GATE EC 2018 Official Paper (Previous Year Paper)

Correct answer: 2.

The sentence has two blanks. The first blank requires a word which is about a part of the cake**.**

Hence the correct word in this blank is 'piece'. Peas and Peace cannot fill the second blank.

Correct answer: Option 1.

The sentence implies that there is immense scope in the area of tourism and that it should be improved upon.

'Improvement' and 'neglected' are two words which are opposite to each other.

Note: Option 3 cannot be correct because tourism in itself cannot be famous. A country or place can be famous due to its tourism policies.

$715~_\ 423$

Let the missing number be ‘x’

For divisibility by 3, the sum of all the digits should be divisible by 3.

The sum of all the digits will be:

= 7 + 1 + 5 + x + 4 + 2 + 3 + 13 + x + 9

= 22 + x

Now, the nearest multiple of ‘3’ is 24. We can, therefore, write:

22 + x = 24

x = 24 – 22

x = 2

Cross check:

For x = 2,

The sum of the digits is, therefore:

= 22 + 2 = 24

Since 24 is divisible by 3, the smallest whole number in the place of _ will be 2

Concept:

For a GP of the form:

a + ar + ar2 + ar3 +…..

The sum is given by:

$Sum= \frac{a}{{1 - r}}$

Calculation:

The given sequence is

$x\left( n \right) = 1 + \frac{1}{4} + \frac{1}{{16}} + \frac{1}{{64}} + \frac{1}{{256}} + \ldots$

$= 1 + \frac{1}{{{2^2}}} + \frac{1}{{{2^4}}} + \frac{1}{{{2^6}}} + \frac{1}{{{2^8}}} + \ldots$

The series is GP series with:

a = 1

$r = \frac{1}{{{2^2}}} = \frac{1}{4}$

Sum of the infinite series will be:

$Sum= \frac{a}{{1 - r}}$

$= \frac{1}{{1 - \frac{1}{4}}} = \frac{1}{{\frac{3}{4}}}$

$= \frac{4}{3}$

The situation can be drawn as:

Since the length of the shadow is twice the height of the girl, we can write:

CQ = 2(1.5) = 3 m

Δ PQC ~ Δ ABC

$\frac{{PQ}}{{AB}} = \frac{{QC}}{{BC}}$

$\frac{{1.5}}{x} = \frac{3}{6}$

$x = \frac{{6 \times 1.5}}{3} = 3;m$

Concept:

The amount in compound interest is given by:

$A = P{\left( {1 + \frac{r}{{100}}} \right)^n}$

P = Principle amount

r = Rate

n = Time (in years)

Calculation:

Putting on the respective given values, we can write:

$10,00,000 = P{\left( {1 + \frac{{10}}{{100}}} \right)^5}$

${10^6} = P{\left( {\frac{{11}}{{10}}} \right)^5}$

$P = \frac{{{{10}^6} \times {{10}^5}}}{{{{11}^5}}}$

= 620921.32

Now, from the given options, the minimum value greater than the above value is:

≈ 6,21,000

A → Gold : Copper 2 : 3

B → Gold : Copper 3 : 7

Since equal of alloys A and B are melted to make the alloy C, we multiply the ratio of Gold : Copper in A by 2 to get:

A = 4 : 6 

Now, 

A = 4 : 6 Weight = 4 + 6 = 10 units

B = 3 : 7 Weight = 3 + 7 = 10 units

Weight in the mixture will be:

(4 + 3) : (6 + 7)

= 7 : 13

• Option (i) which says 'Even as a junior cricketer, John was considered a good captain.' is incorrect as there is no information provided in the passage to support this inference. We do not know if John was a captain as a junior cricketer.
• Option (iii) which says 'Fans and the Cricket Board have differing views on what they want in a captain.' is incorrect because while we know that fans love John's aggression from the phrase 'While this aggression has filled stadiums with die-hard fans', but there is no information provided in the passage to support the Cricket Board's view of John.
• Option (ii) which says 'Finding a complete captain is a challenge.' is correct as the passage talks about the difficulties John faces while he was the captain. It is very difficult to find a captain who can be 'complete' throughout the captainship.
• Option (iv) which says 'Over the past three seasons, John has accumulated big scores.' is correct as it is given in the passage that while earlier John 'found it difficult to convert his aggression into big scores', but 'Over the past three seasons though' he became a 'cerebral, calculative and successful batsman-captain'. From this, inference (iv) can be safely inferred.

