GATE EE 2011

-1 is one of the root since
(-1)^3+(-1)^2+(-1)+1=0
By polynomial division
\frac{x^3+x^2+x+1}{(x-(-1))}=x^2+1 \Rightarrow \;\; x^3+x^2+x+1=(x^2+1)(x+1)
So root are (-1,+j,-j)

\begin{aligned} \frac{dy}{dx}&=e^{-3x}\\ \int dy&=\int e^{-3x}dx\\ y&=\frac{e^{-3x}}{-3}+K\\ y&=-\frac{1}{3}e^{-3x}+K \end{aligned}

V_s=1 \sin t\equiv V_m \sin \omega t
V_m=1 V and \omega =1 rad/sec
Impedance of the branch containing inductor and capacitor
Z=j(X_L-X_C)
\;\;=j\left ( \omega L-\frac{1}{\omega C} \right )
\;\;=j\left ( 1 times 1 -\frac{1}{1 times 1} \right )=0
So, this branch is short circuit and the whole current flow through it
i(t)=\frac{1.0 \sin t}{1}=1.0 \sin t
rms value of the current =\frac{1}{\sqrt{2}}A

Let, x(t)= Even and Hws

Fourier series expansion of x(t) contains cos terms with odd harmonics.

NOw, f(t)=1+x(t)
Fourier series of f(t) contains dc and cos terms with odd harmonics.

4-point starter is used to control the speed of shunt motor in field weakening region, i.e. above rates speeds.
In field weakening region field current will reduce in 3-point starter holding coil unable to hold the plunger in ON position.

As field current reduces, the flux will start reducing, to keep this flux constant, the armature draws high current from bus.
Ar zero field current the motor acts as synchronous reluctance motor in this case the magnetizing current fully taken from bus. Ie we increase field current in reverse direction to keep flux constant, motor draws more current. During this process the load angle increases at one point the reversed field force dominates reluctance torque and rotor slip one pole pitch anf align to opposite pole. The instant align to opposite pole the flux will be very high, to reduce this flux current drops steeply t synchronous motor value. Here torque is reluctance + synchronous motor torque.

For transmission of bulk power of very long distance high voltage (400 kV) is used.
To increase reliability, double circuit is used.

At gain crossover frequency (\omega_{gc}), magnitude of G(j\omega) is 1.
|G(j\omega_{gc})|=1
Phase of G(j\omega )=\angle G(j\omega_{gc})=-150^{\circ}
Phase margin=180^{\circ}+\angle G(j\omega_{gc})
\;\;\; =180^{\circ}-150^{\circ}=30^{\circ}
At phase cross frequency (\omega_{pc}), phase of G(j\omega ) is -180^{\circ},
\angle G(j\omega_{pc})=-180^{\circ}
M=magnitude of G(j\omega ) at \omega _{pc}
=|G(j\omega_{pc})|=0.5
Gain margin
=20log\frac{1}{M}
=20 log \frac{1}{0.5}=6 dB

Let the system is represnted as

\frac{Y(s)}{X(s)}=\frac{G(s)}{1+G(s)H(s)}
H(s)=1 (unity \; feedback)
Error=E(s)=\frac{X(s)}{1+G(s)}
Steady state error for X(s)=1/s, \; e_{ss}=0.1
e_{ss}=\lim_{s \to 0}sE(s)
\;\; =\lim_{s \to 0}\frac{sX(s)}{1+G(s)}
\Rightarrow \; 0.1=\lim_{s \to 0}\frac{s\cdot \frac{1}{s}}{1+G(s)}
\Rightarrow \; 0.1=\lim_{s \to 0}\frac{1}{1+G(s)}
When input,

X(t)=10[u(t)-u(t-1)]
X(s)=10\left ( \frac{1}{s}-\frac{e^{-s}}{s} \right )
\;\;=\frac{10}{s}(1-e^{-s})
e_{ss}=\lim_{s \to 0}sE(s)
\;\;=\lim_{s \to 0}\frac{sX(s)}{1+G(s)}
\;\;=\lim_{s \to 0}\frac{s \times \frac{10}{s}[1-e^{-s}]}{1+G(s)}
e_{ss}=\lim_{s \to 0}\frac{10(1-e^{-s})}{1+G(s)}=0

The current coil carries a current of I +I_P and produces a filed corresponding to this current. The compensating coil is connected in series with the pressure coil circuit and is made as nearly as possible identical and coincident with the current coil. It is so connected that it opposes the field of the current coil. The compensating coil carries a current I_P and produces a field corresponding to this current. This field acts as in opposition to the current coil field. Thus the resultant field is due to current I only. Hence the error caused by the pressure coil current flowing in the current coil is neutralized.