GATE EC 2011 Official Paper (Previous Year Paper)

The Gladiator fights in the Arena to win battles. 

Similarly; 

A lawyer fights in the court to win cases.

Hence, "Lawyer ∶ Courtroom" is the correct answer.

Additional Information1) Dancer ∶ Stage →

Dancer dances on the Stage.

2) Commuter ∶ Train →

Commuter travels in train or Traveller travels in Train. (commuter means traveler)

3) Teacher ∶ Classroom →

Teacher teaches in Classroom.

Explanation:

Let; Total number of voters = X

Number of voters P was supposed to get = 0.4X

Number of voters Q was supposed to get =  0.6X

According to the given information;

Number of voters P got = 85/100 (0.4X) + 25/100 (0.6X)

Number of voters Q got = 75/100 (0.6X) + 15/100 (0.4X)

Q - P = 2

⇒ 75/100 (0.6X) + 15/100 (0.4X) - 85/100 (0.4X) - 25/100 (0.6X) = 2

⇒ 50/100 (0.6X) - 70/100 (0.4X) = 2

⇒ 30X - 28X = 200

⇒ 2X = 200

⇒ X = 100

Hence, total number of voters are 100.

The correct answer is 'exacerbated'.

Key Points

• Let's explore options;
• identified: establish or indicate who or what (someone or something) is.
• Example: Even the smallest baby can identify its mother by her voice.
• ascertained: find (something) out for certain; make sure of.
• Example: The police have so far been unable to ascertain the cause of the explosion.
• exacerbated: make (a problem, bad situation, or negative feeling) worse.
• Example: This attack will exacerbate the already tense relations between the two communities.
• analysed: discover or reveal (something) through detailed examination.
• Example: Researchers analysed the purchases of 6,300 households.
• Thus, from above, we can conclude that 'exacerbated' will be used in the blank because of the keyword 'counterproductive' which suggests technocrats must have done something to make the situation worse.
• Therefore, the correct answer is option 3.

The correct answer is 'rarity'.

Key Points

• Frequency: the fact of being frequent or happening often.
• Example: Complaints about the frequency of buses rose in the last year.
• Rarity: the state or quality of being rare.
• Example: Snow in Florida is a rarity.
• Thus, the correct answer is option 2.

Additional Information

• Let's explore the other options:
• periodicity: the quality or character of being periodic; the tendency to recur at intervals.
• gradualness: Occurring or developing slowly or by small increments
• persistency: the continued or prolonged existence of something.

The correct answer is 'available'.

Key Points

• Let's explore options:
• similar: a person or thing similar to another.
• Example: My father and I have similar views on politics.
• most: a large number of.
• Example: What's the most you've ever won at poker?
• uncommon: out of the ordinary; unusual.
• Example: Accidents due to failure of safety equipment are uncommon nowadays.
• available: able to be used or obtained.
• Example: Is this dress available in a larger size?
• Thus, from above, we can conclude that the correct answer is option 4.

Therefore, the correct sentence is: Under ethical guidelines recently adopted by the Indian Medical Association, human genes are to be manipulated only to correct diseases for which available treatments are unsatisfactory.

The correct answer is 'generally quite immune to diseases'.

Key Points

• Let's refer to the following lines of the passage:
• Horses were injected with toxins of diseases until their blood built up immunities. Then a serum was made from their blood.
• From above it can be inferred that the horses are quite immune to diseases as they themselves build the immunities in their blood to fight the toxins.
• Thus, the correct answer is option 2.

Calculation:

Consumption (km/liter) means distance covered by the car in 1 liter of fuel

⇒ Fuel consumption per liter = $\frac{1}{Consumption\ (km/liter)}$

 

From the above table, fuel consumption per km was least during the lap Q.

∴ The correct answer is Q.

Let the total number of toffees in bowl be x

R took 1/3 of toffees and returned 4 to the bowl

∴ Number of toffees with R = x/3 - 4

Remaining of toffees in bowl = 2x/3 + 4

Number of toffees with $S = \frac{1}{4}[\frac{2}{3}x+4]-3$

Remaining toffees in bowl = $\frac{3}{4}[\frac{2}{3}x+4]+3$

Number of toffees with T = $\frac{1}{2}[\frac{3}{4}(\frac{2}{3}x+4)+3]-2$

Remaining toffees in bowl = $\frac{1}{2}[\frac{3}{4}(\frac{2}{3}x+4)+3]+2$

Given: $\frac{1}{2}[\frac{3}{4}(\frac{2}{3}x+4)+3]+2=17$

⇒ x = 48

∴ There were 48 toffees in the bowl.

