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Given: W = [1, 2, 3]

\(W_x^T=1\)

\( \left[\begin{array}{lll} 1 & 2 & 3 \end{array}\right]\left[\begin{array}{l} x \ y \ z \end{array}\right]=1 \)

ϕ : x + 2y + 3z = 1

Normal vector or gradient of ϕ = î + 2ĵ + 3k̂

Normal vector = [1, 2, 3]^T

Hence, the correct option is (D).

Signal flow graph:

Forward paths,

P₁ = 3, Δ₁ = 1

P2 = \(\frac{2}{S}\), Δ₂ = 1

Loops : L₁ = \(\frac{1}{S}\)

Using Masson's graph formula,

\(\frac{\mathrm{Y}(\mathrm{s})}{\mathrm{R}(\mathrm{s})}=\frac{\mathrm{P}_1 \Delta_1+\mathrm{P}_2 \Delta_2}{1-\mathrm{L}_1}\)

\(=\frac{3+\frac{2}{\mathrm{~S}}}{1-\frac{1}{\mathrm{~S}}} \)

\(=\frac{3 \mathrm{~S}+2}{\mathrm{~S}-1}\)

Nyquist Contour:

Given:

\( \mathrm{G}(\mathrm{s}) \mathrm{H}(\mathrm{s}) =\frac{3 \mathrm{~s}+5}{\mathrm{~s}-1} \)

Put s = Re^jθ

\(\mathrm{G}(\mathrm{s}) \mathrm{H}(\mathrm{s}) =\operatorname{Lim}_{\mathrm{R} \rightarrow \infty} \frac{3 \operatorname{Re}^{j \theta}+5}{\operatorname{Re}^{j \theta}-1}\)

G(s)H(s) = 3

Given system can be drawn as shown below,

Here,

X(s) → Controller output

C(s) → Plant output

\( C(s)=R(s) \frac{\left(3+\frac{1}{s}\right)\left(\frac{1}{s-1}\right)}{1+\left(3+\frac{1}{s}\right) \frac{1}{s-1}}\) = \(R(s) \frac{(3 s+1)}{s(s-1)+3 s+1}\) = \(\frac{(3 s+1)}{s(s-1)+3 s+1} \times \frac{1}{s} \)

∴ c(∞) = \(\lim _{s \rightarrow 0} s C(s)=1 \)

\(X(s)=[R(s)-C(s)]\left(3+\frac{1}{s}\right) \)

\(X(s)=[R(s)-C(s)]\left(\frac{3 s+1}{s}\right) \)

\(\rm x(\infty) =\lim s_{s \rightarrow 0}X(s)=\lim _{s \rightarrow 0}R(s)-C(s)\)

\(x(\infty) =\lim _{s \rightarrow 0}\left(\frac{1}{s}-\frac{1}{s} \frac{(3 s+1)}{\left(s^2+2 s+1\right)}\right)(3 s+1) \)

\(=\lim _{s \rightarrow 0} \frac{1}{s}\left[\frac{s^2+2 s+1-3 s-1}{s^2+2 s+1}\right](3 s+1) \)

\(=\lim _{s \rightarrow 0} \frac{1}{s}\left[\frac{s^2-s}{s^2+2 s+1}\right](3 s+1)\) = \(\lim _{s \rightarrow 0} \frac{(s-1)(3 s+1)}{s^2+2 s+1}=-1\)

The torque slip characteristics are shown in the figure below.

S > 1 ⇒ Plugging mode

0 < S < 1 ⇒ Mtoring mode

S < 0 ⇒ Generating mode

Given : P = 10, f = 50 Hz, V_f = 240 V, N_r = 540 pm

The frequency of induced rotor current due to backward field is given by,

\(N_s=\frac{120 f}{P}=\frac{120 \times 50}{10}=600 \mathrm{rpm} \)

\(S_f=\frac{N_s-N_r}{N_s}\) = \(\frac{600-540}{600}=\frac{60}{600}=0.1\)

As we know, backward field will rotate opposite to forward field so it will try to make the rotor in opposite direction so slip will be

\(S_b=\frac{N_s+N_r}{N_s}\) \(=\frac{600+540}{600}=\frac{1140}{600}=1.9\)

As we know, frequency in the rotor is slip frequency.

So, frequency due to backward slip will be S_bf_s = 1.9 × 50 = 95 Hz

Hence, the correct option is (B).

Given: \(\frac{d y(t)}{d t}+3 y(t)=2 x(t)\)

Taking Laplace transform on both sides, we get

sY(s) + 3Y(s) = 2X(s)

⇒ (s + 3)Y(s) = 2X(s)

⇒ \( \frac{Y(s)}{X(s)}=\frac{2}{s+3} \)

⇒ \(H(s)=\frac{2}{s+3}\)

∴ Impulse response will be, h(t) = L^-1(H(s)) = 2e^-3tu(t)

Hence, the correct option is (C).

Given:

 \(X(ω)= \begin{cases}1, & \text { for }|ω|<ω_0 \ 0, & \text { for }|ω|>ω_0\end{cases}\)

By taking inverse Fourier transform,

\( x(t)=\frac{\sin ω_0 t}{\pi t} \)

\(x\left(\frac{\pi}{2 ω_0}\right)=\frac{2 ω_0}{\pi \times \pi} \sin ω_0 \times \frac{\pi}{2 ω_0}\) \(=\frac{2 ω_0}{\pi^2} \sin \frac{\pi}{2}=\frac{2 ω_0}{\pi^2}\)

So, option (C) and (D) are wrong.

\(x(0)=\underset{t \rightarrow 0}{L t} \frac{\sin ω_0 t}{\pi t}=\) \(\underset{t \rightarrow 0}{L t} \frac{ω_0 \cos ω_0 t}{\pi}=\frac{ω_0}{\pi}\)

So, x(0) ∝ ω₀ Option (B) is wrong.

When ω0​ → ∞, X(ω) will be a D.C signal and inverse Fourier transform of a D.C signal will be impulse signal.

So, option (A) is correct.

Hence, the correct option is (A).

Given:

 \(X(z)=\frac{4 z}{\left(z-\frac{1}{5}\right)\left(z-\frac{2}{3}\right)(z-3)}\)

Poles of X(z) are located at z = \(\frac{1}{5}\), z = \(\frac{2}{3}\) and z = 3.

For DTFT to converge, the ROC of Z-transform of x() should contain unit circle.

If x(n) is a right sided sequence then the ROC is |z|>3 which does not include unit circle. So, option (D) and (A) are wrong.

If R.O.C. is \(\frac{2}{3}\) < |z| < 3, the R.O.C. includes unit circle. So, option (B) is correct.

If x(n) is a left sided then R.O.C will be |z| < \(\frac{1}{5}\) which does not include unit circle. So, option (C) is wrong.

Hence, the correct option is (B).

Hence, R₃ and R₄ are directional over current relays, which only operates when fault occurs in line-1 and line-2 respectively. But blocking faults towards bus-2.

Hence, the correct option is (A).