GATE EC 2022 Official Paper (Previous Year Paper)

The correct answer is 'neither / nor'.

Key Points

• In the given blanks, it can be understood that conjunctions are to be filled.
• We will use the alternative conjunctions 'neither/nor' as we can understand that Mr. X is unable to speak both languages.

The complete sentence will be: Mr. X speaks neither Japanese nor Chinese.

• Hence, option 3 is the correct answer.

Additional Information

• When we have to choose one out of two given alternatives, we use alternative conjunctions such as either/or, neither/nor, else, or otherwise.
• For eg.- Walk slowly on the ice, otherwise, you'll fall.

Given:

The ratio of distributed money among P, Q, R, and S in the proportion 5 ∶ 2 ∶ 4 ∶ 3, respectively.

R gets Rs. 1000 more than S.

Calculation:

The ratio of distributed money,

P : Q : R : S = 5 ∶ 2 ∶ 4 ∶ 3

Suppose, P gets Rs. 5x, Q gets Rs. 2x, R gets Rs. 4x, and S gets Rs. 3x.

According to the question, R gets Rs. 1000 more than S.

Therefore, 4x - 3x = 1000

⇒ x = 1000

∴ Q gets = 2 × 1000 = Rs. 2000

∴ The share of Q (in Rs.) is Rs. 2000

Given:

A trapezium has vertices marked as P, Q, R, and S (in that order anticlockwise), and PQ || RS

PQ = 11 cm, QR = 4 cm, RS = 6 cm and SP = 3 cm

Formula used:

For a right-angle triangle, \(Perpendicular = \sqrt{(Hypotenuse^2 - Base^2)}\)

Calculation:

Let's draw the diagram according to the question,

Here, the trapezium has vertices marked as P, Q, R, and S (in that order anticlockwise), and PQ || RS

PQ = 11 cm, QR = 4 cm, RS = 6 cm and SP = 3 cm

Draw two perpendiculars from point R and point S that connect to PQ at point B and point A respectively are shown in the above diagram.

Therefore, △BCQ and △ASP are both right-angle triangles. [∠QBR = ∠PAS = 90º]

Here, BR = AS = h [PQ || RS]

PQ = 11 cm and RS = 6 cm

Suppose, PA = x

∴ QB = 11 - (6 + x) = (5 - x)

Therefore, \(h = \sqrt{QR^2 - QB^2} = \sqrt{SP^2 - PA^2}\)

⇒ \(h = \sqrt{4^2 - (5 - x)^2} = \sqrt{3^2 - x^2}\)

⇒ \(\sqrt{4^2 - (5 - x)^2} = \sqrt{3^2 - x^2}\)

⇒ 16 - (5 - x)² = 9 - x²

⇒ 16 - (25 - 10x + x2) = 9 - x2

⇒ 16 - 25 + 10x - x2 = 9 - x2

⇒ 10x = 9 - 10 + 25 = 18

⇒ x = 18/10 = 1.8

Now, \(h = \sqrt{3^2 - 1.8^2}\)

⇒ \(h = \sqrt{9 - 3.24} = \sqrt{5.76}\) = 2.4 cm

∴ The shortest distance between PQ and SR (in cm) is 2.4 cm.

According to the question, 

Total number of squares = 16

Number of squares without hole,

4 × 4 - 1 = 16 - 1 = 15,

∴ the total number of 1 × 1 squares (without hole) = 15

Now,

the total number of 2 × 2 squares (without hole) = 5

Total number of squares = 15 + 5 = 20

If a person will stand on ‘P’ then he/she can’t see the visibility side of R and vice versa.

But if the security guard will be stand at Q and S then he will be whole 360°

Security guard posting at Q and S can watch entire space of gallery.

The correct answer is 'Using chemicals to kill mosquitoes may have undesired consequences but it is not clear if using genetically modified mosquitoes has any negative consequence'.

Key Points

• Let's refer to the passage:
• 'In Florida, authorities have used genetically modified mosquitoes to control the overall mosquito population.'
• 'It remains to be seen if this novel approach has unforeseen consequences.'
• From the above-mentioned statements, it is evident that the correct logical inference is that 'using chemicals to kill mosquitoes may have undesired consequences but it is not clear if using genetically modified mosquitoes has any negative consequence'.
• Hence, option 4 is the correct answer.

Additional Information 

• The word 'novel', here, is being used to denote something that is interestingly new or unusual.
• For eg.- He hit on a novel idea to solve his financial problems.

Given:

2x − 1 > 7

2x − 9 < 1

Calculation:

The given inequalities are,

2x − 1 > 7

⇒ 2x > 7 + 1

⇒ 2x > 8

⇒ x > 4     .....(1)

2x − 9 < 1

⇒ 2x < 1 + 9

⇒ 2x < 10

⇒ x < 5     .....(2)

From (1) and (2), we can see that x is greater than 4 but less than 5 means 4 < x < 5

∴ The two inequalities express 4 < x < 5. Hence, option 3 is correct.

Given:

The quadrilateral has four points P(0, 1), Q(0, -3), R(-2, -1), and S(2, -1)

Formula used:

The distance between two points using coordinates = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) where (x₁, x₂) and (y₁, y₂) are coordinates of two points.

Area of a square = (side)²

Calculation:

Four points P(0, 1), Q(0, -3), R(-2, -1), and S(2, -1) represent the vertices of a quadrilateral. Let's draw the diagram,

Length of PS = \(\sqrt{(2 - 0)^2 + (-1 - 1)^2} = \sqrt{8}\)

Length of SQ = \(\sqrt{(0 - 2)^2 + (-3 + 1)^2} = \sqrt{8}\)

Lenght of QR = \(\sqrt{(0 - 2)^2 + (-3 + 1)^2} = \sqrt{8}\)

Length of PR = \(\sqrt{(-2 - 0)^2 + (-1 - 1)^2} = \sqrt{8}\)

Length of PQ = \(\sqrt{(0 - 0)^2 + (-3 - 1)^2} = \sqrt{16} = 4\)

Length of RS = \(\sqrt{(2 + 2)^2 + (-1 + 1)^2} = \sqrt{16} = 4\)

Here, PS = SQ = QR = PR and PQ = RS

Hence, PRQS is a square.

Area of the square = \((\sqrt{8})^2\) = 8

∴​ The area enclosed by the quadrilateral is 8.

Statement of S: Only one of us is telling the truth.

As, given in the question 'S' is telling the truth. 

Let, us suppose that 

T is telling the truth, 

that means R is telling the truth. 

and according to R, we know that P did not copy in the exam.

But if T is telling the truth then that means, R is also telling the truth and then that means 2 of them are telling the truth, which is not possible. Hence T is not telling the truth and hence R is not truthful as well. 

" P did not copy in the exam"  is hence false.

Which means "P did copy in the exam".

1. For XYZZ

and ZZ

2. XY

3. ZZZZ

Concept:

Green theorem:

\(\oint (Mdx+Ndy)= \iint_{R}^{}\left ( \frac{\partial N}{\partial x}-\frac{\partial M}{\partial y} \right )dxdy\)

Where the path of integration along with C is anticlockwise. 

Calculation:

\(\oint_C \vec F(x,y).(dx\ \vec i+dy\ \vec j) \space\)

Given 

\(\vec F(x,y)=x\vec i+y \vec j \space \)

\(\oint_C \vec F(x,y).(dx\ \vec i+dy\ \vec j) \space = \oint_C (x\ \vec i+y\ \vec j).(dx\ \vec i+dy\ \vec j) \space \)

\(\oint_C \vec F(x,y).(dx\ \vec i+dy\ \vec j) \space = \oint_C (xdx+ydy) \space \)

By Green's theorem

\(\oint_C (xdx+ydy) \space = \iint_s {\dfrac{\partial y}{\partial x}-\dfrac{\partial x}{ \partial y}} \space =0\space \)

Applying KCL at junction we get current through middle branch

= (I + 10^-3)A

Applying KVL we get

5 = (2K)I + 2K(I + 10^-3)

5 = 2000 I + 2000 I + 2

solving for I we get

I = 0.75 × 10^-3 A

Concept: 

The current leaving a current/voltage source must be equal to current entering the source in order to justify the law of conservation of charge.

