Solution for Question 8 : IES Conventional 1998 : Electronics & Telecommunication Engineering : Paper II

Q.8. (a): For (6,3) systematic linear block code, the three parity-check bits C4, C5 and C6 are formed from the following equation  C4 = d1 ⊕ d3 ; C5 =  d1 ⊕ d2 ⊕ d3; C6 = d1 ⊕ d2

             (i)    Write the generator matrix G.
             (ii)   Construct all possible code words.
             (iii)  If the received work is 010111 find the location of the error and the transmitted data bits. 

Solution:

(i) The Generator matrix G as defined below -

       |  1    0    0    1    1    1  |
G =  |  0    1    0    0    1    1  |
       |  0    0    1    1    1    0  |

Also,     
       |  1    1    1  |
       |  0    1    1  |
H' = |  1    1    0  |
       |  1    0    0  |
       |  0    1    0  |
       |  0    0    1  |

(ii) Now, all possible code-words are -
C = dG
here,
                  d                                   G                                            C
         |  1    1    1  |                                                     |  1    1    1    0    1    0  |
         |  1    1    0  |                                                     |  1    1    0    1    0    0  |
         |  1    0    1  |     |  1    0    0    1    1    1  |        |  1    0    1    0    0    1  |
         |  1    0    0  |     |  0    1    0    0    1    1  |  =    |  1    0    0    1    1    1  |
         |  0    1    1  |     |  0    0    1    1    1    0  |        |  0    1    1    1    0    1  |
         |  0    1    0  |                                                     |  0    1    0    0    1    1  |
         |  0    0    1  |                                                     |  0    0    1    1    1    0  |
         |  0    0    0  |                                                     |  0    0    0    0    0    0  |

(iii) to find the location of error, we need to calculate decoding table as S = eH'

                 S                                                                              e
         |  1    1    1  |                                                   |  1    0    0    0    0    0  |
         |  1    1    0  |                                                   |  0    1    0    0    0    0  |
         |  1    0    1  |                                                   |  0    0    1    0    0    0  |
         |  1    0    0  |                                                   |  0    0    0    1    0    0  |
         |  0    1    1  |                                                   |  0    0    0    0    1    0  |
         |  0    1    0  |                                                   |  0    0    0    0    0    1  |
                                                                                  |  0    0    0    0    1    1  |
                                                                                    
here, the received work is 010111, hence to decode this code, we will use  s = rH'

                                                       |  1    1    1  | 
                                                       |  1    1    0  | 
s = rH' =  |  0    1    0    1    1    1|  |  1    0    1  |   =  |  1    0    0  |  
                                                       |  1    0    0  | 
                                                       |  0    1    1  |
                                                       |  0    1    0  |  


we can tabulate all the above codes in following fashion - 
                     
      r          |       s    |          e          |          C         |     d
0 1 0 1 1 1 |  1 0 0  |  0 0 0 1 0 0  |  0 1 0 0 1 1  |  0 1 0





Q.8. (b): Calculate the ratio of circular waveguide cross-sectional area to the rectangular waveguide cross-sectional area assuming that both these waveguides have equal cutoff frequency for the dominant mode, if P11 = 1.841

Solution: For the dominant (TE11) mode in the circular waveguide, we have

λo = 2πr / (Kr) = 2πr / (P11) =  2πr /1.841 = 3.41r

The area of a circle with radius r is given by - 

Ac = πr.r

In the rectangular waveguide, for the TE10  mode -

λo = 2a

If the two cut off wavelengths are to be the same, then

2a = 3.41 r

=> a = 1.705 r

The area of a standard rectangular waveguide is -

Ar = ab = a.a/2 = squr (1.705 r) / 2 = 1.45.r.r

Hence, the ratio of circular waveguide cross-sectional area to the rectangular waveguide cross-sectional area will be -

Ac/Ar = πr.r/1.45.r.r = 2.17

                                                  

Solution for Question 2 : IES Conventional 1998 : Electronics & Telecommunication Engineering : Paper II

Q.2. (a):  Calculate Ri, Av and Ai  for the circuit shown.  Use hie = 100 Ohms, hfe = 99, hie = hoe = 0.

