2

we already got a=1 , so above condition is possible only if b=0  => g[1] = 0

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22)   A BPSK scheme operating over an AWGN channel with noise power spectral density of N₀/2, uses equiprobable signals s₁(t) = √(2E/T )sinω_ct and  s₂(t) = - √(2E/T )sinω_ct over the symbol interval (0, T). If the local oscillator in a coherent receiver is ahead in phase by 45° with respect to the received signal, the probability of error in the resulting system is

(A)  Q(√(2E/N₀))                    (B)  Q(√(E/N₀))                                                 (C)  Q(√E/2N₀))                    (D)  Q(√(E/4N₀))

 Solution   

This can be consider as non-coherent decoding of BPSK.

The probability of error is given by,

P_b=Q { √(2E/N₀ ) sinθ}, where θ is the phase shift of local oscillator

P_b=Q { √(2E/N₀ ) sin45}

P_b=Q { √(2E/N₀ ) (1/√2}

P_b=Q { √(E/N₀)}

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The transfer function of a compensator is given as

                            G_c(s) = (s+a) / (s+b)

23)  G_c(s) is a lead compensator if

(A) a =1, b = 2                         (B) a = 3, b = 2

(C) a = –3, b = –1                     (D) a = 3, b = 1

Solution

 G_c(s) is a lead compensator when b > a

So answer is (A)

24)  The phase of the above lead compensator is maximum at

(A) √2 rad/s     (B)  √3 rad/s     (C) √6 rad/s     (D) 1/ √3 rad/s

Solution

ω_max = √(ω_z ω_p)

        = √(1 * 2))
       

        = √2