Hence the correct answer is option 3.

Let,

A = Event that the car accident occurs by a blue car

B = Event that the person correctly identifies the car

The probability that the person correctly identifies the car is:

P (Correct) = $\frac{80}{100}$

Also, P(Incorrect) = $\frac{20}{100}$

The probability that the car is blue in color is:

P (Blue) = $\frac{15}{100}$

The probability that the car is green is:

P (Green) = $\frac{85}{100}$

The probability that the car accident occurs by a blue car will be:

P(A) = P(Correct) × P(Blue) + P(Incorrect) × P(Green)

$P(A)=(\frac{80}{100})(\frac{15}{100})+(\frac{20}{100})(\frac{85}{100})$

 

$P(A)= \frac{{2900}}{{{{10}^4}}}$

The required conditional Probability is:

$P\left( {\frac{A}{B}} \right) = \frac{{\frac{{80}}{{100}} × \frac{{15}}{{100}}}}{{\frac{{2900}}{{{{10}^4}}}}}$

$\Rightarrow \frac{{1200}}{{2900}} × 100 = 41.37$ %

Options 1, 2, and 3 are incorrect as they are assumptions. 

Only option 4 which says 'Helicopter connectivity would lead to an increase in tourists coming to the region.' can be logically inferred from the passage. This is because the passage mentions in the beginning that the coastal region has remained 'inaccessible to tourists due to poor connectivity' which would not be the case once a 'helicopter service to the nearest major city airport' is established.

Hence option 4 is the correct answer.

AC equivalent model will be:

Applying KVL from V_g2 to r₀₁ ground, we get:

-V_g2 - I_x r₀₁ = 0

V_g2 = -I_x r₀₁   ---(1)

V_x = r₀₂ (I_x - g_m2V_g2) + I_xr₀₁

V_x = I_x (r₀₁ + r₀₂) + g_mr₀₁r₀₂I_x

∴ The output resistance is:

\({{\rm{R}}{\rm{0}}} = \frac{{{{\rm{V}}{\rm{x}}}}}{{{{\rm{I}}{\rm{x}}}}} = {{\rm{r}}{{\rm{0}}1}} + {{\rm{r}}{{\rm{0}}2}} + {{\rm{g}}{\rm{m}}}{{\rm{r}}{{\rm{0}}1}}{{\rm{r}}{{\rm{0}}2}}\)

(r₀₁ + r₀₂) will be very small in comparison with g_mr₀₁r₀₂. We can, therefore, write:

R₀ ≅ g_mr₀₁r₀₂

Since the M₂ NMOS transistor is acting as a current buffer, i.e. I_D2 = I_D1, the transconductance will be:

\(g_{m2}= \frac{{\partial {\rm{I}}{{\rm{D}}2}}}{{\partial {{\rm{V}}{{\rm{gs}}}}}} \cong \frac{{\partial {{\rm{I}}{{\rm{D}}1}}}}{{\partial {{\rm{V}}{{\rm{gs}}}}}} = {{\rm{g}}_{{\rm{m}}1}}\)

Concept:

A capacitor does not allow a sudden change in voltage, i.e.

V_C(0⁺) = V_C(0^-)

Application:

Initially, both the capacitor voltages are zero, i.e.

V_C1 (0^-) = V_C2 (0^-) = 0 V

At t = 0, when the switch is closed, initially the Zener will be open-circuited and remains the same till the voltage across C₂ becomes greater than its breakdown voltage i.e. 2.5 volts.