Given

f(y) = |y|/y

f(q) = |q|/q      ----(1)

f(-q) = |-q|/(-q)

f(-q) = |q|/-(q) 

f(-q) = - |q|/q     ----(2)

From (1) and (2)

|f(q) - f(-q)| = 2|q|/q = 2

∴ The correct answer is 2.

Let S = 4(1 + 11 + 111 + ...) = $\frac{4}{9}(9 + 99 + 999 + ...)$

$\Rightarrow\frac{4}{9}[(10 - 1) + (10^2-1)+(10^3-1)+...]$

$\Rightarrow\frac{4}{9}[(10 + 10^2 +...10^n)-n]=\frac{4}{9}[10\frac{(10^n-1)}{9}-n]$

$\Rightarrow \frac{4}{81}[10^{n+1}-9n-10]$

Poynting vector:

It states that the cross product of electric field vector (E) and magnetic field vector (H) at any point is a measure of the rate of flow of electromagnetic energy per unit area at that point that is 

$\vec S = \vec E \times \vec H$

S = Poynting vector

E = Electric field and

H = Magnetic field

The Poynting vector describes the magnitude and direction of the flow of energy in electromagnetic waves per unit volume.

$∯_s Re\left( {\vec P } \right)\cdot \hat n\ d\vec s$,  gives average power and it decreases with the increasing radial distance from the source.

Concept:

Phase velocity: 

v = f.λ 

Calculation:

Phase difference for the distance,

 $2mm = \frac{\pi }{4}$

Distance for phase difference,

 $2\pi = 2mm \times \left( 8 \right)$ 

$\Rightarrow ;λ ; = ;16mm$

Excitation frequency f = 10 GHz

Phase velocity,

 $v = f.;λ = \left( {10 \times {{10}^9}} \right) \times \left( {16 \times ;{{10}^{ - 3}}} \right) = 1.6 \times {10^8}~m/s$

Entropy for equiprobable and independent bits.

$H = {\log _2}N = {\log _2}4 = 2~bits/symbol$

Now, two symbols are transmitted per second.

So,Information rate = rH = 2 × 2 = 4 bit/s.

Concept:

Every branch of a root locus diagram starts at a pole (K = 0) and terminates at a zero (K = ∞) of the open-loop transfer function.

Application:

Zero = -1

Notice that there are two-locus lines going out from the pole at -3, which implies that there are two poles at s = -3,

Pole = 0, -2, -3, -3

On the real axis to the right side of any section, if the sum of total number of poles and zeros are odd, root locus diagram exists in that section.

∴ 

$G\left( s \right)H\left( s \right) = \frac{{k\left( {s + 1} \right)}}{{s\left( {s + 2} \right){{\left( {s + 3} \right)}^2}}}$

Important Points

1. Root locus diagram is symmetrical with respect to the real axis. 2. Number of branches of the root locus diagram are:

N = P if P ≥ Z

= Z, if P ≤ Z

3. Number of asymptotes in a root locus diagram = |P – Z| 4. Centroid: It is the intersection of the asymptotes and always lies on the real axis. It is denoted by σ.

$\sigma = \frac{{\sum {P_i} - \sum {Z_i}}}{{\left| {P - Z} \right|}}$

ΣPi is the sum of real parts of finite poles of G(s)H(s)

ΣZi is the sum of real parts of finite zeros of G(s)H(s)

5. Angle of asymptotes: ${\theta _l} = \frac{{\left( {2l + 1} \right)\pi }}{{P - Z}}$

l = 0, 1, 2, … |P – Z| – 1

6. Break-in/away points: These exist when there are multiple roots on the root locus diagram.

At the breakpoints gain K is either maximum and/or minimum.