Solution:  The ac base current through ri (hie) is

Ib = (Vi - V0). ri

Since, V0 = (1 + hfe).Ib.RE

Substituting V0 in Ib,

Ib.ri = Vi - V0 = Vi - (1 + hfe).Ib.RE

=> Vi = Ib (ri + (1 + hfe).RE) ~= Ib (ri + hfe.RE)

hence, the ac input impedance looking in to the transistor base is -

 Vi/Ib = (ri + hfe.RE)

=>  Ri  = RB || (ri + hfe.RE)

Here RB  = 0,

Ri  = (ri + hfe.RE) = 1000 + (100)(1K) = 101 KOhms

Now, 

Av = V0/Vs = 1  (approximate)

and  

Ai = hfe1 . hfe2 . (Zin.rE / Zin(base).RL)

=>   Ai = hfe1 . hfe2       ....(approximate)

=>      Ai = 10e+6

Q.2. (b):  An amplifier with open loop voltage gain A = 1000 ± 100 is available. It is required to have an amplifier whose gain varies by no more than ± 0.2 %. Find (i) reverse transmission factor B of the feedback network (ii) gain with feedback. Derive the formula you used.

Solution: The given data is-
            open loop voltage gain, A = 1000 ± 100 
            variation in gain = ± 0.2 %
i.e.

We know, the relation between the variation in gain after and before application of the feedback can be expressed as-

from given data, it emplies-

hence, we can obtain value of  reverse transmission factor as - 

Now, to find the gain with the effect of the feedback as -

Solution for Question 1 : IES Conventional 1998 : Electronics & Telecommunication Engineering : Paper II

Q.1. (a): A silicon transistor with VBE (sat) = 0.8 V,  hFE = 100, VCE = 0.2 V is used in the circuit shown. Find the minimum value of RC(sat) for which the transistor remains in saturation. 

Sol: By applying the KVL in the lower arm of the circuit,

                          5 - (200 K)Ib - 0.8 = 0
                  =>                         4.2 = (200 K) Ib
                  =>                           Ib = 0.021 mA

               As, Ic = hfe . Ib = 2.1 mA

     Now, by applying the KVL on the output side of the circuit,

                                 10 - IcRc - 0.2 = 0
                  =>                          9.8 = IcRc
                  =>                            Rc = 4.66 KOhm




Q.1. (b): A silicon single phase full wave bridge rectifier circuit is shown. Explain what happens if the transformer and the load positions are interchanged.

Sol: The transformers, that may be step up or step down type, are typically used to raise or to lay down the voltage level, respectively. Generally, the input applied to transformers are in AC form, say 230 V (ac). If the transformer is of step down type, it generate the DC voltage, say 12 V (dc).

But, as discussed in question, if we interchange the transformer and the load positions, nothing noticeable will happen. DC power is not transmitted between the coils of the transformer. There would be no current on the other side of the transformer, unless the power of source was constantly modulated, because flux does not change its state. After some time, excessive heat is produced and winding may burn. 

Q.1. (d): The truth-table for A-B flip-flop is shown. Draw schematic diagram using J-K flip-flop and any additional logic to implement it. Show the design steps.

   An      Bn        Qn+1 

   0        0           Qn

   1        0           Qn

 0        1           1

 1        1           0

Sol: 

Step 1: Truth table for AB Flip-flop.

    A        B         Qn          Qn+1

0        0          0             0

0        0          1             1

0        1          0             1

0        1          1             1

1        0          0             0

1        0          1             1

1        1          0             0

1        1          1             0

Step 2: Now, lets draw the transition table for JK Flip-flop considering Qn and Qn+1 as inputs.

Qn        Qn+1        J          K

0           0             0          X

0           1             1          X

1           0             X          1

 1           1             X          0 

Step 3: A and B have to express in order of J, K and Qn. Hence, use the column J, K, Qn, A and B and use K map method to obtain the expression for A and B. 

