GATE EC 2012 Official Paper (Previous Year Paper)

Given:

(1.001)1259 = 3.52 and (1.001)2062 = 7.85

Concept used:

X(a + b) = X(a) × X(b)

Calculation:

⇒ (1.001)3321 = (1.001)1259 + 2062 

= (1.001)1259 × (1.001)2062

= 3.52 × 7.85

= 27.64

Hence, the correct answer is "27.64".

The correct answer is 'should have taken'.

Key Points

• Mixed Conditional Sentence: Mixed conditionals are conditional sentences that mix two different times in one sentence or we can say that the if-clause is not the same as the time in the result.
• Structure for Mixed conditional (for present condition, past result):
• If/when + past simple, would have + verb infinitive.
• Example: 
• If Sam spoke Russian, he would have translated the letter for you.
• (But Sam doesn't speak Russian and that is why he didn't translate the letter.)
• Note: Would can be replaced with should/might in the sentence.

Therefore, the correct sentence is: If the tired soldier wanted to lie down, he should have taken the mattress out on the balcony. 

Additional Information

• Conditional sentence:- As the name suggests, these sentences express conditions.
• In conditional sentences, there are two clauses of the sentences. One is a conditional clause and another one is main clause.
• Other Conditional sentences are-

The correct answer is 'nonchalance'.

Key Points

• Let's explore the options:
• beggary: a state of extreme poverty.
• Example: They have no benefits to stand between them and beggary.
• nomenclature: the devising or choosing of names for things, especially in a science or other discipline.
• Example: The Linnean system of zoological nomenclature
• jealousy: the state or feeling of being jealous.
• Example: He broke his brother's new bike in a fit of jealousy.
• nonchalance: the trait of remaining calm and seeming not to care.
• Example: He leaned back in his chair with apparent nonchalance.
• Thus, from above we can refer that the correct answer is option 4.

Therefore, the correct sentence is: 'Given the seriousness of the situation that he had to face, his nonchalance was impressive.'

The correct answer is 'Freedom'.

Key Points

• Latitude: scope for freedom of action or thought.
• Example: Journalists have considerable latitude in criticizing public figures.
• Freedom: the power or right to act, speak, or think as one wants without hindrance or restraint.
• Example: We do have some freedom of choice.
• Thus, the correct answer is option 2.

Additional Information

• Let's explore options:
• Eligibility: the state of having the right to do or obtain something through satisfaction of the appropriate conditions.
• Coercion: the practice of persuading someone to do something by using force or threats.
• Meticulousness: showing great attention to detail; very careful and precise.

The correct answer is 'should be given'; i.e. the error lies in this part of the sentence.

Key Points

• The given sentence is in active voice which represents a form or set of forms of a verb in which the subject is typically the person or thing performing the action and which can take a direct object.
• Thus, the usage of 'be given' is incorrect which is used with a passive voice which is used when we want to emphasize the action (the verb) and the object of a sentence rather than subject.
• Thus, 'should be given' needs to be replaced with 'should give'.

Therefore, the correct sentence is: 'I requested that he should give the driving test today instead of tomorrow.'

The correct answer is 'Thorough regimentation was the main reason for the efficiency of the Roman legions even in adverse circumstances.'

Key Points

• Let's refer to the following lines of the passage:
• In the legions, military law prevailed and discipline was brutal. Discipline on the battlefield kept units obedient, intact and fighting, even when the odds and conditions were against them.
• Thus, from above we can infer that the correct answer is option 1.

Additional Information

• Regimentation: organize according to a strict system or pattern.

Let's consider number of 10 rupee notes as 'x'

And number of 20 rupees notes as 'y'

Then total number of notes = x + y = 14 -------(1)

Total money = 10x + 20y = 230  ---------(2)

By solving (1) and (2), we get

x = 5 and y = 9

Explanation:

Divide 9 bags into three parts.

2,2 and 5 respectively.

If we compare the 2,2 bags on pans of a balance.

We can identify which side is the lighter bag placed.

And then we will need only one more weighing for identifying the faulty bag.

So only two weighings are required.

Alternative solution:

In the case of a weighing balance (i.e. beam balance) the following model can be observed.

1 - 3 → 1 weighing required

4 - 9 → 2 weighings required

10 - 27 → 3 weighings required

28 - 81 → 4 weighings required

And the process will go so on.

