GATE EC 2016 Official Paper: Shift 1 (Previous Year Paper)

The sentence is stating the highest peak in the world.

Since it is a specific thing, we need to use the definite article ‘the’ before it.

Also, the sentence is using the superlative degree and so we say ‘the highest peak in the world’ making option 1 the correct answer.

There cannot be many highest peaks in the world and so options 3 and 4 are incorrect.

• The context of the sentence is asking a person who has been deprived of something because of theft.
• The word to be used to fill the blank is ‘lose’ which means to be deprived of something.
• ‘Loose’ means something that is not fitted. ‘Louse’ is the singular form of the word ‘lice’ that is a parasite that lives in the skin of mammals and birds.
• ‘Loss’ is a noun that means the process of losing someone or something. Eg: He suffered tremendous loss in his business.

The sentence is looking for contrast as it is joined by the conjunction ‘despite’.

The best pair of words that can fit the context of the sentence is ‘effectiveness…prescribed’.

Though medicine is ‘effective’ in treating diabetes, it is not being ‘prescribed’ widely.

A new medicine cannot have a ‘prescription’ or ‘availability’ for treating a disease.

Proscribed’ means forbidden by law. In case we use ‘acceptance…proscribed’ the sentence will not make any sense because it will mean that though the medicine is accepted widely, it is not forbidden by law.

Let T = total no of fruits = 5692000

R = Ripe fruits

U = Unripe fruits

A = Apple        

O = Oranges

Given U = 15% of T :  $\frac{{15}}{{100}} \times 5692000$ = 853800

R = T – U= 4838200

A(U) = 45% of U:  $\frac{{45}}{{100}} \times 853800 = 384210$

A(R) = (100-66)% of R: $\frac{{34}}{{100}} \times 4838200 = 1644988$

∴ A(U) + A(R) = 2029198

In question, it is given that Ahmed is 5Km away and Susan is 7Km away from where I live.

Further, it is given that Arun is farther away than Ahmed from where I live and not as far as Susan.

That means Arun must be living at a distance of more than 5Km but less than 7Km from my house which is according to given options can be 6.02Km. Note: Information about Michal is unnecessary and just given to confuse.

Percentage probability of being infected :

${\rm{;P}}\left( {\rm{A}} \right){\rm{;}} = {\rm{;}}50{\rm{\% }}$

Percentage probability of an infected person developing the disease is having system:

${\rm{;P}}\left( {\rm{B}} \right){\rm{;}} = {\rm{;}}30{\rm{\% }}$

∴ Percentage probability of infected person not showing symptoms:

${\rm{P}}\left( {{\rm{\bar B}}} \right) = 70{\rm{\% ;}}$

∴ Percentage probability of person moving through a TB prone zone remaining infected but not showing symptoms:

${\rm{P}}\left( {\rm{A}} \right).{\rm{;P}}\left( {{\rm{\bar B}}} \right) = \frac{{50}}{{100}} \times \frac{{70}}{{100}} = {\rm{;}}35{\rm{\% }}$

The paragraph states that the subject was very confident about his good friends helping him in his times of need because he had always helped them before in their time. Thus, inference III follows. Since the events of the last week proved him wrong, this means that his confidence was broken and his friends had not helped him. Thus inference 4 also follows.

Hence option 2 is the correct answer.

According to given information the points we got are

1. Shiva is brother of Pavithra
2. Shiva and Pavithra are cousin’s of Leela
3. According to their ages Shiva > Leela > Pavithra
4. They all live play chess
5. Pavithra wins more often than Leela but information about winning cases of Shiva is not given.

So from the given options statement which is clearly true is that Pavithra is the youngest of all.

${{\rm{q}}^{ - {\rm{a}}}} = {\rm{;}}\frac{1}{{\rm{r}}}$ ;   ${{\rm{r}}^{ - {\rm{b}}}} = {\rm{;}}\frac{1}{{\rm{s}}}$  and ${{\rm{s}}^{ - {\rm{c}}}} = {\rm{;}}\frac{1}{{\rm{q}}}$

∴ ${{\rm{q}}^{\rm{a}}} = {\rm{;r}}$  ; ${{\rm{r}}^{\rm{b}}} = {\rm{;s}}$ and ${{\rm{s}}^{\rm{c}}} = {\rm{;q}}$

∴ ${\rm{a}}\log {\rm{q}} = \log {\rm{r;}}$-------(1)

And   ${\rm{b}}\log {\rm{r}} = \log {\rm{s;}}$------(2)

And ${\rm{c}}\log {\rm{s}} = \log {\rm{q;}}$-----(3)

Multiplying equation  (1),(2) and (3)

$\begin{array}{l} {\rm{abc;}}(\log {\rm{q}}){\rm{;}}(\log {\rm{r}})(\log {\rm{s}}) = (\log {\rm{r}})(\log {\rm{s}})(\log {\rm{q}})\ \therefore {\rm{;abc;}} = {\rm{;}}1 \end{array}$

After 7 days from start of project:

Q took sick leave on first 2 days

∴ Man hours by Q = 5×12

∴ work done by ${\rm{Q;}} = {\rm{;}}5 \times 12 \times \frac{1}{{25 \times 12}} = \frac{1}{5}$

Man hours by R = 7×18

∴ work done by R = $\frac{1}{{50 \times 12}} \times 7 \times 18{\rm{;}} = \frac{{21}}{{100}}$

∴ ratio of work done by Q to work done by ${\rm{R;}} = \frac{1}{5}:\frac{{21}}{{100}} = \frac{{100}}{{5 \times 21}} = \frac{{20}}{{21}}$

Explanation:

Let M⁴ = I      ---(i)

Multiplying by M^-1 on both sides

M⁴ M^-1 = I – M^-1

M³ = M^-1      ---(ii)

Multiplying M⁴ on both sides in equation (i)

M⁴⋅ M⁴ = I ⋅ M⁴

M⁸ = M⁴      ---(iii)

Multiplying M^-1 on both sides, we get:

M⁸ – M^-1 = M⁴ M^-1

M⁷ = M³      ---(iv)

From (ii) and (iv)

M^-1 = M³ = M⁷

Now again multiplying M⁴ on both sides in equation (iii)

M⁸⋅ M⁴ = M⁴ M⁴

M¹² = M⁸

M¹² = M⁴ [from (iii)]

M¹² = I      ---(v) [from (i)]

Multiplying M^-1 on both sides we get

M¹¹ = M^-1      ---(vi)

From (ii), (iv) and (vi)

M^-1 = M³ = M⁷ = M¹¹ = …. & So on

3, 7, 11 ….. from general term 4k + 3 where k = 0, 1, 2

Hence, M^-1 = M^{4k + 3}

Option C is correct.

The following properties are true in calculus:

• If a function is differentiable at any point, then it is necessarily continuous at the point.
• But the converse of this statement is not true i.e. continuity is a necessary but not sufficient condition for the Existence of a finite derivative.
• Differentiability implies Continuity.
• Continuity does not necessarily imply differentiability.

Hence,

Statement P is wrong.

Statement Q is right. 

Statement R is right.

Properties of Laplace equation:

∇²f = 0 (Laplace equation)

1. The solutions have neither maxima nor minima anywhere except at the boundaries.

2. The solution is separable in co-ordinates

∇²f = 0

$\frac{{{\partial ^2}{f_x}}}{{\partial {x^2}}} + \frac{{{\partial ^2}{f_y}}}{{\partial {y^2}}} + \frac{{{\partial ^2}{f_z}}}{{\partial {z^2}}} = 0$ 

Where f = f_x a_x + f_y a_y + f_z a_z

3. The solutions are continuous.

4. The solutions are dependent on the boundary conditions.

So, only option A is correct.

Concept:

The integration of Ramp is parabolic, i.e.

$\smallint \frac{1}{t}dt = \frac{1}{{{t^2}}}$

Similarly, the integration of a step function is a Ramp, i.e.

$\smallint Au\left( t \right)dt = \smallint Adt = At$

Application:

The resultant plot will be as shown:

Concept:

if output y(t) is scalar multiple of input x(t), i.e.

y(t) = H(s) . x(t) then the input function is called as the eigen function.

where H(s) is called the eigenvalue.

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

Let , x(t) = e^st

So, x(t ⋅ τ) e^{s(t - τ)}

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

\(= {e^{st}}\mathop \smallint \limits_{ - \infty }^\infty {e^{ - s\tau }}h\left( \tau \right) \cdot d\tau \)

So, y(t) = H(s)⋅ x(t)

If the input to the system is an eigen signal, the output will also be the same eigen signal.

Analysis:

When,

 $\cos {ω _0}t = \frac{{{e^{j{ω _0}t}} + {e^{ - j{ω _0}t}}}}{2}$ is passed through an LTI system, we get the output as:

$\frac{{{e^{j{ω _0}t}}H\left( {j{ω _0}} \right) + {e^{ - j{ω _0}t}}H\left( { - j{ω _0}} \right)}}{2}$

Since Output can't be written in the form of H(jω₀) x(t) (because H(jω₀) ≠ H(-jω₀)), so it is not eigenfunction.

When ${\rm{sin}}{ω _0}t = \frac{{{e^{j{ω _0}t}} + {e^{ - j{ω _0}t}}}}{{2i}}$ is passed through an LTI system, we get:

$\frac{{{e^{j{ω _0}t}}H\left( {j{ω _0}} \right) + {e^{ - j{ω _0}t}}H\left( { - j{ω _0}} \right)}}{{2i}}$

∴ The output can’t be written in form of H(jω₀)x(t) because H(jω₀) ≠ H(-jω₀), So it is not eigenfunction as well.

 Now, when e^jω₀t is passed through an LTI system, the output can be written as:

$H\left( {j{\omega _0}} \right){e^{j{\omega _0}t}}$

Since the Output is the same as the Eigen signal, we conclude that ejω0t is an eigen value.

Consider x(t) = cos (ω₀t)

If x(t) is sampled with a sampling period T_s

x(n) = cos Ω₀n is obtained

$\frac{{{{\rm{\Omega }}_0}}}{{{T_s}}} = {\omega _0}$

Here $\frac{{2\pi ;m}}{{{N_0};{T_s}}} = \frac{{2\pi }}{{{T_0}}}$ 

$\therefore \frac{{{T_0}}}{{{T_s}}} = \frac{{{N_0}}}{m}$

For the sampled signal to be periodic, ‘m’ should be such that an above expression is a rational number.

Thus $\frac{T}{{{T_s}}} = \frac{{12}}{{10}} = \frac{6}{5}$

T = 1.2 T_s

Concept:

Properties of the region of convergence:

The properties of the ROC depend on the nature of the signal. Assuming that the signal has a finite amplitude and that the z-transform is a rational function.

• The ROC is a ring or disk in the z-plane, centred on the origin (0 < rR < |z| < rL ≤ ∞)
• The Fourier transform of x[n] converges absolutely if and only if the ROC of the z-transform includes the unit circle.
• The ROC cannot contain any poles.
• If x[n] is finite duration, then the ROC is the entire z-plane except perhaps at z = 0 or z = ∞
• If x[n] is a right-sided sequence then the ROC extends outward from the outermost finite pole to infinity.
• If x[n] is left-sided then the ROC extends inward from the innermost nonzero pole to z = 0
• A two-sided sequence (neither left nor right-sided) has a ROC consisting of a ring in the z-plane, bounded on the interior and exterior by a pole (and not containing any poles).
• The ROC is a connected region.

Calculation:

Given  ${\rm{x}}\left( {\rm{n}} \right){\rm{;}} = {\rm{;}}{{\rm{a}}^{\rm{n}}}{\rm{u}}\left( {\rm{n}} \right){\rm{;}} + {\rm{;}}{{\rm{b}}^{\rm{n}}}{\rm{u}}\left( {\rm{n}} \right)$

$\begin{array}{l} {\rm{X}}\left( {\rm{z}} \right) = \frac{{\rm{z}}}{{{\rm{z}} - {\rm{a}}}} + \frac{{\rm{z}}}{{{\rm{z}} - {\rm{b}}}}\ {\rm{ROC}}:{\rm{;}}\left| {\rm{z}} \right| > {\rm{;a;and;}}\left| {\rm{z}} \right|{\rm{;}} > {\rm{;b}} \end{array}$

But given that, $\rm 0< |a| < |b| < 1$

∴ Overall ROC is $\left| {\rm{z}} \right|{\rm{;}} > {\rm{;b}}$

Concept:

Application:

Given \({\rm{T}} = \left[ {\begin{array}{*{20}{c}} {\rm{A}}&{\rm{B}}\ {\rm{C}}&{\rm{D}} \end{array}} \right]\)

The two-port network is symmetric if A = D and reciprocal if AD – BC = 1

∴ Determinant (T) = 1

The new energy level is close to the edge of the conduction band (E_c).

Thus it is an n-type semiconductor that is formed by doping a pure or intrinsic semiconductor with a pentavalent atom.

Note:

Fermi level of an intrinsic semiconductor is as shown:

Concept:

For VDS < VGS – VT, n-channel enhancement type MOSFET will operate in triode/linear 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 VDS ≥ VGS – VT, the MOSFET will be in saturation with the current given by:

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

L = Channel length

Conclusion: Since the Current is inversely proportional to Channel length, the current will increase as the length will decrease. ∴ Statement P and S are correct.

Resistance:

The current in an NMOS in the linear region is:

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

The transistor operates in the linear region for small values of VDS, i.e. neglecting higher powers of V_DS, the above expression can be approximated by:

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

$R_d=\frac{V_{DS}}{I_{DS}}$

$R_d=\frac{1}{\frac{{{\rm{W\mu_n }}{C_{ox}}}}{{L}}\left[ {\left( {{V_{GS}} - {V_T}} \right){V_{DS}}} \right]}$

$R_d=\frac{L}{W\mu_nC_{ox}{{{}}}{}\left[ {\left( {{V_{GS}} - {V_T}} \right){V_{DS}}} \right]}$

Conclusion: As the output resistance is directly proportional to the channel length, it reduces with a decrease in channel length. ∴ Statement Q is incorrect.

Threshold Voltage:

The threshold voltage of a MOSFET is given as:

${V_T} = 2{\phi _F} - \frac{{{Q_d}}}{{{C_i}}} + {\phi _{ms}} - \frac{{{Q_i}}}{{{C_i}}}$

From the above expression, we can conclude that the threshold voltage is also dependent on the channel length. ∴ Statement R is incorrect.

Due to virtual Ground Concept,

V_A = V_B       ---(1)

From figure,

V_A = V_CC - V_ref       ---(2)

${I_E} = \frac{{{V_{CC}} - {V_B}}}{R}$

From equation (1) & (2)

${I_E} = \frac{{{V_{CC}} - \left( {{V_{CC}} - {V_{ref}}} \right)}}{R} = \frac{{{V_{Ref}}}}{R}$

Load current (I₀) = I_C

_{${I_C} = \alpha {I_E} = \left( {\frac{\beta }{{1 + \beta }}} \right)\frac{{{V_{ref}}}}{R}$}

Analytical Approach:

In the combination of two Diode circuits at a time, only one Diode will turn ON.

When V₀ = + V_sat, then D₁ is ON and D₂ is OFF. When V0 = - Vsat then D₂ is ON and D₁ is OFF.

So the given circuit introduces amplitude stabilization by preventing the Op-Amp from saturating and thus producing sinusoidal oscillations of fixed amplitude.

Alternate Approach:

The circuit shown is a Wein bridge oscillator. The amplitude of oscillations can be determined and stabilized by using a nonlinear control network. As the oscillations grow, the diodes start to conduct causing the effective resistance in the feedback to decrease. Equilibrium will be reached at the output amplitude that causes the loop gain to be exactly unity.

Concept: PLL is used in FM demodulation. In a frequency synthesizer the VCO output frequency is Nfin

Application:        

${\rm{Y}} = {\rm{ABC}} \oplus \left( {{\rm{AB}} \oplus {\rm{BC}}} \right) = \left( {{\rm{ABC}}} \right)\left( {{\rm{AB}} \odot {\rm{BC}}} \right) + \left( {{\rm{AB}} \oplus {\rm{BC}}} \right)\left( {\overline {{\rm{ABC}}} } \right)$

Consider${\rm{;AB}} \odot {\rm{BC}} = \left( {{\rm{AB}}} \right)\left( {{\rm{BC}}} \right) + \left( {\overline {{\rm{AB}}} } \right)\left( {\overline {{\rm{BC}}} } \right)$

$\begin{array}{l} = {\rm{ABC}} + \left( {{\rm{\bar A}} + {\rm{\bar B}}} \right)\left( {{\rm{\bar B}} + {\rm{\bar C}}} \right) = {\rm{ABC}} + {\rm{\bar A\bar B}} + {\rm{\bar B}} + {\rm{\bar A\bar C}} + {\rm{\bar B\bar C}}\ = {\rm{ABC}} + {\rm{\bar A\bar C}} + {\rm{\bar B}} = {\rm{\bar A\bar C}} + \left( {{\rm{B}} + {\rm{\bar B}}} \right)\left( {{\rm{AC}} + {\rm{\bar B}}} \right) = {\rm{\bar A\bar C}} + {\rm{AC}} + {\rm{\bar B;;}}\ \left( {{\rm{ABC}}} \right)\left( {{\rm{AB}} \odot {\rm{BC}}} \right) = \left( {{\rm{ABC}}} \right)\left( {{\rm{\bar A\bar C}} + {\rm{AC}} + {\rm{\bar B}}} \right) = {\rm{ABC}} \end{array}$

Consider $\overline {{\rm{ABC}}} \left( {{\rm{AB}} \oplus {\rm{BC}}} \right) = {\rm{AB}}\left( {\overline {{\rm{BC}}} } \right) + \overline {{\rm{AB}}} \left( {{\rm{BC}}} \right) = {\rm{AB}}\left( {{\rm{\bar B}} + {\rm{\bar C}}} \right) + \left( {{\rm{\bar A}} + {\rm{\bar B}}} \right){\rm{BC}} = {\rm{\bar ABC}} + {\rm{AB\bar C}}$

$\begin{array}{l} \therefore {\rm{Y}} = {\rm{ABC}} + {\rm{\bar ABC}} + {\rm{AB\bar C}} = {\rm{BC}} + {\rm{AB\bar C}} = {\rm{B}}\left( {{\rm{C}} + {\rm{\bar CA}}} \right)\ = {\rm{B}}\left( {{\rm{C}} + {\rm{\bar C}}} \right)\left( {{\rm{C}} + {\rm{A}}} \right) = {\rm{B}}\left( {{\rm{C}} + {\rm{A}}} \right) \end{array}$

Analysis:

Transfer characteristic of CMOS Inverter is as follows:

Since the CMOS Inverter is connected in the feedback loop formed by connecting a 10 kΩ resistor between the output and input, the output goes and stays at the middle of the characteristic:

${V_a} = \frac{{{V_{IR}} + {V_{IH}}}}{2}$

V_a Switching threshold of the inverter.

Concept:

To find the closed system stability by using RH criteria we require a  characteristic equation. Whereas in remaining all stability techniques we require open-loop transfer function.

The nth order general form of CE is

a0 sn + a1 sn-1 + a2sn-2 + __________an-1 s1 + an

RH table shown below

Necessary condition: 

All the coefficients of the characteristic equation should be positive and real.

Sufficient Conditions for stability:

1. All the coefficients in the first column should have the same sign and no coefficient should be zero.
2. If any sign changes in the first column, the system is unstable.

And the number of sign changes = Number of poles in right of s-plane.

Analysis:

i) If all the elements present in the first column of the RH table have a positive (+ve) sign then we can say that system is stable.

ii) If there is any change in the sign of the first column then we can say that system is Unstable and the number of sign changes indicates the Number of poles present on the right side of the S – plane.

iii) If any element is zero in the first column then it says the test breaks down.

From the principal of argument theorem the number of encirclements about (-1, 0) is ${\rm{N;}} = {\rm{;P;}}-{\rm{;Z}}$ 

Where P = Number of open loop poles on Right – Half of s – plane

Z = Number of Closed Loop Poles on Right Half of s – plane

Given that closed loop system is stable means ${\rm{z;}} = {\rm{;}}0 \Rightarrow {\rm{;N;}} = {\rm{;P}}$

∴ So the Nyquist encircles the s – plane point (-1 +j0) in the counter clockwise direction as many times as the number of right half s – plane poles

Concept:

If the characteristic impedance is purely real and the propagation constant is complex, the transmission line is distortionless and for this line  ${\rm{\alpha }} = \sqrt {{\rm{RG}}}$ and ${\rm{\beta }} = {\rm{\omega }}\sqrt {{\rm{LC}}}$. and ${{\rm{Z}}_{\rm{o}}} = \sqrt {\frac{{\rm{R}}}{{\rm{G}}}}$

Application: 

As ${\rm{\alpha }} = \sqrt {{\rm{RG}}}$ and ${{\rm{Z}}_{\rm{o}}} = \sqrt {\frac{{\rm{R}}}{{\rm{G}}}}$

\(\begin{array}{l} \Rightarrow {\rm{\alpha }}{{\rm{Z}}{\rm{o}}} = {\rm{R}}\ {{\rm{Z}}{\rm{o}}}{\rm{R}} = 2 \times 50 = 100{\rm{;\Omega }}/{\rm{m}} \end{array}\)

${\rm{G}} = \frac{{\rm{R}}}{{{\rm{Z}}_0^2}} = \frac{{100}}{{2500}} = 0.04\frac{{\rm{S}}}{{\rm{m}}}$ . We see only option b fits these two requirements. Confirming ${\rm{\beta }}$ from option b we see,

$\begin{array}{l} {\rm{\omega }}\sqrt {{\rm{LC}}} = {10^6}\sqrt {250 \times {{10}^{ - 6}} \times 0.1 \times {{10}^{ - 6}}} \ = {10^6} \times {10^{ - 6}} \times 5\ = 5 = {\rm{\beta }} \end{array}$

Thus, option b is true

Concept:

Laplace Transform of Causal Periodic Signals

Consider a causal Periodic signal as shown in the figure:

Let x₁(t), x₂(t), x₃(t) …. Be the signal representing 1^st, 2^nd, 3^rd, …. Cycles of a causal periodic signal with fundamental time-period T.

Therefore,

x₂(t) = x₁(t - T)

x₃(t) = x₂(t – 2T)

Similarly:

x(t) = x₁(t) + x₁(t - T) + x₁(t – 2T) + …

Taking Laplace transform, using time shifting properly

X(s) = X₁(s) + e^-sT X₁(s) + e^-2sT X₁(s) + …

= (1 + e^-sT + e^-2sT + …) X₁(s)

$X\left( s \right) = \frac{{{X_1}\left( s \right)}}{{1 - {e^{ - sT}}}}$

Where:

\({X_1}\left( s \right) = \mathop \smallint \limits_0^T x\left( t \right){e^{ - st}}dt\)

Calculation:

Laplace transform for a periodic signal is given as:

\({\rm{F}}\left( {\rm{s}} \right) = \frac{1}{{1 - {{\rm{e}}^{ - {\rm{sT}}}}}}\mathop \smallint \limits_0^{\rm{T}} {\rm{f}}\left( {\rm{t}} \right){{\rm{e}}^{ - {\rm{st}}}}{\rm{dt}}\)

\(= \frac{1}{{1 - {{\rm{e}}^{ - {\rm{sT}}}}}}\left[ {\mathop \smallint \limits_0^{\frac{{\rm{T}}}{2}} \left( 1 \right).{{\rm{e}}^{ - {\rm{st}}}}{\rm{dt}} + 0} \right] \)

$= \frac{1}{{1 - {{\rm{e}}^{{\rm{sT}}}}}}\left[ {\frac{{{{\rm{e}}^{ - {\rm{st}}}}}}{{ - {\rm{s}}}}} \right]_0^{\frac{{\rm{T}}}{2}}$

$\ = \frac{1}{{\left( {1 - {{\rm{e}}^{ - {\rm{sT}}}}} \right)}}.\frac{1}{{\rm{s}}}.\left[ {1 - {{\rm{e}}^{ - \frac{{{\rm{sT}}}}{2}}}} \right]$

$= \left( {\frac{{1 - {{\rm{e}}^{ - \frac{{{\rm{sT}}}}{2}}}}}{{{\rm{s}}\left( {1 + {{\rm{e}}^{ - \frac{{{\rm{sT}}}}{2}}}} \right)\left( {1 - {{\rm{e}}^{ - \frac{{{\rm{ST}}}}{2}}}} \right)}}} \right)$

$\ {\rm{F}}\left( {\rm{s}} \right) = \frac{1}{{{\rm{s}}\left( {1 + {{\rm{e}}^{ - \frac{{{\rm{sT}}}}{2}}}} \right)}}$

Given input \({\rm{x}}\left( {\rm{t}} \right) = \mathop \sum \limits_{{\rm{k}} = 1}^3 {{\rm{a}}{\rm{k}}}\cos \left( {{\rm{k}}{{\rm{\omega }}{\rm{o}}}{\rm{t}}} \right)\)

Suppose ${\rm{R;}} = {\rm{;}}1,{\rm{;L;}} = {\rm{;}}1,{\rm{;C;}} = {\rm{;}}1$ and they are connected in series then transfer function of system

$\rm H(s) = R + sL + 1/sC$

$\Rightarrow {\rm{H}}\left( {\rm{s}} \right) = {\rm{R}} + {\rm{sL}} + \frac{1}{{{\rm{sC}}}} = 1 + {\rm{s}} + \frac{1}{{\rm{s}}} = \frac{{{{\rm{s}}^2} + {\rm{s}} + 1}}{{\rm{s}}}$

At ${\rm{\omega ;}} = {\rm{;k}}{{\rm{\omega }}_{\rm{o}}} \Rightarrow {\rm{;H}}\left( {{\rm{j\omega }}} \right) = {\rm{;M}}\angle \phi$

\(\therefore {\rm{y}}\left( {\rm{t}} \right) = \mathop \sum \limits_{{\rm{k}} = 1}^3 {{\rm{a}}{\rm{k}}}.{\rm{M}}.\cos \left( {{\rm{k}}{{\rm{\omega }}{\rm{o}}}{\rm{t}} + \phi } \right) = \mathop \sum \limits_{{\rm{k}} = 1}^3 {{\rm{b}}{\rm{k}}}.\cos \left( {{\rm{k}}{{\rm{\omega }}{\rm{o}}}{\rm{t}} + \phi } \right)\)Where \({{\rm{b}}{\rm{k}}}{\rm{;}} \ne {\rm{;}}{{\rm{a}}{\rm{k}}}\)

Let open-loop transfer function is:

${\rm{G}}\left( {\rm{s}} \right) = \frac{1}{{{\rm{ST}}}}$

then,

Closed-loop transfer function is:

${\rm{H}}\left( {\rm{s}} \right) = \frac{1}{{1 + {\rm{sT}}}}$ (assume unity negative feedback)

${\rm{Y}}\left( {\rm{s}} \right){\rm{;}} = {\rm{;X}}\left( {\rm{s}} \right).{\rm{;H}}\left( {\rm{s}} \right)$

Case i :

if input = impulse

${\rm{X}}\left( {\rm{s}} \right){\rm{;}} = {\rm{;}}1$

$\begin{array}{l} {\rm{Y}}\left( {\rm{s}} \right) = \frac{1}{{1 + {\rm{sT}}}}.\frac{1}{{\rm{T}}}\left[ {\frac{1}{{\frac{1}{{\rm{T}}} + {\rm{s}}}}} \right]\ {\rm{y}}\left( {\rm{t}} \right) = \frac{1}{{\rm{T}}}{{\rm{e}}^{ - \frac{{\rm{t}}}{{\rm{T}}}}} \end{array}$

Case ii:

If input = unit step

${\rm{;X}}\left( {\rm{s}} \right) = \frac{1}{{\rm{s}}}$

\(\begin{array}{l} {\rm{Y}}\left( {\rm{s}} \right) = \frac{1}{{1 + {\rm{sT;}}}}.\frac{1}{{\rm{s}}} \= \frac{1}{{\rm{s}}} + \frac{{ - {\rm{T}}}}{{\left( {1 + {\rm{sT}}} \right)}} \= \frac{1}{{\rm{s}}} - \frac{1}{{\frac{1}{{\rm{T}}} + {\rm{s}}}}\ {\rm{y}}\left( {\rm{t}} \right) = 1 - {{\rm{e}}^{ - \frac{{\rm{t}}}{{\rm{T}}}}} \end{array}\)

Case iii:

 If input = Ramp $\Rightarrow {\rm{;x}}\left( {\rm{s}} \right) = \frac{1}{{{{\rm{s}}^2}}}$

\(\begin{array}{l} {\rm{Y}}\left( {\rm{s}} \right) = \frac{1}{{1 + {\rm{sT}}}}.\frac{1}{{{{\rm{s}}^2}}}\ = \frac{1}{{{{\rm{s}}^2}}} + \frac{{{{\rm{T}}^2}}}{{\left( {1 + {\rm{sT}}} \right)}} + \frac{{\left( { - {\rm{T}}} \right)}}{{\rm{s}}}\ {\rm{y}}\left( {\rm{t}} \right) = {\rm{t}} + {\rm{T}}{{\rm{e}}^{ - {\rm{t}}/{\rm{T}}}} - {\rm{T}} \= {\rm{t}} - {\rm{T}}\left( {1 - {{\rm{e}}^{ - \frac{{\rm{t}}}{{\rm{T}}}}}} \right) \end{array}\)

Considering low level injection the excess electron generated can be neglected. The excess holes generated at time ${\rm{t}} = 0$ is \({\rm{\delta p}}\left( 0 \right) = {{\rm{G}}{{\rm{opt}}}}.{{\rm{\tau }}{{\rm{po}}}} \Rightarrow {\rm{\delta p}}\left( 0 \right) = 1.5 \times {10^{20}} \times {\rm{;}}0.1{\rm{;\mu sec}}\)

$\Rightarrow {\rm{\delta p}}\left( 0 \right) = 1.5 \times {10^{13}}{\rm{holes}}/{\rm{c}}{{\rm{m}}^3}$

Now hole concentration as a function of time is

\(\begin{array}{l} {\rm{\delta p}}\left( {\rm{t}} \right) = {{\rm{\delta }}{\rm{p}}}\left( {\rm{O}} \right){{\rm{e}}^{ - {\rm{t}}/{{\rm{\tau }}{{\rm{po}}}}{\rm{;}}}}\ {\rm{\delta p}}\left( {0.3{\rm{;\mu sec}}} \right) = 1.5 \times {10^{13}}.{{\rm{e}}^{ - \frac{{0.3 \times {{10}^{ - 6}}}}{{0.1 \times {{10}^{ - 6}}}}}}\ = 1.5 \times {10^{13}} \times {{\rm{e}}^{ - 3}}\ \Rightarrow {{\rm{\delta }}_{\rm{p}}}\left( {0.3{\rm{\mu sec}}} \right) = 7.47 \times {10^{11}}{\rm{;hole}}/{\rm{c}}{{\rm{m}}^3} \end{array}\)

Once the generating source as removed, the generated excess holes recombine with excess electrons and deplete exponentially as a function of time. This conclusion can be verified by solving the ambipolar transport equation.

Concept:

Gray to Binary

Gray codes are widely used to facilitate error correction in digital communications such as digital terrestrial television and some cable TV systems.

Conversion from Gray Code to Binary Code:

Let Gray Code be G3  G2  G1  G0 and Binary Code be B3 B2 B1 B0. Then the respective Grey Code can be obtained as follows:

Binary to Gray

It is given by

y1 = x1

y2 = x1 ⊕ x2

y3 = x2 ⊕ x3

y4 = x3 ⊕ x4

Analysis:

Assume X₂X₁X₀ = (111)₂ = 7₁₀

Then the output of the first decoder = OP₇  which is connected to IP₅ of the encoder

(5)₁₀ = (101)₂

∴ output of encoder Y₂Y₁Y₀ = 101

∴ input to the circuit = 1 1 1 and output = 1 0 1

Consider Gray to Binary converter if input = 111 then output is 101.

∴ So the given circuit acts like a Gray to Binary converter

Hence option (3) is correct

CONCEPT:

A tri-state buffer is a non-inverting device that produces output when input to EN (enable pin) is high else it goes to a high impedance (HI-Z) state.

It is similar to a normal buffer but it adds an Enable (EN) buffer input that decides whether primary input is passed to its output or not.

If EN signal input is true, the tri-state buffer will produce output.

If it is false, the Tri-state buffer goes to a high impedance state such that it disconnects its output from the circuit.

When EN = 0 then, It will go into a high impedance state or disconnected state irrespective of the input value.

When EN = 1 then, It will produce output the same as the input value.

Calculation:

When

C₁ = 0 and C₀ = 0 then O₀ is high → Y = P

C₁ = 0 and C₀ = 1 then O₁ is high → Y = Q

C₁ = 1 and C₀ = 0 then O₂ is high → Y = R

C₁ = 0 and C₀ = 1 then O₃ is high → Y = S

∴ So the given circuit acts like 4 : 1 multiplexer with inputs P, Q, R and S  and  selection inputs C₁ and C₀

PUSH  is  pre decremented operation

Since it is 16 bit operation it takes 6 clock cycles for Fetch and decoding operation.

For PUSH operation:

Total no of clock cycles = 6+3+3= 12 clock cycles

Whereas POP is a post incremented operation and it takes only 4 clock cycles in fetch and decode operation.

For POP:

Total no of clock cycles = 4+3+3= 10 clock cycles

Concept: If  the signal-to-noise ratio (SNR) is maximized at the output of the filter then the filter is matched fi

Analysis: The SNR is maximum when the filter is matched to the input signal. And when the filter is matched to the input the energies of both the input and matched filter responses are the same. Thus \({{\rm{E}}{\rm{s}}} = {{\rm{E}}{\rm{h}}}\). And in that case the SNR \(= \frac{{2{{\rm{E}}_{\rm{s}}}}}{{{{\rm{N}}0}}} = \frac{{2{{\rm{E}}{\rm{h}}}}}{{{{\rm{N}}_0}}}{\rm{;}}\)

Concept: 

The current through any surface is related to current density as ${\rm{I}} = \int!!!\int {\rm{\vec J}}.{\rm{d\vec a}}$

Application:

The current density at ${\rm{r}} = 0.8{\rm{m}}$ is

\(\begin{array}{l} {\rm{\vec J}} = \frac{{400{\rm{sin\theta }}}}{{2{\rm{\pi }}\left( {0.64 + 4} \right)}}{{{\rm{\hat a}}}{\rm{r}}}{\rm{;A}}/{{\rm{m}}^2}\ \Rightarrow {\rm{\vec J}} = \frac{{400{\rm{sin\theta }}}}{{2{\rm{\pi }}\left( {464} \right)}}{{{\rm{\hat a}}}{\rm{r}}}{\rm{;A}}/{{\rm{m}}^2} \end{array}\)

So total current I through portion ${\rm{r}} = 0.8{\rm{m}}$,  $\frac{{\rm{\pi }}}{{12}} \le {\rm{\theta }} \le \frac{{\rm{\pi }}}{4}$

And $0 \le \phi \le 2{\rm{\pi }}$ is ${\rm{I}} = \int!!!\int {\rm{\vec J}}.{\rm{d\vec a}}$

\(\begin{array}{l} {\rm{I}} = \mathop \smallint \limits_{\phi = 0}^{2{\rm{\pi }}} \mathop \smallint \limits_{{\rm{\theta }} = \frac{{\rm{\pi }}}{{12}}}^{\frac{5}{4}{\rm{;;}}} \frac{{400\sin {\rm{\theta }}}}{{2{\rm{\pi }} \times 4.64}}{{{\rm({\hat {dA}}}}{\rm{r}}} = {{\rm{r}}^2}{\rm{sin\theta d\theta d}}\phi {{{\rm{\hat a}}}{\rm{r}}})\ {\rm{I}} = \frac{{400{\rm{;}}}}{{2{\rm{\pi }} \times 4.64}}{\left. {\mathop \smallint \limits_{\phi = 0}^{2{\rm{\pi }}} \mathop \smallint \limits_{{\rm{\theta }} = \frac{{\rm{\pi }}}{{12}}}^{\frac{5}{4}{\rm{;;}}} {{\rm{r}}^2}{\rm{sin^2\theta d\theta d}}\phi } \right|_{{\rm{r}} = 0.8}}\ = \left. {\left. {\frac{{400 \times 0.64}}{{2{\rm{\pi }} \times 4.64}}\left( \phi \right)} \right|0^{2{\rm{\pi }}}\left[ {\frac{{\rm{\theta }}}{2} - \frac{{{\rm{sin}}2{\rm{\theta }}}}{4}} \right]} \right|{\frac{{\rm{\pi }}}{{12}}}^{\frac{{\rm{\pi }}}{2}}\ \Rightarrow {\rm{I}} = \frac{{400 \times 0.64}}{{2{\rm{\pi }} \times 4.64}} \times 2{\rm{\pi }}\left[ {\left( {\frac{{\rm{\pi }}}{8} - \frac{{\rm{\pi }}}{{24}}} \right) - \frac{1}{4}\left( {\sin \frac{{\rm{\pi }}}{2} - \frac{{{\rm{sin\pi }}}}{6}} \right)} \right]\ \Rightarrow {\rm{I}} = \frac{{400 \times 0.64}}{{2{\rm{\pi }} \times 4.64}} \times 2{\rm{\pi }} \times \left[ {\frac{{\rm{\pi }}}{{12}} - \frac{1}{4}\left( {1 - \frac{1}{2}} \right)} \right]\ = \frac{{400 \times 0.64}}{{4.64}} \times \left[ {\frac{{\rm{\pi }}}{{12}} - \frac{1}{8}} \right] \end{array}\)

$\Rightarrow {\rm{I}} = 7.55{\rm{;A}}$ (No option matching)

Now, area of the portion of the spherical surface

\(\begin{array}{l} {\rm{S}} = {\left. {\mathop \smallint \limits_{\phi = 0}^{2{\rm{\pi }}} \mathop \smallint \limits_{{\rm{\theta }} = \frac{{\rm{\pi }}}{{12}}}^{\frac{{\rm{\pi }}}{4}} {{\rm{r}}^2}\sin {\rm{\theta d\theta d}}\phi } \right|_{{\rm{r}} = 0.8}}\ \Rightarrow {\rm{S}} = 0.64 \times \left. {\left( \phi \right)} \right|0^{2{\rm{\pi }}} \times \left. {\left( { - \cos {\rm{\theta }}} \right)} \right|{\frac{{\rm{\pi }}}{{12}}}^{\frac{{\rm{\pi }}}{4}}\ \Rightarrow {\rm{S}} = 0.64 \times 2{\rm{\pi }} \times \left( {0.26} \right)\ \Rightarrow {\rm{S}} = 1.04 \end{array}\)

Now average current density ${{\rm{J}}_{{\rm{avg}}}} = \frac{{7.55}}{{1.04}} = 7.25\frac{{\rm{A}}}{{{{\rm{m}}^2}}}$

Antenna’s noise temperature ${{\rm{T}}_{{\rm{AN}}}} = 50{\rm{K}}$

Ambient temperature ${{\rm{T}}_{\rm{A}}} = 290{\rm{K}}$

Preamplifier noise figure ${{\rm{F}}_{\rm{P}}} = 2{\rm{dB}}$

Amplifier gain ${{\rm{G}}_{\rm{P}}} = 40{\rm{dB}}$

Now, effective noise temperature of amplifier input \({{\rm{T}}{\rm{e}}} = \left( {{\rm{F}} - 1} \right){{\rm{T}}{\rm{A}}}\)

Noise figure of antenna ${\rm{F}} = {10^{\frac{2}{{10}}}} = {10^{\frac{1}{5}}} = 1.585$

Thus, ${{\rm{T}}_{\rm{e}}} = 290\left( {1.585 - 1} \right) = 290 \times 0.585 = 169.7{\rm{;K}}$

Now, noise power at the preamplifier output, \({{\rm{P}}{{\rm{ao}}}} = {\rm{GK}}{{\rm{T}}{{\rm{IN}}}}{\rm{B}}\)

Gain , ${\rm{G}} = {10^{\frac{{40}}{{10}}}} = {10^4}$ and ${{\rm{T}}_{{\rm{IN}}}}$ is the input noise temperature at amplifier input

\(\begin{array}{l} {{\rm{T}}{{\rm{IN}}}} = {{\rm{T}}{{\rm{AN}}}} + {{\rm{T}}{\rm{e}}} = 50{\rm{K}} + 169.6{\rm{K}}\ \Rightarrow {{\rm{T}}{{\rm{IN}}}} = 219.6{\rm{K}} \end{array}\) 

Amplifier output noise, \({{\rm{P}}{{\rm{ao}}}} = {\rm{GK}}{{\rm{T}}{{\rm{IN}}}}{\rm{B}}\)

\(\begin{array}{l} {{\rm{P}}{{\rm{ao}}}} = {10^4} \times \left( {1.38 \times {{10}^{ - 23}}} \right) \times \left( {219.6} \right) \times \left( {12 \times {{10}^6}} \right)\ {{\rm{P}}{{\rm{ao}}}} = 3.63 \times {10^{ - 10}}{\rm{W}} \end{array}\) 

Thus, option a is correct.

Given

Electric field of the incident wave is:

$E^i_W = \left(\hat a_x + j\hat a_y\right) E_0e^{jkz}$

at z = 0;

(in time-varying form)

$\vec E^i_w = E_0 \cos ω t;\hat a_x - E_0 \sin ω t; \hat a_y$

at ωt = 0

$\vec E^i_w = E_0 \hat a_x$

at $\omega t = \frac {\pi}{2}$

$\vec E^i_w = E_0 \left(-\hat a_y\right)$

As a tip of electric field intensity is tracing a circle when time varies, hence the wave is said to be circularly polarized in clockwise direction (or) RHCP. Polarizing vector of incident wave is given by,

$\vec P_i = \frac {\hat a_x + j\hat a_y}{\sqrt 2}$

radiated electric field from the antenna is

$\vec E_a = \left(\hat a_x + 2\hat a_y\right)E_l \frac l \gamma e^{-jk\gamma}$

at r = 0

$\vec E_a = E_l \cos \omega t;\hat a_x + 2E_l \cos \omega t;\hat a_y$ (in time varying form)

As both x & y components are in-phase, hence the wave is said to be linear polarized. Polarizing vector of radiated field is $\hat P_a = \frac {\left(\hat a_x + 2\hat a_y\right)}{\sqrt 5}$ polarizing mismatch; The polarizing mismatch is said to have, if the polarization of receiving antenna is not same on the polarization of the incident wave. The polarization loss factor (PLF) characterizes the loss of EM power due to polarization mismatch.

$PLF = \left|\hat P_i.\hat P_a\right|^2$

in dB; PLF (dB) = 10 log (PLF)

$PLF = \left|\left(\frac {\hat a_x + j\hat a_y}{\sqrt{2}}\right).\left(\frac {\hat a_x + 2\hat a_y}{\sqrt 5}\right)\right|^2 = \left|\frac {1 + j2}{\sqrt 2\sqrt 5}\right|^2 = \frac 1 2$ (or) 0.5

PLF (dB) = 10 log 0.5 = -3.0102

Concept: 

Power Radiated can be calculated from average power density by integrating it over the given surface.

Application:

Given that \({{\rm{W}}{{\rm{rad}}}} = {{\rm{W}}{{\rm{avg}}}} = \frac{1}{{{{\rm{r}}^2}}}{\rm{co}}{{\rm{s}}^4}{\rm{\theta }}.{{\rm{C}}0}.\overrightarrow {{{\rm{a}}{\rm{r}}}}\)

Power radiated $= \smallint {{\rm{W}}_{{\rm{rad}}}}.{\rm{d\vec S}}$

But ${\rm{d\vec S}} = {{\rm{r}}^2}{\rm{sin;\theta ;d\theta d}}\phi .\overrightarrow {{{\rm{a}}_{\rm{r}}}}$

Power radiated = $\smallint \smallint \frac{1}{{{{\rm{r}}^2}}}{\rm{co}}{{\rm{s}}^4}{\rm{\theta }}.{{\rm{C}}_0}.{{\rm{r}}^2}{\rm{sin;\theta ;d\theta d}}\phi$

Where $0 \le {\rm{\theta }} \le \frac{{\rm{\pi }}}{2}{\rm{;and;}}0 \le \phi \le 2{\rm{\pi }}$

$\begin{array}{l} \Rightarrow {\rm{Power;Radiated}},{{\rm{P}}_{{\rm{rad}}}} = {\rm{;}}\smallint \smallint {\rm{co}}{{\rm{s}}^4}{\rm{\theta }}.{{\rm{C}}_0}.{\rm{sin;\theta ;d\theta d}}\phi \ = {{\rm{C}}_0}\left( {2{\rm{\pi }}} \right)\smallint {\rm{co}}{{\rm{s}}^4}{\rm{\theta }}.{\rm{sin\theta }}.{\rm{d\theta }}:{\rm{where;}}0 \le {\rm{\theta }} \le \frac{{\rm{\pi }}}{2}\ = {{\rm{C}}_0}\left( {2{\rm{\pi }}} \right) \times \frac{{3 \times 1 \times 1}}{{5 \times 3 \times 1}} = 1.256{{\rm{C}}_0}. \end{array}$

${\rm{Radiation;intensity}} = {\rm{U}} = {{\rm{r}}^2}.{{\rm{W}}_{{\rm{rad}}}} = {\rm{co}}{{\rm{s}}^4}{\rm{\theta }}.{{\rm{C}}_0}$

Directivity \({{\rm{D}}0} = 4{\rm{\pi }} \times \frac{{{{\rm{U}}{{\rm{max}}}}}}{{{{\rm{P}}_{{\rm{rad}}}}}} = 4{\rm{\pi }} \times \frac{{{{\rm{C}}_0}}}{{1.256{{\rm{C}}_0}}} = 10.005\)

Directivity in $\left( {{\rm{dB}}} \right) = {\rm{;}}10{\rm{;lo}}{{\rm{g}}_{10}}10.005 = 10$