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Given signal y(t) and x(t) is shown in figure below,

Given: z(t) = x(t)*y(t) = \(\int_{-\infty}^{\infty} y(τ) x(t-τ) d τ\)

x(t - τ) can be plotted as shown below,

Maximum value of convolution will be obtained when positive portion of y(τ) is multiplied with x(t - τ)

∴ 1 + t = 5

t = 4 sec

At t = 4 sec, maximum value of convolution will be obtained.

Circuit can be redrawn

By applying voltage division rule, we can get

\(V_{a b}=8 \times \frac{1.5}{1.5+0.5}=6 \mathrm{~V}\)

For nominal π model:

B = Z = [R + j(2π + L)]l

= [0.035 + j(2π × 50 × 1 × 10^-3)] × 400

= 126.44 ∠83.642° = |B|∠β

\(P_R=\left|\frac{V_S V_R}{B}\right| \cos (β-δ)-\left|\frac{A V_R^2}{B}\right| \cos (β-\alpha)\)

\(\frac{2 P_R}{2 δ}=0\) ⇒ δ = β

At δ = β ⇒ \(\left(P_R\right)_{\max }=\left|\frac{V_S V_R}{B}\right|-\left|\frac{A V_R^2}{B}\right| \cos (\beta-\alpha)\)

For maximum PR occur at

δ = β = 83.64° = θ

Given: \(\frac{d y}{d t}=\frac{e^{-α t}}{2+α t}\)    ...(i)

α = 0.01

y(0) = 0 (initial)

\(f(t, y)=\frac{d y}{d t}=\frac{e^{-0.01 t}}{2+0.01 t}\)

Euler's method: Step size h = t (given)

y₁ = y₀ + hf(t, y) (at t₀ = 0, y₀ = 0)

\(=0+1\left[\frac{e^{-0.01 \times 0}}{2+0.01 \times 0}\right] \)

\(=0+1 \times \frac{1}{2}=\frac{1}{2}=0.5\)

So, at t = 1, y₁ = 0.5

Given that:

Best fit of straight line (y = ax + b) by least square.

Approximation given by normal equations

∑y_i = a ∑x_i + bn   ...(i)

∑x_iy_i = a ∑\(x_i^2\) + b ∑x_i   ...(ii)

Substituting values in eqn.

(1) and (2)

5.8 = a(0) + 3b   ...(iii)

2 = a(2) + 0(b)    ...(iv)

Solving we get, a = 1, \(b = \frac{5.8}{3}\) = 1.93

 ∴ Slope of best fit y = ax + b is 'a' = 1

The transfer function, TF = Ke^-sT_d (tranportation lag)

Given magnitude, M = 8 dB = 20 log [K]

K = 2.511

and angle at ω = 1 rad/sec = -60°

Angle, \(\phi=-\omega T_\alpha \times \frac{180^{\circ}}{\pi}\)

\(60^{\circ}=-1 \times T_d \times \frac{180^{\circ}}{\pi}\)

T_d = 1.047

So, required transfer function

TF = 2.511e^-1.047s

For positive half cycle the diodes D1 and Dz are forward bias and D2 is reverse biased and the circuit is shown below,

When V_in = 10V,

\(V_0=\frac{-R}{R} \times 10=-10 \mathrm{~V}\)

∴ V₀ will be -10 V.

For negative half cycle, diode D₂ is forward bias and D₁, D_z is reverse bias. Zenor diode is in breakdown region and the circuit is shown below,

∴ V₀ = 5 V

∴ \(V_{0_{\max }}\) = 5 V and \(V_{0_{\max }}\) = -10 V

Hence, the correct option is (D).

\( \mathrm{TF}=\frac{1+\frac{s}{α}}{1+\frac{s}{α β}}, α>0, β>1 \)

\( \mathrm{TF}=\frac{s+α}{(s+α β)} \)

Given: \( ω_m=√{α(α β)}=4\)

α√β = 4   ...(1)

Given at ω = ω_m

\( x =\frac{1}{β} \)

Amplification, \(M =10 \log _{10} \frac{1}{x}=6\)

10 log₁₀ (β) = 6

β = 4

and \(\alpha = \frac{4}{\sqrt 4} = 2\)

Given: V_line = 400 Volt ⇒ Y - connected

V_phase = \(\frac{V_{\text {ine }}}{√{3}}=\frac{400}{√{3}}\)

I_line = I_phase = 2 Amp

Wattmeter reading = -400 Watt

This Wattmeter connection is related to reactive power measurement by single wattmeter method.

So, Wattmeter reading = \(=\sqrt{3} \cdot V_{\mathrm{ph}} \cdot I_{\mathrm{ph}} \sin (ϕ) \)

\(-400 \text { Watt } =\sqrt{3} \times \frac{400}{\sqrt{3}} \times 2 \times \sin (ϕ) \)

\(\sin (ϕ) =-\frac{1}{2} \)

⇒ \(ϕ =\sin ^{-1}\left(-\frac{1}{2}\right)=-30^{\circ} \) ⇒ Leading

P.f. of load = cos(ϕ) = cos (-30°)

P.f. = 0.866 leading

Vin = 1.992 V, 

n = 8

Vfs = 5 V

We know, from graph

Step size = \( \frac{V_{f s}}{2^n-1} \ \)

Step size = \( \frac{5}{2^8-1}=\frac{5}{255}\)

Analog input = Step size × (decimal equivalent of binary code)

\( 1.992 =\frac{5}{255} \times D \)

\(D =\frac{1.992 \times 255}{5}=(101.592)_{10}\)

For decimal equivalent of 101, 

Analog input = \(\frac{5}{255} \times 101=1.980\)

Here 1.98 is less than 1.992 so, we have to take decimal equivalent as 102

For (102)₁₀ Analog input = \(\frac{5}{255} \times 102=2\)

2 is near about 1.992 so, decimal equivalent will 102 in hexadecimal, (102)₁₀ = 66 H