So, the roots of $\frac{{dK}}{{ds}}$ are the break points.

h(n) = 2ⁿu(n - 2)

h(n) is existing for n > 2; so that h(n) = 0; n < 0 ⇒ causal

$\sum_{n = -\infty} ^{\infty} |h(n) = \sum_{n = \infty} ^{\infty}2^n = \infty \Rightarrow$ System is unstable

Concept:

• The derivative of a unit-step function [u(t)] is an impulse function [δ(t)]
• Impulse function = $\frac{d}{{dt}}$ (Unit step response)
• Unit step function = $\frac{d}{{dt}}$ (Response of ramp function)
• Ramp function = $\frac{d}{{dt}}$ (Response of parabolic function)

 

Calculation:

The unit step response of network is:

c(t) = 1 - e^-αt​​​​

$\frac{d}{{dt}}\left( {unit;step;response} \right) = impulse;response$

$\frac{d}{{dt}}\left( {1 - {e^{ -α t}}} \right) = α {e^{ -α t}}$

From the figure,

$\rm X = \overline {PQ} .\overline {QR} = \left( {\overline {PQ + QR} } \right)$

$\rm \therefore Y = \overline {\left( {\overline {PQ + QR} } \right)\overline {PR} } = PQ + QR + PR$

if two (or) more inputs are zero,

→ Y = 0

if two (or) more inputs are one,

→ Y = 1

For the parallel RLC circuit, the resonant frequency is

${w_r} = \frac{1}{{\sqrt {LC} }} = \frac{1}{{\sqrt {10 \times {{10}^{ - 6}} \times 1 \times {{10}^{ - 9}}} }} = 10\ M\ rad/sec$

∴ given circuit is in resonance

∴ impedance of parallel RLC = Z_max = R = 2KΩ

∴ gain = - g_m (Z_C || R_L)

= - g_m (2K || 2K)

∴ gain is maximum at resonance

∴ option (1) is correct

Drift currently density in a material can be given by​

$\rm{J=σE}$

$\rm = e\left( {n{\mu _n} + p{\mu _p}} \right)E$

So, Drift current in semiconductors depends upon

1) Electric charge ( 1.6 x 10-19  C)

2. Mobility of charge carriers 

3) Electric Field

4) Carrier concentration

• The forward voltage rating of Zener diode is the same as a normal diode
• A zener diode is used as a voltage regulator in reversed biased where voltage across it remains constant as shown in the figure
• The slope (V/I) is positive
• At reverse breakdown voltage (typically around 5V), there is a sharp breakdown at a reverse voltage as shown in the figure

We can see from the circuit,

$\begin{array}{l} \frac{{{V_{out}}}}{{{V_{in}}}} = \frac{{(R||{X_c})}}{{(R||{X_c}) + R + {X_c}}} \end{array}$

Solving the above by putting $X_c=\frac{1}{j\omega C}$ we get,

$\frac{{{V_{out}}}}{{{V_{in}}}} = \frac{{jω RC}}{{jω RC + {{\left( {1 + jω RC} \right)}^2}}}$

Given:

$V_{in}=\cos \left( {\frac{t}{{RC}}} \right)$   ---(1)

By comparing (1) with cos(ωt), we get

  $ω = \frac{1}{{RC}}$ 

∴ $\frac{{{V_{out}}}}{{{V_{in}}}} = \frac{j}{{j + {{\left( {1 + j} \right)}^2}}} = \frac{1}{3}$

$∴ {V_{out}} = \frac{1}{3}{V_i}$

$= \frac{{{V_p}}}{3}~{\rm{cos}}( {\frac{t}{{RC}}})$

From Divergence theorem, we have $\int \int \int \vec \nabla \cdot {\rm{\vec Adv}} = \mathop \oint \limits_{\rm{s}}^{\rm{;}} {\rm{\vec A}} \cdot {\rm{\hat n;ds}}$

The position vector  \({\rm{\vec r}} = \left( {{{{\rm{\hat u}}}{\rm{x}}}{\rm{x}} + {{{\rm{\hat u}}}{\rm{y}}}{\rm{y}} + {{{\rm{\hat u}}}_{\rm{z}}}{\rm{z}}} \right)\)

${\rm{Here}},{\rm{;\vec A}} = {\rm{;}}5{\rm{\vec r}},{\rm{;thus}}$

\(\nabla \cdot {\rm{\vec A}} = \left( {{{{\rm{\hat u}}}{\rm{x}}}\frac{\partial }{{\partial {\rm{;x}}}} + {{{\rm{\hat u}}}{\rm{y}}}\frac{\partial }{{\partial {\rm{;y}}}} + {{{\rm{\hat u}}}{\rm{z}}}\frac{\partial }{{\partial {\rm{;z}}}}} \right) \cdot 5\left( {{{{\rm{\hat u}}}{\rm{x}}}{\rm{x}} + {{{\rm{\hat u}}}{\rm{y}}}{\rm{y}} + {{{\rm{\hat u}}}{\rm{z}}}{\rm{z}}} \right)\)