Applying KCL at node A, we get:

${\rm{I_{C}}} =I= \frac{{{{\rm{1-0}}}}}{{{{\rm{1k}}_{{}}}}} = 1{\rm{mA}}$

We observe that the current through both the capacitors is constant, and hence it is a case of linear charging of capacitors. As the initial voltage is zero, hence the voltage across C₁ and C₂ for t > 0 is given as:

\({V_{{C_1}}}\left( t \right) = \frac{1}{{{C_1}}}\mathop \smallint \limits_0^t {I_C}dt = 1000\left| t \right|_0^t\)   ---(1)

Since V_C1 = V_C2, we get:

\({V_{{C_2}}}\left( t \right) = \frac{1}{{{C_2}}}\mathop \smallint \limits_0^t {I_C}~dt = 1000\left| t \right|_0^t\)

Applying KVL starting from point A and ending at output V₀, we get:

• V_A + V_C1 + V_C2 + V₀ = 0 (where V_A = 0)

V₀ = - (V_C1 + V_C2)

Cut in voltage of Zener diode = 2.5 V

∴ When V_C2(t)­ = 2.5 V, breakdown of the Zener diode will occur and the voltage will be fixed to the Zener voltage of  2.5 V, i.e.

When V_C1(t) = V_C2(t) = 2.5 V(Zener turns on)

Using Equation (1), the time required to reach 2.5 V is calculated as:

$2.5 = 1000\times {{\rm{t}}_1}$ 

∴ t₁ = 2.5 ms (Zener : ON)

But we need V_o = -10 V

∴ C₁ must continue to charge up to 7.5 V. Since the charging is linear, the time required to charge to another 5 V will be calculated using the same Equation (1) as:

$5 =1000 \times {{\rm{t}}_2}$ 

${{\rm{t}}_2} = 5{\rm{;msec}}$

∴ the total times at which the output Voltage (V_out) crosses (- 10) V will be:

T = t₁ + t₂ = (2.5 + 5) msec = 7.5 ms

Transresistance or trans-impedance amplifier is normally used as a current to voltage converter which requires low input and low output impedances for its proper operation.

Requirements of input and output impedances for different amplifiers are listed below:

Concept:

A system is said to be linear if it satisfies to properties:

Additivity:

If states that, if an input x₁(t) produces output y₁(t) and another input x₂(t) also acting along produces output y₂(t), then, when path inputs acting on the system simultaneously, produce output y₁(t) + y₂(t).

If x₁(t) → y₁(t)

x₂(t) → y₂(t)

then x₁(t) + x₂(t) → y₁(t) + y₂(t)

Homogeneity:

It states that if input is scaled by c, then output also scaled by same amount.

if x₁(t) → y₁(t)

then cx₁(t) → cy₁(t): c is any real or Imaginary number

Solution:

y = au + b

y(t) = au(t) + b

u(t) = u₁(t) + u₂(t)

y(t) = au₁(t) + au₂(t) + b

y(t) ≠ [au₁(t) + b] + [au­₂(t) + b]

y(t) ≠ y₁(t) + y₂(t)

Hence the system is non linear

y = au + b is a nonlinear system

• Both Nyquist and RH criteria can be used to find the range of gain (k) for the system to be stable
• Bode plot can be used to find the transfer function which further can be used to determine the Nyquist Plot.
• RH criteria do not give accurate terms in case of exponential terms, an approximation can be obtained using exponential expansion up to 1st degree
• Nyquist plots are the continuation of polar plots for finding the stability of the closed-loop control systems by varying ω from −∞ to ∞.

 

Derivation:

Let us consider O.L.T.F of a unity feedback system as:

$G\left( s \right) = \frac{K}{{s\left( {s + a} \right)}}$

This represents a type = 1 and order – 2 system for which the Nyquist plot will be as shown below:

The closed-loop transfer function for the system is given as:

$T\left( s \right) = \frac{{G\left( s \right)}}{{1 + G\left( s \right)}} = \frac{{K/s\left( {s + a} \right)}}{{1 + \left[ {\frac{K}{{s\left( {s + a} \right)}}} \right]}}$

$T\left( s \right) = \frac{K}{{{s^2} + as + K}}$

We can draw its Nyquist plot by simply substituting s = jω and obtain polar coordinates of T(jω) for different values of ω as:

$T\left( {j\omega } \right) = \frac{K}{{{{\left( {j\omega } \right)}^2} + ja\omega + K}}$

$T\left( 0 \right) = \frac{K}{{0 + 0 + K}} = 1$

T(∞) = 0

So, the statement given in option (D) is false.

