(a) Consider the Boolean function: F(A, B, C, D) = Σ m (1, 3, 4, 11, 12, 13, 14, 15). Implement it with a 4-to-1 multiplexer and external gates. Connect inputs A and B to the selection lines. Input to the four data lines is a function of the variable

Question

(a) Consider the Boolean function: F(A, B, C, D) = Σ m (1, 3, 4, 11, 12, 13, 14, 15). Implement it with a 4-to-1 multiplexer and external gates. Connect inputs A and B to the selection lines. Input to the four data lines is a function of the variables C and D which are obtained by expressing F as a function of C and D for each of the four cases when AB = 00, 01, 10 and 11. Functions are to be implemented with external gates. (20 marks)

(b) In the circuit given below, transistors T₁ and T₂ are having V_BE = 0.6 V and β = 499.
(i) Calculate small signal ac voltage gain of the amplifier at 20 Hz and 2 kHz.
(ii) Find dc voltages on collectors of transistors T₁ and T₂ respectively. (20 marks)

(c) Impulse response of an LTI system, h(n) is defined in the interval N₀ ≤ n ≤ N₁. If the input x(n) to the LTI system is zero except in the interval N₂ ≤ n ≤ N₃, find the interval for which the output y(n) exists in forms of N₀, N₁, N₂ and N₃. (10 marks)

UPSC Answer Check · Accepted Answer

(a) 
Method: Shannon expansion for a 4-to-1 multiplexer. Take A as the most significant select line and B as the least significant select line. Then
F = A′B′I₀ + A′B I₁ + AB′I₂ + AB I₃.
For AB = 00, the relevant minterms are m₀, m₁, m₂, m₃. From the given list, m₁ and m₃ are 1, while m₀ and m₂ are 0. In C,D order 00, 01, 10, 11 this is 0, 1, 0, 1, so
I₀ = C′D + CD = D.
For AB = 01, the relevant minterms are m₄, m₅, m₆, m₇. Only m₄ is 1. In C,D order this is 1, 0, 0, 0, so
I₁ = C′D′.
For AB = 10, the relevant minterms are m₈, m₉, m₁₀, m₁₁. Only m₁₁ is 1. In C,D order this is 0, 0, 0, 1, so
I₂ = CD.
For AB = 11, the relevant minterms are m₁₂, m₁₃, m₁₄, m₁₅. All four are 1, so
I₃ = 1.
Verification: A′B′D gives m₁ + m₃; A′BC′D′ gives m₄; AB′CD gives m₁₁; AB gives m₁₂ + m₁₃ + m₁₄ + m₁₅. This reproduces the original sum of minterms.
External gates: D goes directly to I₀; C′ and D′ are formed by inverters and ANDed for I₁; C and D are ANDed for I₂; I₃ is tied to logic 1. The MUX select lines are A and B.
Final MUX data inputs: I₀ = D, I₁ = C′D′, I₂ = CD, I₃ = 1.

[(b)] The figure printed on the original question paper for this part could not be recovered from the scan, so this part is not answered here.

(c) 
Method: support of discrete convolution. The output is
y(n) = Σ x(k) h(n-k).
A term can be nonzero only when x(k) is nonzero and h(n-k) is nonzero. The given supports are
N₂ ≤ k ≤ N₃ and N₀ ≤ n-k ≤ N₁.
The second condition is equivalent to n-N₁ ≤ k ≤ n-N₀. Hence a nonzero term requires the intersection
[N₂, N₃] ∩ [n-N₁, n-N₀]
to be non-empty.
Non-emptiness requires both
N₂ ≤ n-N₀ and n-N₁ ≤ N₃.
These give
n ≥ N₀ + N₂ and n ≤ N₁ + N₃.
Conversely, for any integer n satisfying N₀+N₂ ≤ n ≤ N₁+N₃, choose k = max(N₂, n-N₁). This k also satisfies k ≤ N₃ and k ≤ n-N₀, so at least one convolution term lies inside both supports. Thus the output can exist throughout the whole interval.
Therefore, assuming N₀ ≤ N₁ and N₂ ≤ N₃, the output interval is
N₀ + N₂ ≤ n ≤ N₁ + N₃.
If either input or impulse-response interval is empty, the output is empty; if endpoint values are zero, the actual nonzero support may be a subinterval, but the possible output interval is the one above.

(a) Consider the Boolean function: F(A, B, C, D) = Σ m (1, 3, 4, 11, 12, 13, 14, 15). Implement it with a 4-to-1 multiplexer and…

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