$= \left( {\frac{{{\rm{\partial x}}}}{{{\rm{\partial x}}}} + \frac{{{\rm{\partial y}}}}{{{\rm{\partial y}}}} + \frac{{{\rm{\partial z}}}}{{{\rm{\partial z}}}}} \right)5 = 3 \times 5 = 15$

So, $\mathop \int!!!\int \limits_{\rm{s}}^{\rm{;}} 5{\rm{;\vec r}} \cdot {\rm{\hat n;ds}} = \int \int \int {\rm{;}}15{\rm{;dv}} = 15{\rm{;V}}$

For a TE_mn mode to exist, We need to have  m,n = 0,1,2,......

but does not exist for m = n = 0. For all other combinations TE_mn exists.

For a TM_mn mode to exist, We need to have m,n = 1,2,3, ....... i.e m & n both must be non-zero

⇒ TM₀₁, TM₁₀ modes do not exist

Important Points Waveguides only allow frequencies above the cut-off frequency to pass through. It blocks or attenuates the frequencies below the cut-off frequencies.

The cut off frequency is mathematically calculated as:

${{f}_{c\left( min \right)}}=\frac{c}{2}\sqrt{{{\left( \frac{m}{a} \right)}^{2}}+{{\left( \frac{n}{b} \right)}^{2}}}$

Where a and b are the dimensions of the waveguide

m and n are mode numbers TEmn.

The given differential equation is, (dy/dx) = ky

$\frac{{dy}}{y} = kdx$

On integrating both the sides, we get

ln y = kx + ln c

y = ce^kx

Advantages of FM over AM are:

• Improved signal to noise ratio.
• Smaller geographical interference between neighbouring stations.
• Less radiated power.
• Well defined service areas for given transmitter power.
• In SSB (Single Side-band), only one side-band is necessary for the transmission of information.
• This reduces the Bandwidth requirement to half.
• A typical SSB transmission spectrum is as shown:

         

• For SSB transmission, it does not matter whether the upper or lower sideband is used, since the information is contained in both.

Vestigial Side Band (VSB):

VSB (vestigial sideband) transmission transmits one sideband fully and the other sideband partially thus, reducing the bandwidth requirement.

$\frac{100d^2y}{dt^2}-\frac{20dy}{dt}+ y = x(t)$

Apply L.T. both sides

(100s² - 20s + 1) Y(s) = $\frac{1}{s}$  $[\because x(t) \times (s) = \frac{1}{3}]$

$Y(s) = \frac{1}{s(100s^2 - 20s + 1)}$

So we have poles with positive real part ⇒ system is unstable.

G (jω) = 5 + jω

 ω = 0  

 G (jω) = 5 + j 0 

 ω = 10  

 G (jω) = 5 + j 10 

 ω = ∞ 

 G (jω) = 5 + j ∞ 

 ∴ G (jω) is a straight line parallel to jω axis

The Fourier series for the function f(x) in the interval α < x < α + 2π is given by

\(f\left( x \right) = \frac{{{a_o}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}\cos nx + \mathop \sum \limits_{n = 1}^\infty {b_n}\sin nx\)

where

\({a_o} = \frac{1}{\pi }\mathop \smallint \limits_\alpha ^{\alpha + 2\pi } f\left( x \right)dx;;{a_n} = \frac{1}{\pi }\mathop \smallint \limits_\alpha ^{\alpha + 2\pi } f\left( x \right)\cos nxdx;;{b_n} = \frac{1}{\pi }\mathop \smallint \limits_\alpha ^{\alpha + 2\pi } f\left( x \right)\sin nxdx\)

An even function is any function f such that f(-x) = f(x)

Example: cos x, sec x, x2, x4, x6 …….., x-2, x-4 ……..

An odd function is any function f such that f(-x) = -f(x)

Example: sin x, tan x, cosec x, cot x, n, x3 ……., x-1, x-3 ……..

\(\mathop \smallint \limits_{ - L}^L f\left( x \right)dx = \left{ {\begin{array}{*{20}{c}} {2\mathop \smallint \limits_0^L f\left( x \right)dx,;;when;f\left( x \right);is;an;even;function}\ {0,;;when;f\left( x \right);is;an;odd;function} \end{array}} \right.\)

When f is an even periodic function of period 2L, then its Fourier series contains only cosine (include possibly, the constant term) terms.

