GATE EC 2017 Official Paper: Shift 2 (Previous Year Paper)

Let us look at the meanings of the given words:

• Figuratively - not using words to mean something literal; metaphorically
• For e.g.- he man flew out the front door, figuratively.
• Retrospectively - with consideration of past events or situations
• For e.g.-Their actions could only be understood retrospectively.
• Respectively - separately or individually and in the order already mentioned
• For e.g.- Their sons, Ben and Jonathan, were three and six respectively.
• Rightfully - according to a legitimate right or claim to property, position, or status
• For e.g.- She could rightfully claim that she is one of the world's leading authorities on the topic.

In the context of the given sentence, we use an adverb that will help us enumerate two items (Monday and Tuesday) that refer back to a previous part of the statement (ninth and tenth, in that order). Out of the given options, respectively fits the context perfectly.

The complete sentence should be:

The ninth and the tenth of this month are Monday and Tuesday respectively.

Hence, option 3 is correct.

We can see from the context of this sentence that a comparison is being made between two items. 

Hence, we will use the comparative form of the adjective 'easy' i.e. 'easier'.

Also, when comparing two things, the comparative forms of adjectives are paired with the preposition 'than'.

For example: He is taller than his brother.

Out of the given options, only option 1 has both the correct words that fit the context of the blanks.

A rule states that in order to drink beer, one must be over 18 years old.

P is 16 years old – as P’s age is lower than 18, we need to check his drink

Q is 25 years old – as Q’s age is greater than 18 no need to check his drink

R is drinking milkshake – as R is drinking milkshake no need to check his age

S is drinking a beer – as S is drinking beer, we need to check his age

So, we need to check P’s drink and S’s age.

The path of Fatima is shown

From the figure

x² + 6² = 10²

x² = 64

x = 8 km

Hence, Fatima should move 8 km to East.

n(C) = 329

n(P) = 186

n(M) = 295

n(C⋂P) = 83

n(C⋂M) = 217

n(P⋂M) = 63

n(P⋃C⋃M) = 500

n(P⋃C⋃M) = n(P) + n(C) + n(M) - n(P⋂C) -n (C⋂M) - n(M⋂C) + n(P⋂C⋂M)

500 = 186 + 329 + 295 - 83 - 217 - 63 + n(P⋂C⋂M)

n(P⋂C⋂M) = 53

Out of the given options, the sentence in option 1 i.e. 'An intimate association does not allow for the necessary perspective.' best reflects the opinion stated by the author in the final part of the sentence. It is the opinion that the closer we are to a particular topic or subject, the more biased is our perspective towards it.

Option 2 i.e. 'Matters are recorded with an impartial perspective.' is vague and does not convey much information. 

Option 3 i.e. 'An intimate association offers an impartial perspective.' is completely contradictory to the author's opinions, which state that an intimate association offers a partial perspective.

Option 4 i.e. 'Actors are typically associated with the impartial recording of matters.' is factually distorted and also fails at reflecting any part of the author's opinions.

Hence, the correct answer is option 1.

The given information can be represented as shown below.

Z is the grandfather of the daughter S through R. So, P and R are the parents of S and it implies that P and R are married to each other.

So, the statement “P and R are not married to each other” is necessarily false.

Let one man can complete work in x weeks and one woman can complete in y weeks

In 1 day work completed by man \(= \frac{1}{x}\)

In 1 day work completed by woman \(= \frac{1}{y}\)

According to questions

\(\frac{{1200}}{x} + \frac{{500}}{y} = \frac{1}{2}\) 

\(\frac{{900}}{x} + \frac{{250}}{y} = \frac{1}{3}\) 

⇒ x = 3600

1 man can complete in 3600 weeks to complete work in 1 week, 3600 men are required

Total number of 3 digits numbers

= 999 - 99 = 900

Case I:

21 are together as shown

Total cases = 10

Case 2:

21 are together as shown

Total cases where 1 comes to right of 2 = 10 + 9 = 19

Total cases where 1 does not come to immediate right of 2 = 900 - 19 = 881

The steepest path will be the one where there will be maximum number of contour lines.

P to Q → 4 contour lines

P to R → 7 contour lines

P to S → 3 contour lines

P to T → 3 contour lines

Hence, steepest path is from P to R.

Calculation:

Given:

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

Apply now transformation, R₄ → R₄ + R₁, we get,

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

R₂ ↔ R₃

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

R₄ → R₄ + R₂

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

R₄ → R₄ + R₃

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

R₅ → R₅ + R₄

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

This matrix is in triangular form with last row = 0,

So, Rank = 4

Given Differential equation:

\(\frac{{{d^2}y}}{{d{x^2}}} + \frac{{2dy}}{{dx}} - 5y = 0\)

Auxillary equation = D² + 2D – 5 = 0

Solving, we get roots \(= \frac{{ - 2 \pm \sqrt {4 + 20} }}{2}\)

\(=\frac{{ - 2 \pm \sqrt {24} }}{2} = - 1 \pm \sqrt 6\)

Therefore the general solution is;

y = k₁e^(-1+√6)x + k₂e^(-1-√6)x

Concept:

Equation of a plane is given as:

ax + by + cz = d

The unit vector normal to the plane is given by:

\({a_n} = \frac{{a\hat i + b\hat j + c\hat z}}{{\sqrt {{a^2} + {b^2} + {c^2}} }} \)

Calculation:

Given two planes with equation:

x + y + z = 1     … (1)

2x - y + 2z = 0     … (2)

Unit vectors in the direction ⊥ to the respective planes are:

\({a_{{n_1}}} = \frac{{\hat i + \hat j + \hat z}}{{\sqrt 3 }}\) and \({a_{{n_2}}} = \frac{{2\hat i - \hat j + 2\hat z}}{3}\) respectively.

\(\widehat {{a_{{n_1}}}}.\widehat {{a_{{n_2}}}} =\) cos (θ)

[∵ a̅.b̅ = |a̅| |b̅| cos (θ)]

\(\left( {\frac{{\hat i + \hat j + \hat k}}{{\sqrt 3 }}} \right).\left( {\frac{{2\hat i - \hat j + 2\hat k}}{3}} \right) =\) cos (θ)(1)(1)

\(= \frac{{2 - 1 + 2}}{{3\sqrt 3 }} =\) cos(θ)

\(=cos\theta = \frac{1}{{\sqrt 3 }} = 0.577\)

θ = 54.73°

Concept:

For simple poles at z = a, b, c…

Residue of \({{ f\left( z \right)} _{z = a}} = \mathop {\lim }\limits_{z \to a} \left{ {\left( {z - a} \right) \times f\left( z \right)} \right}\)

For multiple poles at z = a, a, a … n times

{Residue of f(z)}_z=a \(= \frac{1}{{\left( {n - 1} \right)!}}{\left{ {\frac{{{d^{n - 1}}}}{{d_z^{n - 1}{}}}{\left( {z - a} \right)}^n f\left( z \right)} \right}_{z = a}}\)

Calculation:

Given, \(f\left( z \right) = \frac{1}{{\left( {z - 4} \right){{\left( {z + 1} \right)}^3}}}\)

For a simple pole at z = 4

Residue of {f(z)}_z=4 \(= \mathop {\lim }\limits_{z \to 4} \left{ {\left( {z - 4} \right).\frac{1}{{\left( {z - 4} \right){{\left( {z + 1} \right)}^3}}}} \right}\)

\( = \mathop {\lim }\limits_{z \to 4} \frac {1}{\left( {z + 1} \right)^3} \Rightarrow \frac{1}{{{5^3}}} = \frac{1}{{125}}\)

For multiple pole (n = 3) at z = -1

Residue will be  

\(\Rightarrow \frac{1}{{\left( {3 - 1} \right)!}}\left{ {\frac{{{d^2}}}{{d{z^2}}}{{\left( {z + 1} \right)}^3}.\frac{1}{{\left( {z - 4} \right){{\left( {z + 1} \right)}^3}}}} \right} \)

\(\Rightarrow {z_{ \to - 1}}\frac{{1\left( { - 1} \right)\left( { - 2} \right)}}{2}\frac{1}{{{{\left( {z - 4} \right)}^3}}}\)

\(\Rightarrow {z_{ \to - 1}}\frac{1}{{{{\left( {z - 4} \right)}^3}}} \Rightarrow \frac{1}{{{{\left( { - 5} \right)}^3}}} = - \frac{1}{{125}}\)

Concept:

In a series RLC circuit; if the voltage and current are in phase, then this indicates that the inductive and capacitive reactance are both equal and the circuit is in resonance.

Given, I̅ and V̅ are in phase.

i.e. X_C = X_L.

Calculation:

At resonance,

\({\omega _o} = \frac{1}{{\sqrt {LC} }}\)    ---(1)

The voltage across the capacitor is:

\(|V_c|=|X_c|\times|I|\)    ---(2)

Similarly, the voltage across the resistor is:

 \( \left| {{V_R}} \right| =|I|\times{R}\)    ---(3)

Using Equations 1,2 and 3 we get:

\(\frac{{\left| {{V_C}} \right|}}{{\left| {{V_R}} \right|}} = \frac{{{X_C}}}{R} \)

\(\Rightarrow\frac{1}{{{\omega _o}C \times R}} = \frac{{\sqrt {LC} }}{{C \times R}} = \sqrt {\frac{L}{C}} .\frac{1}{R}\)

Given,

R = 5Ω

L = 5 H

and C = 5 F

So, \(\frac{{\left| {{V_C}} \right|}}{{\left| {{V_R}} \right|}} = \frac{1}{5} = 0.20\)

Concept:

The Net resistance when two resistances are in parallel is smaller than the smallest resistance in parallel.

The parallel combination of resistances R₁ & R₂ is given by:

\({R_{net}} = \frac{{{R_1}{R_2}}}{{{R_1} + {R_2}}}\)

The series combination of resistances of R₁ and R₂ is given by:

R_net = R₁ + R_{2 .}

Calculation:

The given configuration is as shown:

Given resistances: 10 Ω, 5 Ω, and 2 Ω.

To get the maximum value of net resistance, A needs to be maximum, and 5 and 2 ohms will be in parallel.

Taking A = 10 Ω, B = 5 Ω and C = 2Ω

(R_net) _max = 10 + (5||2)

=10 + 10/7 = 10 + 1.4 = 11.42 Ω

To get the minimum value of net resistance, A needs to be minimum.

Taking  A = 2 Ω, B = 10 Ω and C = 5 Ω

(R_net) _min = 2 + (5||10)

\(= 2 + \frac{{50}}{{18}}\)

\(=2 + \frac{{10}}{3} = 2 + 3.33\)

= 5.33 Ω

\(\frac{{{{\left( {{{\text{R}}_{{\text{net}}}}} \right)}_{{\text{max}}}}}}{{{{\left( {{{\text{R}}_{{\text{net}}}}} \right)}_{{\text{min}}}}}} = \frac{{11.428}}{{5.33}} = 2.14\)

Bandpass Filter:

This filter passes a certain band of frequencies and blocks low and high frequencies.

It has two corner frequencies, smaller (fc1) and larger (fc2).

It attenuates signals with frequencies below fc1 and above fc2.

Calculation:

x(n) = 5δ (n) – 7δ (n - 1) + 7δ (n - 3) – 5δ (n - 4)

Since  δ(n) ↔ 1

δ(n - to) ↔ e-jωto

Taking the Fourier transform of x[n], we get:

H(e^jω ) = 5 – 7e^-jω  + 7e^-3jω  – 5e^-4jω               

= 5(1 – e^-4jω) + 7(e^-j3ω – e^-jω)

= 5 e^-j2ω (e^j2ω – e^-j2ω) – 7e^-j2ω (e^jω – e^-jω )

= 5e^-j2ω (2j sin (2ω)) – 7e^-j2ω (2j sin ω )

= e^-j2ω [10j sin 2ω – 14j sin ω]

When ω  = 0; |H(e^jω)| = 0 (Low frequency)

When \({\rm{\omega }} = \frac{\pi }{2},\left| {H\left( {{e^{j{\rm{\omega }}}}} \right)} \right| = 14\) (medium frequency)

When \({\rm{\omega }} = \pi ;\left| {H\left( {{e^{j{\rm{\omega }}}}} \right)} \right| = 0\) (High frequency)

So, the given filter is a band-pass filter.

Concept:

To prove that a given system is a linear, it has to satisfy both the homogeneity and superposition property.

And to prove a given system to be Time Invariant, the input is first delayed and the output is noted. Then, the output is delayed by the same amount, and is noted.

If both the output; i.e. the output when the input is delayed and delayed output are equal, then the system is said to be time invariant.

Calculation:

Given system \( \Rightarrow y\left( t \right) = \mathop \smallint \limits_{t - T}^t x\left( u \right)dx\)

Delaying the input first;

x(u) → x(u – t₀)

let,  u – t₀ = λ

du = dλ

When, u = t – T

λ = t – t₀ – T

and when u = t,

λ = t – t₀

So, \(y\left( t \right) = \mathop \smallint \limits_{t - {t_0} - T}^{t - {t_0}} x\left( \lambda \right)d\lambda ;;; \ldots \left( i \right)\)

Now, when the output is delayed by the same amount we get;

\(y\left( {t - {t_0}} \right) = \mathop \smallint \limits_{t - {t_0} - T}^{t - {t_0}} x\left( u \right)dx;;;; \ldots \left( {ii} \right)\)

We observe that both equation (i) and equation (ii) are equal,

So the system is Time-Invariant.

The system is also linear because integral of an input is a linear operation.

Ideal enhancement MOSFET is a MOSFET which saturates when V_DS ≥ V_GS - V_T and allows a constant current to flow across it even after a further increase in V_DS.

But for a non-ideal MOSFET (i.e. MOSFET in which channel length modulation effect is significant), a change in V_DS beyond V_GS - V_T, will result in a further increase in current beyond saturation. This effect is known as channel length modulation.

The output characteristics will now become:

now is no longer infinite.

• As the reverse-biased base-collector voltage is increased, the transition (or) the depletion region of the base-collector junction is increased resulting in the extension of this region more towards P-side (or base side), thereby reducing the effective base width.
• This reduction in the effective base-width results in the increase in current gain as well because of the increase in the slope of concentration gradient variation from the emitter side to the collector side.

Concept:

For VDS < VGS – VT, n-channel enhancement type MOSFET will operate in triode region:

\({I_D} = \frac{{{\rm{W\mu_n }}{C_{ox}}}}{{2L}}\left[ {2\left( {{V_{GS}} - {V_T}} \right){V_{DS}} - V_{DS}^2} \right]\)

For V_DS ≥ V_GS – V_T, the MOSFET will be in saturation with current given by:

\({I_D} = \frac{{{\rm{W\mu_n }}{C_{ox}}}}{{2L}}\left[ {{{\left( {{V_{GS}} - {V_T}} \right)}^2}} \right]\)

Also, the transconductance is defined as:

\({g_m} = \frac{{\partial{I_D}}}{{\partial{V_{GS}}}}\)

Calculation:

Given,

V_DS = 0.1 V

V_GS = 0.7 V

V_th = 0.3 V

V_GS – V_th = 0.4

V_DS = 0.1 < 0.4;

So, the MOSFET is in the triode region:

\({g_m} = \frac{{d{i_D}}}{{d{V_{GS}}}} = \frac{{{\rm{W\mu }}{C_{ox}}}}{{2L}}.2{V_{DS}} = \frac{{{\rm{W\mu }}{C_{ox}}{V_{DS}}}}{L}\)

Putting values, we get,

g_m = 50 × 100μ × 0.1

= 0.5 mA/V

Concept:

DC-voltmeter gives the average value of the varying voltage.

The average value of a varying voltage is given by:

\(V_{avg}=\frac{1}{T}\mathop \smallint \limits_0^t V(t).dt\)

Also, an ideal diode acts as a short circuit in the forward direction and as an open circuit in reverse direction.

Calculation:

The time-varying nature of the output voltage (V₀) is as shown in the figure:

\({V_0}\left( {avg} \right) = {V_0}\left( {;DC;} \right) = \frac{1}{{2\pi }}\mathop \smallint \limits_0^\pi 10~sin(\omega{t}).d(\omega{t})\)

\(= - \frac{{10}}{{2\pi }}{\rm{cos}}(\left. { \propto )} \right|\begin{array}{*{20}{c}} \pi \ 0 \end{array}\)

\(= - \frac{{10}}{{2\pi }}\left[ {\cos \left( \pi \right) - {\rm{cos}}\left( 0 \right)} \right] = \frac{{20}}{{2\pi }} = \frac{{10}}{\pi } = 3.18V\)

Concept:

When forward biased, a p-n junction diode will allow current to flow with a voltage drop of V_R (Voltage drop) and acts as on open circuit when reverse biased.

Also, a Zener diode is a device used for voltage regulation; i.e. to maintain a constant voltage at the load.

This voltage is called the Zener voltage (or breakdown Zener Voltage V_Z .)

Also, A capacitor doesn’t allow a sudden change in the voltage (i.e. V₀⁺ = V₀^- )

Calculation:

During the positive half-cycle, the circuit will be as shown:

So, the output voltage for the positive half-cycle

⇒ V_out + 6.5 V – 14 V = 0 (Applying KVL)

⇒ V_out = 14 – 6.5 V = 7.5 V

When the V_in switches now to the negative cycle the capacitor will still be at the same voltage of 6.5 V.

V₀⁺(C) = 6.5 V only

The circuit for negative half-cycle in as shown:

So, at t = 0⁺ (i.e. just after switching from positive half cycle to negative half cycle)

V_out + 6.5 + 14 = V_out (Applying KVL)

V­_out = -14 – 6.5 = - 20.5 V

As the time increases, the capacitor voltage will start decreasing resulting in less output voltage.

So, the maximum V_out will be 7.5 V

And the minimum V_out will be -20.5 V

Concept:

When ∝ ≈ 1 (i.e. very close to unity)

β ≈ ∞ ( approaches ∞ )

which results in I_C ≈ I_E

and I_B ≈ 0

Calculation:

From the given figure:

As I_B ≈ 0, V_B will simply be:

\({V_B} = \frac{{18 \times 16}}{{16 + 44}} = \frac{{288}}{{60}} = 4.8;V\)

V_B – V_BE = V_E

V_E = 4.8 – 0.8 V = 4V

\({i_E} = \frac{4}{2} = 2mA \approx {i_C}\)

18 – 4 × 2 = V_C

V_C = 18 – 8 = 10 V

Now, V_CE = V_C – V_E

= 10 – 4 V

= 6 V

⇒ If P = high:

PMOS is OFF

NMOS is ON 

then y = Q̅

⇒If P = low

PMOS is ON

NMOS is OFF 

then  y = Q

The truth table of the above will be as shown:

\(\left. {\begin{array}{*{20}{c}} P&Q&Y\ 0&0&0\ 0&1&1\ 1&0&1\ 1&1&0 \end{array}} \right}\) So, Y = P̅ Q + P Q̅

= P ⊕ Q

So, Ex – OR Operation is being performed.

F' = X̅.Y + X.0 = X̅.Y

F = Z̅.F' + Z.F̅'

= Z̅.(X̅.Y) + Z.(X̅.Y)'   ---(1)

\(\overline{XY}=\overline{X}+\overline{Y}\)

Using the above in Equation (1):

= Z̅.X̅.Y + Z.(X + Y̅)

= X̅YZ̅ + XZ + Y̅Z

In a DRAM, a capacitor to store a bit of data is used along with a MOSFET (transfer device) which acts as a switch. The circuit is as shown –

In a DRAM:

• Periodic refreshing is required.
• The information is stored in a capacitor.
• Both read and write operations cannot be performed simultaneously.

The circuit is redrawn as shown:

[X(s) – Y(s)] G(s) + X(s) = Y(s)

Given G(s) = 2

(X(s) – Y(s)) 2 + X(s) = Y(s)

= 2 X(s) + X(s) = Y(s) + 2Y(s)

= 3 X(s) = 3 Y(s)

\(\frac{{Y\left( s \right)}}{{X\left( s \right)}} = 1\)

The total response of the system is given by

X(t) = ZIR + ZSR

Where,

ZIR = Zero input response

ZSR = Zero state response

ZIR = e^t x(0)

ZSR = L^-1 [ϕ(s) B U(s)]

Where

Φ(s) = (SI - A)^-1

Calculations:

ZIR = e^at x(0)

X(0) = \(\left[ {\begin{array}{*{20}{c}} 0\ 0 \end{array}} \right];\)

\(\Rightarrow ZIR = \left[ {\begin{array}{*{20}{c}} 0\ 0 \end{array}} \right]\)

ZSR = L^-1 [ϕ (s) B U(s)]

\(SI;A = \left[ {\begin{array}{*{20}{c}} s&0\ 0&{s + 9} \end{array}} \right]\)

\(Adj;\left( {sI - A} \right) = \left[ {\begin{array}{*{20}{c}} {s+9}&0\ 0&{s} \end{array}} \right]\)

\(\phi \left( S \right) = {\left( {sI - A} \right)^{ - 1}} = \frac{{Adj;\left[ {sI - A} \right]}}{{\left| {sI - a} \right|}}\)

\(= \left[ {\begin{array}{*{20}{c}} {\frac{1}{s}}&0\ 0&{\frac{1}{{s + 9}}} \end{array}} \right]\)

\(ZSR = {L^{ - 1}}\left[ {\left( {\begin{array}{*{20}{c}} {\frac{1}{s}}&0\ 0&{\frac{1}{{s + 9}}} \end{array}} \right)\left( {\begin{array}{*{20}{c}} 0\ {45} \end{array}} \right)\left( {\frac{1}{s}} \right)} \right]\)

CI \(= {L^{ - 1}}\left[ {\begin{array}{*{20}{c}} 0\ {\frac{{45}}{{s\left( {s + 9} \right)}};} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0\ {5\left( {1 - {e^{ - 9t}}} \right)} \end{array}} \right]\)

x1(t) = 0

x2 (t) = 5 (1 – e-9t)

\(\mathop {\lim }\limits_{t \to \infty } \left| {\sqrt {x_1^2\left( t \right) + x_2^2\left( t \right)} } \right|\)

\(= \sqrt {0 + {5^2}}\)

= 5

Concept:

• Lag Compensation adds a pole at origin (or for low frequencies).
• Lag Compensation to reduce the steady-state error of the system.
• An unstable system is a system that has at least one pole at the right side of the s-plane.
• So, even if we add a lag compensator to an unstable system, it will remain unstable.

Explanation:

If an already unstable system has a CLTP =\(\frac{K}{{\left( {s - 3} \right)\left( {s + 2} \right)}}\)

After using a lag compensator the CLTP = \(\frac{K}{{s\left( {s - 3} \right)\left( {s + 2} \right)}}\)

The system is still unstable.

Concept:

Binary symmetric channel:

The channel capacity per manage symbol for which the unit is bits/symbol is defined as the maximum value of mutual information and can be written as;

C_m = max [I(x, y)]

When I(x, y) = Mutual information

C_m = 1 + plog₂ P + (1 - p) log₂ (1 - p) … (1)

Calculation:

From Equation-(1) the capacity is varying logarithmically.

Let, p = cross over probability ‘or’ probability of error.

For p = 0;

From Equation - (1),

C_m = 1;

For p \(= \frac{1}{2}\)

\({C_m} = 1 + \frac{1}{2}{\log _2}\left( {\frac{1}{2}} \right) + \left( {1 - \frac{1}{2}} \right){\log _2}\left( {1 - \frac{1}{2}} \right)\)

= 0

And for p = 1;

C_m = 1

Since the capacity varies logarithmically, the complete graph will be as shown:

Option (C) →

Concept: Zero mean Gaussian random variable is represented as shown:

Mean of a random process also called as Expectation (E) of a random process in defined as;

⇒ E [X (t)] = Mean of Random Process

⇒ E (x) = Mean \(= \smallint x.f\left( x \right);dx\)

Where

(n) is the probability density function of a random variable.

Calculation:

Given; X (t) = U + Vt

E (X (t)) = E (U + Vt)

= E (u) + E (Vt).

At t = 2 sec.

E[x (t)] = E (U) + E (2V)

= E (U) + 2E (v)

⇒ E (U) = 0 (Given zero mean Random Process).

And E (V) \(= \mathop \smallint \limits_0^2 V.\frac{1}{2}dv\)

\(\Rightarrow {\left. {\frac{{{V^2}}}{4}} \right|_0}^2\)

= 1

PDF of Random Variable V =

2 (x) = 1

\(x = \frac{1}{2}\) (Amplitude of PDF)

So,

E[X(t)] = E (u) + 2E (v)

= 0 + 2

= 2

Concept:

SNR (Signal to Noise Ratio) for a sinusoidal message signal Quantized uniformly with ‘n’ bits in given by;  SNR_AB = 6n + 1.8

Calculation: Given, SNR = 40 dB, required at the output;

So, 40 ≤  6n + 1.8

6n ≥  40 – 1.8

n ≥  6.366

So, the minimum number of bits per sample needed to achieve the derived SQNR is 7 bits

When two conducting bodies are connected by a conducting wire, change will flow from one body with a higher electric potential to the other body with lower electric potential, till their potentials becomes equal i.e.,

V_a = V_b

\(\Rightarrow \frac{{{Q_1}}}{{4\pi \epsilon a}} = \frac{{{Q_2}}}{{4\pi \epsilon b}}\) 

(∵ Potential at the surface of a conducting sphere Q/4πϵr)

So, \(\frac{{{Q_1}}}{{{Q_2}}} = \frac{a}{b}\)      ---- (1)

Now, \({E_a} = \frac{{{Q_1}}}{{4\pi \epsilon {a^2}}}\)

(∵ Electric field at the surface of the conducting sphere = Q/4πϵr²)

And, \({E_a} = \frac{{{Q_1}}}{{4\pi \epsilon {b^2}}}\) 

\(\therefore \frac{{{E_a}}}{{{E_b}}} = \frac{{{Q_1}}}{{{Q_2}}}.\frac{{{b^2}}}{{{a^2}}}\) 

Using equation (1)

\( \Rightarrow \frac{{{E_a}}}{{{E_b}}} = \frac{b}{a}\) 

⇒ Given, b > a

So, E_a > E_b

Concept:

VSWR (Voltage standing wave Ratio) is defined as \( = \frac{{\left( {1 + \left|\Gamma \right|} \right)}}{{\left( {1 - \left| \Gamma \right|} \right)}}\) where Γ = Reflection coefficient.

Case 1:

For Γ = 1; VSWR = ∞

 All the power is reflected back to the transmission line.

Case 2:

For Γ = 0; VSWR = 1

Transmission line is perfectly matched to the Antenna and no power is reflected back to the transmission line

So, 1 ≤ VSWR < ∞ (Range)

Calculation:

Given, VSWR = 5.8

\( \Rightarrow 5.8 = \frac{{\left( {1 + \left| \Gamma \right|} \right)}}{{\left( {1 - \left| \Gamma \right|} \right)}}\)

Solving, we get

|Γ| = 0.7

\(\Gamma = \frac{{{V_ - }}}{{{V_ + }}}\)

(Where V_- = Reflected voltage Amplitude at load)

(Where V₊ = Forward path voltage Amplitude)

% Power reflected \(= \frac{{V_ - ^2}}{{V_ + ^2}} = \Gamma ^2 = {\left( {0.7} \right)^2} = 0.49 \times 100\) 

= 49%

Solving

\(\mathop \smallint \limits_0^1 \left( {\mathop \smallint \limits_0^1 \frac{{x - y}}{{{{\left( {x + y} \right)}^3}}}dy} \right)dx\)     ---- (1)

Solving the inner integral first,

\(\Rightarrow \mathop \smallint \limits_0^1 \frac{{x - y}}{{{{\left( {x + y} \right)}^3}}}dy = \mathop \smallint \limits_0^1 \frac{{x - y + x - x}}{{{{\left( {x + y} \right)}^3}}}dy = \mathop \smallint \limits_0^1 \frac{{2x}}{{{{\left( {x + y} \right)}^3}}}dy - \mathop \smallint \limits_0^1 \frac{1}{{{{\left( {x + y} \right)}^2}}}dy\) 

\(\frac{{2x{{\left( {x + y} \right)}^{ - 2}}}}{{ - 2}} - \left. {\frac{{{{\left( {x + y} \right)}^{ - 1}}}}{{ - 1}}} \right|_0^1\) 

\(\Rightarrow \frac{{ - x}}{{{{\left( {x + y} \right)}^2}}} + \left. {\frac{1}{{x + y}}} \right|_0^1 = \left. {\frac{y}{{{{\left( {x + y} \right)}^2}}}} \right|_0^1 = \frac{1}{{{{\left( {x + 1} \right)}^2}}}\) 

Equation (1) now becomes,

\(\Rightarrow \mathop \smallint \limits_0^1 \frac{1}{{{{\left( {x + 1} \right)}^2}}}dx = = \left. {\frac{{ - 1}}{{x + 1}}} \right|_0^1\)

\(\Rightarrow - \left( {\frac{1}{2} - 1} \right) = - \left( { - \frac{1}{2}} \right) = \frac{1}{2} = 0.5\) 

Similarly, solving for x first now, i.e.

\(\mathop \smallint \limits_0^1 \left( {\mathop \smallint \limits_0^1 \frac{{x - y}}{{{{\left( {x + y} \right)}^3}}}dx} \right)dy\)     ---- (2)

Solving for the inner integral first,

i.e. \(\mathop \smallint \limits_0^1 \frac{{x - y}}{{{{\left( {x + y} \right)}^3}}}dx\)

\(\Rightarrow \mathop \smallint \limits_0^1 \frac{{x - y + y - y}}{{{{\left( {x + y} \right)}^3}}}dx = \mathop \smallint \limits_0^1 \frac{1}{{{{\left( {x + y} \right)}^2}}}dx - \left( {\frac{{2y}}{{x + y}}} \right)dx\) 

\(= \frac{{{{\left( {x + y} \right)}^{ - 1}}}}{1} - \left. {\frac{{2y{{\left( {x + y} \right)}^{ - 2}}}}{{ - 2}}} \right|_0^1\) 

\(\Rightarrow \frac{{ - 1}}{{x + y}} + \left. {\frac{y}{{{{\left( {x + y} \right)}^2}}}} \right|_0^1\) 

\(\Rightarrow \left. {\frac{{y - x - y}}{{{{\left( {x + y} \right)}^2}}}} \right|_0^1 = \left. {\frac{{ - x}}{{{{\left( {x + y} \right)}^2}}}} \right|_0^1 = - \left( {\frac{1}{{{{\left( {y + 1} \right)}^2}}}} \right)\) 

Equation (2) now becomes,

\(\mathop \smallint \limits_0^1 \frac{{ - 1}}{{{{\left( {y + 1} \right)}^2}}}dy\) 

\(\Rightarrow \left. {\frac{{ - {{\left( {y + 1} \right)}^{ - 1}}}}{{ - 1}}} \right|_0^1 = \left. {\frac{1}{{y + 1}}} \right|_0^1\) 

\(\Rightarrow \frac{1}{2} - 1 \Rightarrow - \frac{1}{2} = - 0.5\)

Concept:

For a given complex function with poles, the complex integral \(I = \mathop \oint \limits_C f\left( z \right)dz\) is given by

Residue theorem as;

\(I = \mathop \oint \limits_C f\left( z \right)dz = 2\pi ;i \times \left{ {Sum;of;residue;of;poles;inside;or;on;C} \right}\)

Calculation:

Given C, |Z|= 3

|x + iy| = 3

\(\sqrt {{x^2} + {y^2}} = 3\)

x² + y² = 9

\(f\left( z \right) = \left( {\frac{{{z^2} - 1}}{{{z^2} + 1}}} \right){e^z}\)

Poles are defined by z² + 1 = 0

(z + i)(z - i) = 0

z = i, -i (simple poles lying inside of closed path)

Residue of f(z) at z = i \( \Rightarrow \mathop {\lim }\limits_{z \to i} \left( {z - i} \right).\frac{{{z^2} - 1}}{{\left( {z + i} \right)\left( {z - i} \right)}}{e^z}\)

\(\Rightarrow \frac{{\left( {{i^2} - 1} \right){e^i}}}{{2i}} \to - \frac{{2{e^i}}}{{xi}} = i{e^i}\)

Residue of f(z) at z = -i \(\Rightarrow \mathop {\lim }\limits_{z \to - i} \left( {z + i} \right).\frac{{{z^2} - 1}}{{\left( {z + i} \right)\left( {z - i} \right)}}{e^z}\)

\(\Rightarrow \frac{{\left( {i - 1} \right)}}{{ - 2i}}.{e^{ - i}} \to - i{e^{ - i}}\)

Now, using Cauchy’s Residue Theorem,

\(I = \mathop \oint \limits_c \frac{{{z^2} - 1}}{{{z^2} + 1}}{e^z}dz = 2\pi i;\left( {Sum;of;residue} \right)\)

= 2πi (ieⁱ – ie^-i)

= (2πi) (i) (eⁱ – e^-i)

= 2πi² (2i sin (1))

⇒ -4πi sin (1)

Concept:

If a vector field in irrorational, its curl will be 0.

i.e. \(\nabla \times \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\rightharpoonup}$}} {F} = 0\)

Calculation:

Given \(\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\rightharpoonup}$}} {F} = \left( {3y - {K_1}Z} \right){a_x} + \left( {{K_2}x - 2Z} \right){a_y} - \left( {{K_3}y + Z} \right){a_z}\)

\(\nabla \times \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\rightharpoonup}$}} {F} = 0 \Rightarrow \left| {\begin{array}{*{20}{c}}
{{a_x}}&{{a_y}}&{{a_z}} \
{\frac{\partial }{{{\partial _x}}}}&{\frac{\partial }{{{\partial _y}}}}&{\frac{\partial }{{{\partial _z}}}} \
{3y - {K_1}z}&{{K_2}x - 2z}&{ - {K_3}y - z}
\end{array}} \right| = 0\)

This has to happen for F̅ to be irrotational.

Solving the determinant,

a_x(-K₃ + 2) - a_y(0 + K₁) + a_z(K­₂ - 3).

For the above to be zero,

-K₃ + 2 = 0;

K₁ = 0 and

K₂ - 3 = 0

K₁ = 0

K₂ =  3

K₃ = 2

Concept:

For a discrete probability density distribution mean value ‘or’ Average Value ‘or’ Expected Value is given as;

\({\rm{E}}\left[ x \right] = \mu ;\left( {Mean} \right) = \mathop \sum \limits_{i = 1}^n {X_i}P\left( {{X_i}} \right)\)

Calculation:

With, X = Discrete samples = Number of attempts that the passengers need to make to get a seat reserved.

The probability distribution is as shown:

(Here X = number of attempts)

Now, Average number of attempts that the passenger need to make to get a seat reserved is-

\(E\left( X \right) = \mu = \mathop \sum \limits_{i = 1}^\infty {X_i}P\left( {{X_i}} \right)\)

= 1 × 0.4 + 2 × 0.6 × 0.4 + 3 × 0.6² × 0.4 + . . .

= 0.4 (1 + 2 × 0.6 + 3 × 0.6² + 4 × 0.6³ …)

= 0.4(1 – 0.6)^-2

= 0.4(0.4)^-2

\(\Rightarrow 0.4 \times \frac{1}{{0.4}} = \frac{1}{{0.4}}\)

⇒ 2.5

Concept: 

Maxima and Minima:

Steps to finding maxima and minima

1. Find the stationary point at which f’(x) = 0.

2. Then to find maxima and minima we will check the polarity of the second derivative.

Case 1: If f”(xo) > 0 then x = xo will be minima where xo is stationary point on f(x).

Case 2: If f”(xo) < 0 then x = xo will be maxima where xo is stationary point on f(x).

Calculation:

Given:

\(f\left( x \right) = \frac{1}{3}x\left( {{x^2} - 3} \right) = \frac{{{x^3}}}{3} - x\)

Now, we know that

For relative minimum and maximum;

putting \(\frac{{df}}{{dx}} = 0\) and solving for x, we get;

⇒ x² - 1 = 0

⇒ x = ±1

\(\frac{{{d^2}f}}{{d{x^2}}} = 2x\) which gives \(\frac{{{d^2}f}}{{d{x^2}}} = +~2\)  i.e. + Ve for x = +1

So, x = 1 is a point of relative minimum.

For the range given the minimum value can also take place to x = -100

So, checking at x = 1 and x = -100, we get:

At x = 1,

\(f\left( 1 \right) = \frac{1}{3}\left( {1 - 3} \right) = - \frac{2}{3}\)

at, x = -100,

f(-100) \(= \frac{1}{3}\left( { - 100} \right)\left( {10000 - 3} \right)\)

f(-100) \(= - \frac{{100}}{3}\left( {9997} \right)\)

f(-100) = -333233.33

So, f(-100) is the minimum value in the interval given.

∴ The minimum value occurs at x = -100.

Concept:

• Transients are present in the network when the network is subject to changes either by changing source magnitude or by changing any circuit elements
• the provided circuit should consist of any energy storage elements.
• No transients occur in the resistive element.
• Inductor doesn’t allow sudden change in current i.e. I_L(0^-) = I_L(0⁺).
• Capacitor doesn’t allow sudden change in voltage i.e. V_C(0^-) = V_C(0⁺)

Calculation:

At, t = 0^-, inductor is short circuited

\({i_L}\left( {{0^ - }} \right) = {i_L}\left( {{0^ + }} \right) = \frac{{10}}{2} = 5;Amp\)

At, t = 0⁺ switch is closed

At, t = ∞

i_L(∞) = 0 amp

Current through inductor is given by

\({i_L}\left( t \right) = {i_L}\left( \infty \right) + \left( {{i_L}\left( {0) - {i_L}(\infty } \right)} \right){e^{ - \frac{R}{L}t}}\)

\({i_L}\left( t \right) = 0 + \left( {5 - 0} \right){e^{ - \frac{5}{{2.5}}t}}\)

= 5 e^-2t

At, t = 0.5

i_L(t = 0.5) = 5e^-2×0.5 = 5e^-1

After the switch is closed, the whole current of the current source passes through Short Circuit. i.e.,

i(t) = 10 - i_L(t)

i(t = 0.5) = 10 – 5e^-1

= 8.16 Amp

Concept:

Thevenin’s Theorem states that to get the equivalent resistance between two points, we short circuit the voltage source and open circuit the current source and if we have dependent sources as well, we put a voltage source or a current source at the terminal where the equivalent resistance is to be obtained.

Calculation:

Given circuit:

For Thevenin’s equivalent resistance independent sources are replaced by their internal resistance and connected a 1 V source across PQ.

i₀ = 1/1 = 1 amp

Applying super node analysis method,

-I + i₀ + V_B/1 = 0

I = 1 + V_B

With V_A = 1 V

V_A - V_B = 3i₀

1 - V_B = 3

V_B = -2V

I = 1 – 2 = -1

Thevenin’s Resistance will be:

R_th = 1/1 = 1/(-1) = -1Ω

Concept:

When the input to an LTI system with impulse response H (f) = A sin (ωt + ϕ₁),

the output is - B sin (ωt + ϕ₂) 

With B = A |H (f)|

And Phase = ∠ H (f) + ϕ₁

Calculation:

\(Given;x\left( t \right) = 8\cos \left( {20\pi t + \frac{\pi }{4}} \right) + 16\sin \left( {40\pi t + ;\frac{\pi }{8}} \right) + 24\cos \left( {80\pi t + \frac{\pi }{{16}}} \right)\)

\(and;;\left| {H\left( f \right)} \right| = \begin{array}{*{20}{c}} {;;1 - \frac{{\left| f \right|}}{{20}}}&{,\left| f \right| \le 20}\ 0&{,\left| f \right| > 0} \end{array};\)

For an input frequency of \(10;Hz,;\left| {H\left( f \right)} \right| = 1 - \frac{{10}}{{20}} = \frac{1}{2}\)

For an input frequency of \(20;Hz,;\left| {H\left( f \right)} \right| = 1 - \frac{{20}}{{20}} = 0\)

For the input frequency of \(40;Hz,;\left| {H\left( f \right)} \right| = 0\)

So, only the frequency of 10 Hz will pass through the given system,

The output y (t) will, therefore, be  \(B;cos;\left( {20\pi t; + ;\phi ;} \right)\),

\(with;B = 8 \times {\left| {H\left( f \right)} \right|_{10Hz}} = 8 \times \frac{1}{2} = 4\)

y (t) = 4 cos (20πt + ϕ)

Since, the power of a signal depends only on the Amplitude and not on phase,

Therefore, for y (t) = 4 cos (20πt + ϕ),0 

\(Average;Power = \frac{{{4^2}}}{2} = 8\)

Concept:

If x(t) ↔ X(s)

Then \(\frac{{x\left( t \right)}}{t} \leftrightarrow \mathop \smallint \limits_s^\infty X\left( s \right)ds\)

Final value theorem states that:

\(\mathop {\lim }\limits_{t \to \infty } f\left( t \right) = \mathop {\lim }\limits_{s \to 0} sf\left( s \right)\)

Calculation:

Given \(x\left( t \right) = \left( {\frac{{\sin t}}{{\pi t}}} \right)u\left( t \right)\)

\(\sin t \leftrightarrow \frac{1}{{{s^2} + 1}}\)

So, \(\frac{{\sin t}}{t} \leftrightarrow \mathop \smallint \limits_s^\infty \frac{1}{{{s^2} + 1}};ds\)

= tan^-1 (∞) – tan^-1 (s)

\(= \frac{\pi }{2} - {\tan ^{ - 1}}\left( s \right)\)

\(\frac{1}{\pi }\left( {\frac{{\sin t}}{t}} \right) \leftrightarrow \frac{1}{\pi }\left[ {\frac{\pi }{2} - {{\tan }^{ - 1}}\left( s \right)} \right]\)

The above signal x(t) is passed through causal LTI system with:

\(H\left( s \right) = \frac{1}{s}\) ,

Output Y(s) = X(s).H(s)

\( = \frac{1}{s}\left[ {\frac{1}{\pi }\left( {\frac{\pi }{2} - {{\tan }^{ - 1}}\left( s \right)} \right)} \right]\)

\(\mathop {\lim }\limits_{{\rm{t}} \to {\rm{;}}\infty ;{\rm{;}}} {\rm{y;}}\left( {\rm{t}} \right){\rm{;}} = ;\mathop {\lim }\limits_{{\rm{s}} \to 0} {\rm{;s;Y}}\left( {\rm{s}} \right){\rm{;}}\)

\(\mathop {{\rm{y}}\left( {\rm{t}} \right)}\limits_{t \to \infty } = {;_{\mathop {\lim }\limits_{s \to 0} \frac{s}{s}\left( {\frac{1}{\pi }} \right)\left( {\frac{\pi }{2} - {{\tan }^{ - 1}}\left( s \right)} \right)}}\)

\(\Rightarrow {\rm{li}}{{\rm{m}}_{s \to 0}}\left( {\frac{1}{\pi }.\frac{\pi }{2} - \frac{1}{\pi }.{{\tan }^{ - 1}}\left( s \right)} \right) = \frac{1}{2} = 0.5\)

Concept:

The parallel combination of two systems with impulse responses h(t) and h(t) is simply the sum of the two impulse responses as shown:

The equivalent of the above circuit is;

Normalized energy of a signal is given as \(\mathop \smallint \limits_0^T {n^2}\left( t \right)dt\)

Calculation:

Given h₁(t) = 2δ(t + 2) - 3δ(t + 1)

and h₂(t) = δ (t - 2)

h₁(t) + h₂(t) = 2δ(t + 2) - 3δ(t + 1) + δ(t - 2)

The equivalent system in as shown;

y(t) = x(t)⊗(h₁(t) h₂(t))

= n(t)⊗(2δ(t + 2) - 3 δ(t + 1) + δ(t - 2))

y(t) ⇒ 2x(t + 2) - 3 × (t + 1) + x(t - 2)

[∵ x(t)⊗ δ(t - n) = n(t - n)]

Given x(t) = Unit step signal,

y(t) is as shown below,

Energy of \(y\left( t \right) = \smallint {y^2}\left( t \right)dt\)

\(= \mathop \smallint \limits_{ - 2}^{ - 1} {y^2}\left( t \right)dt + \mathop \smallint \limits_{ - 1}^{2;} {y^2}\left( t \right)dt\)

= 4 × 1 + 1 × 3 = 7J

Concept:

Flat-band voltage is the voltage at which the surface-potential becomes zero or ϕ_s = 0.

Flat-band voltage results because of the difference in the Work function of the metal, oxide, and semiconductors and also because of the presence of oxide charges.

It is given by:

\({V_{fb}} = {ϕ _{ms}} - {\frac{{{Q_{ss}}}}{{{C_{OX}}}}}\)

Q_ss = Oxide charge

C_OX = Oxide capacitance

ϕms = Work function difference between metal and semiconductor, defined as:

\({ϕ _{ms}} = {ϕ _m} - \left( {X + \frac{{{E_g}}}{2} + {ϕ _{fp}}} \right)\)    ---(1)

where ϕ_m = Metal work function

X = Electron affinity

E_g = Energy-band gap.

And ϕ_fp = difference (in volts) between E_fi and E_f, and is given by:

\({\phi_{fp}} = \frac{{KT}}{q}\ln \left( {\frac{{Na}}{{n_i}}} \right)\)

Calculation:

For the given p-type substrate, we can write:

\(\frac{{{{\text{E}}_g}}}{2} + {ϕ _{fp}} = {E_c} - {{\text{E}}_f} = 0.9~eV\)   ---(2)

Given ϕ_m = 4.1 eV and X = 4 eV

Usinig Equaiton (1) and (2), we get:

\({ϕ _{ms}} = {4.1} - \left( {4 + 0.9} \right)\)

ϕ_ms = - 0.8 eV

Now,

\({V_{fb}} = {ϕ _{ms}} - {\frac{{{Q_{ss}}}}{{{C_{OX}}}}}\)

\(-1 = -0.8 - {\frac{{{Q_{ss}}}}{{{C_{OX}}}}}\)

\(0.2 = \frac{{{Q_{ss}}}}{{{C_{OX}}}}\)

With \(C_{ox}=\frac{{{\text{}}{\epsilon_{ox}}}}{{{t_{ox}}}}\), we can write:

\({Q_{ss}} = 0.2 \times {C_{ox}} = 0.2 \times \frac{{{\text{}}{\epsilon_{ox}}}}{{{t_{ox}}}}\)

\(Q_{ss} =\frac{{0.2 \times 3.9 \times 8.85 \times {{10}^{ - 14}}}}{{0.1 \times {{10}^{ - 4}}}} ~nC/cm\)

\(Q_{ss}= 6.9 \times {10^{ - 9}}~nC/cm\)

Concept:

V_oc = (Open circuit voltage) of a solar cell

\({V_{oc}} = \eta {V_T}\ln \left( {1 + \frac{{{J_L}}}{{{J_S}}}} \right)\)

Where η = 2 for Si p-n diode.

J_L = Photo current density

Js = Saturation (Reverse) current density.

Calculation:

Given V_oc = 0.451 V

For J_L = 2.5 mA /cm², n = 2 and V_T2 = 25 mV.

\(0.451 = 2 \times 26;m\ln \left( {2 + \frac{{2.5m}}{{{J_s}}}} \right)\)

\( 9.02 = \ln \left( {1 + \frac{{2.5m}}{{{J_s}}}} \right)\)

\( 8266.77 = 1 + \frac{{2.8m}}{{{J_s}}}\)

Solving, we get J_s = 3.024 × 10^-7

If the intensity of the incident light is increased by 20 times, J_L (Photocurrent density) also increases by 20 times

i.e. J_L2 = 20 J_L1 = 20 × 2.5 m

= 50 mA/cm²

So, \({V_{oc2}} = \eta {V_T}\ln \left( {1 + \frac{{{J_{L2}}}}{{{J_S}}}} \right)\)

Putting values, we get

\({V_{oc2}} = - 2 \times 2.5;m\ln \left( {1 + \frac{{50m}}{{3.024 \times {{10}^{ - 7}}}}} \right) \)

\(= 0.6;V\)

Concept:

n-channel MOSFET biased in saturation region will result in a current given by:

\(\Rightarrow {I_D} = \frac{{W\mu {C_{ox}}}}{{2L}}{\left( {{V_{GS}} - {V_{th}}} \right)^2}\)

And the transconductance is defined as \({g_m} = \frac{{\partial {I_D}}}{{\partial {V_{GS}}}}\)

\(\Rightarrow {g_m} = \frac{{W\mu {C_{ox}}}}{{2L}}\left( {{V_{GS}} - {V_{th}}} \right)\)

Calculation:

Given, two transistors T₁ and T₂ which are identical in all respects except for width and gate overdrive voltage (V_GS - V_th).

For T₁ MOSFET,

g_m1 ∝ W₁ (V_GS - V_th)₁  and  I_D1 ∝ W₁ (V_GS - V_th)²₁

Similarly for T₂, g_m2 ∝ W₂ (V_GS - V_th)₂  and  I_D2 ∝ W₂ (V_GS - V_th)²₂

\(\Rightarrow \frac{{{I_{D1}}}}{{{I_{D2}}}} = \frac{{{W_1}\left( {{V_{GS}} - {V_{th}}} \right)_1^2}}{{{W_2}\left( {{V_{GS}} - {V_{th}}} \right)_2^2}};;;; \ldots \left( 1 \right)\)

Given W₂ = 2W₁,

And (V_GS - V­_th)₂ = 2 (V_GS - V_th)₁

Putting these in Equation (1), we get,

\(\Rightarrow \frac{{{I_{D1}}}}{{{I_{D2}}}} = \frac{{{W_1}\left( {{V_{GS}} - {V_{th}}} \right)_1^2}}{{2{W_1}{{\left( 2 \right)}^2}\left( {{V_{GS}} - {V_{th}}} \right)_1^2}}\)

\(\Rightarrow \frac{{{I_{D1}}}}{{{I_{D2}}}} = \frac{{{W_1}\left( {{V_{GS}} - {V_{th}}} \right)_1^2}}{{8{W_1}\left( {{V_{GS}} - {V_{th}}} \right)_1^2}} \Rightarrow {I_{D2}} = 8{I_{D1}}\)

Similarly,

\(\frac{{{g_{m1}}}}{{{g_{m2}}}} = \frac{{{W_1}{{\left( {{V_{GS}} - {V_{th}}} \right)}_1}}}{{{W_2}{{\left( {{V_{GS}} - {V_{th}}} \right)}_2}}}\)

\(\frac{{{g_{m1}}}}{{{g_{m2}}}} = \frac{{{W_1}{{\left( {{V_{GS}} - {V_{th}}} \right)}_1}}}{{2{W_1}\left( 2 \right){{\left( {{V_{GS}} - {V_{th}}} \right)}_1}}}\)

g_m2 = 4g_m

Concept:

When two ‘p’ type and ‘n’ type semiconductors come into contact to form a p-n junction diode as shown in the figure, there is a diffusion of majority carries which takes place, leaving behind uncompensated ions behind, as shown;

Application:

The figure for charge density within the transition region is as shown below:

According to Poisson’s equation, the electric field is given as

\(\frac{{d{\text{E}}\left( x \right)}}{{dx}} = \frac{q}{\epsilon}\left( {N_d^ + - N_a^ - } \right)\)

This shows that E(x) will vary linearly with distance.

So, the electric field distribution is as shown:

For the given figure,

Let x be the voltage at node A.

Now, \({V_{D{S_2}}} = 3 - x\)

\({V_{GS2}} - {V_t} = 2.5 - x - 1\)

= 1.5 - x

We observe that, \({V_{D{S_2}}}\) is always > \({V_{G{S_2}}} - {V_t}\)

So, M₂ will always be in saturation.

Assuming M₁ to be in saturation, we calculate the saturation current as shown.

(I_DSat)₁ = (I_DSat)₂

⇒ K_n(V_GS - V_t)²₁ = K_n (V_GS - V_t)²₂

⇒ (2 - 1)² = (2.5 - x - 1)²

⇒ 1 = 1.5 - x

x = 0.5 V

If ‘x’ were to be 0.5 V then for M₁,

V_DS = x - 0 = 0.5 V

and V_GS - V_t = 1 V

So, V_DS < V_GS - V_t

This shows that our assumption is wrong, and M₁ cannot be in saturation.

So, M₁ is in cut off.

Concept:

Small signal Analysis can be used to simplify circuits to find gain and other parameters. For a fixed bias circuit with emitter resistance the small signal model is shown:

The AC equivalent circuit is drawn as:

Applying KVL in 1^st loop, we get:

\({V_i} - \left( {\beta {r_e}} \right){i_b} - \left( {\beta + 1} \right){i_b}{R_e} = 0\)

KVL at loop 2,

V_o = - i_c R_c

since i_c = βi_b ,  

\(;{V_0} = ;-\beta {i_b}{R_c}\)

⇒ So,  \({i_b} = \frac{{ - {V_o}}}{{\beta {R_c}}}\)                         (1)

\({V_i} = \left( {{\beta _r}_e + \left( {\beta + 1} \right);{R_E}} \right);{i_b}\)

From eq. (1), we can write:

\({V_i} = \left( {{\beta _{re}} + \left( {\beta + 1} \right){R_E}} \right).\frac{{ - {V_o}}}{{\beta {R_C}}}\)

Since \(\left( {\beta + 1} \right) \approx \beta ,\) if β is large

\(So,\frac{{{V_o}}}{{{V_i}}} = \frac{{ - \beta {R_C}}}{{\beta \left( {{r_e} + {R_E}} \right)}} = \frac{{ - {R_C}}}{{{r_e} + {R_E}}}\)             

Calculation:

The given circuit’s small-signal model is drawn as shown:

\( \frac{{{V_o}}}{{{V_s}}} = \frac{{ - {R_c}}}{{{r_{{e_1}}} + {r_{{e_2}}}}} = \frac{{ - 1k}}{{{r_{{e_1}}} + {r_{{e_2}}}}}\)

r_e1 = r_e2 (because the collector current for both the transmitters is equal), i.e.

\({r_{e1}} = {r_{e2}} = \frac{{{V_t}}}{I} \)

\(r_{e1}= \frac{{26m}}{{2.6m}} = 10;{\rm{\Omega }}\)

\(So,;\frac{{{V_o}}}{{{V_s}}} = \frac{{ - 1000}}{{20}} = - 50\)

\(\left| {\frac{{{V_o}}}{{{V_s}}}} \right| = 50\)

For an ideal op-amp, V₊ = V^- (Virtual ground)

From the given figure:

Current through Q₂, Q₃…..Q₃₂ will be equal because

As V_BE2 = VBE3 = VBE4 = .....VBE32 = Vp = 0.7     (Given)

And i_s is same (became the transistors are all identical)

Using KCL, we can state that,

\(\Rightarrow \frac{{{V_{out}} - {V_ - }}}{{20;K}} = \frac{{{V_ - } - {V_P}}}{{5K}}\)       ---- (1)

Since, V_- = V₊ = \({V_{BE1}}\) (where \({V_{B{E_1}}}\) in the base to emitter voltage of Q₁)

Solution equation (1) becomes,

\(\frac{{{V_{out}} - V_{-}}}{{20;K}} = \frac{{{V_ - } - {V_P}}}{{5K}} = 31;{i_s};{e^{\frac{{q{V_P}}}{{KT}}}}\)       ---- (2)

Also, \({i_{{c_1}}}\) (current though Q₁) \(= {i_s}{e^{\frac{{q{V_{BE1}}}}{{KT}}}} = \frac{{{V_{out}} - {V_{B{E_1}}}}}{{20;K}}\)      ---- (3)

From (2) and (3)

Solving, \(e^{\frac{qV_{BE1}}{KT}}= 31\times e^{\frac{0.7}{0.026}} = 1.52 \times 10^{13}\) 

Given,

V_p = 0.7 V and \({V_ + } = \frac{{KT}}{q} = 26;mV\)

\(\Rightarrow \frac{{{V_{B{E_1}}}}}{{V_T}} = \ln \left( {1.52 \times {{10}^{13}}} \right)\)

\({V_{B{E_1}}} = 0.789;V\)

Now, using equation (2)

\(\frac{{{V_{out}} - {V_ - }}}{{20;K}} = \frac{{{V_ - } - {V_P}}}{{5;K}}\)

As V_- = V₊ = \({V_{BE1}} = 0.789\)

So,

\(\Rightarrow \frac{{{V_{out}} - 0.789}}{{20K}} = \frac{{0.789 - 0.7}}{{5K}}\)

⇒ V_out - 0.789 = 4(0.089)

V_out = 1.14 V

Calculation:

The initial state of FSM is S_o, and the given sequence of the input = 10101101001101

From the given FSM, the output will be one (1) whenever the input sequence is 101 (i.e. after every 101 the output changes to 1)

In the given input sequence, 101 appears 4 times as shown,

So, the number of times ‘Out’ will be 1 is 4.

In the given question, the inputs of X and Y are given as,

The first two bits from LSB does not generate any carry and does not involve in addition hence, the last two bits operation will be the main factor for calculating delay,

Now,

When X₂ = 1, Y₂ = 1, Z₂ = 0

Step - 1)

After first XOR (20 ns), output = 0 is transferred to second XOR gate (20 ns) hence, the total sum delay = 20 ns + 20 ns = 40 ns.

Step - 2) Carry generation

After first AND (15 ns), output = 1 is transferred to the OR gate (10 ns) as we know that OR GARE if one input is high output will be high, so we will not add the delay of second AND gate hence, the total carry delay = 15 ns + 10 ns = 25 ns

When X₃ = 1, Y₃ = 0, Z₃ = 1

Step - 3) Sum (S₁) generation

After first XOR (20 ns), output = 1 is transferred, and carry Z₁ = 1 is fed after 25 ns,

Total sum delay = max (20 ns, 25 ns) + second XOR gate (20 ns)

Hence, the total sum delay = 25 ns + 20 ns = 45 ns.

Step - 4) Carry (Z₃) generation

After first AND (15 ns), output = 0 is transferred to the OR gate (10 ns) but in second AND gate will receive input from 1^st XOR gate and Z₂ = 1 (after 25 ns), we know that OR GATE if one input is high output will be high t

the total carry delay = AND₂ gate delay + OR gate delay

= max (25 ns, 20 ns) + 15 ns + 10 ns = 50 ns

Fromthe given figure

A = P̅ Q̅ R

B = P̅ Q R

C = P Q̅ R

F = A + B + C

= P̅ Q̅ R + P̅ Q R + P Q̅ R

Concept:

For a unity feedback control system the characteristic equation in given as 1 + G (s) = 0.

Routh Array can be used to check for a marginally stable system (oscillatory nature).

Calculation:

Given \(G\left( s \right) = \frac{{2\left( {s + 1} \right)}}{{{s^3} + K{s^2} + 2s + 1}}\)

q(s) = Characteristic equation = 1 + G(s) = 0

\( 1 + \frac{{2\left( {s + 1} \right)}}{{{s^3} + K{s^2} + 2s + 1}}\)

s³ + Ks² + s(2 + 2) + 1 + 2 = 0

s³ + Ks² + 4s + 3 = 0

Forming Routh Array,

\(\begin{array}{*{20}{c}} {{s^3}}&1&4\ {{s^2}}&K&3\ {{s^1}}&{\frac{{4k - 3}}{k}}&0\ {{s^0}}&3&; \end{array}\)

For marginally stable (or oscillations), coefficients of s¹ has to be zero, i.e.,

\(\frac{{4k - 3}}{k} = 0\)

\(k = \frac{3}{4} = 0.75\)

The number of the system can be determined by finding its transfer function.

T. F = C [sI- A]^-1 B + D

Calculation

\(\frac{d}{{dt}}{x_1}\left( t \right) - {x_2}\left( t \right) = 0\)

x₁ – x₂ = 0

x₁ = x₂         --------(1)

\(\frac{d}{{dt}}{x_2} + 2{x_1}\left( t \right) + 3{x_2}\left( t \right) = r\left( t \right)\)

x₂ (t) + 2x₁ (t) 3x₂ (t) = r (t)

x₂ = -2x, -3x₂ + r          ------(2)

From (1) and (2)

\(\left[ {\begin{array}{*{20}{c}} {{x_1}}\ {{x_2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&1\ { - 2}&{ - 3} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}}\ {{x_2}} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 0\ 1 \end{array}} \right]r\)

c(t) = x₁ (t) [output]

\(\left[ C \right] = \left[ {1;0} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}}\ {{x_2}} \end{array}} \right]\)

Transfer function:

TF = C (SI – A)^-1 B + D

D = 0

\(\left( {sI - A} \right) = \left( {\begin{array}{*{20}{c}} s&{ - 1}\ 2&{s + 3} \end{array}} \right)\)

\(\frac{{Adj\left( {sI - A} \right)}}{{\left| {sI - A} \right|}} = \frac{{\left( {\begin{array}{*{20}{c}} {s + 3}&{ + 1}\ { - 2}&s \end{array}} \right)}}{{s\left( {s + 3} \right) + 2}}\)

\(T.;F = ;\left[ {1;0} \right]\frac{{\left( {\begin{array}{*{20}{c}} {s + 3}&1\ { - 2}&s \end{array}} \right)\left( {\begin{array}{*{20}{c}} 0\ 1 \end{array}} \right)}}{{s\left( {s + 3} \right) + 2}}\)

\(= \frac{{\left[ {1;0} \right]\left[ {\begin{array}{*{20}{c}} 1\ s \end{array}} \right]}}{{{s^2} + 3s + 2}} = \frac{1}{{{s^2} + 3s + 2}}\)

C. E = s² + 3s + 2

s² + 2 ζ ω_ns + ω_n²

\(w_n^2 = Z\)

\({\omega _n} = \sqrt 2\)

2ζ ω_ns = 35

\(\zeta = \frac{3}{{2\sqrt 2 }} > 1\)

The system is overdamped.

Concept

From Nyquist criteria we have

N = P

Where,

N = Number of encirclements of point (-1, j0) by G(s) H(s) plot, the positive direction of encirclement being Anticlockwise.

P = Number of open loop poles in RHS of s-plane.

Z = Number of closed loop poles in RHS of s-plane

Calculations:

P = open loop RHS poles

C.E = s³ + 3s² + 0s +10

From RH Criteria

Number of sign changes = P = 2

For 0 < K < 1

N = 0

From Nyquist criteria

N = P – Z

= 2 – Z

Z = 2

Number of poles of the closed loop transfer function that lie in RHS = 2

Given signal x(t) = sin (1400 πt)

When sampled at a frequency of 9K samples second. The frequency spectrum shifts to the right by nf_s and to the left by -nf_s. (where, n = ±0, ±1 ….)

The output spectrum will be as follows:

When this is passed through an ideal low pass filter with H(f) given as

\(H\left( f \right) = \left{ {\begin{array}{*{20}{c}} {1,}&{\left| f \right| \leqslant 12kHz} \ {0,}&{\left| f \right| > 12kHz} \end{array}} \right..\)

The output will be then

So, the number of Sinusoids at the output will be = 3,

With frequencies = 2, 7, 11.

