(a) (i) Consider the shift register shown in the figure below, which is implemented using D flip-flops and 2 : 1…

(a)(i) Shift register: truth table and timing diagram

Each 2:1 multiplexer selects its D input: with Load = 1, D_i = X_i (parallel load); with Load = 0, D₀ = X_L (serial input) and D₁ = Q₀, D₂ = Q₁, D₃ = Q₂ (shift towards Q₃). The flip-flops are cleared asynchronously when CLR̄ = 0.

Truth table (next state Q₃Q₂Q₁Q₀):

(i) CK = X, CLR̄ = 0, Load = X: 0 0 0 0.

(ii) CK = ↑, CLR̄ = 1, Load = 0: Q₃Q₂Q₁Q₀ = Q₂ Q₁ Q₀ X_L, a left shift with X_L entering at Q₀.

(iii) CK = ↑, CLR̄ = 1, Load = 1: Q₃Q₂Q₁Q₀ = X₃ X₂ X₁ X₀, a parallel load.

Timing with X₃X₂X₁X₀ = 0101: at the first rising edge, Load = 1, so Q₃Q₂Q₁Q₀ = 0101 (Q₀ = 1, Q₁ = 0, Q₂ = 1, Q₃ = 0). At the following edges (Load = 0, X_L = 0) the word shifts left: 1010, 0100, 1000, 0000 and stays 0000. When X_L is high at the edges that follow, a 1 enters Q₀ and moves along: 0001, then 0011 at the second such edge. With X_L low again the word continues 0110, 1100, 1000. When CLR̄ falls, all outputs go to 0 at once, without waiting for a clock edge.

(a)(ii) Implementing F = Σm(3, 4, 5, 6, 7, 9, 10, 12, 14, 15) with a 4:1 multiplexer

Use A and B as select lines (S₁ = A, S₀ = B) and find each data input as a function of C and D.

(i) AB = 00 (m0 to m3): only m3 is present, so I₀ = C·D.

(ii) AB = 01 (m4 to m7): all four are present, so I₁ = 1.

(iii) AB = 10 (m8 to m11): m9 and m10 are present, so I₂ = C'D + CD' = C ⊕ D.

(iv) AB = 11 (m12 to m15): m12, m14 and m15 are present (m13 is absent), so I₃ = C + D'.

Hardware: one 4:1 multiplexer, an AND gate for I₀, logic 1 for I₁, an XOR gate for I₂, and an inverter with an OR gate for I₃.

(b)(i) Fourier transform of the triangular pulse

The pulse is f(t) = V₀(1 − 2|t|/a) for |t| ≤ a/2, and zero elsewhere. It is even, so

F(ω) = 2∫₀^(a/2) V₀(1 − 2t/a) cos ωt dt.

Integrating by parts, ∫₀^(a/2) (1 − 2t/a) cos ωt dt = 2(1 − cos(ωa/2))/(aω²), so

F(ω) = (4V₀/(aω²)) (1 − cos(ωa/2)) = (8V₀/(aω²)) sin²(ωa/4) = (V₀a/2) [sin(ωa/4)/(ωa/4)]².

Check: the triangle is the convolution of two rectangles of width a/2, so its transform is the square of a sinc. At ω = 0, F = V₀a/2, the area of the pulse.

(II) Amplitude spectrum. |F(ω)| = (V₀a/2) sinc²(ωa/4). It is a real, non-negative curve, with its maximum V₀a/2 at ω = 0, zeros at ω = ±4πn/a (n = 1, 2, 3…), and side lobes that fall as 1/ω², so it is concentrated within |ω| < 4π/a.

(b)(ii) Inverse Laplace transform by convolution

F₁(s) = s/(s + 1) = 1 − 1/(s + 1), so f₁(t) = δ(t) − e^(−t); F₂(s) = 1/(s² + 1), so f₂(t) = sin t.

f(t) = f₁ * f₂ = sin t − ∫₀^t e^(−τ) sin(t − τ) dτ.

The integral is ∫₀^t e^(−τ) sin(t − τ) dτ = (1/2)(sin t − cos t + e^(−t)).

Hence f(t) = sin t − (1/2)(sin t − cos t + e^(−t)) = (1/2)(sin t + cos t − e^(−t)) for t ≥ 0.

Check by partial fractions: s/[(s + 1)(s² + 1)] = −(1/2)/(s + 1) + (1/2)(s + 1)/(s² + 1), which gives the same result.

(c) Inverting summer design

For the inverting summer, v₀ = −(R_f/R₁)v₁ − (R_f/R₂)v₂, so R_f/R₁ = 1 and R_f/R₂ = 4.

The feedback current at the maximum output is I_f = |v₀|/R_f = 4 V/R_f ≤ 1 mA, so R_f ≥ 4 kΩ.

Choose R_f = 4 kΩ, which gives R₁ = R_f/1 = 4 kΩ and R₂ = R_f/4 = 1 kΩ. Check: with R_f = 4 kΩ the current at 4 V is exactly 1 mA. Any larger R_f with R₁ = R_f and R₂ = R_f/4 also meets the limit.