\(f\left( x \right) = \frac{{{a_o}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}\frac{{\cos n\pi x}}{L}\)

\({a_o} = \frac{1}{L}\mathop \smallint \limits_{ - L}^L f\left( x \right)dx = \frac{2}{L}\mathop \smallint \limits_0^L f\left( x \right)dx\)

\({a_n} = \frac{1}{L}\mathop \smallint \limits_{ - L}^L f\left( x \right)\cos \frac{{n\pi x}}{L}dx = \frac{2}{L}\mathop \smallint \limits_0^L f\left( x \right)\cos \frac{{n\pi x}}{L}dx\)

When f is an odd periodic function of period 2L, then its Fourier series contains only sine terms.

\(f\left( x \right) = \mathop \sum \limits_{n = 1}^\infty {b_n}\sin \frac{{n\pi x}}{L}\)

\({b_n} = \frac{1}{L}\mathop \smallint \limits_{ - L}^L f\left( x \right)\sin \frac{{n\pi x}}{L}dx = \frac{2}{L}\mathop \smallint \limits_0^L f\left( x \right)\sin \frac{{n\pi x}}{L}dx\)

 

Concept:

D flip-flop can be built using NAND gate or with NOR gate. Whenever the clock signal is LOW, the input is never going to affect the output state. The clock has to be high for the inputs to get active. Thus, D flip-flop is a controlled Bi-stable latch where the clock signal is the control signal. Again, this gets divided into positive edge triggered D flip flop and negative edge triggered D flip-flop

Truth table of D Flip-Flop:

 

New,

 ${y_n} = {Q_{{1_n}}}{Q_{{2_n}}}$

$= {Q_{{1_n}}}{D_{{2_n}}}$

$= {Q_{{1_n}}}\overline {{Q_{{1_{\left( {n - 1} \right)}}}}}$

$= {D_{{1_n}}}\overline {{D_{{1_{\left( {n - 1} \right)}}}}}$

The output will be high when,

 ${D_{{1_{\left( {n - 1} \right)}}}} = 0\ and\ {D_{{1_n}}} = 1$

i.e when the data changes from ‘0’ to ‘1’

Concept:

In a 4 × 1 MUX

Truth-Table

 

Y = Output = S̅₁ S̅₀ I₀ + S̅₁ S₀ I₁ + S₁ S̅₀ I₂ + S₁ S₀ I₃

MUX contains AND gate followed by OR gate

Calculation:

By re-drawing circuit diagram

∴ I₀ = 0, I₁ = 1, I₂ = 1, I₃ = 0 & (P = S₁, Q = S₀)

Now output of 4 × 1 MUX is

Y = F = (P̅ Q̅) 0 + (P̅ Q)1 + (P Q̅) 1 + (P Q)0

∴ F = P Q̅ + P̅ Q = P ⊕ Q

∴ F = XOR (P, Q)

At low frequencies,

 $i.e. \ w=0 : \ L\rightarrow\ S.C$

$\therefore\ V_0\ =0V$

similarly at higher frequencies,

 $i.e. w=\infty:L\rightarrow\ O.C$

$\therefore\ V_0=i_iR_1$  

So, the given circuit passes only high frequencies so it acts like a high pass filter.

For Si diode:

The forward bias voltage change by – 2.5mV/°C.

Thus, for 10°C change in temperature change in forward bias.

${\rm{\Delta }}V = - 2.5\frac{{mV}}{{^\circ C}} \times 10^\circ C;$

$=-25 mV$

Note : It may be noted here that reverse saturation current doubles for every 10 °C rise in temperature

Converting current source into voltage source form, we get

The capacitor is connected across the short circuited branch hence it can be removed.

$\begin{array}{l} \therefore {I_{SC}} = \frac{{400\angle {0^\circ }}}{{40 + j30}}\ = 6.4 - j4.8\ A \end{array}$

(1) is correct.

Concept:

From LMS

Calculation:

R_TH = (10 || 10) + 10 = 15 Ω

Concept:

Cauchy’s Theorem:

If f(z) is an analytic function and f’(z) is continuous at each point within and on a closed curve C, then

$\mathop \oint \limits_C f\left( z \right)dz = 0$

Calculation:

$\mathop \oint \limits_c^; \frac{{ - 3z + 5}}{{{z^2} + 4z + 5}}dz$

Where c is circle |z| = 1

Here poles are at,

z = -2 ± j which are outside the unit circle

Hence,

 $\mathop \oint \limits_c^; \frac{{ - 3z + 5}}{{{z^2} + 4z + 5}}dz = 0$

Boundary conditions for magnetic field are

${B_{{n_1}}} = {B_{{n_2}}}$ and ${H_{{t_1}}} - {H_{{t_2}}} = - {\vec J_s} \times {\hat a_n}$