As there are nine objects. The minimum number of weighings required will be only 2.

Total budget = 10500 Rs

Expenditure  other than savings = 9000

The approximate percentage of the monthly budget NOT spent on savings is = (9000/ 10500) × 100

= 85.71 % ≈ 86%

A meeting occurs if the person arrives between 1:00 PM and 1:45 PM and the second person arrives in the next 15 minutes or if both the persons arrive between 1:45 and 2:00.

Case 1:

45/60 are favorable cases and hence the probability of first-person arriving between 1:00 and 1:45 is 3/4.

Probability of second person arriving in the next 15 min = 15/60 = 1/4

So, the probability of one person arriving between 1:00 and 1:45 and meeting the other = 3/4 × 1/4 × 2 = 3/8 (2 for choosing the first arriving friend)

Case 2: 

Both friends must arrive between 1:45 and 2:00

Probability = 1/4 × 1/4 = 1/16

So, probability of a meet = 3/8 + 1/16 = 7/16

Hence, the correct answer is 7/16.

Given $\rm i_b(dc) = 1$

We know $\rm {r_{\rm{\pi }}} = \frac{\beta }{{{g_m}}} = \frac{\beta \ V_T }{{{I_c{\left( {dc} \right)}}}} = \frac{{\beta {V_T}}}{{\beta\ {i_b{\left( {dc} \right)}}}}$

$\rm {r_{\rm{\pi }}} = \frac{{{V_T}}}{{{i_b{\left( {dc} \right)}}}} = \frac{{25\ mV}}{{1\ mA}} = 25\ {\rm{\Omega }}$

\(\rm E\left[ {{X^2}\left( t \right)} \right] = \frac{1}{{2\pi }}\mathop \smallint \limits_{ - \infty }^\infty {S_x}\left( \omega \right)d\omega \)

Since x(t) is real process the PSD is even function hence above equation can be written as

\(\rm = \frac{2}{{2\pi }}\mathop \smallint \limits_0^\infty {S_x}\left( \omega \right)d\omega \)

\(\rm = \frac{1}{\pi }\mathop \smallint \limits_0^\infty {S_x}\left( \omega \right)d\omega \)

$\rm = \frac{1}{\pi }\left[ {400 + \frac{1}{2} \times 2 \times {{10}^3} \times 6} \right]$

$\rm = \frac{{6400}}{\pi }$

$\rm E\left[ {X\left( t \right)} \right] = 0$ as \(\rm \frac{1}{{2\pi }}\mathop \smallint \limits_{{0^ - }}^{{0^ + }} {S_X}\left( \omega \right)d\omega = 0\)

Concept:

B_T = 0.5 R_s(β + 1)

R_s → Symbol rate

Rs = (2 × β_T)/(β + 1)

Calculations:

β = 0.75

Rs = (2 × 3500)/(0.75 + 1)

= 4000 symbols/sec

The given waveform can be represented as:

$\vec E = \left( {8{{\hat a}_x} + 6{{\hat a}_y} + 5{{\hat a}_z}} \right){e^{j\left( {\omega t + 3x - 4y} \right)}}$

A perfect conductor will reflect  $\vec E$ completely

${\vec E_t} = 6{\hat a_y} + 5{\hat a_z},;{\vec E_n} = 8{\hat a_x}$

For reflected wave $\vec E$ tangential component will cancel out tangential component of $\vec E$ so that net tangential field is zero.

i.e. ${\vec E_{{r_t}}} = - \left( {{{\vec E}_t}} \right) = - \left( {6{{\hat a}_y} + 5{{\hat a}_z}} \right)$

Similarly for normal component, ${\vec E_{{r_n}}} = - {\vec E_n} = - \vec 8{a_x}$

Since, the wave travelling is ‘-x’ & ‘+y’ direction and gets reflected at boundary x ≤ 0, the wave will travel along ‘+x’ & ‘+y’ direction.

$\Rightarrow {\vec E_r} = \left( { - 8{{\hat a}_x} - 6{{\hat a}_y} - 5{{\hat a}_z}} \right){e^{j\left( {\omega t - 3x - 4y} \right)}}V/m$

Note: It is an Official GATE Question, asked in GATE EC 2012 Paper. Please understand that Marks to All was allotted for this particular Question. This is because, in any TEM wave, Poynting Theorem must be satisfied according to which the propagation direction and field direction should be perpendicular.