∴ For loop 2:

y = y + I

I = 0

Similarly, for Loop 1:

i = i - I

I = 0

∴ The current 'I' though resister 10 Ω will be:

I = 0 A

Concept:

Let us assume the Fourier transform of x(t) is X(ω)

then as per property of duality,

x(t) \(\rightleftharpoons\) X(ω)

X(t) \(\rightleftharpoons\) 2π x(-ω)

Calculation:

We know that,

\(\begin{aligned} &e^{-a t} u(t) \stackrel{F \cdot T \cdot}{\rightleftharpoons} \frac{1}{a+j ω} \ &e^{-a|t|} u(t) \stackrel{F T}{\rightleftharpoons} \frac{2 a}{a^{2}+ω^{2}} \end{aligned}\)

for, a = 1

e^-|t| \(\stackrel{F \cdot T \cdot}{\rightleftharpoons}\) \(\frac{2}{1+ω^2}\)      ....(1)

applying differentiation in frequency property,

x(t) \(\rightleftharpoons\) X(ω)

Now,

t x(t) \(\rightleftharpoons\) j \(\rm \frac{d}{dω}X(ω)\)

\(\rm te^{-|t|}u(t)\rightleftharpoons(j)\frac{d}{dω}\left[\frac{2}{1+ω^2}\right]\)

\(\rm te^{-|t|}u(t)\rightleftharpoons\frac{-4jω}{(1+ω^2)^2}\)       .....(2)

applying property of duality in eqn (2) we get,

\(\rm te^{-|t|}u(t)\rightleftharpoons\frac{-4jω}{(1+ω^2)^2}\)

t = -ω

\(\frac{-4jt}{(1+t^2)^2}\rightleftharpoons-2\pi \omega e^{-|\omega|}\)

\(\frac{t}{(1+t^2)^2}\rightleftharpoons\frac{\pi}{2j} \omega e^{-|\omega|}\)

Concept:

Extrinsic p-type semiconductors are formed when a trivalent impurity is added to a pure semiconductor. Examples of trivalent impurity are Boron, Gallium, and Indium.

Also, the law of mass action is used to determine the minority carriers in a doped semiconductor.

According to law:

n₀.p₀ = ni2

n₀ = concentration of electrons

p₀ = concentration of holes

For a steady-state carrier injection, the excess carrier concentration dies out exponentially for a distance ‘x’ due to recombination.

Mathematically this is defined as:

δp(x) = δp(0) e-x/Lp

δn(x) = δn(0) e-x/Ln

Ln and Lp are the diffusion length of electrons and holes respectively.

Calculation:

Given, P = 10¹⁷ cm^-3 = P₀

\(\rm n_{0}=\frac{n_{i}^{2}}{P_{0}}=\frac{10^{20}}{10^{17}}=10^{3} \mathrm{~cm}^{-3}\)

L_n = 2 μm

n'_PO = 10¹⁴ cm^-3

Excess electron concentration at any distance x is

δn_p(x) = n'_P0 e^-x/L_n

= 10¹⁴ e^-2/2

= 10¹⁴ e^-1

= 3.67 × 10¹³ cm^-3

= 0.367 × 10¹⁴ cm^-3

Calculation:

P(s) = s⁴ + 5s² + (4 + K)

Given: p(s) = s⁴ + 5s² + (4 + K)

Routh's Table:

\(\begin{array}{c|ccc} s^{4} & 1 & 5 & (4+K) \ s^{3} & 0 & 0 \ s^{2} && \ s^{1} & & & \ s^{0} & & & \end{array}\)

As all row of s³ are zero,

A(s) = s⁴ + 5s² + (4 + K)

\(\begin{array}{c|ccc} s^{4} & 1 & 5 & (4+K) \ s^{3} & 4 & 10 & 0 \ s^{2} & \frac{10}{4}-\frac{5}{2} & (4+K) & \ s^{1} & \frac{25-4(4+K)}{5 / 2} & & \ s^{0} & 4+K & & \end{array}\)

1^st column of R-H table must be all positive.

i.e., \(\frac{25-4(4 + K)}{5/2} > 0 ⇒ K < \frac{9}{4}\)

4 + K > 0 ⇒ K > -4

Range of K: \(-4 < K < \frac{9}{4}\)

Concept:

Fermi level in n-type Semiconductor:

The position of the Fermi level in the n-type semiconductor is given as:

\( {E_F} = {E_C} -KT\ln \left[ {\frac{{{N_C}}}{{{N_D}}}} \right]\)

Fermi level in p-type Semiconductor:

The position of the Fermi level in the p-type semiconductor is given as:

\( {E_F} ={E_V} - KT\ln \left[ {\frac{{{N_V}}}{{{N_A}}}} \right]\)

Calculation:

Given:

ND1 = 1016 cm-3

ND2 = 0.5 × 1016 cm-3

\(\rm (E_{C}-E_{F n})_{1}=k T \ln \left(\frac{N_{C}}{N_{D 1}} \right)=200 \) meV

\(\rm (E_{C}-E_{F n})_{2}=k T \ln \left(\frac{N_{C}}{N_{D 2}} \right)\)

\(\rm \left(E_{C}-E_{F n}\right)_{1}-\left(E_{C}-E_{F n}\right)_{2}=k T \ln \left(\frac{N_{C}}{N_{D 1}}\right)-k T \ln \left(\frac{N_{C}}{N_{D 2}}\right)\)

\(\rm 200 \ meV -\rm (E_{C}-E_{F n})_{2}=k T \ln \left(\frac{N_{D2}}{N_{D 1}} \right)\)

\(\rm 200 \ meV -\rm (E_{C}-E_{F n})_{2}=0.026 \ln \left( \frac{10^{16} \times 0.5}{10^{16}} \right)\)

\(\rm 200 \ meV -\rm (E_{C}-E_{F n})_{2}\) = 0.026 ln(0.5)

\(\rm 200 \ meV -\rm (E_{C}-E_{F n})_{2}\) = -0.01802 volt

\(\rm 200 \ meV -\rm (E_{C}-E_{F n})_{2}\) = -18.02 meV

(E_C - E_Fn)₂ = 200 meV + 18.02 meV

= 218.02 meV

Concept:

MOSFET is in saturation region:

VGS > Vth

VDS > VGS – Vth

The current equation for a MOSFET in the saturation region is given by:

\({{I}_{D}}=\frac{1}{2}{{\mu }_{n}}{{C}_{OX}}\frac{W}{L}{{\left( {{V}_{GS}}-{{V}_{th}} \right)}^{2}}\)

\({{I}_{D}}=\frac{K_n}{2}{{\left( {{V}_{GS}}-{{V}_{th}} \right)}^{2}}\)

Calculation:

Both MOSFETs are in saturation because drain is shorted to Gate.

I_DSN = I_SDP

\(\frac{\mu_{n} C_{o x}}{2} ×\left(\frac{W}{L}\right)_{N}\left(V_{G S N}-V_{T N}\right)^{2}=\frac{\mu_{p} C_{o x}}{2}\left(\frac{W}{L}\right)_{P}\left(V_{S G P}-\left|V_{T P}\right|\right)^{2}\)

300 × 1 (V₀ - 1)² = 40 × 5(4 - V₀ - 1)²

\(3(V_0^2 + 1 - 2V_0) = 2(9 + V_0^2 - 6 V_0)\)

⇒ \(V_0^2 + 6V_0 - 15 = 0\)

\(V_0 = \frac{-6 \pm \sqrt{36 + 4 \times 15}}{2} = \frac{-6 \pm \sqrt{96}}{2}\)

V₀ = 1.898 V, -7.89 V

V₀ cannot be negative because V₀ should lie between 0 and 4 V.

V₀ = 1.898 V (Less than 2V)

The correct answer is option 3.

Concept:

The multiplexer abbreviated "MUX" or "MPX," is a combinational logic circuit that uses a control signal to switch one of the numerous input lines to a single common output line.

S_1, S₀ are the select lines, I₁, I₂, I₃, and I₄ are the input lines and produce the output Y.

Y= (S₁' S₀' )I₀+( S₁'S₀ )I1+(S₁ S₀' )I2 +( S₁S₀ )I3

 

Explanation:

Option 1: A0 = 0, A1 = 0, A2 = 1, A3 = 1

These input lines A0 = 1, A1 = 0, A2 = 1, A3 = 0, C and D are the select lines and produce the output 

Y= (C' D' ) 0+( C' D ) 0+(C D' ) 1+( CD) 1

Y=0+0+CD'+CD

Y=C(D'+D)

Y=C

Option 2: A0 = 1, A1 = 0, A2 = 1, A3 = 0

These input lines A0 = 0, A1 = 0, A2 = 1, A3 = 1, C and D are the select lines and produce the output 

Y= (C'D') 1+( C' D ) 0+(C D' ) 1+( CD) 0

Y= C'D'+0+CD'+0

Y=D'(C'+C)

Y=D'

Option 3: A0 = 0, A1 = 1, A2 = 1, A3 = 0

These input lines A0 = 0, A1 = 1, A2 = 1, A3 = 0, C and D are the select lines and produce the output 

Y= (C'D') 0+( C' D) 1+(CD') 1+( CD) 0

Y= 0 + C'D + CD' + 0

Y=C'D + CD'

Option 4: A0 = 1, A1 = 1, A2 = 0, A3 = 0

These input lines A0 = 1, A1 = 1, A2 = 0, A3 = 0, C and D are the select lines and produce the output 

Y= (C'D') 1 + ( C' D ) 1 + (C D' ) 0 + ( CD) 0

Y= C'D '+ C'D + 0 + 0

Y=C'D '+ C'D

Y=C'(D' + D)

Y = C'

Hence the correct answer is A0 = 0, A1 = 1, A2 = 1, A3 = 0.