$\Rightarrow {B_{{x_1}}} = {B_{{x_2}}}$ .......(x is normal direction) 

$\Rightarrow {H_x}_2 = \frac{{1 \times 3{{\hat u}_x}}}{2}$

${H_x}_2 = 1.5{\hat u_x}$

Now, 

$\left( {{H_t}_1 - {H_t}_2} \right) = ; - \left( {10{{\hat u}_y}} \right) \times \left( {{{\hat u}_x}} \right)$

$\ = {\rm{;}}10{\rm{;}}{\hat u_z}$

${H_{{t_2}}} = {H_{{t_1}}} - 10{\hat u_z} = 30{\hat u_y} - 10{\hat u_z}$

$\Rightarrow {H_t}_2 = 30{\hat u_y} - 10{\hat u_z}$

\(\Rightarrow {\vec H_2} = {\vec H_x}2 + {\vec H{{t_2}}} = 1.5{\hat u_x} + 30{\hat u_y} - 10{\hat u_z}\)

VSWR $= ;5; = ;\frac{{1 + \left| \Gamma \right|}}{{1 - \left| \Gamma \right|}} \Rightarrow \left| \Gamma \right| = \frac{2}{3} \Rightarrow \Gamma = \pm \frac{2}{3}$

Maxima at distance  $\frac{\lambda }{4}$ from load

⇒ minima at load itself

$\Rightarrow {Z_L} < Z \Rightarrow \Gamma = - \frac{2}{3}$

 

$\Rightarrow \Gamma = \frac{{{Z_L} - {Z_o}}}{{{Z_L} + {Z_o}}} = - \frac{2}{3} \Rightarrow \frac{{{Z_L} - 50}}{{{Z_L} + 50}} = - \frac{2}{3} \Rightarrow {Z_L} = 10;\Omega$

The total transfer function H(f) = (j2πf - 1)

S_x(f) = |H(f)|² S_x(f) R_x(τ) $\overset{F}{\large\leftrightarrow}$ S_x(f)

= (4π²f² + 1)e^-πt2 (∵ e^-πt2 $\overset{F}{\large\leftrightarrow}$ e^-πt2)

The initial state of the Johnson counter is as follows –

                                    D₂        D₁        D₀

            State -1            0          0          0

            State-2             1          0          0

            State-3             1          1          0

            State-4             1          1          1

            State-5             0          1          1

            State-6             0          0          1

            State-7             0          0          0                      (it is equivalent to state 1)

            So the corresponding digital output assuming LSB bit contribute unit Analog output is –

<br>

So, option (1) is correct.

Concept:

Synchronous Counters are so-called because the clock input of all the individual flip-flops within the counter are all clocked together at the same time by the same clock signal.

With the Synchronous Counter, the external clock signal is connected to the clock input of EVERY individual flip-flop within the counter so that all of the flip-flops are clocked together simultaneously (in parallel) at the same time giving a fixed time relationship.

Example of the Synchronous up counter with the waveforms is shown below:

 

Calculation:

Given counting sequence is 

The construction of the truth table is explained below:

Excitation table of D flip flop

Output equations for the flip flops are

For D_B

<sub>

</sub>

_{\({{\bf{D}}{\bf{B}}} = \overline {{{\bf{Q}}{\bf{B}}}} \)}

For D_A

<sub>

</sub>

_{${D_A} = \overline {{Q_B}} ;\overline {{Q_A}} + {Q_B}{Q_A}$}

_{\({{\bf{D}}{\bf{A}}} = {{\bf{Q}}{\bf{B}}} \odot {{\bf{Q}}_{\bf{A}}}\)}

Since the Gate and Drain of pull-down transistor is shorted. So it has operated in the saturation region, let pull up transistor operated in the saturation region.

Then I_D1 = I_D2

\(\frac{1}{2}{\mu n}{C{ox}}{\left( {\frac{\omega }{L}} \right)1}{\left( {{V{GS1}} - V_{T1}} \right)^2}\)

\( = \frac{1}{2}{\mu n}{C{ox}}{\left( {\frac{\omega }{L}} \right)2}{\left( {{V{Gs2}} - V_{T2}} \right)^2}\)

4 [5 – V_x - 1]² = 1[V_x - 1]²

2(4 - V_x) = V_x – 1

9 = 3 V_x → V_x = 3V

V_GS1 = 5 – 3 = 2, V_DS1 = 6 – 3 = 3V

∴ V_DS1 > (V_GS1 – VT₁)

So our assumption is correct

∴ V_x = 3 V

So option (3) is correct.