$\vec E = 10\left( {{{\hat a}_y} + j{{\hat a}_z}} \right){e^{ - j25x}}$

⇒ since $\left| {{{\hat a}_y}} \right| = \left| {{{\hat a}_z}} \right|$

⇒ Circular, due to phase difference of 90° between y and z, here phase difference between y component and x component is + π/2 hence the wave is left circular.

The x direction is into the plane 

to go in the direction of propogation while curling fingers in direction of rotation in circle, you have to use left hand.

Thus given wave is left hand circularly polarised

$\beta ; = ;25; \Rightarrow ;f; = \frac{c}{\lambda } = \frac{{c \times \beta }}{{2\pi }} = \frac{{3 \times {{10}^8} \times 25}}{{2\pi }}$

⇒ f ≈ 1.2 × 10 ⁹ Hz

The circuit is a SR flip-flop with input A = S and B = R. In SR flip-flop there is no race around condition for any combination of input.

Note :

11 is a not allowed state because the output Q and Q'  will be 1. 

For race around Q should toggle.

It occurs in JK flip flop when J=K=1 and there is unequal propagation delay.

The only possible combinations are

A = 01 and B = 0 0

A = 10 and B = 00, 01

A = 11 and B = 00, 01, 10

So there are only 6 combinations

Tips and Tricks:

$\frac{2^{2n} - 2^n}{2}$

where n = 2 bit

Let's consider v >.7 and the diode is forward biased. 

By applying KVL, we get

10 - I × 1k - v = 0

10 -[(v-.7)/(500)] × 1000 - v = 0

⇒ v = 3.8 V 

And v > .7 and our assumption is correct,

⇒ i =[(v-.7)/(500)] = [3.8 - 0.7]/500 = 6.2 mA

The S – domain equivalent circuit is

$I\left( S \right) = \frac{{\frac{{{V_c}\left( 0 \right)}}{S}}}{{\frac{1}{{{C_1}S}} + \frac{1}{{{C_2}S}}}} = \frac{{{V_c}\left( 0 \right)}}{{\left( {\frac{1}{{{C_1}}} + \frac{1}{{{C_2}}}} \right)}}\% MathType!End!2!1!$

V_c(0) = 12 V

$I\left( S \right) = \left( {\frac{{{C_1}{C_2}}}{{{C_1} + {C_2}}}} \right)\left( {12} \right)\% MathType!End!2!1!$

I(S) = 12 C_eq

Taking Inverse Laplace transform for the current in the time domain.

i(t) = 12 C_eq δ(t) 

The above equation shows that it's an impulse.

In phasor form:

Z = 4 – j3

Z = 5 ∠-36.86° Ω

I = 5 ∠100° A

Average power delivered will be:

${P_{avg}} = \frac{1}{2}{\left| I \right|^2}Zcos\theta \% MathType!End!2!1!$

= (½) × 25 × 5 cos 36.86°

= 50 W

Alternate Approach:

The power delivered to the load will simply be the power dissipated by the resistor R, i.e.

P_delivered = I_RMS² R

With I_m = 5, I_RMS will be:

$I_{RMS} = \frac{5}{\sqrt 2}$

$P_{delivered} = (\frac{5}{\sqrt 2})^2\times 4$

$P_{delivered} =\frac{25}{2}\times 4 = 50~W$

Concept

Multiplication of function in one domain corresponds to the differentiation in another domain

If f(t) having the Laplace transform F(s) then t ⋅ f(t) will have the transform as

  t f(t) ↔$- \frac{{dF\left( s \right)}}{{ds}}$

Calculation:

Given function f(t) is . And  g(t) = t ⋅ f(t)

Laplace transform of g(t) is:

$L[g(t)] = - \frac{d}{{ds}}\left( {\frac{1}{{{s^2}s + 1}}} \right)$

$= - \frac{{\frac{{d\left( 1 \right)}}{{ds}}\left( {{s^2} + s + 1} \right) - 1 \times \frac{{d\left( {{S^2} + S + 1} \right)}}{{ds}}}}{{{{\left( {{S^2} + S + 1} \right)}^2}}}$

$= - \left( {\frac{{0 - 1\left( {2s + 1} \right)}}{{{{\left( {{s^2} + s + 1} \right)}^2}}}} \right)$

$= \frac{{2s + 1}}{{{{\left( {{s^2} + s + 1} \right)}^2}}}$

Given differential equation is $t\frac{{dx}}{{dt}} + x; = ;{t}$

$\Rightarrow ;\frac{{dx}}{{dt}} + \frac{1}{t}x; = ;1$

This is a linear differential equation in t.