V_c1 = min(V_G - V_T, V_D)

= min(4, 5)

Vc1 = 4V

   

V_SG  > |V_TP|

V_C2 + 4 > 1

VC2 > -4

VSD > 0

VC2 > 5V

VC2 = 5V

Concept:

Nyquist stability criterion:

N = P – Z

N is the number of encirclements of (-1+j0) point by the Nyquist contour in an anticlockwise direction.

P is the open-loop RHP poles

Z is the closed-loop RHP poles

Analysis:

For K = 1,

For K = 2, the plot will be

N = No. of encirclement about (-1, 0) in anticlockwise.

P = Total number of open loop poles, in R.H.S.

Z = P - N

N = -2, P = 0

Z = 0 - (-2) = 2

Z = 2

Two poles in right side.

Concept:

The number of branches of the root locus diagram is:

N = P if P ≥ Z

= Z, if P ≤ Z

Application:

Here 5 Root Locus branches are diverging from the same point, this can possible only when if we have 5 poles in the system at the same point because the Root Locus branch departs from the open-loop pole and

The Number of Root Locus branches = Number of open-loop poles or Number of zero (Whichever is greater).

Here, there are 5 multiple Real poles, matching with option (A). 

So, correct option is \(\frac{1}{(s + 1)^5}\)

Additional Information

1. 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.
2. The root locus diagram is symmetrical with respect to the real axis.
3. The number of branches of the root locus diagram are:

N = P if P ≥ Z

= Z, if P ≤ Z

4. Number of asymptotes in a root locus diagram = |P – Z| 5. 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)

6. Angle of asymptotes: 

l = 0, 1, 2, … |P – Z| – 1

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

Analysis(1):

Let x(t) signal is delayed by time T_d. If y(t) is the response of delayed system-

y(t) = x(t-Td) ..............(1)

Let take the Fourier transform of equation (1)

Y(f) = X(f).\(e^{-2j\pi fT_d}\)

H(f) = \(Y(f)/X(f) = e^{-2j\pi fT_d}\)

| H(f) | = 1 and

\(\angle H(f) ={-2j\pi fT_d}\)

It shows that system has magnitude 1 and system is bandlimited to −α ≤ f ≤ α, hence statement 1 is correct.

Analysis(2):

For a wide-sense stationary input process:

Power spectral density (PSD)_o = SX(f)

Output power spectral density (PSD)_i = 𝑆Y(f) 

\(\begin{aligned} (\mathrm{PSD})_{o} &=(\mathrm{PSD})_{i} \cdot|H(f)|^{2} \ \end{aligned}\)

\(\begin{aligned} S_{Y}(f) &=S_{X}(f) \quad ; \quad-\alpha \leq f \leq \alpha \ \end{aligned}\)

Phase response of the system is not given, hence statement 2 is correct.

Concept:

Conduction current density:

• Conduction current comes from the movement of charges such as electrons present in conductors or and semiconductors.
• The electrons present in the conduction band of conductors or semiconductors are free and they respond to the applied electric field.
• The current passing through due to such free election per unit cross-sectional area per unit time is commonly referred to as conduction current density.

 

\({J_c} = \frac{{{I_c}}}{A} = \sigma E\left( {\frac{{Amp}}{{{m^2}}}} \right)\)

Displacement current density:

• Displacement current comes from the change in the electric field caused by the displacement of the center of positive charge and center of negative charge of a dielectric or insulating medium.
• Dielectric does not contain free electrons for conduction, rather they contain immobile charges and are displaced slightly from their equilibrium position in the presence of the electric field.
• Consequently, they produce a change in the electric field present in the dielectric medium and hence displacement current.
• The rate of change of such electric field with respect to time is commonly referred to as displacement current density.

 

\({J_D} = \frac{{{J_D}}}{A} = \frac{{\frac{{\partial \mu }}{{\partial t}}}}{A} = \frac{\partial }{{\partial t}}\left( {\frac{\psi }{A}} \right) = \frac{{\partial D}}{{\partial t}}\)

\({J_D} = \frac{{\partial D}}{{\partial t}} = \omega D = 2\pi f{\varepsilon _o}{\varepsilon _r}E\)

Analysis:

In a series connection, the current through each element remains the same, Hence i_c = i_d

\(\rm i_d = 2 \sin \left( t + \frac{\pi}{2} \right)\)

Concept:

Relationship between de-emphasis filter and pre-emphasis filter is-

\(\left|H_{p e}(ω)\right|=\frac{1}{\left|H_{d e}(ω)\right|}\)

Where,

H_pe(ω) - Frequency response of pre-emphasis filter

H_de(ω) - Frequency response of de-emphasis filter

Given:

Frequency response of pre-emphasis filter-

\(\rm H_{pe}(ω)=1+\frac{jω}{ω_0}\)

ω0 = 104 rad/sec

and

de-emphasis filter

Solution:

\(\left|H_{p e}(\omega)\right|=\frac{1}{\left|H_{d e}(\omega)\right|}\)

\(\left|H_{p e}(\omega)\right|^{2}=\frac{1}{\left|H_{d e}(\omega)\right|^{2}}\)

So,

\(1+\left(\frac{\omega}{\omega_{0}}\right)^{2}=1+(\omega R C)^{2}\)

\(\frac{1}{\omega_{0}}=R C\)

RC = 10^-4

Only Option (1) satisfies the above condition.

Hence option (1) is correct.

Analysis:

Given

f(x, y) = e(ξ x + n y)

\(\frac{\partial^2f(x,y)}{\partial x^2}=ξ ^2\space f(x,y)\space \) →  (1)

\(\frac{\partial^2f(x,y)}{\partial y^2}=n^2\space f(x,y)\space \) → (2)

By the given partial equation and equation 1 and 2 

a ξ 2 f(x, y) + b n2 f(x, y) = f(x, y)

a ξ 2  + b n2  = 1

Now go by options

From option 1

\(ξ=\frac{1}{\sqrt{2a}}, \eta=\frac{1}{\sqrt{2b}} \space \)

a ξ 2  + b n2  = 1   ⇒  a × (1/2a) + b (1/2b)  = 1      (Option 1 is correct)

From option 2 

\(ξ=\frac{1}{\sqrt{a}}, \eta=0 \space \)

a ξ 2  + b n2  = 1  ⇒  a × (1/a) + 0 = 1    (Option 2 is correct)

From option 3

ξ = 0, η = 0

a ξ 2  + b n2  = 1  ⇒  a × (0) + 0 = 1   

a ξ 2  + b n2  ≠ 1        (Option 3  is incorrect)

From option 4

\(\xi=\frac{1}{\sqrt{a}}, \eta=\frac{1}{\sqrt{b}} \space \)

a ξ 2  + b n2  = 1  ⇒  a × (1/a) + b× (1/b) =  2 

a ξ 2  + b n2  ≠ 1        (Option 4  is incorrect)

Concept:

Probability of correct decoding (Pc) = Probability of at most 1 bit error

If n bit transmitted then probability of getting error in r bits \({ }^{n} C_{p} P^{r}(1-P)^{n-r}\)

Given:

Probability of bit error = ∈

∈ = 0.1 ⇒ Probability of correct decoding

Hamming code = (7, 4) 

Solution:

Probability of correct decoding (P_c) = Probability of atmost 1 bit error 

P_c = No error (or) Probability of 1 error

\(P_{c}={ }^{7} C_{0}(0.1)^{0}(1-0.1)^{7-0}+{ }^{7} C_{1}(0.1)^{1}(1-0.1)^{7-1}\)

P_c = 0.9⁷ + 7 × 0.9⁶ × 0.1 = 0.8503

The given op-amp is inverting type, hence the gain is given as:

\(\frac{V_o}{V_{in}}=-\frac{2000}{1000+\frac{1}{jω\times10^{-6}}}\)

\(\frac{V_0}{V_{in}}=\frac{-2}{1+\frac{1}{\frac{jω}{1000}}}\)

At ω = ∞

\(Gain=\frac{V_0}{V_{in}}=-2\)

At ω = 0

\(Gain=\frac{V_0}{V_{in}}=0\)

Hence the given op-amp is High Pass Filter.