Concept:

If p(s) has 'n' real roots, then p'(s) will have at least 'n - 1' real roots.

Application:

Given p(s) = s³ + a₂ s² + a₁s + a₀

p'(s) = 3s² + 2a₂s + a₁

Also given that p'(s) has no real roots, therefore, we can write:

n - 1 = 0

n = Number of real roots in p(s)

∴ n = 1

In a p-n junction at equilibrium, i.e. no bias

The hole electrons flow (drift, diffusion) directions are as follows (convention based on the above PN diagram)

 

• The concentration of holes on the p-side is much greater than that in the n-side, a very large hole diffusion current tends to flow across the junction from the p to the n material.
• Hence an electric field must build up across the junction in such a direction that a hole drift current will tend to flow across the junction from n-side to p-side in order to counterbalance the diffusion current.
• After the depletion region is formed and an electric field is a set-up; holes on the p-side and electrons on the n-side move away from the junction.
• However, diffusion current components (before the formation of the depletion region) and drift current components (after the formation of the depletion region) are in the same direction.
• Electron drift and diffuse currents are in the opposite direction.

​Option (d) is not true.

Concept:

CMOS logic circuit is an extension of a CMOS inverter. It consists of two network transistors, a pull-down network (PDN) constructed of an n-MOS and Pull-up Network (PUN) constructed of P-MOS.

PDN: Since nMOS conducts when the signal gate is high, PDN is activated when the inputs are high.

PUN: It comprises PMOS and conducts when the input signal gate is low.

The PDN and PUN are connected in parallel to form OR logic function and they are connected in series to form AND logic as shown:

 

 

 

Application:

$\overline {{\rm{\bar x}}.{\rm{\bar y}} + {\rm{xy}}} = \overline {{\rm{x}} \odot {\rm{y}}}$ = x ⊕ y = XOR

F = A̅.B̅.C̅ + A̅.B.C̅ + A.B̅.C̅

In terms of minterms, this can be represented as:

F = ∑m (0, 2, 4)

The equivalent maxterm will contain the terms not present in the minterm representation, i.e.

F = ∑m (0, 2, 4) = π(1, 3, 5, 6, 7) = M₁. M₃. M₅. M₆. M₇

⇒ (A + B + C̅) (A + B̅ + C̅) (A̅ + B + C̅) (A̅ + B̅ + C) (A̅ + B̅ + C̅)

Smith chart consists of constant resistance and constant reactance circle representing normalized impedance. (Z_L/Z_O)

The various points on the smith chart are:

Point P: Short circuit (r = 0, x = 0)

${{\rm{\Gamma }}_L} = \frac{{{Z_L} - {Z_0}}}{{{Z_L} + {Z_0}}}$

Z_L = 0

Γ_L = -1 = 1∠180°

Point Q: Matched load (r = 1, x = 0)

${{\rm{\Gamma }}_L} = \frac{{{Z_L} - {Z_0}}}{{{Z_L} + {Z_0}}}$

Matched load Z_L = Z₀

Γ_L = 1

Point R: Open circuit (r = ∞, x = ∞) (Z_L = ∞)

${{\rm{\Gamma }}_L} = \frac{{{Z_L} - {Z_0}}}{{{Z_L} + {Z_0}}}$

$= \frac{{1 - \frac{{{Z_0}}}{{{Z_L}}}}}{{1 + \frac{{{Z_0}}}{{{Z_L}}}}}$

Z_L = ∞

Γ_L = 1 ∠0

The upper half of the smith chart represents the inductive load.

The lower half of the smith chart represents capacitive.

The given values are given by the solution of the characteristic equation, i.e.

|A - λI| = 0

λ₁, λ₂ ….. are different eigenvalues.

If Eigenvalues are distinct, then corresponding eigenvectors are linearly independent.

Vectors V₁, V₂ …. V_n are linearly independent.

If C₁V₁ + C₂V₂ + C₃V₃ …… + C_nV_n = 0

Implies C₁ = C₂ = ……. C_n = 0

|A| ≠ 0 implies that ‘O’ is not an eigenvalues

Other eigenvalues may be repeated, i.e. 1, 2, 2

The two same eigenvalues give linearly dependent eigenvectors.

$f\left( {x,y} \right) = \frac{{a{x^2} + b{y^2}}}{{xy}} = \frac{{ax}}{y} + \frac{{by}}{x}$

$\frac{{\partial f}}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {\frac{{ax}}{y} + \frac{{by}}{x}} \right]$

$\frac{\partial f}{\partial x}=\frac{a}{y} - \frac{{by}}{{{x^2}}}$

At, x = 1 and y = 2, we get:

$\frac{\partial f}{\partial x}|_{x=1,y=2}=\frac{a}{2} - \frac{{b\left( 2 \right)}}{{{1^2}}}$

$= \frac{a}{2} - 2b$   ----(1)

Similarly,

$\frac{{\partial f}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {\frac{{ax}}{y} + \frac{{by}}{x}} \right]$

$= - \frac{{ax}}{{{y^2}}} + \frac{b}{x}$

At, x = 1 and y = 2

$\frac{\partial f}{\partial y}|_{x=1,y=2}= \frac{{ - a}}{4} + b$     ----(2)

For the given condition, equating the value of $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ at x = 1 and y = 2, we get:

$\frac{a}{2} - 2b = \frac{{ - a}}{4} + b$

$\frac{a}{2} + \frac{a}{4} = 3b$

$\Rightarrow \frac{{2a + a}}{4} = 3b$

3a = 12b

a = 4b

Concept:

Properties of discrete-time all pass system:

(1) Magnitude of |H(e^jω)| = constant

(2) Location of ${\rm{Zeroes}} = \frac{1}{{{\rm{Pol}}{{\rm{e}}^{\rm{*}}}}}$

(3) Complex pole exist in conjugate pair

Application:

Now pole one ${{\rm{P}}_1} = 0.25\angle 0{\rm{;}}$

${{\rm{Z}}_1} = \frac{1}{{{{\left( {0.25\angle 0} \right)}^{\rm{*}}}}} = 4\angle 0$

${{\rm{P}}_2} = 2\angle 30{\rm{;}}$  

${{\rm{Z}}_2} = \frac{1}{{{{\left( {2\angle - 30} \right)}^{\rm{*}}}}} = 0.5\angle 30^\circ$

${{\rm{P}}_3} = 2\angle - 30{\rm{;}}$

${{\rm{Z}}_3} = \frac{1}{{{{\left( {2\angle - 30^\circ } \right)}^{\rm{*}}}}} = 0.5\angle - 30^\circ$

Pole P₁ lies inside the unit circle whereas P₂ and P₃ lie outside the unit circle.

For stability: ROC must include in the unit circle

∴ It is stable only when the impulse response is two-sided

Given a fundamental period = 10

According to the question, the function is the same but you can consider another period

If the function is the same, the Fourier series coefficient will also be the same, i.e.

\(\mathop \sum \limits_{{\rm{k}} = - \infty }^\infty \left| {{{\rm{b}}{\rm{k}}}} \right| = \mathop \sum \limits{{\rm{k}} = - \infty }^\infty \left| {{{\rm{a}}_{\rm{k}}}} \right| = 16\)

V_in = 6V ± 5 % indicates a variation from  5.7 V to 6.3 V

If the load voltage is 5 V, the load current = 5 mA

The source current Is will be:

I_s = 2 + 5 = 7 mA

V_i(min) = 5.7 V

${\rm{R}} = \frac{{5.7 - 5}}{{7.5}} = 100{\rm{;\Omega }}$

${{\rm{V}}_{{\rm{i}}\left( {{\rm{max}}} \right)}} = 6.3~{\rm{V}}$

${\rm{R}} = \frac{{6.3 - 5}}{{7{\rm{mA}}}} \cong 186{\rm{;\Omega }}$

For maximum power dissipation, we take R = 100 Ω

${{\rm{I}}_{\rm{s}}} = \frac{{6.3 - 5}}{{100}} = 13{\rm{;mA}}$ 

I_Z = 13 -5 = 8 mA

P_z = 5 × 8 = 40 mW

∴ The minimum power rating of the device means higher the maximum power dissipation in the circuit.

c(t) = A_c cos 2πf_ct

m(t) = cos 2π f_m(t)

signal: c(t) + m(t)

v_i = A_c cos 2π f_ct + cos 2π f_mt

output of non-linear device:

v₀(t) = av_i(t) + bv_i²(t)

v₀(t) = a[A_ccos2π f_ct + cos 2π f_mt] + b[A_c cos2π f_ct + cos2π f_mt]

v₀(t) = aA_ccos2π f_ct + acos2π f_mt + b[A_c²cos²2π f_ct + bcos²2π f_mt + 2bA_ccos2πf_ct cos2π f_mt]

When the above signal is passed through the band pass filter with center frequency f_c and bandwidth 3 f_m, the output will contain:

y(t) = aA_c cos(2πf_ct) + 2bA_c cos(2πf_c) cos(2πf_mt)

The carrier power will be:

${P_c} = \frac{{{{\left( {aAc} \right)}^2}}}{2} = \frac{{{a^2}A{c^2}}}{2}$

And the side band power will be:

${P_{SB}} = \frac{{2{{\left( {bAc} \right)}^2}}}{2} = {b^2}A{c^2}$

${P_{SB}} = \frac{1}{2}{P_c}$

${b^2}A{c^2} = \frac{1}{2}\left( {\frac{{{a^2}A{c^2}}}{2}} \right)$

$\frac{{{a^2}}}{{{b^2}}} = 4$

$\frac{a}{b} = 2$

Concept: 1. PSD of output = |H(f)|2  PSD of input. 

                 2. Fourier transform of Gaussian Pulse is also Gaussian in nature i.e ${e^{ - a{t^2}}}\mathop \leftrightarrow \limits^{F.T.} \sqrt {\frac{π }{a}} {e^{ - {π ^2}{f^2}/a}}$                  3. Average Power from PSD can be calculated as \(P = \mathop \smallint \limits_{ - \infty }^{ + \infty } Sy\left( f \right)df \)

Application: Given $\frac{{No}}{2} = 0.5;W/Hz$

$h\left( t \right) = 0.5;{e^{ - {t^2}/2}}$

The output  P_SD  is related to the input PSD as:

Sy(f) = |H(f)|² S_N(f)    ----(1)

With $h\left( t \right) = 0.5{e^{ - {t^2}/2}}$, the Fourier transform is calculated as:

${e^{ - a{t^2}}}\mathop \leftrightarrow \limits^{F.T.} \sqrt {\frac{π }{a}} {e^{ - {π ^2}{f^2}/a}}$

The Fourier transform of h(t) will be:

$0.5{e^{ - {t^2}/2}}\mathop \leftrightarrow \limits^{F.T.} 0.5\sqrt {\frac{π }{{{0.5}}}} {e^{ - \frac{{{π ^2}{f^2}}}{{{0.5}}}}}$

$0.5{e^{ - \frac{{{t^2}}}{2}}}\mathop \leftrightarrow \limits^{F.T.} 0.5\sqrt {2π } {e^{ - 2{π ^2}{f^2}}}$

Substituting this in Equation (1), we get:

$Sy\left( f \right) = {\left| {0.5\sqrt {2π } {e^{ - 2{π ^2}{f^2}}}} \right|^2}\left( {0.5} \right)$

$Sy\left( f \right) = 0.125\left( {2π } \right){e^{ - 4{π ^2}{f^2}}}$

$Sy\left( f \right) = \frac{π }{4}{e^{ - 4{π ^2}{f^2}}}$

The integration of PSD gives power, i.e. the area under the PSD gives power.

Total power P will therefore be:

\(P = \mathop \smallint \limits_{ - \infty }^{ + \infty } Sy\left( f \right)df \)

\(P=\frac{π }{4} \mathop \smallint \limits_{ - \infty }^{ + \infty } {e^{ - 4{π ^2}{f^2}df}}\)   ---(1)

The area of $e^{-\pi f^2}$ is always equal to 1, i.e.

\(\mathop \smallint \limits_{ - \infty }^{ + \infty } {e^{ - {π }{f^2}df}} = 1\)

Using the area scaling property, we get:

\(\mathop \smallint \limits_{ - \infty }^{ + \infty } x\left( t \right)dt = A\)

\(\mathop \smallint \limits_{ - \infty }^{ + \infty } x\left( kt \right)dt = A/k\)

Applying this property to Equation (1), we can write:

\(P=\frac{π }{4}\mathop \smallint \limits_{ - \infty }^{ + \infty } {e^{ - π {{\left( {2\sqrt π f} \right)}^2}}}df = \frac{1}{{2\sqrt π }}\)

$\Rightarrow \frac{π }{4} \times \frac{1}{{2\sqrt π }} = \frac{{\sqrt π }}{8} = 0.2215$

The transfer function is calculated as:

${\rm{T}}\left( {\rm{s}} \right) = \frac{{{\rm{C}}\left( {\rm{s}} \right)}}{{{\rm{R}}\left( {\rm{s}} \right)}}={\rm{C}}{\left[ {{\rm{sI}} - {\rm{A}}} \right]^{ - 1}}{\rm{B}}$

\(\left[ {{\rm{sI}} - {\rm{A}}} \right] = \left[ {\begin{array}{*{20}{c}} {{\rm{s}} + 4}&{1.5}\ { - 4}&{\rm{S}} \end{array}} \right]\)

\({\left[ {{\rm{sI}} - {\rm{A}}} \right]^{ - 1}} = \left[ {\begin{array}{*{20}{c}} {\rm{s}}&{ - 1.5}\ 4&{{\rm{s}} + 4} \end{array}} \right]\frac{1}{{{{\rm{s}}^2} + 4{\rm{s}} + 6}}\)

\({\left[ {{\rm{sI}} - {\rm{A}}} \right]^{ - 1}}{\rm{B}} = \left[ {\begin{array}{*{20}{c}} {2{\rm{s}}}\ 8 \end{array}} \right]\frac{1}{{{{\rm{s}}^2} + 4{\rm{s}} + 6}}\)

${\rm{C}}{\left[ {{\rm{sI}} - {\rm{A}}} \right]^{ - 1}}{\rm{B}} = \frac{{3{\rm{s}} + 5}}{{{{\rm{s}}^2} + 4{\rm{s}} + 6}}$

The bandgap of LED is related to its wavelength of light as:

${E_g} = \frac{{hc}}{λ }$

i.e. ${E_g}~or;{V_{bi}};\propto \frac{1}{λ }$

Hence, as λ increases, the bandgap, and the build-in potential decreases, i.e.

For λ_R > λ_G > λ_B, the bandgap of the respective diodes will be:

⇒ V_R < V_G < V_B

The output of the first multiplexer will be:

A = U̅V̅0 + U̅V.1 + UV̅.1 + UV.0

A = (U̅V + UV̅)   ---(1)

Now, the output of second multiplexer will be:

F = A.(W̅ X̅) + A(W̅X) + 0.(WX̅) + 0.(WX)

F = AW̅ (X̅ + X) = AW̅

Subsitutin A from Equation (1), we get:

F = (U̅V + UV̅) W̅

When W₀ = 1

B₀ =1 and B₁ = 0

When 

When W1 = 1

B0 =0 and B1 = 1

Hence the matrix is 

\(\left[ {\begin{array}{{20}{c}} {{{\rm{D}}{00}}}&{{{\rm{D}}{01}}}\ {{{\rm{D}}{10}}}&{{{\rm{D}}{11}}} \end{array}} \right] = \left[ {\begin{array}{{20}{c}} 1&0\ 0&1 \end{array}} \right]\)

Concept:

The amplitude of E.M wave when it propagates through a lossy medium is given by:

$E = {E_0}{e^{ - \alpha z}}$

α = attention factor which is the inverse of the skin depth (δ), i.e.