$I.F.; = ;{e^{\smallint \frac{1}{t}dt}}; = ;{e^{\log t}}; = ;t$

The solution is,

$\begin{array}{l} x.(IF); = ;\smallint 1.(IF)dt + c\Rightarrow x.t; = ;\smallint tdt + c\ \Rightarrow ;xt; = ;\frac{{{t^2}}}{2} + c \end{array}$

Given x(1) = .5 ⇒ c = 0

The solution is $x; = ;\frac{{{t}}}{2}$

At t = 0⁺, D₁ will be ON and D₂ will be OFF, and the capacitor will be instantly charged to 1 V as the current offered will be infinite.

This 1 V at the capacitor will reverse bias both the diodes, i.e. both D₁ and D₂ will be OFF and the voltage v(t) will be calculated as shown:

– cosωt + 1 + V(t) = 0

V(t) = cosωt - 1 

Hence option (1) is correct

Concept:

MOS logic circuit consists of two network transistors, a pull-down network (PDN) and a Pull-up Network (PUN) as shown:

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:

Since we have the two MOS transistors connected in series, the resultant output will be the AND operation, with inverted output, i.e.

$\overline Y= {AB}$

$Y=\overline {AB}$

Concept:

Information associated with the event is “inversely” proportional to the probability of occurrence.

Entropy: The average amount of information is called the “Entropy”.

\(H = ;\mathop \sum \limits_i {P_i}{\log _2}\left( {\frac{1}{{{P_i}}}} \right);bits/symbol\)

Rate of information = r.H

Calculation:

Given:

Three symbols with a probability of 0.25, 0.25, and 0.50 at the rate of 3000 symbols per second.

Entropy is given as;

 $H = 0.25{\log _2}\left( {\frac{1}{{0.25}}} \right) + 0.25{\log _2}\frac{1}{{0.25}} + 0.5{\log _2}\frac{1}{{0.5}};$

$= 0.25 \times 2 + 0.25 \times 2 + 0.5 \times 1$

= 1.5

Rate of information = r.H

= 1.5 × 3000

= 4500 bits/sec

Inner diameter of outer conductor, D = 1 cm

Diameter of inner conductor, d = 0.01 cm

Characteristic impedance of coaxial cable, 

${Z_0} = \frac{1}{2\pi }\sqrt {\frac{\mu }{\epsilon}} \times ln\left( {\frac{D}{d}} \right) = \frac{{138}}{{\sqrt {{\varepsilon _r}} }}\log \frac{D}{d}$

$\Rightarrow {Z_o} = \frac{{138}}{2}\log \frac{1}{{0.01}} = 138;{\rm{\Omega }}$

Radiation pattern $F\left( \theta \right) = {\cos ^2}\theta ,0 \le \theta \le ;\pi ;/2$

U(θ) = F²(θ)

Directivity of antenna is given as

$D = \frac{{4\pi {U_{max}}}}{{{P_{rad}}}}$

\({P_{rad}} = \mathop \smallint \limits_{\phi = 0}^{2\pi } \mathop \smallint \limits_{\theta = 0}^{\frac{\pi }{2}} \left( {{{\cos }^4}{\rm{\theta }}} \right)\left( {sin\theta } \right)d\theta .d\phi = 2\pi \mathop \smallint \limits_{\theta = 0}^{\frac{\pi }{2}} \left( {{{\cos }^4}{\rm{\theta }}} \right)\left( {\sin {\rm{\theta).(d\theta }}}\right)\)

\(= 2\pi \mathop \smallint \limits_{ - 1}^0 {t^4}dt = 2\pi \left. {\frac{{{t^5}}}{5}} \right|_{ - 1}^0 = \frac{{2\pi }}{5}\)

${U_{max}} = {\left. {{{\cos }^4}\theta } \right|_{max}} = 1$

⇒ Directivity $D = \frac{{4\pi \left( 1 \right)}}{{\left( {\frac{{2\pi }}{5}} \right)}} = 10 \Rightarrow 10\log 10 = 10;dB$