Now, the  3 dB Cut-off frequency is given as:

\(ω_c = \frac{1}{R_1 C_1} = \frac{1}{10^3 \times 10^{-6} } \)

ωc = 1000 rad/sec

Hence, option (b) and (c) is correct.

The correct answer is option 2 and option 3.

Concept:

Boolean function:

The variables used in Boolean Algebra only have one of two possible values, a logic “0” and a logic “1” but an expression can have an infinite number of variables all labeled individually to represent inputs to the expression.

For example, variables A, B, C, etc, give us a logical expression of A + B = C, but each variable can ONLY be a 0 or a 1.

Explanation:

The given expression is = x + yz the equivalent expression is,

Option 1:  x + z + xy

= x + z + xy

= x(1+y)+ z  (∴ 1 + x = x + 1 = 1)

= x + z it is not equivalent x + yz.

Option 2:  (x + y)(x + z)

= (x + y)(x + z)

=x + xz + yx + yz

=x(1+ z+ y) + yz  (∴1 + x = x + 1 = 1)

= x + yz it is equivalent x + yz.

Option 3:  x + xy + yz

= x + xy + yz

= x(1 + y) +yz

= x + yz  it is equivalent x + yz.

Option 4:  x + xz + xy

= x + xz + xy

= x(1+z+y) (∴1 + x = x + 1 = 1)

= x it is not equivalent x + yz.

Hence the correct answer is option 2 and option 3.

Correct answer is option (3)

Explanation:

Option (1):

• Noise is an undesirable signal that is superimposed upon the normal operating signal.
• Noise margin the maximum tolerable noise voltage that can be added so that the circuit that does not cause an undesirable change in the circuit 's output.

​

       

NML = VIL - VOL 

NM_H = V_OH - V_IH

If the size of both transistors ( pMOS and nMOS) are equal

Under the condition:

V_TN = V_TP

K_n = K_P 

V_in = \(\frac{V_{DD}}{2}\)

in such case : N_ML = NMH 

If the size of both transistor are unequal, then \(K_{N}\neq K_{P}\)

When ; \(\uparrow \frac{K_{p}}{K_{N}}>1, \text{than } V_{IL}\uparrow, V_{IH}\downarrow\)

N_MH\(\downarrow\) and \(N_{ML}\uparrow\)

Now,

When ; \(\uparrow \frac{K_{p}}{K_{N}}<1, \text{than } V_{IL}\downarrow, V_{IH}\uparrow\)

\(N_{MH}\uparrow and \quad N_{ML}\downarrow\)

Hence, option (1) is incorrect.

Option (2):

Dynamic power consumption during switching is non zero due to capacitive loading of next stage.

Hence, option(2) is incorrect.

Option (3):

   

Logic High input: \(V_{DD}- |V_{TP}|<V_{in}<V_{DD}\)

In this region : pMOS → Cutoff 

nMOS → Linear 

Hence, option (3) is correct.

Option (4)

• For fast charging and discharging capacitor depends upon the mobility of the charge carrier.
• Mobility influences the switching speed of the NOT gate.

Hence, option (4) is incorrect.

Concept:

Channel capacity (Cs)- 

\(C_{s} = Max [I(X ; Y)] \)

Information(I) and Entropy(H) relationship - 

\(\begin{aligned} I(X ; Y) &=H(Y)-H\left(\frac{Y}{X}\right)~ \end{aligned}\)

Where,

\(H\left(\frac{Y}{X}\right)=-\sum_{i=1}^{3} \sum_{j=1}^{3} P\left(x_{i}, y_{j}\right) \log _{2} P\left(\frac{y_{j}}{x_{i}}\right)\)

Explanation:

\([P(X, Y)]=[P(X)]_{d} \cdot\left[P\left(\frac{Y}{X}\right)\right]\)

\([P(X, Y)]=\left[\begin{array}{ccc} 1-α & α & 0 \ 0 & 0 & 0 \ 0 & 0 & 0~ ~\end{array}\right]\)

So,

\(H\left(\frac{Y}{X}\right)=\left{(1-α) \log _{2}(1-α)+α \log _{2} α\right}\)

Now,

\(I(X ; Y)=H(Y)+(1-α) \log _{2}(1-α)+\log _{2} α\)

Channel capacity will be

\(C_{s} = Max [H(Y)+(1-α) \log _{2}(1-α)+\log _{2} α] \)

\(C_{s}=\log _{2} 3+\underbrace{(1-α) \log _{2}(1-α)+α \log _{2} α}\)

 

At, α = 0, 1 ⇒ H (x, y) is minimum and I(x ; y) is maximum.

So C_s will be maximum for  α interval is [0.25, 1]

Hence α = 1 is the correct answer.

Calculation:

When the diode is ON

 

Write KVL for Loop 1 

V_in - 2k I -0.7 - 2k I = 0

V_in = 4k I + 0.7   

\(I=\dfrac{V_{in}-0.7}{4k}\space \) →  (1)

Write KVL for Loop 2

Vin - 2k I -V_out  = 0

V_out = V_in - 2k I

From equation 1

\(V_{out} = V_{in} - 2k \times \dfrac{V_{in}-0.7}{4k}\space =\ \dfrac{V_{in}-0.7}{2k}\space \)

\(\dfrac {\partial{V_{out}}}{\partial V_{in}} = \dfrac {1}{2}\space \)

When diode is off

By voltage division rule

V_out = V_in (4k/6k) = (2/3) V_in

\(\dfrac {\partial V_{out}}{\partial V_{in}} = \dfrac {2}{3}\space \)

The ratio of the minimum to the maximum small-signal voltage gain = \(\dfrac{(\frac {1}{2})} {(\frac {2}{3})} = 0.75 \space \)

Given:

X is a discrete random variable , K possible distinct real values.

Analysis:

If 'K' symbols having equal probability:

Maximum Entropy  H(X)max = log2 (K).

If 'K' symbols having different probabilities:

H(X) < log2 (K)

So, H(X) ≤ log2 (K) Option (1) is correct.

As per given option (2):

H(X) ≤ H(2X)

\(\begin{aligned} &\begin{array}{|c|ccc|} \hline X \in\left{x_{i}\right} & -1 & 0 & 1 \ \hline P_{X}\left(x_{i}\right) & \frac{1}{4} & \frac{1}{2} & \frac{1}{4} \ \hline \end{array}\ \end{aligned}\)

\(H(X)=\sum_{i} P_{X}\left(x_{i}\right) \log _{2} \frac{1}{P_{X}\left(x_{i}\right)}\)

\(H(f) = \frac{1}{4} \log _{2} 4+\frac{1}{2} \log _{2} 2+\frac{1}{4} \log _{2} 4\)

\(H(X)=1.5 \frac{\text { bits }}{\text { symbol }}\)

Now , Y = 2X

 

 

\(H(2 X)=H(Y)=\sum_{i} P_{Y}\left(y_{i}\right) \log _{2} \frac{1}{P_{Y}\left(y_{i}\right)}=1.5 \frac{\text { bits }}{\text { symbol }}\)

For Y = 2X, distant X values results in distinct ‘Y’ values so that H(X) = H(Y). So, option (2) is correct.

As per given option (3):

H(X) ≤ H(X2)

Now , Y = X2

\(H\left(X^{2}\right)=H(Y)=\sum_{i} P_{Y}\left(y_{i}\right) \log _{2} \frac{1}{P_{Y}\left(y_{i}\right)}\)

\(H(f)=\frac{1}{2} \log _{2} 2+\frac{1}{2} \log _{2} 2=1 \frac{\text { bit }}{\text { symbol }}\)

Whereas,

\(\mathrm{H}(\mathrm{X})=1.5 \frac{\text { bits }}{\text { symbol }}\)

Hence option (3) is incorrect.

As per given option (4) :

H(X) ≤ H(2X)

Now , Y = 2X

Here distinct ‘X’ values results in distinct ‘Y’ values. So that,

H(X) = H(Y) i.e. H(X) = H(2X) so, Option (4) is correct.

NOTE: Option 1, 2, 4 are correct.

Concept:

One dimensional wave equation:

 \(\frac{{{\partial ^2}u}}{{\partial {t^2}}} = {C^2}\frac{{{\partial ^2}u}}{{\partial {x^2}}}\)

Two-dimensional wave equation:

 \(\frac{{{\partial ^2}u}}{{\partial {t^2}}} = \frac{{{\partial ^2}u}}{{\partial {x^2}}} + \frac{{{\partial ^2}u}}{{\partial {y^2}}}\)

Analysis:

The solution is given as:

F = f(x ± ct)

option 1 and 3 satisfies the given conditions.

Calculation:

Total matches = 5 + 7+ 8 + 25 + 20 + 17 + 8 + 4 + 3 + 2 + 1 = 100

Median is given by average of 50^th  and 51^th matches

50^th matches = 4 

51^th matches = 4 

Median = (4 + 4) / 2 

Median = 4

Concept:

Cauchy's residue theorem: Contour integral is given by 

2πi (sum of residue inside the contour)

Analysis:

Poles are given as 

z² - 1 = 0

z =  ± 1

z = 1 lies outside the contour 

Residue at z = -1

\(\frac{2^z}{(z-1)(z+1)}\times (z+1) |_{z = -1} \implies \dfrac {-1}{4} \space \)

\(\oint \frac{2^z}{z^2-1}dz=2\pi i(residue\space at\space z= -1)\space \)

\(\oint \frac{2^z}{z^2-1}dz=2\pi i(-\dfrac {1}{4})=-\dfrac {\pi i}{2}\space \)

Therefore 

A = 0.5

Calculation:

Given 

D̅ (x, y, z) = ( x³ î + y³ ĵ + xy² k̂ ) c\m²

Charge enclose given by

Q_(enclose) = \(\int_v ρ _v \ \ dv \ \ \)

By gauss law   ∇ . D = ρ_v

Q_(enclose) = \(\int _v \nabla. D\ \ dv \ \ \space \)

\(\nabla .D =\dfrac { \partial { x^3} }{\partial x} +\dfrac { \partial { y^3} }{\partial y}+ \dfrac { \partial { xy^2} }{\partial z \ \ }\space \)

 ∇ . D = 3x² + 3y² 

dv = dx dy dz 

Q(enclose) = \(\int _v (3x^2 + 3y^2) dx \ \ dy\ \ dz\ \ \space \)

Q(enclose) = \(\int _{x=-0.5}^{0.5}\ \ \int _{y=-0.5}^{0.5}\ \ \int _{z=-0.5}^{0.5} (3x^2 + 3y^2) dx \ \ dy\ \ dz\ \ \space \)

Q(enclose) = \(3\times{ ({\dfrac {x^3}{3}} )}_{-0.5}^{0.5}\times (y)_{-0.5}^{0.5}\times (z)_{-0.5}^{0.5}\ \ + 3 \times (x)_{-0.5}^{0.5} \times (\dfrac {y^3}{3})_{-0.5}^{0.5} \times (z)_{-0.5}^{0.5}\ \ \space \)

Q(enclose) = 0.25 + 0.25 

Q(enclose) = 0.5 C

Correct answer of the question is 0.

Solution:

Given that,       

 \(x_{1}(t)= e^{-t}u(t)\ x_{2}(t) = u(t)-u(t-2)\ y(t)= x_{1}(t)\ast x_{2}(t) \)                         

By applying Laplace Transform: 

\(x_{1}(t)= e^{-t}u(t)\leftrightarrow X_{1}(s)= \frac{1}{s+1}\ x_{2}(t) = u(t)-u(t-2)\leftrightarrow X_{2}(s) = \frac{1}{s}-\frac{e^{-2s}}{s}=\frac{1}{s}(1-e^{2s})\ y(t)= x_{1}(t)\ast x_{2}(t)\leftrightarrow Y(s)= X_{1}(s)\times X_{2}(s)\ \ \ Y(s)= \frac{1-e^{-2s}}{s(s+1)}\)                          

By applying Final Value Theorem:

\(\begin{align*} \lim _{t\rightarrow \infty }f(t)=\lim _{s\rightarrow 0}s.F(s)\ =\lim _{s\rightarrow 0}\frac{s(1-e^{2s})}{s(s+1)}\ =\frac{(1-1)}{1} = 0 \end{align*} \)                                                                  

Hence, the answer to the question is 0.

Solution:

Quantization noise power for positive values of x,

(QNP)_P = E[(x-α)²]

\(\begin{aligned} &=\left.\frac{(x-α)^{3}}{3} \cdot \frac{1}{2}\right|_{0} ^{1} \ \end{aligned}\)

\(\begin{aligned} &=\frac{1}{6}\left[(1-α)^{3}-(-α)^{3}\right] \ \end{aligned}\)

\(\begin{aligned} &=\frac{1}{6}\left[(1-α)^{3}+α^{3}\right] \ \end{aligned}\)

\(\begin{aligned} &=\frac{1}{6}\left[1-α^{3}+3 α^{2}-3 α+α^{3}\right] \ \end{aligned}\)

\(\begin{aligned} &=\frac{1}{6}\left[1+3 α^{2}-3 α\right] \ \end{aligned}\)

Now , taking differentiation on both side with respect to α .

\(\frac{d}{d α} (Q N P)_P=0+6 α-3=0\)

α = 1/2

α = 1/2 =0.5*

Quantization noise power for negative values of x,

(QNP)_N = E[(x-β)2]

\(\begin{aligned} &=\int_{-2}^{0}(x-β)^{2}\left(\frac{1}{4} x+\frac{1}{2}\right) d x \ \end{aligned}\)

\(\begin{aligned} &=\frac{1}{4} \int_{-2}^{0}(x-β)^{2} x d x+\frac{1}{2} \int_{-2}^{0}(x-β)^{2} d x \ \end{aligned}\)

\(\begin{aligned} &=\frac{1}{4}\left[\int_{-2}^{0}\left(x^{2}+β^{2}-2 β x\right) x d x\right] +\frac{1}{2}\left[\int_{-2}^{0}\left(x^{2}+β^{2}-2 β x\right) d x\right] \ \end{aligned}\)

\(\begin{aligned} &=\frac{1}{4}\left[\frac{\left(x^{4}\right)_{-2}^{0}}{4}+\frac{β^{2}}{2}\left(x^{2}\right)_{-2}^{0}-\frac{2}{3} β\left(x^{3}\right)_{-2}^{0}\right] +\frac{1}{2}\left[\frac{8}{3}+2 β^{2}+4 β\right] \end{aligned}\)

\(=\frac{1}{4}\left[-4-2 β^{2}-\frac{16}{3} β\right]+\frac{1}{2}\left[\frac{8}{3}+2 β^{2}+4 β\right]\)

Now , taking differentiation on both side with respect to β .

\(\frac{d}{d β} (Q N P)_N=-4 β-\frac{16}{3}+2 β+4=0 ~\)

2β = -4/3

β = -2/3

β = -2/3 = - 0.667*

So,

α* -β* = 0.5 - (- 0.667)

α -β = 1.167**

Calculation:

Given 

V_G = 2, C_Ox = 1.7 μF/cm², V_c = 4v, Q_in = 2.2  μC/cm2

Q_in can be given as

Q_in = C_ox( V_GS - V_T) 

2.2 μ  = 1.7 μ  ( 2 - V_T) 

V_T = 0.706 v

Q_in at V_G = 4 v 

Qin = Cox ( VGS - VT) 

Qin = 1.7 × 10^-6 ×(  4 - 0.706) 

Q_in = 5.5998 μ C/cm²

Q_in ≈ 5.6 μC/cm²

Concept:

Bit Rate = = symbol rate × Number of bits per symbol

Given:

QPSK and 16 QAM

Symbol rate = 1 mega-symbol/sec

Solution:

If QPSK-

Bit-rate = symbol rate × No. of bits per symbol

Bit-rate = 1 M × 2 = 2 Mbps

If 16-QAM-

Bit rate = 1 M × 4 = 4 Mbps

∴ QPSK and 16-QAM are used alternately-

Effective data rate (Avg.) = (2+4)/2 = 3 Mbps

Calculation:

f(x) = 8 loge x - x2 + 3

f ' (x) = 8/x - 2 x 

For Stationary points  f ' (x) = 0

\(\dfrac {8}{x} - 2x = 0 \ \ \)

x = ± 2 

\(x =\ \ 2 \in [1 \ \ e] \ \ \)

\(f '(x)\ \ =\ \ {( -\dfrac {8}{x^2}-2} )\ |_{at\ \ x = 2}\ \ = -4 \ \ \)

f '' (x) < 0 

f(x) is maximum at x = 2

f(1) = 2

f(e) = 8 - e² + 3 = 3.61

Minimum of f(x) in [1 , e] = min { f (1) , f (e) }

Minimum of f(x) in [1 , e] = min { 2 , 3.61 }

f(x) minimum  at x = 1

Analysis:

Given 

v₁^Tv₁ = 1,  v₁^Tv₂ = 0,  v₂^Tv₂ = 1

A = α v₁ v₁^T + β v₂ v₂^T

Post multiply by v₂ on both side 

A v2 = α v1 v₁^Tv2 + β v2 v₂^T v₂

A v₂ = β v₂

β  is a eigenvalue of A

Post multiply by v₁ on both side 

A v₁ = α v1 v1Tv1 + β v2 v2T v1

A v1 = α v1

α  is a eigenvalue of A

Let v₁ = [ a  b  c ]

\(v_1v_1^T=\begin{bmatrix}a\b\c\end{bmatrix} \begin{bmatrix}a&b&c\end{bmatrix}= \begin{bmatrix}a^2&ba&ca\ab&b^2&cb\ac&bc&c^2\end{bmatrix} \ \ \)

| v₁ v₁^T | = 0

similarly | v₂ v₂^T | = 0

v1 v₁^T and v₂ v₂^T both are singular matrices

A is also a singular matrix

| A | = 0 

Product of eigenvalues is zero

eigenvalues of A are 0, α ,β

Concept:

To find the system of equation for Ax = B

We define augmented matrix [A: B] 

For

Unique solution →  Rank [A] = Rank [A:B] = number of unknown

Infinite many solution →   Rank [A] = Rank [A:B]  <  number of unknown

No solution →  Rank [A]  ≠  Rank [A : B]

Analysis:

Define augmented matrix

[A : B] = \(\begin{bmatrix}1&-\sqrt2&3&1\\ -1&\sqrt2&-3&3\end{bmatrix}_,\)

R₂ = R₂ + R₁ 

[A:B] = \(\begin{bmatrix}1&-\sqrt2&3&1\\ 0&0&0&4\end{bmatrix}_,\)

Rank [A:B] = 2

Rank [A] = 1 

Rank [A:B] ≠ Rank [A]

Therefore No solution for x

Correct answer is option (1): R₁ = 2K , R₂ = 3k

Solution:

At ω= 0 rad/sec

• Capacitor will be open circuited ; \(X_{C}= \frac{1}{ω C}= ∞\)

Impedance, \(Z(j0)= R_{1}+ R_{2}\)

From the graph ; 

\(Z(j0)= 5K Ω\ R_{1}+R_{2}=5K Ω \)                                  

At ω= ∞ rad/sec:

Capacitor will be short circuited : \(X_{c}= \frac{1}{\omega C}= 0\) 

Impedance, 

\(Z(j\infty)= R_{1}= 2 KΩ\ \ R_{1}+R{2}= 5 KΩ\ \rightarrow R_{2}= 3KΩ\)

Hence, R₁ = 2 K Ω and R₂ = 3 K Ω

Correct answer is option (2): an inductor of 1H.

Explanation:

As per the given graph, the current through the element I(t) is lagging, it implies that the element is inductor.

\(I(t)= \frac{V(t)}{Z_{o}}\ \ I(t)= \frac{sin(t)}{Z_{o}}\)          

The maximum value of the current   \(I(t)_{max} = \frac{1}{\sqrt{2}}\)

\(|Z_{o}|=\frac{|V(t)|}{|I(t)|}\ =\frac{1}{\frac{1}{\sqrt{2}}}\ =\sqrt{2}\)

As per the circuit given,

Z_o = R + jωL

\(|Z_{o}|=\sqrt{R^{2}+(j ω L)^{2}}\)

\(\sqrt{2}=\sqrt{1+ ω^{2}L^{2}}\)

Given, ω = 1 rad/sec

\( \sqrt{2} = \sqrt{1+L^{2}}\)

2 = 1 + L²

L = 1 H

Correct answer is option (c)

**Concept:**​

Mode of operations of nMOSFET

• If, V_DS > (V_GS - V_T)   nMOSFET is operating in Saturation mode.
• If, VDS < (VGS - VT)   nMOSFET is operating in Linear mode.
• Here, ​ VDS  is Drain to source voltage , VGS  is Gate to Source voltage and V_T is the required Threshold voltage.

​Expressions of Drain to Source current in different mode of operation.

​Linear mode:

 \(I_{D} =μ_{n}C_{ox}\frac{W}{L}{(V_{GS}-V_{T})V_{DS}-\frac{1}{2}V_{DS}^{2}}\)

Saturation mode:

 \(I_{D} =μ_{n}C_{ox}\frac{W}{2L}(V_{GS}-V_{T})^{2}\)

 

Calculation:

Given,

L= 10 μm ,

W= 100 μm , 

μ_nC_ox= 1 mA/V²,

V_T= 1V,

V_DS = 1V

V_GS= [2 - sin(2t)]

V_GS(max) = 3 v

V_GS(min)= 1 V

here, VDS > (VGS - VT) condition will not be satisfied

Hence, nMOSFET is in Linear region

\(I_{D_{max}} =\mu_{n}C_{ox}\frac{W}{L}{(V_{GS}-V_{T})V_{DS}-\frac{1}{2}V_{DS}^{2}}\)

\(=1 mA/V^{2}\times 10[(3-1)1- \frac{1}{2}\times1)]\)

= 10[1.5] mA = 15 mA

Hence, the maximum value of the drain-to-source current is 15 mA

Correct answer is option (3)

Solution:

If, V_O = 15V

D₁ is ON

D₂ is OFF

\(V_{UT}= \frac{15 \times R}{R + 2R}=5V\)

Capacitor will charge up to V_UT 

\(V_{Cmax}= V_{UT}= 5 V\)

If, VO = -12V

D1 is OFF

D2 is ON

\(V_{UT}= \frac{-12 \times 2R}{R + 2R}=-8V\)

Capacitor will discharge up to V_LT 

\(V_{Cmin}= V_{LT}= -8V\)

 Difference between the maximum and the minimum values of the capacitor voltage (Vc):

 

\(V_{Cmax}-V_{Cmin}= 5-(-8)=13V\)

                                                                    

Hence, answer to the question is 13 V.

Solution:

Correct answer is option(1)

At ω=0

Gain of the amplifier is maximum,

20log₁₀(|A_Vmax|) = 12 dB

|A_Vmax| = 3.98

|A_Vmax| ≈ 4                                 .

Capacitor will be open circuited

​\( Gain=(1+\frac{R}{1k})\ (1+\frac{R}{1k})=4\ R=3K \ \)                                                   

As per the Bode plot corner frequency will be at,

log₁₀(ω_c) = 3

ωc = 1000 rad/sec

ω_c = 1/R₃C

C = 1/(R₃ × ω_c) 

= 1/10⁶

= 1 μF                                                

Hence, R = 3K and C= 1 μF.

The correct answer is option 1.

Concept:

2-bit Counter:

The 2-bit Counter With two T flip-flops in a chain you build a 2-bit counter. The output of the first flip-flop is connected to the input of the second flip-flop.

Explanation:

The given 2 bit counter,

The 2-bit count 

LSB 0, 1, 0, 1

For JK flip-flop (FF), 00 will not change the state.

So, output frequency = f₀/2  

Two-time change of state and duty cycle = 50%.

The output frequency is:

The output frequency is= (f₀/2 )/2

The output frequency is= f0/4.

Hence the correct answer is frequency is f0/4 and duty cycle is 50%.

It is a MSQ question.

Correct answers would be option(2) and option (4)

Concept:

To check whether the given expression converges or not we will apply Ration test and P test.

Ratio Test:

Let an is the given expression.

Let \(L=\lim_{n \mapsto \infty }|\frac{a_{n+1}}{a_{n}}|\)

if L<1, then the series Converges

if L>1, then the series Diverges

if L<1, then the test is Inconclusive.

P series test:

 \(\sum_{n=1}^{\infty }\frac{1}{n^{p}}= \frac{1}{1^{p}}+\frac{1}{2^{p}}+\frac{1}{3^{p}}+..... \) where p> 0 by definition

If p > 1 , then the series converges. If 0 < p <= 1 , then the series diverges.

Calculation:

Option (1): Substitute c= 1, d= -1

Perform  test:

\(\sum_{n=1}^{\infty }\frac{n^{d}}{c^{n}}\ =\sum_{n=1}^{\infty }\frac{n^{-1}}{1^{n}}\ =\sum_{n=1}^{\infty }\frac{1}{n}\)   

As, p=1 , hence series will diverge.

Hence, option (1) is incorrect.

Option (2):

Substitute c=2, d= 1.

Perform Ration test

\( \lim_{n \mapsto \infty }|\frac{a_{n+1}}{a_{n}}|=\lim_{n \mapsto \infty}\frac{n+1}{2^{n+1}}\times \frac{2^{n}}{n}=\frac{1}{2}\)

​​​​As, L < 1 , the series converges.

Hence, option (2) is correct 

Option (3): 

​Substitute c= 0.5, d= -10

Perform Ratio test;

\( \sum a_{n}= \sum \frac{n^{-10}}{(0.5)^{n}}\ \lim_{n \mapsto \infty }|\frac{a_{n+1}}{a_{n}}|=\lim_{n \mapsto \infty}\frac{(n+1)^{-10}}{(0.5)^{n+1}}\times \frac{0.5^{n}}{n^{10}}=\frac{1}{0.5}=2\)

As L>1, the series will diverge.

Hence, option (3) is incorrect.

Option (4):

​Substitute c= 1, d= -2

Perform P test:

• \(\sum a_{n}= \sum \frac{n^{-2}}{(1)^{n}}=\sum\frac{1}{n^{2}}\)
• Here, p > 1 , therefore the series converges.
• Hence, option (4) is correct.

Concept:

For an LTI system, the following results hold true:

• The same form of the sinusoidal signal is produced at the output.
• The only change will be in magnitude and phase.
• ∴ If a system produces frequencies in the output other than those which are in input, then that system can’t be called as the ‘Linear shift-invariant’ system.

 

Analysis:

\(\sin t \rightarrow \boxed{S_1} \rightarrow \sin (-t) = - \sin t\)

\(\sin t \rightarrow \boxed{S_2} \rightarrow \sin (t + 1)\)

\(\sin t \rightarrow \boxed{S_3} \rightarrow \sin (2t )\)

\(\sin t \rightarrow \boxed{S_4} \rightarrow \sin^2 (t ) = \frac{1 - \cos 2t}{2}\)

Since, LTI system does not change the frequency of sinusoidal input,

So S₃ and S₄ are definitely not LTI as input and output sinusoidal frequencies are different.

Remark: It is a MSQ question.

Correct answers are option (1) and option (3)

Explanation:

Option(1):

• An intrinsic semiconductor is an undoped semiconductor.
• For an intrinsic semiconductor, the concentration of electrons in the conduction band is equal to the concentration of holes in the valence band at thermal equilibrium.
• Hence, option (1) is correct.

Option(2):

• ​Collector region is generally lightly doped than base region in BJT
• Hence, option (2) is incorrect.

Option(3):

• The total current due to injected minority carrier ( diffusion current ) and majority carrier ( drift current ) is spatially constant in two terminal electronic device, however they individually will be spatially different under dark and steady state condition.

• Hence, Option(3) is correct.

• ​Option(4):

​Mobility of electrons always decreases with temperature in Silicon beyond 300 K.

Hence, option(4) is incorrect​

The correct answer is option 1 and option 3.

Concept:

The given data, 

A state transition diagram with states A, B, and C, and transition probabilities p1, p2, …, p7

From state A outgoing transitions are,

P7 A, P1 B, P4 C are equal to 1.

P1 + P4 + P7 =1

From state B outgoing transitions are,

P2 A, P3 B are equal to 1.

P2+P3=1

From state C outgoing transitions are,

P6A, P5C are equal to 1.

P5+P6=1

Option 1: p2 + p3 = p5 + p6

True, P2+P3=P5+P6

From State B and State, C are the same number of transitions and those are similar.

Option 2: p1 + p3 = p4 + p6

False, P1+P3=P4+P6

We can not equalize those probabilities and are not similar or not equal.

Option 3: p1 + p4 + p7 = 1

True, p1 + p4 + p7 = 1 

From state A we can find the sum of all probabilities is equal to 1.

Option 4: p2 + p5 + p7​ = 1

False, p2 + p5 + p7​ = 1

We can not equalize those probabilities and are not similar or not equal.

Hence the correct answer is option 1 and option 3.

The correct answer is option 1 and option 3.

Concept:

The given function is,

y=A+B' = (A'B)'

f(A,B)= A+B'

As 0 and 1 are available.

f(0,B)= A+B'= B'= NOT Gate.

f(A,B')= A+(B')' = A+B= OR Gate.

f(o,f(A,B'))=0 +(A+B)'= NOR Gate.

f(A',B)=A'+B'= (A.B)' =NAND Gate.

f(0,f(A',B)) = 0+((A.B)')'= A.B= AND Gate.

With the combination of OR and NOT, NOR gate can be implemented. Since NOR gate is universal logic gate, so all the functions can be implemented. 

Hence, we can implement NOT Gate we can also implement OR, AND, NOR and NAND Gate.

Hence the correct answer is option 1 and option 3.

Calculation:

Given 

\(\rm G_1(s)=\frac{10}{s^2+s+1} \ \ \)  &  \(\rm G_2(s)=\frac{10}{s^2+s√{10}+10} \ \ \)

Comparing standard second-order equation 

S2 + 2 ζ wn S + Wn2 = 0

Wn1 = 1,  2 ζ1 Wn1  = 1

ζ1 = 0.5

Wn2 = √ 10

2 ζ2 Wn2 = √ 10

ζ2 = 0.5

Percentage peak overshoot:

% Mp = \(\dfrac {e^{-\pi ζ }}{\sqrt {1-ζ ^2}}\ \ \)

Percentage peak overshoot is depend only on ζ 

ζ1 = ζ2 

Therefore % Mp is the same as for both y1(t) and y2(t)         (option 1 is correct)

Steady-state value: 

  \(lim _{s→ 0}~~sY(s)\ \ \)

\(\rm Y_1(s)=\frac{10}{s(s^2+s+1)} \ \ \) and  \(\rm Y_2(s)=\frac{10}{s(s^2+s√{10}+10)} \ \ \)

Steady-state value of y1(t) →  10

Steady-state value of y1(t) →  1 

Therefore the steady-state values of both systems are different          (option 2 is incorrect)

 

Damped frequency of oscillation (Wd): 

  \(W_d=W_n\sqrt {(1-ζ ^2}\ \ \ \ \)

\(W_{d1}=W_{n1} \sqrt {1-ζ_1 ^2}\ \ \ \ \)

\(W_{d1}=1 \sqrt {1-\dfrac {1}{4}}\ \ =\dfrac {\sqrt 3}{2}= 0.86\ \ \ \ \)

\(W_{d2}=\sqrt{10} \sqrt {1-\dfrac {1}{4}}\ \ =\sqrt {10} \times \dfrac {\sqrt 3}{2}= 2.73\ \ \ \ \)

Wd1 ≠ Wd2        (option 2 is incorrect)

Settling time (ts):

For 2% tolerance band 

\(t_s=\dfrac{4}{ζ W_n}\ \ \)

By putting values of ζ and Wn

\(t_{s1}= \dfrac {4}{\dfrac {1}{2}\times 1}=8 \ \ sec\ \ \ \ \)

\(t_{s2}= \dfrac {4}{\dfrac {1}{2}\times \sqrt {10}}=2.53\ \ sec\ \ \space \)

ts1 ≠ ts2     (option 4 is incorrect)

Concept:

Cut off frequency: Every waveguide has a cut off frequency below which there is no propagation.

For parallel plate cut off frequency is given as:

\(f_c= \dfrac {m c}{2 a} \ \ Hz \space \space \)

Calculation:

The mode that propagates at the lowest cut off frequency is TM1 mode

\(f_{{c}\ \ _{TM_1}}=\dfrac {c}{2a}=\dfrac {3× {10}^8}{2× {10}^{-4}}=1.5× {10}^{14}\ \ Hz \space \)

For propagation,

f > fc 

Therefore the frequency is greater than 1.5 × 1014 Hz can propagate in the waveguide

In the given option 6× 1015 Hz and 8 × 1014 Hz can propagate in waveguide.

Calculation:

To cover shaded region 

x → 0 to 4 

y → -x to x 

\(I = \rm \iint_D 3(x^2+y^2)\ dxdy \ \ \)

\(I=\int_{x=0}^{4} \int_{y=-x}^{y=x}3(x^2+y^2)dx\ \ dy\ \ \)

\(I=\int_{x=0}^{4} | ({3x^2}y+\dfrac{3y^3}{3})|_{y=-x}^{y=x} \ \ dx\ \ \)

\(I=\int_{x=0}^{4} (3x^2(2x)+2x^3)\ \ dx\ \ \)

\(I=\int_{x=0}^{4} 8x^3\ \ dx\ \ \)

\(I= \left|\dfrac{8x^4}{4}\right|_{x=0}^{4}\ \ \)

I= 2 × 4⁴ = 512

Calculation:

First, find V_th and  R_th From the graph 

From the graph

\(V_2= \dfrac {-1}{4m}I_2+5\space \)

When I₂ = 0 

V₂ = V_osc = V_th = 5

Replace network with its Thevenin's equivalent and variable resistance with load

\(I_2=\dfrac {10-5}{1250}=4\ \ mA\space \)

Concept:

By duality property

 \(x(n)\xrightarrow [ ] {DFT}\ \ \xrightarrow [] {DFT} Nx(-k)\space \)

Calculation:

Given 

X̅  = DFT x(n) =  X[k] = [1, 0, 0, 0, 2, 0, 0, 0] 

\(\rm X[k]=\sum_{n=0}^{7}x[n]\ exp\left(-j\frac{2\pi}{8}nk\right) \space \)

Therefore we hane to find 

y̅ = (DFT (x̅ )) = DFT (DFT x(n))

By duality property

\(x(n)\xrightarrow [ ] {DFT}\ \ \xrightarrow [] {DFT} Nx(-k)\space \) 

DFT (DFT x(n)) = y̅  = N x(-k)

X(k) = [1, 0, 0, 0, 2, 0, 0, 0] 

X(-k) = [1, 0, 0, 0, 2, 0, 0, 0] 

y(n) =  N X(-k) = 8 [1, 0, 0, 0, 2, 0, 0, 0] 

y(0) = 8

Solution:

We know; \(Generation ,rate =Recombination ,rate = \frac{\delta}{\tau} \)

\(\delta_{n}(x) = R\tau_{n}=10^{20}e^{-\frac{|x|}{l_{n}}}\tau_{n}\ at ,x=l_{n}\ \delta_{n}(l_{n}) = R\tau_{n}=10^{20}e^{-1}\tau_{n} ; ; ......(i)\)

          

For \(l_{n}\leq x\leq 2l_{n}\) :

Continuity equation in the steady state is; 

\(D_{n}\frac{\partial^2 \delta_{n}}{\partial x^2}+ G-R=0\)

Here, G = R = 0               

\(D_{n}\frac{\partial^2 \delta_{n}}{\partial x^2}=0\)​                                 

Solution of the differential equation will be in the form of 

\(\delta_{n}(x) =Ax+B\)                                      

Since;

 \(At,x = 2l_{n}: \quad \delta_{n}(2l_{n})=0 (given)\ \quad \quad \quad \quad \quad \quad \quad 0=A(2l_{n})+B\ \quad \quad \quad \quad \quad \quad \quad A=- \frac{B}{2l_{n}}\ \quad \quad \quad \quad \quad \quad \therefore \delta_(n)= -\frac{B}{2l_{n}}x+B=B(1- \frac{x}{2l_{n}}) \quad \quad \quad \cdots(ii)\ \therefore At,x=l_{n}: \quad \quad equation(i)=equation(ii)\ \ \quad \quad \quad 10^{20}e^{-1}\tau_{n}=B(1- \frac{l_{n}}{2l_{n}})\ \therefore B= 2 \times 10^{20}e^{-1}\tau_{n}\ \therefore \delta_{n}(x)= 2 \times 10^{20}e^{-1}\tau_{n}(1-\frac{x}{2l_{n}})\ \ \ \therefore Electron , diffusion , current , density:\ |J_{n}|_{diff}= qD_{n}\frac{\mathrm{d} n}{\mathrm{d} x}=qD_{n}\times 2 \times10^{20}\times e^{-1} \times \tau_{n}(0- \frac{1}{2l_{n}})\ =\frac{1.6 \times 10^{-19}\times l_{n}^{2} \times 2 \times 10^{20} \times e^{-1}}{2l_{n}}\quad (l_{n}=10^{-4}cm)\ =0.588 mA/cm^{2}\ \approx 0.59 mA/ cm^{2} \)

Hence, answer to the question is 0.59 mA/cm²

Concept:

The analog output of a DAC is given as 

Analog output = R× D× Gain 

Where

D → decimal equivalent of a binary number

Resolution of a DAC (R): \(Resolution \ \ = \dfrac {V_r}{2^n}\space \)

V_r → Reference voltage, n → number of bits

Calculation 

Given 

V_r = 1.6

n = 4 

\(Resolution \ \ = \dfrac {V_r}{2^n}\space = \dfrac {1.6}{2^4}\space =0.1\space \)

From the DAC 

Binary input = 0101

The decimal equivalent of 0101 is 5

Replacing circuit with R_th and V_th 

 

Gain = -1

Analog output (V₀ ) = R× D× Gain  =  (0.1) ×  5 × (-1) = -0.5

Calculation:

Given

First, take V_A as output 

This is a common source amplifier with R_s

Therefore

Voltage gain  →  \(\dfrac {V_A}{V_{in}}=\dfrac {-g_mR_D}{1+g_mR_s} \space \)→  (1)

Taking V_B as an output 

We are applying input through the gate and taking output through the source therefore it is a common drain amplifier

The voltage gain of a common drain amplifier is given as

Voltage gain →  \(\dfrac {V_B}{V_{in}}=\dfrac {g_mR_s}{1+g_mR_s} \space \) →  (2)

\(\dfrac{V_A}{V_B}={\dfrac {V_A}{V_{in}}}\times \dfrac {V_{in}}{V_B}\space \)

From 1 and 2 

\(\dfrac {V_A}{V_B}=\dfrac {-g_mR_D}{1+g_mR_S}\times \dfrac{1+g_mR_S} {g_mR_S} \space = \dfrac {-R_D}{R_S}\space \)

\(\dfrac {V_A}{V_B}=\dfrac {-4}{2}= -2\space \)

Calculation:

Given 

r(t) = 0

e(t) = r(t) - y(t)

e(t) = - y(t)

E(s) =  - Y(s) →  (1)

Circuit can be redraw as 

Y(s) = [ 10 E(s) + D(s) ] × \(\dfrac {1}{s(s+10)}\ \ \)

From  (1)

• E(s) =  [ 10 E(s) + D(s) ] × \(\dfrac {1}{s(s+10)}\ \ \)

For unit step disturbance d(t) 

D(s) = 1/ s 

E(s) =  - [ 10 E(s) + \(\dfrac {1}{s} \ \ \) ] × \(\dfrac {1}{s(s+10)}\ \ \)

\(E(s) = \dfrac {-1}{10s+s^2(s+10)}\ \ \)

steady-state error is given as

\(lim_{s\rightarrow 0 } \ \ sE(s)=lim_{s\rightarrow 0} \dfrac {-s}{10s+s^2(s+10)}=\ \ - 0.1\ \ \space \)

Calculation:

Given

\(\rm f_{Y|x_A}(y)=e^{-(y+1)}u(y+1)\ \ \)

\(\rm f_{Y|x_B}(y)=e^{(y-1)}(1-u(y-1))\ \ \)

The conditional Pdf's can be drawn as 

From ML  decoding

If 

\(\rm f_{Y|X_A}(y)>\rm f_{Y|X_B}(y)\rightarrow decision \ \ in \ \ favour \ \ of \ \ X_A\)

\(\rm f_{Y|X_A}(y)<\rm f_{Y|X_B}(y)\rightarrow decision \ \ in \ \ favour \ \ of \ \ X_B \ \ \)

For region

-1 < y < 0  & 1 < y < - \(\infty \)  

\(\rm f_{Y|X_A}(y)>\rm f_{Y|X_B}(y)\rightarrow decision \ \ in \ \ favour \ \ of \ \ X_A\)

• \(\infty \)< y < -1 & 0 < y < 1 

\(\rm f_{Y|X_A}(y)<\rm f_{Y|X_B}(y)\rightarrow decision \ \ in \ \ favour \ \ of \ \ X_B \ \ \)

The probability of error can be given as

P_e = P(X_A) . PeXA   + P(X_B) PeX_B 

When X_A is transmitted error occurred in 0 < y < 1 only

Probability of error when X_A is transmitted PeXA 

\(P_{e_{X_A}}= \int _0^1f_{Y/X_A}(y)dy=\int _0^1e^{-(y+1)}u(y+1)dy\ \ \)

\(P_{e_{X_A}} = \int _0^1e^{-(y+1)}dy\ \ =| -e^{-(y+1)} |_0^1=e^{-1}-e^{-2}=0.23\ \ \)

When X Bis transmitted error occurred in -1 < y < 0 only

Probability of error when X_B is transmitted PeXB

\(P_{e_{X_B}}= \int _{-1}^0f_{Y/X_A}(y)dy=\int _{-1}^0e^{(y-1)}[1-u(y-1)]dy\ \ \)

\(P_{e_{X_B}} = \int _{-1}^0e^{(y-1)}dy\ \ =| e^{(y-1)} |_{-1}^0=e^{-1}-e^{-2}=0.23\ \ \)

Pe = P(XA) . PeXA   +  P(XB) PeXB 

P_e = P(XA) × 0.23 + P(X_B) × 0.23

P_e = 0.23  [ P(XA) + P(XB) ]

P_e = 0.23