$X(s) = \frac{1}{s + 2} + e^{-6s} \text {and H}(s) = \frac{1}{s}$

$Y(s) = H(s) \times (s) = \frac{1}{s(s+2)}+\frac{e^{-6s}}{s}=\frac{1}{2}\frac{1}{s}-\frac{1}{2(s+2)}+\frac{e^{-6s}}{s}$

⇒ y(t) = 0.5(1 - e^-2t) u(t) + u(t - 6)

Concept:

Current gain β is given as:

$\beta = \frac{\alpha }{{1 - \alpha }}$

Collector current in terms of reverse bias saturation current is given as:

${I_c}={β{I_B}}+{\left(1+β\right)}I_s$

Calculation:

Given:

Is = 0.6μA

α=0.98

IB= 20µA

$\beta = \frac{0.98}{{1 - 0.98}}=49$

putting values,

${I_c}={49\times{20μ{A}}}+{\left(50\right)}\times0.6μ{A}$

${I_c}=0.98{m}A+0.03{m}A=1.01mA$

Concept:

By initial value theorem,

Initial value of f(t) is given by,

$\mathop {\lim }\limits_{t \to 0} f\left( t \right) = \mathop {\lim }\limits_{s \to \infty } sF\left( s \right)$      ---(1)

Final value of f(t) is given by,

$\mathop {\lim }\limits_{t \to \infty } f\left( t \right) = \mathop {\lim }\limits_{s \to 0} ;sF\left( s \right)$      ---(2)

Calculation:

$\displaystyle\lim_{t \rightarrow o}f(t) = \displaystyle\lim_{(s \rightarrow \infty)} \frac{s(2s + 1)}{s^2 + 4s + 7}=2$

$\displaystyle\lim_{t \rightarrow \infty}f(t) = \displaystyle\lim_{(s \rightarrow 0)} \frac{s(2s + 1)}{s^2 + 4s + 7}=0$

From star to delta conversion

${R_1} = \frac{{{R_a}{R_b}}}{{{R_a} + {R_b} + {R_c}}} = 2\ \Omega$

Similarly,   ${R_2} = {R_3} = {R_1} = 2\ \Omega$

∴The circuit will be like

$\begin{array}{l} \therefore Z = \left( {\left( {2 + j4} \right)} \right)||\left( {2 - j4} \right)) + 2\ Z=\frac{20}{4}+2 = 7; \Omega \ \therefore I =\frac{V}{Z}= \frac{{14\angle {0^\circ }}}{7} = 2\angle {0^\circ }A \end{array}$

Concept:

Using Newton-Raphson method

$\mathop x\nolimits_1 = \mathop x\nolimits_0 - \frac{{f(\mathop x\nolimits_0 )}}{{\frac{d}{{dx}}f(\mathop x\nolimits_0 )}}$

Calculation:

$\mathop x\nolimits_o = 2$

$\frac{d}{{dx}}f(x) = 1 + \frac{1}{{2\sqrt x }}$

at x₀ = 2

$\frac{d}{{dx}}f(2) = 1 + \frac{1}{{2\sqrt 2 }}$

$\begin{array}{l} \mathop x\nolimits_1 = 2 - \frac{{f(2)}}{{\frac{d}{{dx}}f(2)}} = 2 - \frac{{\left[ {2 + \sqrt 2 - 3} \right]}}{{\left[ {1 + \frac{1}{{2\sqrt 2 }}} \right]}} = 1.693\ \end{array}$

$\vec E = {E_p}.{e^{j\left( {\omega t - 280\pi y} \right)}};;{\hat u_z}~\frac{v}{m}$

\(H = 3~{e^{j\left( {\omega t - 280\pi y} \right)}}{\hat u_x}\frac{A}{m}\)

$c = 3 \times {10^8}~m/s$  (speed of light)

$\beta = 280;\pi = \frac{{2\pi }}{\lambda }$

 $\lambda = \frac{1}{{140}}m$

$v = f\lambda$

$v= 14 \times {10^9} \times \frac{1}{{140}}$

$v = 1 \times {10^8}$

$\Rightarrow v = \frac{c}{{\sqrt {{\epsilon_r}} }}$

${\epsilon_r} = {\left( {\frac{c}{v}} \right)^2}$

$\epsilon= {\left( {\frac{{3 \times {{10}^8}}}{{1 \times {{10}^8}}}} \right)^2} =9$

${E_p} = \left| {\vec H} \right|\eta = \frac{{\left| {\vec H} \right|.{\eta _o}}}{{\sqrt {{\epsilon_r}} }}$

$= \frac{{3 \times 120\pi }}{{\sqrt 9 }} = \textbf{120};\pi$

For faithful demodulation, the time constant should satisfy the following condition.