$\begin{array}{l} x\left[ n \right] = {\left( {\frac{1}{3}} \right)^{\left| n \right|}} - {\left( {\frac{1}{2}} \right)^n}u\left[ n \right]\ = {\left( {\frac{1}{3}} \right)^n}u\left[ n \right] + {\left( {\frac{1}{3}} \right)^{n - 1}}u\left[ { - n - 1} \right] - {\left( {\frac{1}{2}} \right)^n}u\left[ n \right] \end{array}$

Taking Z-transform

\(\begin{array}{l} X\left[ z \right] = \mathop \sum \limits_{n = - \infty }^\infty {\left( {\frac{1}{3}} \right)^n}{z^{ - n}}u\left[ n \right] + \mathop \sum \limits_{n = - \infty }^\infty {\left( {\frac{1}{3}} \right)^{ - n}}{z^{ - n}}u\left[ { - n - 1} \right] - \mathop \sum \limits_{n = - \infty }^\infty {\left( {\frac{1}{2}} \right)^n}{z^{ - n}}u\left[ n \right]\ = \mathop \sum \limits_{n = 0}^\infty {\left( {\frac{1}{3}} \right)^n}{z^{ - n}} + \mathop \sum \limits_{n = - \infty }^{ - 1} {\left( {\frac{1}{3}} \right)^{ - n}}{z^{ - n}} - \mathop \sum \limits_{n = 0}^\infty {\left( {\frac{1}{2}} \right)^n}{z^{ - n}}\ = \underbrace {\mathop \sum \limits_{n = 0}^\infty {{\left( {\frac{1}{{3Z}}} \right)}^n}}I + \underbrace {\mathop \sum \limits{m = 1}^\infty {{\left( {\frac{Z}{3}} \right)}^m}}{II} - \underbrace {\mathop \sum \limits{n = 0}^\infty {{\left( {\frac{1}{{2Z}}} \right)}^n}}_{III} \end{array}\)

Series I converges if $\left| {\frac{1}{{3Z}}} \right| < 1or\left| Z \right| > \frac{1}{3}$

Series II converges if $\left| {\frac{1}{3}z} \right| < 1\ or\ \left| Z \right| < 3$

Series III converges if $\left| {\frac{1}{{2Z}}} \right| < 1\ or\ \left| Z \right| > \frac{1}{2}$

Region of convergence of X(Z) will be intersection of above three

So, ROC  $\frac{1}{2}< |Z| < 3$.

Prime implicant is a minterm, which are obtained by combining maximum possible adjacent cell in k-map

$F\left( {x,y,z} \right) = \bar x\bar y + xy$

r(t) = sin(ωt) ⇒ R(s) is not zero for all time duration for any value of ω

C(S) = G(S) R(S)

For C(S) at steady state to be zero

|(GS)| = 0

∵ R(S) ≠ 0

$\Rightarrow \left| {G\left( {j\omega } \right)} \right| = \left| {\frac{{\left( {16 - {\omega ^2}} \right)\left( {j\omega + 2} \right)}}{{\left( {j\omega + 1} \right)\left( {j\omega + 3} \right)\left( {j\omega + 5} \right)}}} \right| = 0$

Only possible if

16 – ω² = 0

⇒ ω = 4 rad/s

Assume a voltage source V between 1 and 2 ,

then $\frac{V}{I} = impendance$ at 1 and 2

${i_b} = \frac{{0 - V}}{{10\ K{\rm{\Omega }}}}$         ___________(1)

Apply Nodal analysis then

$- {i_b} + \frac{V}{{100}} - 99{i_b} - I = 0$

V = 10⁴ i_b + 100 I        ___________(2)

From (1) and (2) we get

$V = {10^4}\left( {\frac{{ - V}}{{{{10}^4}}}} \right) + 100\ I$

2V = 100 I

Impendence $= \frac{V}{I} = 50\ {\rm{\Omega }}$

∴ Option (1) is correct

Applying nodal analysis at top node

$\frac{{{V_1} + 1\angle 0^\circ }}{1} + \frac{{{V_1} + 1\angle 0^\circ }}{{j1}} = 1\angle 0^\circ$

V₁ = (j1 + 1) + j1 + 1 ∠ 0° = j1

$\begin{array}{l} {V_1} = \frac{{ - 1}}{{1 + j1}}\ {I_1} = {V_1} = \frac{{{V_1} + 1\angle 0^\circ }}{{j1}} = \frac{{\frac{{ - 1}}{{j + 1}} + 1}}{{j1}} = \frac{j}{{\left( {1 + j} \right)j}} = \frac{1}{{\left( {1 + j} \right)}}A \end{array}$

$f\left( z \right); = ;\frac{1}{{z + 1}} - \frac{2}{{z + 3}}; = ;\frac{{ - z + 1}}{{\left( {z + 1} \right)\left( {z +3} \right)}}$

The poles of f(z) are - 1 and -3 and -3 is outside of the circle |z + 1| = 1.

By Cauchy’s formula,

$\begin{array}{l} \smallint \frac{{f\left( z \right)}}{{z - a}}dz; = ;2\pi if\left( a \right)\ \smallint \frac{{ - z + 1}}{{\left( {z + 3} \right)\left( {z +1} \right)}}dz; = ;\smallint \frac{{\frac{{ - z +1}}{{\left( {z +3} \right)}}}}{{\left( {z + 1} \right)}}dz \end{array}$

By Cauchy’s formula,

$\begin{array}{l} \Rightarrow ;2\pi if\left( { - 1} \right); = ;2\pi i\left( {\frac{{-(-1) + 1}}{{ - 1 + 3}}} \right); = ;2\pi i\ \Rightarrow ;\frac{1}{{2\pi i}}\smallint f\left( z \right)dz; = ;1 \end{array}$

P(max(X,Y) < .5) = P(X <.5, Y < 0.5)

Since X and Y are independent random variable, therefore

P(X <.5, Y < 0.5) = P(X < .5) P(Y < .5)

X and Y are uniformly distributed in [-1,1]

PDF of X and Y are

Thus,

P(X > 2) P(Y > 2)

Let A = xx

Taking logarithm both the sides.

Log A = x log x = i log (0 + x)

As |x| = 1 and arg (x) = π/2,

Log A = x(log1 + xπ/2) = x(0 + x π/2)

Log A = -π/2

Hence, A = e- π/2

The volume of the region is calculated as:

$= 1\mu m^2 × 1\mu m$( = {10^{ - 18}}{m^3}\)

For an n-channel MOS capacitor, the type of substrate is p-type.

Also, the type of doping for the source and the drain of an n-channel MOSFET is n-type.

The dopant density at the source is the donor density only as the source is n-type.

The concentration of holes in the source region will be calculated from the mass action law by:

$\frac{{n_i^2}}{{{N_D}}} = \frac{{{{\left( {{{10}^{10}}} \right)}^2}}}{{{{10}^{19}}}}=10\ cm^{-3}=10^{7} m^{3}$(\)

∴ The number of holes in that volume will be:

=10^-18 × 10⁷ = 10^-11 ≈ 0

Based on the information given for the two symbols waveforms, we see that the basis function is $\rm \sqrt {\frac{2}{T}} \sin \left( {{\omega _c}t} \right)$. Now the oscillator frequency is

$\rm \sqrt {\frac{2}{T}} \sin \left( {{\omega _c}t + {{45}^\circ }} \right)$

Using this, we calculate the vectors of demodulated symbols of  $\rm {s_1}\left( t \right)$ and $\rm {s_2}\left( t \right)$ along the basis function and get the following result.

$\rm {S_1} = \sqrt {\frac{E}{2}}$and $\rm {S_2} = - \sqrt {\frac{E}{2}}$

Now, distance between vectors $\rm d = 2\sqrt {\frac{E}{2}}$

Now, using the relation that $\rm {P_e} = Q\left( {\sqrt {\frac{{{d^2}}}{{2{N_o}}}} } \right) = Q\left( {\sqrt {\frac{{4 \times \frac{E}{2}}}{{2{N_o}}}} } \right) = Q\left( {\sqrt {\frac{E}{{{N_o}}}} } \right)$

Alternate Solution:

For BPSK

For coherent detection when there is phase synchronization between the carrier and received signal:

$P_e= Q\left( {\sqrt {\frac{2E_b}{{{N_o}}}} } \right)$

For coherent detection when there is phase shift of ϕ  between the carrier and received signal:

$P_e= Q\left( {\sqrt {\frac{2E_bCos^2\phi}{{{N_o}}}} } \right)$

Substituting the values Cos² ϕ = 1/2

$P_e = Q\left( {\sqrt {\frac{E}{{{N_o}}}} } \right)$

For 429 MHz, $;{l_1} = \frac{{{\lambda _1}}}{4} = \frac{c}{{4{f_1}}}$

$\Rightarrow {l_1} = 0.175;m$

for 1GHz, ${l_2} = \frac{{{\lambda _2}}}{4} = \frac{c}{{4{f_2}}} = 0.075;m$

The length of the transmission line l should be an odd integral multiple of both $l_1$and $l_2$

$\Rightarrow l = LCM\left( {{l_1},{l_2}} \right) = 0.525m$ .....$(0.525 = 3 × 0.175 = 7 × 0.075)$

Since $l$ is not in options we have take integral of l:

l' = 3l = 1.575 m ≈ 1.58 m

1. Time Invariant and Time Variant System:

If in a system, a delay in the input leads to same delay in the output, then such system is known as time invariant system. Otherwise It will be time variant system.

2. Stable and Unstable system:

The system is sold to be stable only when the bounded output for the input.

For a bounded input if output is unbounded then system is said to be UNSTABLE.

Condition:

\(\mathop \smallint \nolimits_{ - \infty }^\infty \left| {h\left( t \right)} \right|dt < \infty \) or finite.

Where h(t) is impulse response of system.

Solution:

Given, \(y\left( t \right) = \mathop \smallint \nolimits_{ - \infty }^t x\left( \tau \right)\cos \left( {3\tau} \right)d\tau\) 

Check for time-invariance:

Since y’(t) ≠ y (t – t0) so it is time-variant

Stability:

Suppose input x(t) = cos 3t

\(y\left( t \right) = \mathop \smallint \nolimits_{ - \infty }^t {\cos ^2}3\tau dt = \mathop \smallint \nolimits_{ - \infty }^t \left( {\frac{{1 + \cos 6\tau }}{2}} \right)d\tau ;\)

\(y\left( t \right) = [\left. {\frac{1}{2}t\tau } \right|{ - \infty }^t + \frac{1}{2}\sin \left. {\frac{{6\tau }}{6}} \right|{ - \infty }^t\)

Since y(t) → ∞, it is unbounded of P so this is an unstable system.

TRICK:

→ check for stability: try to find at least one input value for which output will be unbounded.

e.g. $y\left( t \right) = \frac{{{e^x}\left( {t - 5} \right)}}{{t - 5}}$ at t = 5 y(t) → ∞ - unbounded.

→ check for variance: If any coefficient if input is function of time (t) then it will be time variant.

e.g. tx(t), etx(t), sin t x(t)

The characteristic equation is s³ + as² + (k + 2) s + (k + 1) = 0

Routh array

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System oscillates if row of zero occurs

That occurs when $\frac{{a\left( {k + 2} \right) - \left( {k + 1} \right)}}{a} = 0$

ak + 2a = k + 1

$a = \frac{{k + 1}}{{k + 2}}$

System oscillates at ω = 2 rad/sec

as² + k + 1 = 0

$\frac{{k + 1}}{{k + 2}}{s^2} + k + 1 = 0$

s² = – (k + 2)

$s = \pm \sqrt {k + 2} ;j$

Comparing

k = 2

$a = \frac{{k + 1}}{{k + 2}} = \frac{{2 + 1}}{{2 + 2}} = 0.75$

$\begin{array}{l} H\left( {j\omega } \right) = \frac{{\left( {2\cos \omega } \right)\left( {\sin 2\omega } \right)}}{\omega }\ = \frac{{\sin 3\omega }}{\omega } + \frac{{sin\omega }}{\omega } \end{array}$

We know that the inverse Fourier transform of sinc function is a rectangular function.

