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Electronics & Communications Test - 4

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Electronics & Communications Test - 4
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  • Question 1
    2 / -0.66

    For the transistor circuit shown in the figure, VBE = 0.7 V, if β = 80, the value of VCE is (the VCE at saturation is 0.2 V)  _____

    Solution

    Concept:

    First of all, we can calculate VCE by assuming that the BJT is in the active region.

    If VCE is coming out to be less than 0.2, then our assumption that BJT is in active region is wrong.

    BJT is in saturation and VCE = 0.2 V

    Analysis:

  • Question 2
    2 / -0.66

    Solution

    As the limit is between 0 to π/4, hence the greatest integer of x, will be always 0,

    [x] = 0

    Hence, the problem reduces to,

  • Question 3
    2 / -0.66

    For a wide band pass filter, if the cut off frequencies are 200 Hz and 1 kHz respectively, evaluate centre frequency and quality factor.

    Solution

    Concept:

    For a wide band pass filter if fL and fH are the lower and higher cut-off frequencies respectively then

    Center frequency (fR) = √(fL × fH)

    Bandwidth (BW) = (fH - fL)

    Quality factor (Q) = fR/ BW 

    Calculation:

    Given; fL = 200 Hz , fH = 1KHz 

    Center frequency (fR) = √(200 × 1000) = 447.2 Hz

    Bandwidth (BW) = (1000 - 200) = 800 Hz

    Quality factor (Q) = 447.2/ 800 = 0.559

  • Question 4
    2 / -0.66

    The statistically independent random variables ‘X’ and ‘Y’ have mean value Ux = 2 and Uy = 4. They have second order moments E[X2] = 8 and E[Y2] = 25. What is E[W2] for the random variable W = 3X – Y?

    Solution

    Since the random variables X and Y are independent:

    E[W2] = E[(3X-Y)2]  

    = E [9X2-6XY+Y2]

    = 9E [X2] – 6UxUy+ E [Y2]

    Since the two random variables are independent

    E[W2] = 72 – 48 +25 = 49

  • Question 5
    2 / -0.66

    Given the homogenous state-space equation  The steady-state value of  given the initial state value of x(0) = [10 - 10]T is

    Solution


  • Question 6
    2 / -0.66

    Solution

    We are looking for orthogonal vectors having a span that contains P, Q and R

    Take option (1)

  • Question 7
    2 / -0.66

    A micro-instruction format has micro-ops field which is divided into three subfields F1, F2, F3 each having seven distinct micro-operations, condition field CD for four status bits, branch field BR having four options used in conjunction with address field ADF. The address space is of 128 memory locations. The size of micro-instruction is :

    Solution

    The correct answer is option 2

    The size of the micro-instruction is 20.

    Explanation:

    The given micro-instruction consists of the following three subfields - F1, F2, F3, one condition field - CD, one branch field - BR, one address field - ADF.

    So, 3 bits are required for each.

    Since each sub-field has seven distinct operations, a minimum of 3 bits is required to indicate all of them.

    CD-Condition field has four status bits, it needs 4 bits for four different conditions. 
    BF-Branch field have four option so, it needs 2 bits for four option. 
    Now there are 128 different memory location, So, there 7 bits are required for 128 different location.

    But BR field is used in conjunction with the address field therefore the size of the address field is 7 – 2  = 5.

    Total number of bits required for this micro-instruction = 3+3+3+4+2+5 = 20

    F1

    F2

    F3

    CD

    BR

    ADF

    3 bits

    3 bits

    3 bits

    4 bits

    2 bits

    5 bits

  • Question 8
    2 / -0.66

    A 5-bit D/A converter has a current output. If an output current Iout = 10 mA is produced for a digital input of 10100, the value of Iout for a digital input of 11101 will be

    Solution

    Concept:

    For a DAC

    Analog output = Resolution × Digital equivalent in decimal

    Analysis:

    Given,

    For digital input 10100, analog output is 10 mA

    101002 = (20)10

    Analog output = Resolution × Digital equivalent in decimal

    10 × 10-3 = Resolution ×  20

    Resolution = 0.5 × 10-3

    For digital input 11101

    Decimal equivalent = 29

    Analog output = Resolution × Digital equivalent in decimal

    = 0.5 × 10-3 × 29

    =14.5 × 10-3

    =14.5 mA

  • Question 9
    2 / -0.66

    For the circuit in figure, the values of i1 and i4, are respectively,

    Solution

    Writing kVL equation for loop ABCDEFGHIJ we get

    2 i1 + 4 i3 + 2 i4 +10 + 6 i2 = 0

    i1 + 3i2 + 2i3 + i4 = -5      ...... (1)

    Writing kVl eqn for loop DEFG we get 8(i4 - i3) + 2 i4 + 10 = 0

    5 i4 - 4 i3 + 5 = 0      .......(2)

    Writing kcL eqn for branch JC

    i1 - i2 = -5      ......(3)

    Writing KcL eqn for branch CH i2 - i3 = -3 i4      ......(4) [I0 = -i4]

    Adding eq 1 and 3 we get

    i1 - i2 = -5

    i2 + i3 = -3i4

    i1 - i3 = -5 - 3i4

    Calculating value of i1 in term of i3, i4 ⇒ i1 = -5 - 3 i4 + i3      ......(5)

    from eq 4 we get value of i2 in term of i3, i4 ⇒ i2 = i3 - 3i4      ......(6)

    From eq 5, 6, 1 we get

    (-5 - 3i4 + i3) + 3(i3 - 3i4) + 2i3 + i4 = -5

    6i3 - 11i4 = 0      ........(7)

    solving eq 7 and eq 2 we get

    i4 = 2.142 A

    i3 = 3.928 A

    From eq 5 we get

    i1 = -7.498 A

  • Question 10
    2 / -0.66

    The fermi-level in n-type silicon at T = 300 K is 230 meV below the conduction band and 200 meV below the donor level. The probability of finding an electron in the donor level is

    Solution

    Concept:

    For an n-type semiconductor the energy diagram can be shown as follow:

    Fermi-dirac function:

    It gives us the fraction of the available states within an energy interval dE which are occupied at particular energy and temperature.

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