$\alpha= \frac{1}{{skin;depth;\left( δ \right)}}$

Application:

$\alpha = \frac{1}{{0.1}} = 10;{m^{ - 1}}$

E = E₀ e ^-αz

$\frac{E}{{{E_0}}} = {e^{ - \alpha z}}$

$20\log \left( {\frac{E}{{{E_0}}}} \right) = 20\log \left( {{e^{ - \alpha z}}} \right)$

For the wave to attenuate by 20 dB, we can write:

$-20= 20\log \left( {{e^{ - \alpha z}}} \right)$

-1 = log(e^-αz)

e^-αz = (10)^-1

e-αz = 0.1

αz = 2.302

$z = \frac{{2.302}}{\alpha } = \frac{{2.302}}{{10}}$

= 0.2302 m

$\frac{{dy}}{{dx}} = \frac{{{x^2} + {y^2}}}{{2y}} + \frac{y}{x}$

$\frac{{dy}}{{dx}} = \frac{{{x^2}}}{{2y}} + \frac{y}{2} + \frac{y}{x}$

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

y² = t

$2y\frac{{dy}}{x} = \frac{{dt}}{{dx}}$

$\frac{1}{2}\frac{{dt}}{{dx}} - \left( {\frac{1}{x} + \frac{1}{2}} \right)t = \frac{{{x^2}}}{2}$

$\frac{{dt}}{{dx}} - 2\left( {\frac{1}{x} + \frac{1}{2}} \right)t = {x^2}$

The above is linear differential equation of the form

$\frac{{dt}}{{dx}} = + pt = Q$

Integrating factor $= {e^{ - 2\smallint \left( {\frac{1}{x} + \frac{1}{2}} \right)dx}}$

= e^{-2ln x}. e^-x

$= {e^{\ln {x^{ - 2}}}}.{e^{ - x}}$

$= \frac{{{e^{ - x}}}}{{{x^2}}}$

Solution is given by:

$\frac{{t;{e^{ - x}}}}{{{x^2}}} = \smallint \frac{{{x^2}{e^{ - x}}}}{{{x^2}}}dx + c$

Substitute t = y²

$\frac{{{y^2}{e^{ - x}}}}{{{x^2}}} = - {e^{ - x}} + c$

At, x = 1, y = 0

0 = -e^-1 + c

c = e^-1 = y­_e

substituting in the solution

${y^2}.\frac{{{e^{ - x}}}}{{{x^2}}} = - {e^{ - x}} + {e^{ - 1}}$

$\left( {\frac{{{y^2}}}{{{x^2}}} + 1} \right){e^{ - x}} = {e^{ - 1}}$

$\left( {1 + \frac{{{y^2}}}{{{x^2}}}} \right) = \frac{{{e^{ - 1}}}}{{{e^{ - x}}}} = {e^{x - 1}}$

Taking log both sides

$\ln \left( {1 + \frac{{{y^2}}}{{{x^2}}}} \right) = x - 1$

Concept:

Z = R + Xc

Where Z is the total impedance

R is the resistance of the circuit

X_c is the capacitive reactance of the circuit

${X_c} = \frac{1}{{j2\pi fC}}$

Where,

f is the frequency of the supply

C is the capacitance of the circuit.

Phasor

A phasor is a vector representation of a sinusoidal function that has both a magnitude and a phase angle.

Representations:

Rectangular form: z = x + jy

Polar form: z = r ∠ ϕ

Exponential form: z = z ejϕ 

Calculation:

The capacitive impedance is given by:

$X_C=\frac{1}{ω C}$

With ω = 5 and C = 1 μF, we can write:

${{\rm{X}}_{\rm{C}}} = \frac{1}{{5 \times {{10}^{ - 6}}}}Ω$

X_C = 200 kΩ 

The voltage across the capacitor will be:

${{\rm{V}}_{\rm{C}}} = 5\angle 0 \times \frac{{\left( { - {\rm{j}}200} \right)}}{{200 - {\rm{j}}200}}$

$V_C= 5\angle 0 \times \frac{{ - {\rm{j}}\left( {1 + {\rm{j}}} \right)}}{2}$

$V_C= 2.5\sqrt 2 \angle - 45^\circ$

In terms of π, the above can be written as:

$V_C= 2.5\sqrt 2 \angle - 0.25{\rm{\pi }}$

Converting the phasor representation back to the time domain representation, we can write:

$\ {{\rm{V}}_{\rm{c}}}\left( {\rm{t}} \right) = 2.5\sqrt 2 \sin \left( {5{\rm{t}} - 0.25{\rm{t}}} \right)$