       A        B         Qn          Qn+1    J        K       

0        0          0             0        0        X

0        0          1             1        X        0

0        1          0             1        1        X

0        1          1             1        X        0

1        0          0             0        0        X

1        0          1             1        X        0

1        1          0             0        0        X

1        1          1             0        X        1

               K map for J:                                                 K map for K:  

                     AB                                                                   AB

           00      01      11     10                                   00       01       11       10

         ____________________                                  ______________________

      0|            X       X       1                               0| X                              X

Qn  1|            X       1                                  Qn   1| X                   1         X

                 

                     J = Qn'.A                                                           K = Qn.A

Step 4: Draw the AB flip-flop

                  

Q.1. (e): For open loop transfer function 

 a negative feedback is applied with a feedback factor B. Find the value of A1, 
(i) corresponding to the breakaway point, 
(ii) for which the system becomes unstable. 

Sol: (1) At breakaway point,

the characteristics equation for given circuit is-

by differentiating A1 with respect to S and equating with zero, we can get the breakaway point.

by solving the above equation, we get two values for S, those are S = -2/3 and S = -2.

But only S = -2/3 is acceptable as it lies on root locus.

At S = -2/3,

(2) To determine value of A1 for which system becomes unstable:

by using above characteristics equation and applying Routh-Hurwitz criterion,

Here, we can conclude that,

Q.1. (f): According to CCIR system B standard for TV given the values of the following parameters: (i) Channel B.W (ii) Number of lines per picture (iii) Aspect ratio (iv) Line period (v) Field period.

Sol:  (i) Channel B.W                           : 7 MHz
       (ii) Number of lines per picture    : 625 
       (iii) Aspect ratio                          : (4:3)
       (iv) Line period                            : 15625
       (v) Field period                            : 50


Q.1. (g): Calculate the efficiency of a system which selects one message out of 13 equiprobable messages in (i) binary systems and (ii) decimal systems.

Sol: For equiprobable messages, the efficiency is given by formula as mentioned below:
As here, number of messages are 13;
1) for binary system,
                
                       efficiency = log2(n) = log2(13) = 3.7

2) for decimal system,

                       efficiency = log10(n) = log10(13) = 1.114 



Q.1. (h): The terminating load of an HF transmission line with Z0 = 50 ohms working at 300 MHz is (50 +j50) ohms. Calculate the VSWR and the position of voltage minima nearest to the load.

Sol: To calculate the VSWR, we need to calculate the value of transmission coefficient (k) as follows:

now the VSWR is calculated as :

now, to calculate the position of voltage minima nearest to the load, we need to calculate the position at voltage maxima,

the position at voltage minima:

Q.1. (i): An optical fibre has a core refractive index of 1.5 and a cladding refractive index of 1.47. Find

    (i)    Critical angle at core-cladding interface.

    (ii)   Numerical aperture NA of the fibre.

    (iii)  The acceptance angle in air for the fibre .

Sol: (i)    Critical angle at core-cladding interface is given by:

        (ii)   to find the Numerical aperture (NA) of the fibre

       (iii)  to find the acceptance angle in air for the fibre,

Q.1. (j): Define the following terms used in microprocessors:

                        (i) Instruction Cycle (ii) Machine Cycle (iii) T-State 

Sol: (i) Instruction cycle: is the time required for completely execution of the instruction.

    (ii) Machine Cycle: is the time required for completion of operation of the accessing memory, IO or acknowledging an external request.  

       (iii) T-State: is the one subdivision of operation of performed in one clock period. 

Solution for Question1 (cont.) : IES Conventional 1998 : Electronics & Telecommunication Engineering : Paper I

Q.1 (f) : Synthesize the two Foster networks for 

Sol:  

             

                                   

                                               Foster I form

Foster II form

Q.1 (g):  If the magnetic flux density of a point in a region is 250 sin 120 xt az mWb/m, what is the curl of the electric field intensity?  

Sol: 

Q.1 (h): With the input of

to an amplifier, the measured output amplitude is 1 volt at 1 kHz and 1 mV at 600 Hz. If the amplifier input-output characteristics is given by 

Determine the out- put amplitudes at the other frequencies.

Sol: