GATE EC 2021 Official Paper (Previous Year Paper)

Given:

The present population of the city = 1,10,250

Increasing rate, (R) = 5% P.a.

Time, (n) = 2 years

Formula used:

Amount, (A) = P(1 + R/100)ⁿ

Here A is the present population of the city and P is the population of the city 2 years ago

Concept used:

This problem can be solved by the use of the compound interest formula

Calculation:

(A) = P(1 + R/100)n

⇒ 110250 = P(1 + 5/100)²

⇒ 110250 = P(105/100)²

⇒ 110250 = P(21/20)²

⇒ 110250 = P × (441/400)

⇒ P = (110250 × 400)/441

⇒ P = 100000

∴ The population of the city 2 years ago was 1,00,000

Concept:

(a + b)² = a² + b² + 2ab ---(1)

Calculation:

Given:

\(\frac{p}{q} \ + \ \frac{q}{p}=3\)

Squaring both sides:

\((\frac{p}{q} \ + \ \frac{q}{p})^2=9\)

Using the formula given in equation (1):

\(\frac{p^2}{q^2} \ + \ \frac{q^2}{p^2} \ + \ 2=9\)

\(\frac{p^2}{q^2} \ + \ \frac{q^2}{p^2} \ =7\)

Hence option (1) is the correct answer.

Shows the number of squares added so that P-Q becomes a line of symmetry.

Hence, 6 squares added to the figure.

The correct answer is Past, future.

Key Points

Nostalgia means a feeling of pleasure and also slight sadness when you think about things that happened in the past.

Anticipation means a feeling of excitement about something that is going to happen in the near future.

In short, nostalgia means thinking about the past and anticipation means thinking about the future.

Thus, Nostalgia is to anticipation as past is to future.

The correct answer is (i) and (iii).

Key Points

Wake means to stop sleeping; to make somebody stop sleeping.

The three forms of the verb wake:

• Present - Wake.
• Past - Woke.
• Past participle -  Woken/ waked.

 

​Thus, sentence (i) I woke up from sleep and sentence (iii) I was woken up from sleep is correct.

Hence, option (i) and (iii) follows.

The least possible Venn diagram for the given statements is as follows:

Conclusion I: Some black are purple → False (As there is no direct relation between purple and back)

Conclusion II: No black is purple → False (As there is no direct relation between purple and back)

Both conclusions are complementary to each other.

Hence, Either conclusion I or II follows.

Important Points 

Either and or conditions:

1. One conclusion is positive and one is negative.
2. The elements are the same in both the conclusions.
3. Individually both conclusions must be wrong.

Thus, Either I or II follows.

The correct answer is (i) and (iii).

Key Points

Look at the lines:

Computers are ubiquitous. 

• ​Ubiquitous means present, appearing or found everywhere. ​
• From this line, it can be deduced that Nowadays, computers are present in almost all places.
• ​They are used to improve efficiency in almost all fields from agriculture to space exploration. For humans, sitting in front of a computer for long hours can lead to health issues.

 

​Upon the perusal of the above lines, it can be concluded that computers help in increasing efficiency but can lead to health issues as well. Thus, For humans, there are both positive and negative effects of using computers.

Option (i) and (iii) can be found in the passage.

None of the other options follows.

Concept:

If the side of a square is a unit then its diagonal length will be:

\(D=\sqrt {a^2 \ + \ a^2}\)

D = √2a units  ---(1)

The volume of a cone is given by:

\(V=\frac{\pi r^2h}{3}\)  ---(2)

Where r =radius of the cone

h = height of the cone

Calculation:

When the square is cut along the main diagonal then the square will be divided into two equal triangles.

The length of the main diagonal can be calculated from equation (1)

D = √2 × 1 = √2  

Then one of the cut triangles is revolved about its short edge to form a solid cone:

According to figure

r = 1

h = 1

The volume of come can be calculated using equation (2):

\(V=\frac{\pi \ \times \ 1 \ \times \ 1}{3}\)

\(V=\frac{\pi}{3}\)

Hence option (2) is the correct answer.

Given:

Minimum 10% more

Formula used:

Percentage = (Obtained value/Maximum value) × 100

Concept used:

Here the fraction part is always the smaller number because comparing number is the smaller number

Calculation: 

 

∴ The number of the day in the given week in which one of the students spent a minimum of 10% more than the other students is 6 days

Formula used:

The area of the equilateral triangle = (√3/4) × a²

The area of the regular hexagon = (3√3/2) × b²

Here a → side of the equilateral triangle and b → Side of the regular hexagon

Concept used:

When the corner is cut from the equilateral triangle to produce the regular hexagon then the side of the triangle is divided into 3 equal parts and one part is the side of the hexagon.

Then, b = a/3

Calculation:

The area of the equilateral triangle = (√3/4) × a2

The area of the regular hexagon = (3√3/2) × b²

⇒ (3√3/2) × (a/3)2

⇒ (√3/6) × (a)2

The ratio of the regular hexagon to the equilateral triangle = {(√3/6) × (a)²}/{(√3/4) × a2}

⇒ 4/6 = 2 ∶ 3

∴ The required ratio is 2 ∶ 3

Concept:

∫_C F(r) dr can be solved by putting:

x = r cos θ

y = r sin θ

dx = -r sin θ dθ

dy = r cos θ dθ

Application:

Given r = 1

θ = 0 to 45°

The required integral can be written as:

\(\mathop \smallint \nolimits_C \vec F \cdot dr = \mathop \smallint \nolimits_0^{45^\circ } \left[ {\left( { - r\cos \theta } \right)\left( { - r\sin \theta } \right)d\theta + \left( {r\sin \theta } \right)\left( {r\cos \theta } \right)d\theta } \right]\)

\(\mathop \smallint \nolimits_0^{45^\circ } \left( {{r^2}\cos \theta \sin \theta + {r^2}\sin \theta \cos \theta } \right)d\theta \)

With r = 1, the above integral becomes:

\( = \frac{1}{2}\mathop \smallint \nolimits_0^{45^\circ } \left( {\sin 2\theta + \sin 2\theta } \right)d\theta \)

r = 1

\( = \mathop \smallint \nolimits_0^{45^\circ } \sin 2\theta ;d\theta \)

\( = \frac{{\left( { - \cos 2\theta } \right)_0^{45^\circ }}}{2}\)

\( = \frac{{ - 0 - \left( { - 1} \right)}}{2} = \frac{1}{2}\)

Concept:

\(\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right)\)

The solution of a linear differential equation of a general form shown above is:

y(I.F) = ∫Q(x) (IF) dx + C

Where:

IF = Integrating factor calculated as:

\(I.F = {e^{\smallint Pdx}}\)

Calculation:

Given:

\(\frac{{dy}}{{dx}} + \frac{x}{{1 - {x^2}}}y = x\sqrt y \)

The above differential equation is not in a general form. Converting it first in the general form of a linear differential equation, we divide the equation by √y to get:

\(\frac{1}{{\sqrt y }}\frac{{dy}}{{dx}} + \frac{x}{{1 - {x^2}}};\sqrt y = x\)

Let √y = u

\(\left[ {\frac{1}{{2\sqrt y }};;\frac{{dy}}{{dx}}} \right] = \frac{{du}}{{dx}}\)

\(\frac{1}{{\sqrt y }};\frac{{dy}}{{dx}} = 2\frac{{du}}{{dx}}\)

\(2\frac{{du}}{{dx}} + \frac{x}{{1 - {x^2}}}u = x\)

\(\frac{{du}}{{dx}} + \frac{1}{2}\left( {\frac{x}{{1 - {x^2}}}} \right)u = \frac{x}{2}\)

\(\frac{{du}}{{dx}} + P\left( x \right)u = Q\left( x \right)\)

\(IF = {e^{\smallint Pdx}}\)

\(IF = {e^{\smallint \frac{x}{{2\left( {1 - {x^2}} \right)}}dx}}\)

Let 1 – x2 = t

-2xdx = dt

\(xdt = \frac{{ - dt}}{2}\)

\(IF = {e^{\smallint - \frac{1}{t};\frac{{dt}}{4}}}\)

\( = {e^{\frac{{ - 1}}{4}\ln t}}\)

\( = {e^{\ln {t^{\frac{{ - 1}}{4}}}}}\)

\( = {t^{\frac{{ - 1}}{4}}}\)

\(IF = {\left( {1 - {x^2}} \right)^{\frac{{ - 1}}{4}}}\)

Concept:

Variance of a random variable ‘y’ is given by:

Var[y] = E[y²] – E²[y]

Properties of mean:

1. E[K] = K, Where K is some constant

2. E[c X] = c. E[X], Where c is some constant

3. E[a X + b] = a E[X] + b, Where a and b are constants

4. E[X + Y] = E[X] + E[Y]

Application:

Variance of y = E[(2x + 3)²] – (E[2x + 3])²

= E[4x² + 12x + 9] – (E[2x + 3])²

= 4E[x²] + 12E[x] + 9 – (E[2x] + E[3])²

= 4E[x2] + 12E[x] + 9 – (2E[x] + 3)2

= 4E[x2] + 12E[x] + 9 – (4E²[x] + 9 + 12E[X])

= 4E[x2] + 12E[x] + 9 – 4E2[x] – 9 – 12E[X]

= 4E[x2] – 4E2[x]

= 4[E[x²] – E²[x]]

This can be written as:

= 4 (variance of x), i.e.

The variance of y = 4 times the variance of x

\(\sigma _y^2 = 4;\sigma _x^2\)

Properties of Variance:

1. V[K] = 0, Where K is some constant.

2. V[cX] = c2 V[X]

3. V[aX + b] = a2 V[X]

4. V[aX + bY] = a2 V[X] + b2 V[Y] + 2ab Cov(X,Y)

Cov.(X,Y) = E[XY] - E[X].E[Y]

Concept:

Consider x(t) → f(kHz) signal

Then:

\(x\left( {\frac{t}{2}} \right);\mathop \to \limits^{;;is;;} ;\frac{f}{2}\left( {kHz} \right)\) signal

\(x\left( {\frac{t}{2} + 1} \right) \to \frac{f}{2}\) (kHz) signal; as shifting in the time domain does not affect the frequency spectrum.

\(x\left( t \right)x\left( {\frac{t}{2} + 1} \right) \to \left( {f + \frac{f}{2}} \right)kHz\)

\( = \frac{{3f}}{2}kHz\) signal

Since multiplication in one domain is the convolution in another domain, and in convolution, due to the time-invariant system, the duration of the signal gets added.

Hence the frequency of both signals gets added in the frequency domain.

More clearly, we can understand from the following specturem diagram:

Application:

\(y\left( t \right) = x\left( t \right) \cdot x\left( {\frac{t}{2} + 1} \right);;\mathop \leftrightarrow \limits^{;;F.T;;} ;y\left( f \right) = x\left( f \right)*{x_1}\left( f \right)\)

Given x(t) = 10 kHz signal

Then, \(X\left( {\frac{t}{2} + 1} \right) \Rightarrow 5;kHz\) signal

\(y\left( t \right) = x\left( t \right) \cdot x\left( {\frac{t}{2} + 1} \right)\)

∴ y(t) will be (10 + 5) = 15 kHz signal

Nyquist rate of y(t) = 2 [frequency of y(t)]

= 2 × 15 kHz

= 30 kHz

Concept:

Linear convolution = x[n] * h[n]

y[n] = x[n] * h[n]

Y[k] = X[k] H[k]

Also, the function which is circularly convoled can be written as:

z[n] = IDFT [x(k) y(k)]

i.e. z[n] = x[n] ⊗ h(n)

⊗ → circular convolution

We have 16 point sequence x(n) and h(n) but for the case of understanding and solving, we will

consider 4-point sequence as:

x(n) = {1, 1, 2, 2}

h(n) = {1, 2, 3, 4}

Linear convolution:

y(n) = {1, 3, 7, 13, 16, 14, 8}

Circular convolution:

z[n] = x[n] h(n)

\(\left[ {\begin{array}{*{20}{c}} 1&2&2&1\ 1&1&2&2\ 2&1&1&2\ 2&2&1&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1\ 2\ 3\ 4 \end{array}} \right]\)

z[n] = [15, 17, 15, 13]

z[3] = y[3]

Clearly, if N point sequence is given for x(n) and h(n), then:

\({\left[ {x\left( n \right)*h\left( n \right)} \right]_{at;n = N - 1}} = {\left[ {x\left( n \right)h\left[ n \right]} \right]_{at;n = N - 1}}\)

Application:

The given sequence is a 16 point sequence.

Hence for N = 16 – 1 = 15:

Z[k] = y [k]

Concept:

The number of minority carriers generated because of light is given by the generation rate as:

\(G = \frac{{excess;minority;carrier;generated}}{{minority;carrier;life;time}} \)

\(G= \frac{{{\rm{\Delta }}p}}{{\tau_p}} = \frac{{{\rm{\Delta }}n}}{{\tau_n}}\)

Analysis:

N_A = 10¹⁶ / cm³

G = 10²⁰ / cm³ – sec

τ = 100 μsec

ni = 10¹⁰ / cm³

before shining light:

Hole concertration (p) ≃ N_A = 10¹⁶ / cm³

Electron concentration (n) \(= \frac{{n{i^2}}}{p} = \frac{{{{10}^{20}}}}{{{{10}^{16}}}} = {10^4}/c{m^3}\) 

After illumination of light minority carrier will be generated.

So after light illumination hole conc. (p’) = p + Δp

p’ = 10¹⁶ + 10²⁰ × 10^-6 × 100

= 2 × 10¹⁶ / cm³

After illumination e^- conc. (n’) = n + Δn

= 10⁴ + 10²⁰ × 10^-6 × 100

≃ 10¹⁶ / cm³

Product of e^- and hole concertration = n’ × p’

= 2 × 10³² / cm^-6

Concept:

Electric field

It is defined as the ratio of the applied voltage to the total length of the semiconductor specimen.

\(E = \frac{V}{L}\)

Calculation:

Electric field developed will be calculated as:

\(E = \frac{{ - \partial V}}{{\partial x}}\)

\(V = \frac{{work;done}}{{charge;transferred}}=\frac{dW}{dq}\)

Work done is calculated in terms of energy used. From the energy diagram given the total energy is Δ and the charge is 'q'

\(V=\frac{\Delta}{q}\)

The electric field generated is:

\(\left| E \right| =\frac{V}{L} = \frac{{\bf{\Delta }}}{{qL}}\)

Concept:

Whenever gate and drain are shorted of MOSFET, we can replace directly MOSFET by resistance which is equal to

Analysis:

Since the gate and drain of MOSFET 1 are short-circuited.

It will work like a resistor.

→ drawing AC - equivalent model of MOSFET - 2

\({V_{gs_2}} = {V_{in}}\)

\({V_0} = - {g_{{m_2}}};{V_{g{s_2}}}\left{ {{r_{{0_2}}}||{r_{{0_1}}}||\frac{1}{{{g_{{m_1}}}}}} \right}\)

\(\frac{{{V_0}}}{{{V_{in}}}} = - {g_{{m_2}}}\left{ {{r_{{0_2}}}||{r_{{0_1}}}||\frac{1}{{{g_{{m_1}}}}}} \right}\)

Important Points

When gate and source of MOSFET are short-circuited:

Concept:

When V_IN is active and V_REF = 0, the circuit is drawn as:

\({V_{OUT1}} = - \frac{{{R_F}}}{{{R_{IN}}}}{V_{IN}}\)  

When V_IN = 0 and V_REF is active, the circuit is drawn as:

\({V_{OU{T_2}}} = \left( {\frac{{{V_{REF}} \times {R_2}}}{{{R_1} + {R_2}}}} \right)\left( {1 + \frac{{{R_F}}}{{{R_{IN}}}}} \right)\)

Total V_out = V_out1 + V_out2

\({V_{out}} = \left( {\frac{{{V_{REF}} \times {R_2}}}{{{R_1} + {R_2}}}} \right)\left( {1 + \frac{{{R_F}}}{{{R_{IN}}}}} \right) - \frac{{{R_E}}}{{{R_{IN}}}};{V_{IN}}\)

Given V_REF is fixed then V⁺ is fixed, i.e.

\({V^ + } = \frac{{{V_{REF}} \times {R_2}}}{{{R_1} + {R_2}}}\)

\({V_{OUT}} = \left( {1 + \frac{{{R_F}}}{{{R_{IN}}}}} \right)\left( {\frac{{{V_{REF}};{R_2}}}{{{R_1} + {R_2}}}} \right) - \frac{{{R_F}}}{{{R_{IN}}}}{V_{IN}}\)

For V_IN = 0.1 V, we have V_OUT = 1 V

\(1 = \left( {1 + \frac{{{R_F}}}{{{R_{IN}}}}} \right)\left( {\frac{{{V_{REF}};{R_2}}}{{{R_1} + {R_2}}}} \right) - \frac{{{R_F}}}{{{R_{IN}}}}\left( {0.1} \right)\)        ---(i)

For V_IN = 10 V, we have V_OUT = 6 V

\(6 = \left( {1 + \frac{{{R_F}}}{{{R_{IN}}}}} \right)\left( {\frac{{{V_{REF}};{R_2}}}{{{R_1} + {R_2}}}} \right) - \frac{{{R_F}}}{{{R_{IN}}}}\left( 1 \right)\)      ---(ii)

From (i)

\(\left( {1 + \frac{{{R_F}}}{{{R_{IN}}}}} \right)\left( {\frac{{{V_{REF}} \cdot {R_2}}}{{{R_1} + {R_2}}}} \right) = 1 + \frac{{{R_F}}}{{{R_{IN}}}}\left( {0.1} \right)\)

Putting above value in eq (ii)

\(6 = 1 + \frac{{{R_F}}}{{{R_{IN}}}}\left( {0.1} \right) - \frac{{{R_F}}}{{{R_{IN}}}}\left( 1 \right)\)

\(5 = \frac{{{R_F}}}{{{R_{IN}}}}\left( { - 0.9} \right)\)

\(\frac{{{R_F}}}{{{R_{IN}}}} = - \frac{5}{{0.9}} = - 5.55\)

Since only positive values are given in the options:

\(\frac{{{R_F}}}{{{R_{IN}}}} = 5.55\)

Concept:

In the below circuit  V₁ is connected to the ground, so V₁ = 0.

Thus V₂ also will be at ground potential.

Note:

1. This concept is valid only when negative feedback is applied to op-amp like in inverting amplifiers

2.When op-amp gain has a low value then also this concept is not valid.

Analysis:

Using virtual ground concept:

V_A = V_B = 0 → ground

Applying KVL at node A:

\(\frac{{{V_i} - 0}}{R} + \frac{{ - {V_{Ref}} - 0}}{R} = \frac{{0 - {V_0}}}{{{R_F}}}\)

V₀ = 0

VIN = V_REF

(1235)_x = (3033)_y

Converting both the RHS and LHS into its decimal equivalent, we get:

(x³ + 2x² + 3x + 5)₁₀ = (3y³ + 0y² + 3y + 3)₁₀

Substituting the values given in the options in above equation, we get:

x = 8 and y = 6

Proof:

Putting x = 8 in x³ + 2x² + 3x + 5

= (8)³ + 2(8)² + 3(8) + 5

= (669)₁₀

Putting y = 6 in 3y³ + 0y² + 3y + 3

= 3(6)³ + 3(6) + 3

= 3(216) + 18 + 3

= (669)₁₀

Concept of memory:

Any memory size is given by = 2^K × m

K = address line

m = data line

Eg: 1 KB memory = 2¹⁰ × 8

Concept of decoder:

For n × 2ⁿ decoder no. of AND gates required are 2ⁿ.

Eg: 2:4 decoder,  4 AND gates are required.

Analysis:

Given

For this memory 15 × 2¹⁵ decoders required

Since n = 15

So no. of AND gate are required = 2¹⁵

Concept:

Mason's Gain Formula is used to evaluate an overall transmittance (gain), which can be expressed as,

\(T = \frac{{\sum {P_k}{{\rm{\Delta }}_k}}}{{\rm{\Delta }}}\)

Where

Pk = forward path transmittance of kth path

Δ = graph determinant comprising closed-loop transmittances & mutual interactions between non-touching loops.

ΔK = path factor consisting of all isolated closed loops from the forward path in the graph.

Analysis:

No of forward path = 2

Forward paths:- G₂, G₁

No of Loops = 1

Loop: -G₁H

Using mason gain formula:

\(\frac{{Y\left( s \right)}}{{X\left( s \right)}} = \frac{G_1+ G_2}{{1 + G_1 H }}\)

Note:-

Loop:- G₁H is touching the forward path G₂, hence option 2 is incorrect.

We will observe the encirclement around the origin: (for open loop transfer function → G(s)H(s)= 0 )

N = P - Z

where,

P = the no. of open-loop poles lie in the right half of plane

Z = the no. of closed-loop poles or open-loop zeros lie in the right half of plane

N = no. of encirclement about (-1 + j0) in G(S) H(S) plane

OR,

no. of encirclement about the origin in 1 + G(S) H(S) plane.

Given;

Z = 1 (for open loop system)

N = 2 (at the origin in ACW direction)

then

2 = P - 1

∴ P = 3 

For the analysis of closed loop system:

observe the encirclement around the (-1 , j0) [∵ 1 + GH = 0 → GH = -1]

N = 0

P = 3

Then,

N = P - Z

Z= P - N

∴ Z= 3 - 0 = 3

There are 3 poles in the right half-plane.

Concept:

Normal Incidence is as shown:

The Reflection coefficient is defined as:

\({\rm{\Gamma }} = \frac{{{E_r}}}{{{E_i}}}\)

\({\rm{\Gamma }} = \frac{{{\eta _2} - {\eta _1}}}{{{\eta _2} + {\eta _1}}}\)

Where:

η₁ = intrinsic impedance of medium 1

η₂ = intrinsic impedance of medium 2

Application:

Given medium 1 = free space

ϵ₁ = ϵ₀ and μ₁ = μ₀

Medium 2 = lossless medium

ϵ₂ = ϵ₀ and μ₂ = 4μ₀

Reflection coefficient will be:

\(\eta = \sqrt {\frac{\mu }{\epsilon}} \)

\({\rm{\Gamma }} = \frac{{{\eta _2} - {\eta _1}}}{{{\eta _2} + {\eta _1}}}\)

  \({\rm{\Gamma }} = \frac{{\sqrt {\frac{{{\mu _2}}}{{{\epsilon _2}}}} - \sqrt {\frac{{{\mu _1}}}{{{\epsilon _1}}}} }}{{\sqrt {\frac{{{\mu _2}}}{{{\epsilon _2}}}} ; + \sqrt {\frac{{{\mu _1}}}{{{\epsilon _1}}}} }}\)

\({\rm{\Gamma }} = \frac{{\sqrt {\frac{{4{\mu _0}}}{{{\epsilon_0}}}} - \sqrt {\frac{{{\mu _0}}}{{{\epsilon_0}}}} }}{{\sqrt {\frac{{4{\mu _0}}}{{{\epsilon_0}}}} ; + \sqrt {\frac{{{\mu _0}}}{{{\epsilon_0}}}} }}\)

\({\rm{\Gamma }} = \frac{{2 - 1}}{{2 + 1}}\)

\({\rm{\Gamma }} = \frac{1}{3}\)

Extra Information: The Transmission coefficient is given by:

τ = 1 + Γ

Concept:

Linearly dependent vectors

A set of vectors is said to be linearly dependent if one of the vectors in the set can be defined as the linear combination of the other vectors.

1. If the two vectors are collinear then they are linearly dependent.
2. If a set has a zero vector, then it means that the vector set is linearly dependent.

The determinant of the linearly dependent matrix is zero.

\(\left| {\begin{array}{*{20}{c}} {\overrightarrow {{u_1}} }&{\overrightarrow {{u_2}} }&{\overrightarrow {{u_3}} }\ {\overrightarrow {{v_1}} }&{\overrightarrow {{v_2}} }&{\overrightarrow {{v_3}} }\ {\overrightarrow {{w_1}} }&{\overrightarrow {{w_2}} }&{\overrightarrow {{w_3}} } \end{array}} \right| = 0\)

Calculation:

Given vectors are (1,-1, 2), (7, 3, x) and (2, 3, 1)

\(\left| {\begin{array}{*{20}{c}} 1&{ - 1}&2\ 7&3&x\ 2&3&1 \end{array}} \right| = 0\)

Performing the row and column operations.

C₂ → C₁ + C₂

\(\left| {\begin{array}{*{20}{c}} 1&0&2\ 7&{10}&x\ 2&5&1 \end{array}} \right| = 0\)

C₃ → C₃ – 2C₁

\(\left| {\begin{array}{*{20}{c}} 1&0&0\ 7&{10}&{x - 14}\ 2&5&{ - 3} \end{array}} \right| = 0\)

Calculating the determinant we get

1[10 × (-3)] – 5(x – 14) = 0

-30 - 5x + 70 = 0

5x = 40

x = 8

Explanation:

If a vector F is irrotational, then the curl of F will be zero i.e. \(\nabla\times \vec F = 0\)

The curl of a vector

\(\vec F = \overrightarrow {{F_x}} \widehat {\dot i} + \overrightarrow {{F_y}} \widehat {\dot J} + \overrightarrow {{F_z}} \hat k\)

\(\nabla = \frac{\partial }{{\partial x}}\widehat { i} + \frac{\partial }{{\partial y}}\widehat { j} + \frac{\partial }{{\partial z}}\hat k\)

\(\nabla \times \vec F = Curl\left( {\vec F} \right) \)

\(= \left| {\begin{array}{*{20}{c}} i&{j}&k\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\ {{F_x}}&{{F_y}}&{{F_z}} \end{array}} \right|;\)

Calculation:

F = a_x (4y – c₁z) + a_y (4x + 2z) + a_z (2y + z)

\(\vec \nabla \times \vec F = \left| {\begin{array}{*{20}{c}} {{a_x}}&{{a_y}}&{{a_z}}\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\ {\left( {4y - {c_1}z} \right)}&{\left( {4x + 2z} \right)}&{\left( {2y + z} \right)} \end{array}} \right|\)

\(= {a_x}\left[ {\frac{\partial }{{\partial y}}\left( {2y + z} \right) - \frac{\partial }{{\partial z}}\left( {4x + 2z} \right)} \right] - {a_y}\left[ {\frac{\partial }{{\partial x}}\left( {2y + z} \right) - \frac{\partial }{{\partial z}}\left( {4y - {c_1}z} \right)} \right] + {a_z}\left[ {\frac{\partial }{{\partial x}}\left( {4x + 2z} \right) - \frac{\partial }{{\partial y}}\left( {4y - {c_1}z} \right)} \right]\)

= a_x( 2 – 2) – a_y (0 + c₁) + a_z (4 - 4) = 0

∴  c₁ = 0

Concept:

Current divider rule (CDR)

Whenever current has to be divided among resistors in parallel, use the current divider rule principle.

\({i_1} = \frac{{{R_2}}}{{{R_1} + {R_2}}}{i_s}\)

\({i_2} = \frac{{{R_1}}}{{{R_1} + {R_2}}}{i_s}\)

Calculation:

Redrawing the circuit

 

Using current division rule:

\(I = \frac{{5 ;\times; 1}}{{1; +; 9}} = 0.5;A\)

Concept:

There are two types of Kirchhoff’s Laws:

Kirchhoff’s first law:

• This law is also known as junction rule or current law (KCL). According to it the algebraic sum of currents meeting at a junction is zero i.e. Σ i = 0.

​

• In a circuit, at any junction, the sum of the currents entering the junction must be equal the sum of the currents leaving the junction i.e., i1 + i3 = i2 + i4
• This law is simply a statement of “conservation of charge” as if current reaching a junction is not equal to the current leaving the junction, charge will not be conserved.

 

Kirchhoff’s second law:

• This law is also known as loop rule or voltage law (KVL) and according to it “the algebraic sum of the changes in potential in the complete traversal of a mesh (closed-loop) is zero”, i.e. Σ V = 0.
• This law represents “conservation of energy” as if the sum of potential changes around a closed loop is not zero, unlimited energy could be gained by repeatedly carrying a charge around a loop.
• If there are n meshes in a circuit, the number of independent equations in accordance with loop rule will be (n - 1).

Analysis:

Given circuit diagram is redrawn as:

Consider current I in the branch BC.

Applying KCL at node B:

V_A = V_B = V₀

\(\frac{{{V_0} - 4}}{1} + \frac{{{V_0}}}{1} + I = 6\)

2V0 + I = 10    ---(1)

Applying KCL at node C:

I = 6 + 2

I = 8 mA

Putting value of I in (1), we get:

2V₀ + 8 = 10

V₀ = 1 V

Given:

8 bit unipolar DAC

V_range = 7.68 V

i / p = 10010110

Formula:

DAC o/p = Resolution * decimal equivalent

Where, resolution \( = \frac{V}{{{2^n} - 1}}\) full scale

n = no. of bit

Calculation:

Decimal equivalent of 10010110 = 150₁₀

DAC o / p = Resolution × decimal equivalent

= \(\frac{{7.68}}{{{2^8} - 1}} \times 150\)

DAC o / p = 4.51 V

Concept:

Power spectral density of a wide sense stationary process is a positive real function.

It is related to Auto co-relation function Rx(τ) by:

\({R_x}\left( \tau \right)\begin{array}{*{20}{c}} {Fourier}\ \leftrightarrow \ {transform} \end{array}{S_x}\left( f \right)\)

For wide – sense stationary random process, the power of a signal is given by R_x(0).

Application:

R_x(0) = 2

Average power of signal = R_x(0) = 2

Concept:

The total AM Power is given by:

\({P_t} = {P_c} + \frac{{{P_c}{\mu ^2}}}{2}\)

Expanding the above, we can write:

\( = \begin{array}{*{20}{c}} {{P_c}}\ \downarrow \ {Carrier}\ {Power} \end{array}\begin{array}{*{20}{c}} + \ {}\ {}\ {} \end{array};\begin{array}{*{20}{c}} {\frac{{{P_c}{\mu ^2}}}{4}}\ \downarrow \ {USB}\ {Power} \end{array}\begin{array}{*{20}{c}} + \ {}\ {}\ {} \end{array};\begin{array}{*{20}{c}} {\frac{{{P_c}{\mu ^2}}}{4}}\ \downarrow \ {LSB}\ {Power} \end{array}\)

% Power saving is defined calculated as:

\(% P = \frac{{Power;saved}}{{Total;power}} \times 100\)

When the carrier and one side-band is suppressed, the percent of power-saving will be:

\( = \frac{{{P_c} + \frac{{{P_c}{\mu ^2}}}{4}}}{{{P_c}\left[ {1 + \frac{{{\mu ^2}}}{2}} \right]}} \times 100\)

\( = \frac{1}{2}\left[ {\frac{{4 + {\mu ^2}}}{{2 + {\mu ^2}}}} \right] \times 100\)

Calculation:

Given:

\(\mu = \frac{{50}}{{100}} = 0.5\)

% Power saved will be:

\(% P = \frac{{4 + {{0.5}^2}}}{{2\left( {2 + {{0.5}^2}} \right)}}\)

= 94.44%

Concept:

Shannon’s theorem:

• A given communication system has a maximum rate of information C known as the channel capacity.
• If the information rate R is less than C, then one can approach arbitrarily small error probabilities by using intelligent coding techniques.
• To get lower error probabilities, the encoder has to work on longer blocks of signal data. This entails longer delays and higher computational requirements.

 

Thus, if R ≤ C then transmission may be accomplished without error in the presence of noise, i.e. for small probability of error C ≥ R_b

C = Channel Capacity

R_b = Bit Rate = nf_s

Analysis:

Nyquist Rate = 2 f_m = 2 × 4 = 8 kHz

f_s = 1.25 × N.R. = 1.25 × 8 = 10 kHz

n = 8

from Shanon theorem,

\(B~{\log _2}\left( {1 + \frac{S}{N}} \right) \ge n{f_S}\)

\(B \ge \frac{{n{f_s}}}{{{{\log }_2}\left( {1 + \frac{S}{N}} \right)}} = \frac{{8 \times 10}}{{\log 399.07}}\ \frac{S}{N} = {10^{2.6}} = 398.107\)

B ≥ 9.258 kHz

Concept:

To avoid slope overload:

\(\frac{{\rm{\Delta }}}{{{T_s}}} ≥ 2π ;{A_{max}}{f_m}\)

\(T_s=\frac{1}{f_s}\)

Δ × f_s ≥ 2π × A_max × f_m ---(1)

Where,

Δ = Step size

Ts = Sampling interval

f_s = sampling rate

Amax = Maximum amplitude of message signal

fm = Message signal frequency

Calculation:

Given:

f_m = 4 kHz

f_s = 32 kHz

A_max = 4 V

Using equation (1) we can calculate the minimum step size:

Δ × 32 ≥ 2π × 4 × 4

\({\rm{\Delta }} ≥ \frac{2π}{2}\)

Δ ≥ π

Δ ≥ 3.14

Important Points

Delta Modulation (DM):

It is used to reduce the bandwidth or bit rate.

1. It is considered a 1-bit DPCM. So, the encoder is a 1 bit ADC, i.e. only 1 sample is used in DM. 2. The encoder output is either 1 or 0 3. Two quantization levels are used + Δ and - Δ.

Two types of distortions (Errors) occurs in Delta modulation system i.e.

1. Slope overload error

2. Granular Error

This is explained with the help of the given figure:

Concept:

Critical angle: When a ray of light is going from a denser medium to a rare medium then the angle of incidence at which the refraction angle is 90° is called the critical angle.

The critical angle is given by:

θC = critical angle = Sin-1(n2/n1) ---(1)

η₁ = refractive index of core

η2 = refractive index of cladding

Calculation:

Given:

η₁ = 1.50

η₂ = 1.48

The critical propagation angle can be calculated using equation (1):

\(θ_C=sin^{-1}(\frac{1.48}{1.5})\)

θ_C = sin^-1(0.986)

θ_C = 80.64° 

Here θ_C is the angle that a light ray can make with the wall of the core.

Whereas from the figure θ is the maximum angle that the light beam makes with the axis of the optical fiber to achieve the total internal reflection.

θ = 90° - θ_C

θ = 90° - 80.64° 

θ = 9.4° 

Important Points

1. Total internal reflection (TIR): 

Total internal reflection (TIR) occurs at the interface of two transparent medium when the ray of light travels from a denser medium to a rarer medium.

2. The critical angle is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is 90°.

Conditions for TIR: 

1. The ray must travel from denser medium to rarer medium.

2. The angle of incidence ∠i must be greater than the critical angle.

Concept:

Residue Theorem:

If f(z) is analytic inside and on a closed curve ‘C’ except at a finite number of singularities inside ‘C’

Then:

∮f(z) dz = 2πi (sum of residues)

Singularity: A point where the function f(z) fails to be analytic

Analysis:

Given \(f\left( x \right) = \frac{{\sin \left( x \right)}}{{{x^2}\left( {{x^2} + 4} \right)}}\) 

\(f\left( x \right) = \frac{{\sin \left( x \right)}}{{{x^2}\left( {x + 2i} \right)\left( {x - 2i} \right)}}\) 

x = 0 pole of order 2

x = 2i, -2i

Given curve or region is |x - i| = 2

x = -2i lies outside |x - i| = 2

So the singular points are x = 0 & x = 2i

Residue at x = 0

\(\mathop {\lim }\limits_{x \to 0} \frac{d}{{dx}}\left[ {\frac{{\sin x}}{{\left( {{x^2} + 4} \right)}}} \right]\)

\( = \mathop {\lim }\limits_{x \to 0} \left[ {\frac{{\cos x\left( {{x^2} + y} \right) - \sin x\left( {2x} \right)}}{{{{\left( {{x^2} + 4} \right)}^2}}}} \right]\)

\( = \mathop {\lim }\limits_{x \to 0} \frac{{\cos x}}{{{x^2} + y}} = \frac{1}{4} = 0.25\)

Residue at x = 2i

\( \Rightarrow \mathop {\lim }\limits_{x \to 2i} \left( {x - 2i} \right)\frac{{\sin x}}{{{x^2}\left( {x + 2i} \right)\left( {x - 2i} \right)}}\)

\( \Rightarrow \frac{{\sin \left( {2i} \right)}}{{{{\left( {2i} \right)}^2}\left( {4i} \right)}}\)

\( \Rightarrow \frac{{\sin \left( {2i} \right)}}{{ - 16;i}}\)

 \(\mathop \oint \nolimits_C f\left( x \right)dx = \mathop \oint \nolimits_C \frac{{\sin \left( x \right)}}{{{x^2}\left( {{x^2} + 4} \right)}};dx = 2\pi i\) (sum of residues)

\( = 2\pi i\left[ {\frac{1}{4} - \frac{{\sin \left( {2i} \right)}}{{16i}}} \right]\)

\( = \frac{{\pi i}}{2} - \frac{{\pi \sin \left( {2i} \right)}}{8}\)

Concept:

Conditional probability is defined as:

\(P\left( {\frac{A}{B}} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}}\)

\(P\left( {\frac{B}{A}} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( A \right)}}\)

Application:

Let event A is defined as:

A = Getting head in the first toss

Event B is defined as:

B = Getting head in the second toss

According to the question, we need to find the probability of getting a head in the second toss when

Already a head has occurred in the first toss, i.e.So, the probability of getting head in the first toss will be:

\(P\left( {\frac{B}{A}} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( A \right)}}\)

P(A) = P(fair coin is selected) × P(Getting head) + P(double-headed coin is selected) × P(getting head)

\( = \frac{1}{3} \times \frac{1}{2} + \frac{1}{3} \times 1 = \frac{1}{6} + \frac{1}{3} = \frac{1}{2}\)

\(P\left( A \right) = \frac{1}{2}\)

Now, finding the probability of getting head in both tosses will be: 

\(P\left( {A \cap B} \right) = \mathop {\mathop {\underbrace {\begin{array}{*{20}{c}} {First;toss}\ {\left( {\frac{1}{3} \cdot \frac{1}{2}} \right)} \end{array};;\begin{array}{*{20}{c}} {second;toss}\ {\left( {\frac{1}{2} \cdot 1} \right)} \end{array}}_{\begin{array}{*{20}{c}} {when;first;fair}\ {coin;is;tossed} \end{array}};;}\limits_{} }\limits_; + \underbrace {\begin{array}{*{20}{c}} {First;toss}\ {\left( {\frac{1}{3} \cdot 1} \right)} \end{array};;\begin{array}{*{20}{c}} {second;toss}\ {\left( {\frac{1}{2} \cdot \frac{1}{2}} \right)} \end{array}}_{\begin{array}{*{20}{c}} {When;first;double}\ {headed;coin;is;tossed} \end{array}}\)

\(P\left( {A \cap B} \right) = \frac{1}{{12}} + \frac{1}{{12}} = \frac{1}{6}\)

\(P\left( {\frac{B}{A}} \right) = \frac{{P\left( {A \cap B} \right)}}{{P\left( A \right)}}\)

\(P\left( {\frac{B}{A}} \right) = \frac{{\frac{1}{6}}}{{\frac{1}{2}}} = \frac{1}{3}\)

Concept:

Under steady-state:

• When a capacitor, is present with a D.C supply, it behaves as an open circuit.
• When an inductor is present with D.C supply, it behaves as a short circuit

At t = 0+ :

Inductor replace with current source IL (0+) = IL (0-)

The inductor does NOT allow the sudden change in current”

The capacitor is replaced with the voltage source

Vc (0+) = Vc (0-)

The capacitor does NOT allow the sudden change in voltage.

Calculation:

At t = 0^- → switch is at P position network is in steady-state

Capacitor → open circuit

Inductor → S.C.

i_L(0^-) = 20/20 = 1 mA

i_L(0⁺) = iL(0-) = 1 mA

V_c (0^-) = 10 × 1 = 10 V

V_c (0⁺) = Vc (0-) = 10 V

At t = 0⁺ → switch is at Q position.

Applying KCL at node A

10/5  + i_c(0⁺) + 1 = 0

i_c (0⁺) = -3 mA

\(C{\left. {\frac{{d{V_c}\left( t \right)}}{{dt}}} \right|_{t = {0^ + }}} = - 3mA\)

∵ \({i_c} = C\frac{{d{V_c}}}{{dt}}\)

\({\left. {\frac{{d{V_c}\left( t \right)}}{{dt}}} \right|_{t = {0^ + }}} = - 3V/sec\)

Concept:

Y parameters

These are also called the admittance parameters.

The Y parameters for the two-port network are shown as:

\(\left[ {\begin{array}{*{20}{c}} {{I_1}}\ {{I_2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {{Y_{11}}}&{{Y_{12}}}\ {{Y_{21}}}&{{Y_{22}}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{V_1}}\ {{V_2}} \end{array}} \right]\)

I₁ = Y₁₁V₁ + Y₁₂V₂

I₂ = Y₂₁V₁ + Y₂₂V₂

Calculation:

Consider the circuit with nodes specified as V₁ and V₂

Writing the node equation at the node1

\( - {I_1} + \frac{{{V_1}}}{1} - 3{V_2} + \frac{{{V_1} - {V_2}}}{1} = 0\)

I₁ = 2V₁ – 4V₂  ---- (i)

Writing the node equation at the node2

\( - {I_2} + \frac{{{V_2} - {V_1}}}{1} + \frac{{{V_2}}}{1} = 0\)

I₂ = - V₁ + 2V₂ ---- (ii)

From the equations (i) and (ii) the Y parameters for the given network are:

\(Y = \left[ {\begin{array}{*{20}{c}} 2&{ - 4}\ { - 1}&2 \end{array}} \right]\)

Concept:

The threshold value for the transistor is given by:

\({V_t} = {V_{t0}} + γ \left[ {√ {{V_{SB}} + \left| {2{\phi _F}} \right|} - √ {\left| {2{\phi _F}} \right|} } \right]\)

V_BS for the n channel transistor

Calculation:

Given that substrate, sensitivity is 50 mV/V

\(\frac{{\partial {V_t}}}{{\partial \left| {{V_{BS}}} \right|}} = \frac{γ }{{2√ {{V_{BS}} + 2\left| {{\phi _F}} \right|} }}\)

\(\frac{{50mV}}{V} = \frac{γ }{{2√ 2 }}\left( {\because\left| {{V_{BS}}} \right| \gg 2{\phi _B}} \right)\)

Consider ϕ_B = ϕ_F 

γ = 0.1 √2

γ is defined as:

\(γ = \frac{{√ {2q{N_A}{ \in _{Si}}} }}{{{C_{ox}}}}\)

\({C_{ox}} = \frac{{{ \in _{ox}}}}{{{t_{ox}}}} = \frac{{4{ \in _0}}}{{10nm}}\)

\(0.1√ 2 = \frac{{√ {2q{N_A}{ \in _{Si}}} }}{{{C_{ox}}}}\)

Squaring on both sides, we get:

\(0.01 × \frac{{16 \in _0^2}}{{100 × {{10}^{ - 18}}}} = 1.6 × {10^{ - 19}}{N_A} × 12{ \in _0}\)

\({N_A} = \frac{{16 × 8.85 × {{10}^2}}}{{1.6 × {{10}^{ - 19}} × 12}}\)

\({N_A} = \frac{{8.85 × {{10}^{22}}}}{{12;{m^3}}}\)

N_A = 0.7375 × 10²² m^-3

NA = 7.375 × 10¹⁵ cm-3

Hence option (1) is correct

In the given question both MUX select line is ‘T’ input.

So MUX output depends only on T.

For input ‘T’ there is only two value possible T = 0 or T = 1

At T = 0 for MUX 1 → XOR gate o/p will propagate to next stage.

For MUX 2 → AND gate 1 (A1), o/p will propagate which is available to MUX 2 at t = 2 nsec

So for Y o/p only, AND 1 (A1) + MUX (2) is necessary.

Propagation delay

= 2 + 1

= 3 nsec

Case 2: T = 1

When T = 1 then for o/p Y: 

AND gate (A3) + MUX (1 )+ AND gate (A2) + MUX (2)

Propagation delay (tpd) = 2 + 1 + 2 + 1

= 6 nsec

So maximum propagation delay of the circuit is 6 nsec.

Concept:

PUSH: Store or push the content of the register pair into stock memory. Content is always pushed on top of the stock.

POP: Read or occurs the data from the top of the stock memory into register pair.

Content is always popped from the top of the stack.

Application:

R₁ = 25 H, R₂ = 30 H, R₃ = 40 H

Push R₁:

 

Push R₂:

Push R₃:

POP R₁:

POP R₂:

POP R₃:

 

Therefore

R₁ = 40 H

R₂ = 30 H

R₃ = 25 H

Concept:

For controllability and observability following condition must be satisfied:

Controllability:

Q_c = [B  AB]

If |Q_c| = 0 then system is UNCONTROLLABLE

If |Q_c| ≠ 0 then the system is CONTROLLABLE

Observability:

\({Q_0} = \left[ {\begin{array}{*{20}{c}} C\ {CA} \end{array}} \right]\)

If |Q₀| = 0 then system is UNOBSERVABLE

If |Q₀| ≠ 0 then the system is OBSERVABLE

Application:

Let us first develop the state model of the given circuit:

Let V_L(t) be the voltage across the inductor and V_R(t) be the voltage across resistor 1 Ω.

\({V_R}\left( t \right) = {V_L}\left( t \right) = L\frac{{d{i_L}\left( t \right)}}{{dt}}\)

With L =  1 H:

\({V_L}\left( t \right) = \frac{{d{i_L}\left( t \right)}}{{dt}}\)

i_s(t) = i_L(l) + i_R(t)

\({i_R}\left( t \right) = \frac{{{V_R}\left( t \right)}}{{1;{\rm{\Omega }}}} = {V_L}\left( t \right)\)

\({i_s}\left( t \right) = {i_L}\left( t \right) + \frac{{d{i_L}\left( t \right)}}{{dt}}\)

\(\frac{{d{i_L}\left( t \right)}}{{dt}} = - {i_1}\left( t \right) + {i_s}\left( t \right)\)    ---(1)

Let i_c(t) be the current through the capacitor, then:

\({i_c}\left( t \right) = C\frac{{d{V_c}\left( t \right)}}{{dt}}\)

C = 1 F

\({i_c}\left( t \right) = \frac{{d{V_c}\left( t \right)}}{{dt}}\)

And \({i_c}\left( t \right) = {i_s}\left( t \right) - \frac{{{V_0}\left( t \right)}}{{1;{\rm{\Omega }}}};\)

And V₀(t) = V_c(t)      ---(2)

\(\therefore ;\frac{{d{V_c}\left( t \right)}}{{dt}} = {i_s}\left( t \right) - {V_c}\left( t \right)\)

\(\frac{{d{V_c}\left( t \right)}}{{dt}} = - {V_c}\left( t \right) + {i_s}\left( t \right)\)     ---(3)

From (1) and (3) equation:

\(\frac{{d{i_L}\left( t \right)}}{{dt}} = - {i_L}\left( t \right) + 0.{V_c}\left( t \right) + {i_s}\left( t \right)\)

\(\frac{{d{V_c}\left( t \right)}}{{dt}} = 0;{i_L}\left( t \right) - {V_C}\left( t \right) + {i_s}\left( t \right)\)

\(\left[ {\begin{array}{*{20}{c}} {{i_L}}\ {{{\dot v}_c}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 1}&0\ 0&{ - 1} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{i_L}}\ {{V_c}} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 1\ 1 \end{array}} \right]{i_s}\left( t \right)\)

\({V_0} = \left[ {0;;1} \right]\left[ {\begin{array}{*{20}{c}} {{i_L}}\ {{V_C}} \end{array}} \right]\)

Here \(A = \left[ {\begin{array}{*{20}{c}} { - 1}&0\ 0&{ - 1} \end{array}} \right];;;;;;B = \left[ {\begin{array}{*{20}{c}} 1\ 1 \end{array}} \right]\) 

C = [0  1]

Now, the Controllability matrix will be:

\({Q_c} = \left[ {B;;AB} \right] = \left[ {\begin{array}{*{20}{c}} 1&{ - 1}\ 1&{ - 1} \end{array}} \right]\)

|Q_c| = 0

Hence the system is not controllable.

Observability matrix:

\({Q_0} = \left[ {\begin{array}{*{20}{c}} C\ {CA} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&1\ 0&{ - 1} \end{array}} \right]\)

|Q₀| = 0

Hence the system is Not observable.

∴ The system is neither controllable nor observable.

Concept:

Let d = (d₁, d₂ …. d_k)

Codeword C can be written as:

\(C = \left( {\underbrace {({C_1},;{C_2},; \ldots .{C_k}}_{message;bits},;\underbrace {;{C_{k + 1}} \ldots .{C_{k + m}})}_{parity;bits}} \right)\) 

C_d = d₁

C₂ = d₂

C₃ = d₃

⋮

C_k = d_k

\(\begin{array}{l} {C_{k + 1}} = {P_{11}};{d_1}; \oplus ;;{P_{12}};;{d_2}; \oplus ;; \ldots ..;;; \oplus {P_{1k}};{d_k}\ {C_{k + 1}} = {P_{11}};{d_1}; \oplus ;;{P_{12}};;{d_2}; \oplus ;; \ldots ..;;; \oplus {P_{1k}};{d_k}\ \vdots ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;} ;;;;;;;;;;;;;;;;;;;\ {C_{k + m}} = {P_{{m_1}}};{d_1}; \oplus ;;{P_{{m_2}}};;{d_2}; \oplus ; \ldots .;.;; \oplus {P_{mk}};{d_k}\ \end{array}\) 

Note: The parity bits are generated by a linear combination of message bits.

Application:

Given (7, 4) linear Hamming code.

(m_i, c_i) given are:

(1100 ; 0101100)

(1110 ; 0011110)

(0110 ; 1000110)

Clearly, we can observe that the first 3 bits are parity bits and the last 4 bits are message bits.

 

Observe when P₀, P_1, and P₂ are one, and write their combination, i.e.

P₀ = d₀ ⊕ d₁ ⊕ d₃

P₁ = d₁ ⊕ d₂ ⊕ d₃

P₂ = d₀ ⊕ d₁ ⊕ d₂

Now checking each of the options:

Option (1): 1011010

Message bits are:

_{\(\begin{array}{*{20}{c}} 1&0&1&0\ {{d_0}}&{{d_1}}&{{d_2}}&{{d_3}} \end{array}\)­}

 P₀ = d₀ ⊕ d₁ ⊕ d₃ = 1

P₁ = d₁ ⊕ d₂ ⊕ d₃ = 1

P­₂ = d₀ ⊕ d₁ ⊕ d₂ = 0

So, the codeword is 1101010.

Hence option (1) is invalid codeword.

Option (2): 0110100

Message bits are:

 \(\begin{array}{*{20}{c}} 0&1&0&0\ {{d_0}}&{{d_1}}&{{d_2}}&{{d_3}} \end{array}\)

P₀ = d₀ ⊕ d₁ ⊕ d₃ = 1

P₁ = d₁ ⊕ d₂ ⊕ d₃ = 1

P­₂ = d₀ ⊕ d₁ ⊕ d₂ = 1

So, the valid codeword is 1110100

Hence option (2) is an invalid codeword.

Option (3): 0001011

Message bits are:

\(\begin{array}{*{20}{c}} 1&0&1&1\ {{d_0}}&{{d_1}}&{{d_2}}&{{d_3}} \end{array}\) 

P₀ = d₀ ⊕ d₁ ⊕ d₃ = 0

P₁ = d₁ ⊕ d₂ ⊕ d₃ = 0

P₂ = d₀ ⊕ d₁ ⊕ d₂ = 0

So the valid codeword is 0001011

Hence option (3) is a valid codeword.

Concept:

Consider the figure shown below which shows a lossless transmission line being driven from the left and which is terminated by the Z_L at the end.

Z_in: Input impedance

Z_L: Load impedance

Z_o: Characteristic impedance

The interface between the source and transmission line is located at z = - l.

\({z_{in}}\left( l \right) = {z_0}\left[ {\frac{{1 + {\rm{\Gamma }}{e^{ - j2\beta l}}}}{{1 - {e^{ - j2\beta l}}}}} \right]\)

The result also depends on the length and phase propagation of the line.

Note that Z_in(l) is periodic in l. Since the argument of the complex exponential factors is 2βl, the frequency at which Z_in(l) varies is β/π; and since β=2π/λ, the associated period is λ/2.

\({z_{in}}\left( l \right) = {z_0}\left[ {\frac{{{z_L} + j{z_0}tan\beta l}}{{{z_0} + j{z_L}tan\beta l}}} \right]\)

Calculation:

For the length λ/2, the input impedance will be the same as that of the load.

\({Z_{in1}} = {z_0}\left[ {\frac{{{z_L} + j{z_0}tan\frac{{2\pi }}{\lambda }\frac{\lambda }{2}}}{{{z_0} + j{z_L}tan\frac{{2\pi }}{\lambda }\frac{\lambda }{2}}}} \right]\)

\({z_{in1}} = {z_L}\)

For short circuit stub length is λ/4

\({z_{in2}} = \frac{{z_0^2}}{{{z_L}}}\)

\({z_{in2}} = \frac{{75^2}}{{{0}}}\)

Z_in2 = ∞

The final structure of the transmission line is:

This is similar to that of the quarter-wave transformer. So, the impedance will be

\(z_0' = \sqrt {{z_0}{z_L}} \)

\({z_L} = \frac{{z_0^{'2}}}{{{z_0}}} = \frac{{{{75}^2}}}{{50}}\)

Z_L = 112.5 Ω

Concept:

Non Singular Matrix: It is a matrix whose determinant ≠ 0

1. If A is any square matrix of order n, we can form the matrix [A – λI], where I is the nth order unit matrix. The determinant of this matrix equated to zero i.e. |A – λI| = 0 is called the characteristic equation of A.

2. The roots of the characteristic equation are called Eigenvalues or latent roots or characteristic roots of matrix A.

3. Sum of Eigenvalues is equal to the trace of that matrix.

4. The product of Eigenvalues of a matrix A is equal to the determinant of that matrix A.

Calculation:

\(A = \left[ {\begin{array}{*{20}{c}} x&{ - 3.0}\ {3.0}&{4.0} \end{array}} \right]\)

To find eigen-values:

|A – λI| = 0

\( \left| {\begin{array}{*{20}{c}} x-λ&{ - 3.0}\ {3.0}&{4.0-λ} \end{array}} \right|=0\)

|(4 - λ)(x - λ) - (-3)(3)| = 0

λ² - (4 + x)λ + (4x + 9) = 0

As the roots are repeated (λ₁ = λ₂)

So according to quadratic equations property:

b² - 4ac = 0

(4 + x)² - 4(4x + 9) = 0

16 + x² + 8x - 16x - 36 =0

x² - 8x - 20 = 0

x² + 2x - 10x + 20 = 0

x(x + 2) - 10(x + 2) = 0

x = -2 and x = 10

As x = -2 not possible (given)

So x = 10

Concept:

Gauss’s law

It states that flux leaving any closed surface is equal to the charge enclosed by that surface.

\({\rm{\Psi }} = \oint \vec D.d\vec S\)

The Divergence theorem states that the surface integral can be converted into the volume integral.

\(\oint \vec D.d\vec S = \smallint \left( {\nabla .\vec D} \right)dV\)

Divergence in cylindrical form for the given vector is:

\(\nabla .\vec D = \frac{1}{\rho }\frac{\partial }{{\partial \rho }}\left( {\rho {D_\rho }} \right) + \frac{1}{\rho }\frac{\partial }{{\partial \phi }}\left( {{D_\phi }} \right) + \frac{\partial }{{\partial z}}\left( {{D_z}} \right)\)

Calculation:

A given vector field is:

D = ρ cos²ϕ a_ρ + z²sin²ϕ a_ϕ

From the given vector D_z = 0

\(\nabla .\vec D = \frac{1}{\rho }\frac{\partial }{{\partial \rho }}\left( {{\rho ^2}{{\cos }^2}\phi } \right) + \frac{1}{\rho }\frac{\partial }{{\partial \phi }}\left( {{z^2}{{\sin }^2}\phi } \right) + \frac{\partial }{{\partial z}}\left( 0 \right)\)

\(\nabla .\vec D = 2{\cos ^2}\phi + \frac{{{z^2}}}{\rho }2sin\phi cos\phi \)

\(\smallint \left( {\nabla .\vec D} \right)dV = \)

\(\mathop {\mathop{{\int!!!!!\int!!!!!\int\int\int}\mkern-31.2mu } {} }\limits_{\phi = 0,\rho = 0,z = 0}^{\phi = 2π ,\rho = 3,z = 2} \left( {2{{\cos }^2}\phi + \frac{{{z^2}}}{\rho }2sin\phi cos\phi } \right)\rho d\rho d\phi dz\)

\(= \mathop {\mathop{{\int!!!!!\int!!!!!\int}\mkern-31.2mu } {} }\limits_{\phi = 0,\rho = 0,z = 0}^{\phi = 2π ,\rho = 3,z = 2} \left( {2{{\cos }^2}\phi } \right)\rho d\rho d\phi dz \)

\(+~~ \mathop {\mathop{{\int!!!!!\int!!!!!\int}\mkern-31.2mu } {} }\limits_{\phi = 0,\rho = 0,z = 0}^{\phi = 2π ,\rho = 3,z = 2} \left( {\frac{{{z^2}}}{\rho }2sin\phi cos\phi } \right)\rho d\rho d\phi dz\)

\( = 2\left[ {\frac{{{\rho ^2}}}{2}} \right]_0^3\left[ z \right]_0^2\mathop \smallint \limits_{\phi = 0}^{2π } {\cos ^2}\phi d\phi + \left[ { - \frac{{cos2\phi }}{2}} \right]_0^{2π }\mathop \int!!!\int \limits_{\rho = 0,z = 0}^{\rho = 3,z = 2} {z^2}d\rho dz\)

\( = 9 \times 2\left{ {\left[ {\frac{\phi }{2}} \right]_0^{2π } + \left[ {\frac{{sin2\phi }}{4}} \right]_0^{2π }} \right} \)

\(+ \frac{1}{2}\left[ { - cos4π - \left( { - cos0^\circ } \right)} \right]\mathop \int!!!\int \limits_{\rho = 0,z = 0}^{\rho = 3,z = 2} {z^2}d\rho dz\)

= 18 π = 56.54

Flux leaving through the surface is 56.54

Concept:

Voltage across capacitor is given by

V_c(t) = V_c(∞) + [V_c (0⁺) – V_c (∞)] e^-t/τ 

Analysis:

Given V_c(0^-) = -5 V

V_c(0^-) = V_c (0⁺) = -5 V

At t → ∞ capacitor will act as open circuit

Applying KCL at node A:

\(\frac{{{V_c}\left( \infty \right) - 5}}{{250}} + \frac{{V_R}}{{500}} + \frac{{{V_c}\left( \infty \right)}}{{250}} = 0\)

2 V_c (∞) – 10 + V_R + 2 V_c (∞) = 0     …i)

\(V_R = \left{ {\left( {\frac{{{V_c}\left( \infty \right) - 5}}{{250}}} \right) \times 250} \right}\times - 1\)

V_R = -V_c (∞) + 5     …ii)

Putting V_R from equation ii) to equation i)

2 V_c (∞) – 10 – V_c (∞) + 5 + 2 V_c (∞) = 0

3 V_c (∞) = 5

\({V_c}\left( \infty \right) = \frac{5}{3};V\)

Time constant τ = R_eQ.C

R_eQ = R_th across capacitor C

Applying KCL at node A-

\(I = \frac{V}{{250}} + \frac{{V_R}}{{500}} + \frac{V}{{250}}\) …iii)

\(V_R = - \frac{V}{{250}} \times 250 = - V\)…iv)

Put V_R = -V into equation iii)

\(I = \frac{V}{{250}} - \frac{V}{{500}} + \frac{V}{{250}}\)

\(I = V\left{ {\frac{3}{{500}}} \right}\)

\(\frac{V}{I} = \frac{{500}}{3}{\rm{\Omega }} = {R_{th}} = {R_{eQ}}\)

\(τ = {R_{eQ}}.C = \frac{{500}}{3} \times 6 \times {10^{ - 6}} = {10^{ - 4}};sec\)

\({V_c}\left( t \right) = {V_c}\left( \infty \right) + \left{ {{V_c}\left( {{0^ + }} \right) - {V_c}\left( \infty \right)} \right}{e^{ - \frac{t}{τ }}}\)

\({V_c}\left( t \right) = \frac{5}{3} + \left{ { - 5 - \frac{5}{3}} \right}{e^{ - \frac{t}{{{{10}^{ - 4}}}}}}\)

Let V_c(t) = 0 at t = t₁

\(0 = \frac{5}{3} - \frac{{20}}{3}{e^{ - \frac{{{t_1}}}{{{{10}^{ - 4}}}}}}\)

\(\frac{{{e^{ - {t_1}}}}}{{{{10}^{ - 4}}}} = \frac{1}{4}\)

t₁ = 10^{– 4} ln 4 = 0.1386 msec

Concept:

The Exponential Fourier Series (EFS) of the function g(t) is represented as:

\(g\left( t \right) = \mathop \sum \limits_{n = - \infty }^\infty {C_n}{e^{jn{\omega _0}t}}\)

Where C_n is the EFS coefficient and it is defined by:

\({C_n} = \frac{1}{T}\mathop \smallint \limits_0^T g\left( t \right){e^{ - jn{\omega _0}t}}dt\)

Parseval’s theorem

This tells that the total power in a periodic signal is equal to the sum of the squared amplitudes of each harmonic.

\(\frac{1}{T}\mathop \smallint \limits_0^T {\left| {x\left( t \right)} \right|^2}dt = \mathop \sum \limits_{n = - \infty }^\infty {\left| {{C_n}} \right|^2}\)

Calculation:

The given signal is defined as:

\(x\left( t \right) = \mathop \sum \limits_{n = - \infty }^\infty {a_k}{e^{jk{\omega _0}t}}\)

Where a_k is the Fourier Series coefficient.

Given that the x(t) is real and even. So, this implies that the spectrum also real and even.

\(a_k = \left\lbrace \begin{matrix} k, \ \ 1 \le k \le 3 \\ 0, \ \ \ k > 3 \end{matrix} \right.\)

Average power is:

\(\frac{1}{T}\mathop \smallint \limits_0^T {\left| {x\left( t \right)} \right|^2}dt = \mathop \sum \limits_{n = - 3}^3 {\left| {{a_k}} \right|^2}\)

\({P_{avg}} = {\left| {{a_{ - 3}}} \right|^2} + {\left| {{a_{ - 2}}} \right|^2} + {\left| {{a_{ - 1}}} \right|^2} + {\left| {{a_0}} \right|^2} + {\left| {{a_1}} \right|^2} + {\left| {{a_2}} \right|^2} + {\left| {{a_3}} \right|^2}\)

P_avg = 9 + 4 + 1 + 4 + 1 + 4 + 9

P_avg = 18 + 8 + 2 + 4

P_avg = 32 W

Concept:

Convolution

Convolution is the sliding of one signal over the folded and shifted version of the other signal.

The below figure represents the structures in the continuous and discrete domain.

\(y\left( t \right) = \mathop \smallint \limits_{ - \infty }^\infty x\left( \tau \right)h\left( {t - \tau } \right)d\tau \)

\(y\left[ n \right] = \mathop \sum \limits_{k = - \infty }^\infty x\left[ k \right]h\left[ {n - k} \right]\)

Cascaded signals are convolved to give the output.

NOTE: Convolution in one domain is the multiplication in another domain.

Convolution of signal with del function

x(t) * δ(t – t₀) = x( t – t₀)

Calculation:

Given input is u[n] and the output for u[n] is 2δ[n + 1] + δ[n] + δ[n – 1]

Converting this into the frequency domain we'll get

\(X\left( z \right) = \frac{z}{{z - 1}}\)

Y(z) = 2z + 1 + z^-1

\(H\left[ z \right] = \frac{{Y\left[ z \right]}}{{X\left[ z \right]}} = \frac{{2z + 1 + {z^{ - 1}}}}{{\frac{z}{{z - 1}}}}\)

\(H\left[ z \right] = 2{z} - 1 - {z^{ - 2}}\)

h[n] =  2δ[n + 1] - δ[n] - δ[n – 2]

Now the other input signal applied to the same system.

y[n] = x[n] * h[n]

y[n] = x[n] * [2δ[n + 1] - δ[n] - δ[n – 2]]

y[n] = 2x[n +1] – x[n] – x[n -2]

\(y\left[ n \right] = 2{\left( {\frac{1}{2}} \right)^{n + 1}}u\left[ {n + 1} \right] - {\left( {\frac{1}{2}} \right)^n}u\left[ n \right] - {\left( {\frac{1}{2}} \right)^{n - 2}}u\left[ {n - 2} \right]\)

\(y\left[ 0 \right] = 2{\left( {\frac{1}{2}} \right)^{0 + 1}}u\left[ 1 \right] - {\left( {\frac{1}{2}} \right)^0}u\left[ 0 \right] - {\left( {\frac{1}{2}} \right)^{0 - 2}}u\left[ { - 2} \right]\)

y[0] = 1 – 1 – 0

y[0] = 0

Concept:

\(x\left[ n \right]*y\left[ n \right];\begin{array}{*{20}{c}} {\mathop \leftrightarrow \limits^{Fourier} }\^{Transform} \end{array}X\left( {{e^{j\omega }}} \right)Y\left( {{e^{j\omega }}} \right)\)

Let z(n) = x(n) * y(n)

Then Z(e^jω) = X(e^jω) Y(e^jω)

Also, using central ordinate theorem:

\(z\left[ 0 \right] = \frac{1}{{2\pi }}\mathop \smallint \nolimits_0^{2\pi } z\left( {{e^{j\omega }}} \right)d\omega \)

\( = \frac{1}{{2\pi }}\mathop \smallint \nolimits_0^{2\pi } X\left( {{e^{j\omega }}} \right)Y\left( {{e^{j\omega }}} \right)d\omega \)

Application:

x[n] = 2^n-1 u[-n + 2]

y[n] = 2^-n+2  u[n + 1]

Given:

\(\frac{1}{{2\pi }}\mathop \smallint \nolimits_0^{2\pi } X\left( {{e^{j\omega }}} \right);y\left( {{e^{ - j\omega }}} \right)d\omega \)

If y(n) = y(e^jω)

Then, using time reversal property:

y[-n] = y(e^-jω)

Hence z[n] = x[n] * y[-n]

z[n] = [2^n-1u[-n + 2]] * [2ⁿ⁺²u[-n + 1]]

\(z\left[ n \right] = \mathop \sum \nolimits_{k = - \infty }^\infty {2^{k - 1}};u\left[ { - k + 2} \right] \cdot {2^{n - k + 2}};u\left[ { - n + k + 1} \right]\)

\(z\left[ n \right] = \mathop \sum \nolimits_{k = - \infty }^2 {2^{k - 1}} \cdot {2^{n - k + 2}};u\left[ { - n + k + 1} \right]\)

\(z\left[ n \right] = \mathop \sum \nolimits_{k = - \infty }^2 {2^{k - 1 + n - k + 2}};u\left[ { - n + k + 1} \right]\)

\(z\left[ n \right] = \mathop \sum \nolimits_{k = - \infty }^2 {2^{n + 1}};u\left[ { - n + k + 1} \right];\)

\(\frac{1}{{2\pi }}\mathop \smallint \nolimits_0^{2\pi } X\left( {{e^{j\omega }}} \right);Y\left( {{e^{ - j\omega }}} \right) = z\left[ 0 \right]\)

Putting n = 0, we get:= 2

\(z\left[ 0 \right] = \mathop \sum \nolimits_{k = - \infty }^2 2 \cdot u\left[ {k + 1} \right]\)

\( = \mathop \sum \nolimits_{k = - 1}^2 2 \cdot \left( 1 \right)\)

= 2 × 4

= 8

Concept:

Charge equality equation in diode is given as:

W_N N_D = W_P N_A

Also, the depletion width is given by the formula:

\(\sqrt {\frac{{2 \in }}{q}\left( {\frac{1}{{{N_A}}} + \frac{1}{{{N_D}}}} \right)\left( {{V_{bi}} + {V_{RB}}} \right)} \)

Calculation:

Given: N_A = 5 × 10¹⁶/cm³    N_D = 10 × 10¹⁶/cm³

N_A X_p = N_D X_n → in case of complete depletion 

\(\begin{array}{l} \frac{{{X_p}}}{{{X_n}}} = 2\ \end{array}\)

x_p = 2 × .2 = 0.4 μm

depletion width = X_p + X_n



 = 0.4 + 0.2

= 0.6 μm = 60 μcm

The depletion width will be: 

\({10^{ - 4}} \times0 .6{\rm{;}} = \sqrt {\frac{{2 \times 8.85 \times {{10}^{ - 12}}1 \times 12}}{{1.6 \times {{10}^{ - 19}}}}\left{ {\frac{1}{{5 \times {{10}^{16}}}} + \frac{1}{{{{10}^{17}}}};} \right}\left( {{V_{bi}} + {V_{RB}}} \right)} \)

\({10^{ - 4}} \times 0.6 = \sqrt {3.9825 \times {{10}^{ - 10}}\left( {{V_{bi}} + {V_{RB}}} \right)} \)

V_bi + V_RB = 9.03

V_RB = 8.23 V

Concept:

Clipper

It is a circuit that cuts the portion of the waveform which is given.

Calculation:

Initially assuming that the diode is forward biased for the rising edge of the signal.

τ = 0 when diode forward biased

Capacitor charges to 10V in the positive half.

During negative half

 

V_C + V₀ = 0

V₀ = - V_C

A diode is reverse biased.

Capacitor will discharge through the resistor

τ = RC = 500 × 10³ × 20 × 10^-9

τ = 10^-2 = 10 ms = 10000 μs

Time avaialable is 1 μs.

\({V_C} = 10{e^{ - \frac{t}{{RC}}}}\)

\({V_C} = 10{e^{ - \frac{1}{{10000}}}}\)

V_C ≈ 10V

Capacitor never going to discharge.

In the next cycle, the circuit will be

 

So the swing is 10V.

Concept:

MOSFET is in saturation if V_DS ≥ V_GS – V_T

At the edge of saturation:

V_DS = V_GS – V_T

\({I_D} = {\mu _n}C_{OX}\frac{W }{{2L}}{\left( {{V_{GS}} - {V_T}} \right)^2}\)

Calculation:

Let threshold voltage of MOSFET is V_th

Given: μ_n C_OX = 100 μA / V

\(\frac{W }{L} = 10\)

V_GS = 1

Drain current:

  \({I_{DS}} = \frac{{{\mu _n}COX}}{2}*\frac{W }{L}{\left( {{V_{GS}} - {V_T}} \right)^2}\)

\({I_{DS}} = \frac{{100}}{2} \times 10 \times {10^{ - 6}}{\left( {1 - {V_{th}}} \right)^2};A\)

\({I_{DS}} = \frac{1}{2}{\left( {1 - {V_{th}}} \right)^2};mA\)…1)

Applying KVL at loop 1

V_DS = 3 – 20 I_DS

From equation 1) put I_DS into equation 2)

\({V_{DS}} = 3 - 20 \times \frac{1}{2}{\left( {1 - {V_{th}}} \right)^2}\)

V_DS = 3 – 10 (1 - V_th)²     …3)

Since MOSFET is at edge of saturation

V_DS = V_GS - V_th

1 – V_th = 3 – 10 (1 - V_th)²

As sum 1 – V_th = a

a = 3 – 10 a²

10 a² + a – 3 = 0

\(a = \frac{{ - 1 \pm \sqrt {1 + 120} }}{{20}}\)

\(a = \frac{{ - 1 \pm 11}}{{20}}\)

a = 0.5 V or -0.6 V

when, a = 0.5, 1 – V_th = 0.5, V_th = 0.5

when, a = -0.6 V, 1 – V_th = -0.6, V_th = 1.6 V

But V_GS > V_th

{to flow drain current I_DS}

So V_th = 0.5 is the correct value of threshold voltage.

Note: 

Always try to convert drain current 2^nd order equation in term of voltage because out of two values we can eliminate one value using V_GS > V_th relation.

Method – 1

Since we are applying at t = 0 a short duration pulse.

At t = 0^-; V_in = 0; V_c (0^-) = 0

Since V_c (0⁺) = V_c (0^-) = 0 V

At t = 0⁺ capacitor will behave as a short circuit.

At t = 0⁺; V_in = 5 V

Since it behaves like an inverting amplifier

\({V_{out}} = \frac{{ - {R_2}}}{{{R_1}}}{V_{in}}\)

R₂ = 10 kΩ

R₁ = 0 kΩ

V_out = -∞

Since V_out < -V_sat so V_out = -12 V, hence it will work like a capacitor.

Method – 2

Using the Laplace transform method

_CeQ = 1 + 1 = 2 μF

In s domain –

By virtual ground concept V_A = 0

Applying KCL at node A-

\(\frac{{{V_A}\left( s \right) - {V_{out}}\left( s \right)}}{{10}} + \frac{{{V_A}\left( s \right) - {V_{in}}\left( s \right)}}{{\frac{1}{{2s}}}} = 0\)

V_out = -10 × 25 V_in (s)

\({{V_{in}}\left( s \right) = \frac{5}{s}\left{ {1 - {e^{ - 20 \times 10 ^{- 3}s}}} \right}}\{\begin{array}{*{20}{c}} {}\ {{V_{in}}\left( t \right) = 5\left{ {u\left( t \right) - 4\left( {t - 20 \times {{10}^{ - 3}}} \right)} \right}} \end{array}}\)

\({V_{out}} = \frac{{ - 20s}}{s} \times 5\left{ {1 - {e^{ - 20 \times 10 ^{- 3}s}}} \right}\)

Applying initial value theorem in –

\({V_{out}}\left( {{0^ + }} \right) = \mathop {\lim }\limits_{s \to \infty } s;{V_{out}}\left( s \right)\)

= \(\mathop {\lim }\limits_{s \to \infty } - 100;s\left{ {1 - {e^{ - 20 \times 10 - 3s}}} \right}{e^{ - \infty }} \simeq 0\)

V_out (0⁺) = -∞

Since V_out (0⁺) < -V_sat so V_out = -V_sat = -12 V

Clock frequency = 500 MHz

\({T_{CLK}} = \frac{1}{f} = \frac{1}{{500 \times {{10}^6}}} = 2;nsec\)

Any change / input will occur at FF output only at t = 0, 2, 4, 6, 8 … nsec

XOR gate delay is 3 nsec. So output at XOR will occur after 3 nsec

It is a right shift register.

No. of clock edges to get Q₂ Q₁ Q₀ → 100 is 5.

Concept:

Capacitor do not allow sudden change in voltage i.e. V_c(0^-) = V_c(0⁺)

Inductors do not allow a sudden change in current.

i.e. i_L(0^-) = i_L(0⁺)

Calculation:

Since at t < 0    … switch is open

∴ i_L (0^-) = 0 Amp

V_L (0^-) = 0 V

As we know, when Inductor and capacitor both present, we use the Laplace domain concept:

 

Now, the ckt for t > 0 in the Laplace domain

Apply nodal concept:

\(\frac{1}{s} = \frac{{{V_c}\left( s \right)}}{{\frac{1}{{sC}}}} + \frac{{{V_c}\left( s \right)}}{{R + sL}} = 0\)

\({V_c}\left( s \right) = \frac{1}{s} \cdot \frac{{R + sL}}{{{s^2}LC + RCs + 1}}\)

∵ we need, \({I_L}\left( s \right) = \frac{{{V_c}\left( s \right)}}{{R + sL}}\) 

\({I_L}\left( s \right) = \frac{{1/LC}}{{s\left( {{s^2} + \frac{R}{L}s + \frac{1}{{LC}}} \right)}}\)

From denominator:

\({s^2} + \frac{R}{L}s + \frac{1}{{LC}} = {s^2} + 2\xi {\omega _n}s + \omega _n^2\)

By comparing with std. 2^nd order equation

\({\omega _n} = \frac{1}{{\sqrt {LC} }};;& ;2\xi {\omega _n} = \frac{R}{L}\)

∴ ξ = 0.25

Now

First maxima or peak overshoot \( = {e^{\frac{{ - \xi \pi }}{{\sqrt {1 - {\xi ^2}} }}}}\) 

(m­_p) = 0.4443

Steady-state value of \({I_L}\left( s \right) = \mathop {\lim }\limits_{s \to 0} s \cdot \frac{{\frac{1}{{LC}}}}{{s\left( {{s^2} + \frac{R}{L}s + \frac{1}{{LC}}} \right)}}\) 

= 1

∴ The maximum ammeter reading just after the switch closed is

= 1.444 Amp

Given:

\(G\left( s \right) = \left( {{k_p} + \frac{{{k_I}}}{s}} \right)\left( {\frac{2}{{{s^3} + 4{s^2} + 5s + 2}}} \right)\)  

\( = \left( {\frac{{s{k_p} + {k_I}}}{s}} \right)\left( {\frac{2}{{{s^3} + 4{s^2} + 5s + 2}}} \right)\) 

\(G\left( s \right) = \frac{{2s;{k_p} + 2;{k_I}}}{{{s^4} + 4{s^3} + 5{s^2} + 2s}}\) 

Characteristic equation will be:

 ⇒1 + G(s) = 0

s⁴ + 4s³ + 5s² + (2 + 2k_p)s + 2k_I = 0

From, necessary condition:

k_p > - 1 & k_I > 0

Routh-Array:

 

For stability:

\(\frac{{20 - \left( {2 + 2;{k_p}} \right)}}{4} > 0\) 

18 – 2k_P > 0

k_P < 9

 -1 < k_p < 9

For stability:

\(\frac{{\left( {18 - 2{k_p}} \right)\left( {2 + 2;{k_p}} \right) - 32;{k_I}}}{{\left( {18 - 2;{k_p}} \right)}} > 0\) 

\(36 - 4{k_p} + 36;{k_p} - 4k_p^2 - 32;{k_I} > 0\)  

\(0 < {k_I} < \frac{{ - 4k_p^2 + 32;{k_p} + 36}}{{32}}\) 

For a maximum value of k_I

\(\frac{{d{k_I}}}{{d{k_p}}} = 0\) 

32 – 8 k_p = 0

k_P = 4

For k_p = 4   k_­I is maximum

∴  \({K_{I;max}} = \frac{{ - 4{{\left( 4 \right)}^2} + 32 \times 4 + 36}}{{32}}\)

\( = \frac{{ - 64 + 128 + 36}}{{32}}\) 

K_{I max} = 3.125

Concept:

The general expression for a phase-modulated signal is given by:

\({S_{PM}}\left( t \right) = {A_c}\cos{\mathop {\mathop {}{\underbrace {2\pi {f_c}t + {K_p}m\left( t \right)}_{{\theta _i}\left( t \right)}} }\limits_{} }\)

θ_i is the instantaneous phase

Maximum instantaneous frequency f_i is given by:

\({f_i} = \frac{1}{{2\pi }};\frac{{d{\theta _i}}}{{dt}}\)

Calculation:

S_PM(t) = A_c cos [2πf_ct + k_pm(t)]

K_p = 2 rad/volt

m(t) = A_m sin (2πf_mt)

Given \(\frac{{{A_m}}}{{\sqrt 2 }} = 4;V\) 

And f_m = 10³ Hz

∴ \(m\left( t \right) = 4\sqrt 2 \sin \left( {2\pi ;{{10}^6}t} \right)\) 

\({f_{i,;max}} = \frac{1}{{2\pi }};{\left| {\frac{{d{\theta _i}}}{{dt}}} \right|_{max}}\)

\( = {f_c} + {\left| {\frac{{{k_p}}}{{2\pi }};\frac{{dm\left( t \right)}}{{dt}}} \right|_{max}}\)

\(\frac{{dm\left( t \right)}}{{dt}} = 4\sqrt 2 ;\left( {2\pi \times {{10}^3}} \right)\cos (2\pi \times {10^3}t);;\)  

\({\left| {\frac{{dm\left( t \right)}}{{dt}}} \right|_{max}} = 4\sqrt 2 ;\left( {2\pi \times {{10}^3}} \right)\)    

 \(\therefore {f_{i,;max}} = {f_c} + \frac{2}{{2\pi }}\left( {4\sqrt 2 \times 2\pi \times {{10}^3}} \right)\)

\( = {10^6} + 8\sqrt 2 \times {10^3}\)

= 1011313.7 Hz

Concept:

The image frequency for a super-heterodyne receiver is given by:

fsi = fs + 2IIF ----(1)

fsi = Image frequency

fs = Tuned frequency of the signal

IIF = Intermediate frequency

fLO = Local Oscillator frequency, which is calculated as:

I_IF = fLO -  fs  ----(2)

This is explained with the help of the following spectrum analysis:

Calculation:

Given:

fs = 600 kHz   

fLO = 1000 kHz

From equation (2) I_IF can be calculated as:

I_IF = 1000 - 600 

I_IF = 400 kHz

Also, the image frequency can be calculated using equation (1):

fsi = 600 + (2 × 400)

fsi = 1400 kHz

Concept:

Information (I):

\(I = {\log _2}\frac{1}{P}\) bits

Calculation:

∴ No. of boys and girl students are equal

∴ \(P\left( B \right) = \frac{1}{2};;& ;;P\left( G \right) = \frac{1}{2}\)

∵ 75% students study science and 25% students study commerce.

∴ \(P\left( S \right) = \frac{3}{4};;& ;;P\left( C \right) = \frac{1}{4}\)  

∵ commerce students are two times more likely to be a boy than are science students.

\(P\left( {\frac{B}{C}} \right) = 2P\left( {\frac{B}{S}} \right)\)

Now;

\(P\left( B \right) = \left( {\frac{B}{C}} \right) \cdot P\left( C \right) + P\left( {\frac{B}{S}} \right) \cdot P\left( S \right)\)

\(\frac{1}{2} = P\left( {\frac{B}{C}} \right) \cdot \frac{1}{4} + \frac{1}{2}P\left( {\frac{B}{C}} \right) \cdot \frac{3}{4}\)

\(P\left( {\frac{B}{C}} \right) = \frac{4}{5}\)

Now,

\(P\left( {\frac{C}{B}} \right) = \frac{{P\left( {\frac{B}{C}} \right) \cdot P\left( C \right)}}{{P\left( B \right)}} = \frac{{\frac{4}{5} \times \frac{1}{4}}}{{\frac{1}{2}}} = \frac{2}{5}\)

\(P\left( C \right) = P\left( {\frac{C}{B}} \right)P\left( B \right) + P\left( {\frac{C}{G}} \right) \cdot P\left( G \right)\)

\(\frac{1}{G} = \frac{2}{5} \times + P\left( {\frac{C}{G}} \right)\frac{1}{2}\)

\(P\left( {\frac{C}{G}} \right) = \frac{1}{{10}}\)

The amount of information gained in knowing that a randomly selected girl student studies commerce

\(I = \log_2[ \frac{1}{{P\left[ {\frac{C}{G}} \right]}} ]= - {\log _2}P\left[ {\frac{C}{G}} \right]\)

= log₂10

= 3.333 bits.

Concept:

For PCM, the noise power (N_p) is given by:

\({N_p} = \frac{{{{\rm{\Delta }}^2}}}{{12}}\)

Where \({\rm{\Delta }} = \frac{{{V_{peak - peak}}}}{{{2^n}}}\) 

For calculating n, the bit-rate and the bandwidth of message signal is given as:

Bitrate, R_b = nf_s

f_s = 2f_m (Nyquist criteria)

Signal power = (RMS value of signal)²

Calculation:

Given

RMS value of signal = 0.1 V

Signal power = (0.1)² = 0.01     ---(1)

Bit rate R_b = nf_s

R_b = n (2f_m)

50 = n (2 × 5)

n = 5

Step size will be:

\({\rm{\Delta }} = \frac{{{V_{peak - peak}}}}{{{2^n}}} = \frac{{2V}}{{{2^5}}} = \frac{1}{{{2^4}}}\)

Now, the Noise Power will be:

\(\frac{{{{\rm{\Delta }}^2}}}{{12}} = {\left( {\frac{1}{{{2^4}}}} \right)^2}\frac{1}{{12}} = \frac{1}{{{2^8} \times 12}}\)

∴ The required Signal to Noise Ratio (SNR) will be:

\(\frac{{Signal;power}}{{Noise;Power}} = \frac{{0.01}}{{\left( {\frac{1}{{{2^8} \times 12}}} \right)}}\)

= 30.72

Concept:

Optimum threshold voltage

For an Additive White Gaussian Noise function, the optimum threshold value is given by:

\({v_{th}} = \frac{{{a_1} + {a_2}}}{2} + \frac{{\sigma _n^2}}{{{a_1} - {a_2}}}\ln \left[ {\frac{{P\left( 0 \right)}}{{P\left( 1 \right)}}} \right]\)

P(0): Probability of the error when ‘0’ transmitted

P(1): Probability of the error when ‘1’ transmitted

Calculation:

Given that P(1) = 0.4 then P(0) = 0.6

The variance of the noise is 0.4V²

Given that binary ‘1’ is represented for +2V and binary ‘0’ is represented for -2V

For transmission of binary ‘1’

a₁ = E[X + N]

a₁ = E[X] + E[N]

Given that mean of noise is zero i.e, E[N] = 0

a₁ = E[2] + E[N] = 2 + 0

a₁ = 2

For transmission of binary ‘0’

a₂ = E[X + N]

a₂ = E[- 2] + E[N]

a₂ = - 2

Optimum threshold voltage is

\({v_{th}} = \frac{{2 - 2}}{2} + \frac{{0.4}}{{2 - \left( { - 2} \right)}}\ln \left[ {\frac{{0.6}}{{0.4}}} \right]\)

\({v_{th}} = \frac{{0.4}}{{2 - \left( { - 2} \right)}}\ln \left[ {\frac{{0.6}}{{0.4}}} \right]\)

V_th = 0.04 V

Concept:

Phase velocity is given by:

\({v_p} = \frac{\omega }{\beta }\)

Where, ω = angular frequency

β = phase constant, calculated for a parallel waveguide as:

\(\beta = \sqrt {{\omega ^2}\mu - {{\left( {\frac{{m\pi }}{a}} \right)}^2}} \)         [for parallel waveguide]

The Phase velocity can now be written as:

\({v_p} = \frac{\omega }{{\sqrt {{\omega ^2}\mu - {{\left( {\frac{{m\pi }}{a}} \right)}^2}} }}\)  

\({v_p} = \frac{1}{{\sqrt {\mu - {{\left( {\frac{{m\pi }}{{a\omega }}} \right)}^2}} }}\)

\({v_p} = \frac{{\frac{1}{{\sqrt {\mu } }}}}{{\sqrt {1 - {{\left( {\frac{{m\pi }}{{a\omega \sqrt {\mu } }}} \right)}^2}} }}\)

With \(c = \frac{1}{{\sqrt {\mu } }}\)

\({v_p} = \frac{c}{{\sqrt {1 - {{\left( {\frac{{m\pi c}}{{a\omega }}} \right)}^2}} }};\)

Also, the cut-off frequency is defined as:

\({\omega _c} = \frac{{m{\pi _c}}}{a}\)

\({v_p} = \frac{c}{{\sqrt {1 - {{\left( {\frac{{{\omega _c}}}{\omega }} \right)}^2}} }}\)

\(\sin \theta = \frac{{{\omega _c}}}{\omega }\)

\({v_p} = \frac{c}{{\sqrt {1 - {{\sin }^2}\theta } }}\)

\({v_p} = \frac{C}{{\cos \theta }}\)

\({v_p} = \frac{c}{{\cos \theta }}\)  

Note: This is valid for the Rectangular waveguide also.

Calculation:

Given

a = 8 cm

b = 4 cm

f = 3.4 GHz

The cut-off frequency will be:

\({f_c} = \frac{c}{{2a}} = \frac{{3 \times {{10}^{10}}}}{{2 \times 8}}\)  

\(\sin \theta = \frac{{{f_c}}}{f}\)

 \(\sin \theta = \frac{{1.875}}{{3.4}} = 0.5514\)

∴ cos θ = 0.83419

\(\frac{{{v_p}}}{c} = \frac{1}{{\cos \theta }}\)

\(\frac{{{v_p}}}{c} = 1.198\)

Given:

Directive gain G_d = 6 dB = 3.98

Rotating power P_rad = 16 kW

Distance r = 8 km

Calculation:

We know,

Avg power \({P_{avg}} = \frac{1}{2} \cdot \frac{{E_0^2}}{\eta }\) 

But;

\({P_{avg}} = \frac{{{P_{rod}} \cdot {G_d}}}{{area}} = \frac{{{P_{rad}} \cdot {G_d}}}{{4\pi {r^2}}}\)

∴ \(\frac{{{P_{rad}} \cdot {G_d}}}{{4\pi {r^2}}} = \frac{1}{2} = \frac{{E_0^2}}{{{\eta _0}}};\left{ {\because \eta = {\eta _0}} \right}\)

\(\frac{{{P_{rad}} \cdot {G_d}}}{{4\pi {r^2}}} = \frac{1}{2} \cdot \frac{{E_0^2}}{{120;\pi ;}}\left{ {{\eta _0} = 120\pi = 377;{\rm{\Omega }}} \right}\)

\({E_0} = \frac{{\sqrt {60 \cdot {P_{rad}} \cdot {G_d}} }}{r} = 0.244;V/m\)