Concept: Total power of AM transmitter is given by \({P_t} = {P_c}\left( {1 + \frac{{{\mu ^2}}}{2}} \right)\). Where μ is called modulation index.

Application: Given, P_c = 5 kW (Carrier Power)

μ = 0.5

\({P_t} = {P_c}\left( {1 + \frac{{{\mu ^2}}}{2}} \right)\)

\(= 5k\left( {1 + \frac{{0.25}}{2}} \right)\)

= 5k (1 + 0.125)

P_t = 5k × 1.125

μ is now reduced to μ = 0.4

If the transmitter is not to be overloaded i.e. if P_t is to remain the same,

The maximum unmodulated carrier power which can be used is,

\(5k \times 1.125k = {P_c}\left( {1 + \frac{{{\mu ^2}}}{2}} \right) = {P_c}\left( {1 + \frac{{0.16}}{2}} \right)\)

\(\Rightarrow 5k \times 1.125k = {P_c}\left( {1 + 0.08} \right)\)

\({P_c} = \frac{{5K \times 1.125K}}{{1.08}}\)

⇒ 5.2 KW

Concept:

Phase modulated signal is given as;

⇒ x_p(t) = A cos (ω_ct + K_p x(t))

Where, x(t) is the modulating signal given.

So, x_p(t) = A cos (ω_ct + 5 × 5 sin (4π × 10³ t - 10π cos 2π 10³t))

(∵ K_p = 5 rad/V) and n (t) = 5 sin (4π 10³ t - 10π cos 2π 10³ t)

So, phase modulated wave will be represented as;

x_p(t) = A cos (ω_ct + 25 sin (4π 10³ t - 10π cos 2π 10³ t))

Now, the instantaneous frequency of the above is given as \({\omega _i} = \frac{{d\alpha \left( t \right)}}{{dt}}\)

where α(t) = instantaneous angle;

Calculations:

α(t) = ω_ct + 25 sin [4π 10³ t - 10π as 2π 10³t]

\(\frac{{d\alpha }}{{dt}} = {\omega _i} = {\omega _c} + 25\cos \left[ {4\pi {{10}^3}t - 10\pi \cos 2\pi {{10}^3}t} \right] \times \left( {4\pi {{10}^3} + 10\pi \times 2\pi {{10}^3}\sin 2\pi {{10}^3}t} \right)\)

Instantaneous angular frequency at t = 0.5 ms

⇒ ω_i = ω_c + 25 cos [2π + 10π] × 4π 10³ + 0 → (∵ sin π = 0)

2πf_i = 2πf_c + 25 cos (12π) × 4π 10³

So, instantaneous frequency at t = 0.5 ms

⇒ f_i = f_c + 50 K

Given f_c = 20 K

f_i = 20 K + 50 K

= 70 KHz

Concept: (i) For lossless channel H(X/Y) = 0

                (ii) For the channel to be deterministic;   H(Y/X) = 0

                (iii) For the channel to be noiseless, (H(x/y) = H(y/x) = 0)

Application: From the given channel, the channel matrix will be:

                                y₁             y₂

\(P\left( {Y/X} \right) = \begin{array}{*{20}{c}} {{x_1}}\ {{x_2}} \end{array};\left[ {\begin{array}{*{20}{c}} {;0.25}&{0.75}\ {;0.25}&{0.75} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\frac{1}{4}}&{\frac{3}{4}}\ {\frac{1}{4}}&{\frac{3}{4}} \end{array}} \right]\)

Assuming P(x₁) = P(x₂) = 1/2

\( {\because P\left( {X,Y} \right) = \left[ {P\left( X \right)} \right]\left[ {P\left( {\frac{Y}{X}} \right)} \right];}\)

\(P\left( {X,Y} \right) = \left[ {\begin{array}{*{20}{c}} {\frac{1}{8}}&{\frac{3}{8}}\ {\frac{1}{8}}&{\frac{3}{8}} \end{array}} \right]\)

So, \(P\left( {{y_1}} \right) = \frac{1}{4},;;P\left( {{y_2}} \right) = \frac{3}{4}\)

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

(i) For lossless channel H(X/Y) = 0

\(H\left( {\frac{X}{Y}} \right) = \frac{1}{8};{\log _2}\frac{{2 + 3}}{8}{\log _2}\frac{{2 + 1}}{8}{\log _2}\frac{{2 + 3}}{8}{\log _2}2\)

= 1 ≠ 0 (so, the channel is not lossless)

(ii) For the channel to be deterministic;

H(Y/X) = 0

\(H\left( {\frac{Y}{X}} \right) = \frac{1}{8}{\log _2}\left( 4 \right) + \frac{3}{8}{\log _2}\left( {\frac{4}{3}} \right) + \frac{1}{8}{\log _2}\frac{{4 + 3}}{8}{\log _2}\left( {\frac{4}{3}} \right)\)

So, H(Y/X) ≠ 0; the channel is not determinant as well.

(iii) For the channel to be noiseless,

(H(x/y) = H(y/x) = 0)

But, H(x/y) ≠ H(y/x) ≠ 0

Say it is not noiseless as well.

So, the channel is useless.

Concept:

K. E. of a moving electron \(= \frac{1}{2}m{v^2}\)

And the potential energy between the two particles separated by a distance \(= \frac{{{q_1}{q_2}}}{{4\pi {\epsilon_0}r}}\).

Calculation:

For the given situation, two electrons separated by a distance of exerts a force to each other as they come to each other field.

Eventually, charge q₁ will come to rest and all the kinetic energy that was initially there will get converted to the potential energy when q₁ comes to a halt.

This can be visualized as shown in the figure below,

Equating P.E. = K.E. ,  i.e.,

\(\Rightarrow \frac{1}{2}m{v^2} = \frac{{{q_1}{q_2}}}{{4\pi {\epsilon_0}r}}\)

So, \(r = \frac{{2{q_1}{q_2}}}{{4\pi {\epsilon_0}m{v^2}}};;; \ldots \left( 1 \right)\)

Given m = 9.1 × 10^-31 kg

q₁ = q₂ = -1.6 × 10^-19

\({\epsilon_0} = \frac{1}{{36\pi }} \times {10^{ - 9}}f/m\)

we get r = 5.05 × 10^-8 m

Concept:  

Cut-off frequency of waveguide (rectangular)

\( \Rightarrow {f_c} = \frac{C}{2}\sqrt {{{\left( {\frac{m}{a}} \right)}^2} + {{\left( {\frac{n}{b}} \right)}^2}} ;;;; \ldots \left( 1 \right)\) 

Dominant mode in rectangular waveguide (a > b) is TE₁₀ and the cutoff frequency is defined as;

\({f_{{c_{\left( {10} \right)}}}} = \frac{C}{2}\sqrt {\frac{1}{{{a^2}}}} = \frac{C}{{2a}}\) 

Calculation:

\({f_{{c_{\left( {10} \right)}}}} = \frac{C}{{2a}} = \frac{{3 \times {{10}^{10}}}}{{2 \times 2.29}} = 6.554;Hz\) 

Desired operating frequency = 25% more than \({f_{{c_{\left( {10} \right)}}}}\)

\(\left( {1 + 0.25} \right){f_{{c_{\left( {10} \right)}}}} = 1.25{f_{{c_{\left( {10} \right)}}}} = 1.25 \times 6.55\)

= 8.199 Hz

Using equation (1), we calculate different cut-off frequency for higher modes to find the next dominant mode,

f_c for TE₀₁ mode = 14.74 Hz

f_c for TE₁₁ mode = 16.4 Hz

and

f_c for TE₂₀ mode = 13.144 Hz

So, TE₂₀ is the next higher cut-off frequency,

95% of 13.144 Hz

⇒ 12.484 GHz

Concept:

\( \Rightarrow \frac{{\sin {\theta _2}}}{{\sin {\theta _1}}} = \frac{{{n_2}}}{{{n_1}}}\) (n₂ and n₁ are the refractive index of 2^nd and 1^st Medium respectively)

The critical angle is the angle for which we have no. refracted wave i.e. θ₂ = 90°

Since n = \(\frac{1}{{\sqrt {\epsilon{_r}{\epsilon_0}} }}\), we can write:

\(\frac{{\sin 90^\circ }}{{\sin {\theta _c}}} = \frac{{{n_2}}}{{{n_1}}} = \frac{1}{{\sqrt {{\epsilon_2}{\epsilon_0}} }} \times \sqrt {{\epsilon_1}{\epsilon_0}} \)             

\(\Rightarrow \frac{1}{{\sin {\theta _c}}} = \sqrt {\frac{{{\epsilon_1}}}{{{\epsilon_2}}}} \)

\(\sin {\theta _c} = \sqrt {\frac{{{\epsilon_2}}}{{{\epsilon_1}}}} \)

Calculation:

Given ϵ₂ = 1

ϵ₁ = 1.75

\({\theta _c} = {\sin ^{ - 1}}\sqrt {\frac{1}{{1.75}}} = 49.1^\circ\)

From the given figure:

\(\tan {\theta _c} = \frac{5}{d}\), i.e.

 \(\tan \left( {49.1} \right) = \frac{5}{d}\)

\(d = \frac{5}{{\tan \left( {49.1} \right)}} = \frac{5}{{1.104}} = 4.33\)