$\frac{1}{{{f_c}}} < RC < \frac{1}{{{f_m}}}$

Time constant should be length than $\frac{1}{f_m}$

And time constant should be far greater than $\frac{1}{f_m}$

$f_m = \frac{4000 a}{2a} = 2000$

$\frac{1}{f_c}<< Rc< \frac{1}{2000}$

1 μs << RC << 0.5 ms

x = y₁ & $\dot x = \frac{{d{y_1}}}{{dx}}$

\(\underline y = \left[ {{y_2}} \right] = \left[ {\begin{array}{{20}{c}} 2\ {2x} \end{array}} \right] = \left[ {\begin{array}{{20}{c}} 1\ 2 \end{array}} \right]x\)

${y_1} = \frac{1}{{s + 2}}u$

${\dot y_1} + 2{y_1} = u$

$\dot x + 2x = u$

$\dot x + 2x = u$

$\underline {\dot x} = \left[ { - 2} \right]\underline x + \left[ 1 \right]u$

Alternate solution 

Assume x be a state variable

$[{\rm{X}}\left( {\rm{s}} \right) = \frac{1}{{s + 2}}u\left( s \right)]$

sX(s)+2X(s) = U(S)

taking inverse laplace transform

x(t)’ + 2 x(t) = u(t)

x’(t)= u(t) – 2x(t)

output equation is given by

y₁(t) = x

y₂(t) = 2 x

The overall output y(n) is the same as the input x(n) with a one-unit delay.

y(n) = x(n - 1)

By applying z-transform,

Y(z) = z^-1X(z)

$\Rightarrow H\left( z \right) = \frac{{Y\left( z \right)}}{{X\left( z \right)}} = {z^{ - 1}}$ 

⇒ H₁(z) H₂(z) = z^-1

$\Rightarrow \left( {\frac{{1 - 0.2{z^{ - 1}}}}{{1 - 0.4{z^{ - 1}}}}} \right){H_2}\left( z \right) = {z^{ - 1}}$ 

$\Rightarrow {H_2}\left( z \right) = \frac{{{z^{ - 1}}\left( {1 - 0.4{z^{ - 1}}} \right)}}{{1 - 0.2{z^{ - 1}}}}$

MVI A, 07 H ⇒ 0000 0111 ← The content of 'A'

RLC ⇒ 0000 1110 ← The content of 'A'

MOVT B, A ⇒ 0000 1110 ← The content of 'B'

RLC ⇒ 0001 1100 ← The content of 'B'

RLC ⇒ 0011 1000 ← The content of 'B'

ADD B

\(\frac{A \space 0000 \space 1110\\

• \
  B \space 0011 \space 1000}{0100 \space 0110}\)

$RCC \rightarrow \frac{0010}{2}\frac{0011}{3} 23 H$

Concept:

N-point DFT of a sequence x(n) defined for n = 0, 1, 2 ... N-1 is given by:

$X\left( k \right)=\sum_{n=0}^{N-1}x\left( n \right){{e}^{-jk\omega n}}$

Also If $x(n)\overset{DFT}{\longleftrightarrow} X(k)$

The Conjugate symmetric property of DFT states:

X(k) = X*(N – k) 'or' X(N – k) = X*(k) 

Calculation:

Given, X(k) = {5, 1 - 3j, 0, 3 – 4j, 0, 3 + 4j, A, B]

Where A and B are the missing variable values.

Using the conjugate symmetric property of DFT: X(k) = X*(N – k)

We find X(6) = A = X*(8 – 6) = X*(2)

With X(2) = 0, X*(2) = 0

So, X(6) = A = 0

Similarly,

B = X(7) = X*(8 – 7) = X*(1)

With, X(1) = 1 – 3j

X(7) = X*(1) = 1+3j

So, A = 0 and B = 1 + 3j

Option (2) is therefore correct.