$A~rect(\frac{t}{\tau}) \leftrightarrow A \tau ~Sa(\frac{\omega \tau}{2})$

Now,

$Sa(\frac{\omega \tau}{2}) = \frac{sin~(\frac{\omega \tau}{2})}{\frac{\omega \tau}{2}}$

\(\frac{sin 3\omega}{\omega} = 3 \frac{sin(3\omega)}{3\omega}\)

By comparing, we get

$\tau/2 = 3, A\tau = 3$

hence, A = 0.5

2) 

\(\frac{sin \omega}{\omega} = 1\times \frac{sin(\omega \times 1)}{\omega \times 1}\)

By comparing, we get

$\tau/2 = 1, A\tau = 1$

hence, A = 0.5

 

So, inverse Fourier transform of H (jω)

h(t) = h_1­(t) + h₂(t)

h(0) = h₁(0) + h_2­­(0)

$\frac{1}{2}+\frac{1}{2} = 1$

\({Q_c} = \left[ {\begin{array}{*{20}{c}} B&{AB}&{{A^2}B} \end{array}} \right]\)

\({Q_c} = \left[ {\begin{array}{*{20}{c}} 0&0&{{a_1}{a_2}}\ 0&{{a_2}}&0\ 1&0&0 \end{array}} \right]\)

System is controllable if |Q_C| ≠ 0

a₁a₂ (0 – a₂) ≠ 0

$- {a_1} \cdot a_2^2 \ne 0$

Hence condition for controllability is

a₁ ≠ 0, a₂ ≠ 0, a₃ = 0

We obtain Thevenin equivalent of circuit B

Thevenin impedance:

Thevenin Voltage: V_th = 3∠0° V

Now, circuit becomes as

Current in the circuit,

I₁ = (10-3)/(2 + R)

Power transfer from circuit A to B

P = (I₁)²R + 3 I₁

= (42 + 70 R)/(2 + R)²

dP/dR = 0

⇒ (2 + R) [(2 + R) 70 – (42 + 70 R) 2] = 0

⇒ R = 0.8 Ω

$\frac{{{d^2}y}}{{d{t^2}}} + \frac{{2dy}}{{dt}} + y = \delta \left( t \right)$

Taking Laplace transform with initial conditions,

$\begin{array}{l} \left[ {{s^2}y\left( s \right) - sy\left( 0 \right) - \frac{{dy}}{{dt}}{|_{t = 0}}} \right] + 2\left[ {sy\left( s \right) - y\left( 0 \right)} \right] + y\left( s \right) = 1\ \Rightarrow \left( {{s^2}y\left( s \right) + 2s - 0} \right) + 2\left[ {sy\left( s \right) + 2} \right] + y\left( s \right) = 1\ y\left( s \right)\left( {{s^2} + 2s + 1} \right) = 1 - 2s - 4\ y\left( s \right) = \frac{{ - 2s - 3}}{{{s^2} + 2s + 1}} \end{array}$

We know that, If $y\left( t \right)\mathop \leftrightarrow \limits^Z y\left( s \right)$

Then, $\frac{{dy}}{{dt}}\mathop \leftrightarrow \limits^Z sy\left( s \right) - y\left( 0 \right)$

So, $sy\left( s \right) - y\left( 0 \right) = \frac{{s + 2}}{{{{\left( {s + 1} \right)}^2}}} = \frac{1}{{s + 1}} + \frac{1}{{{{\left( {s + 1} \right)}^2}}}$

Taking inverse Laplace,

$\frac{{dy}}{{dt}} = {e^{ - t}}u\left( t \right) + t{e^{ - t}}u\left( t \right)$

At $t = {0^ + },\frac{{dy}}{{dt}}{|_{t = {0^ + }}} = {e^0} + 0 = 1$

$\left| {\vec A} \right| = k.{r^n},{r^2} = {x^2} + {y^2} + {z^2}$

$\Rightarrow \vec A = k{r^n}.{\hat a_r}$ ........ (Given radial outward) 

In spherical coordinates

$\vec \nabla .{\rm{\vec A}} = \frac{1}{{{{\rm{r}}^2}}}\frac{\partial }{{\partial {\rm{r}}}}\left( {{{\rm{r}}^2}{{\rm{A}}_{\rm{r}}}} \right) + 0 + 0 = \frac{1}{{{{\rm{r}}^2}}}{\rm{;}}\frac{\partial }{{\partial {\rm{r}}}}\left( {{{\rm{r}}^2}.{\rm{k}}{{\rm{r}}^{\rm{n}}}} \right)$

$\vec \nabla .{\rm{\vec A}} = \frac{{\rm{k}}}{{{{\rm{r}}^2}}}\frac{\partial }{{\partial {\rm{r}}}}\left( {{r^{\left( {n + 2} \right)}}} \right)=\frac{{\rm{k}}}{{{{\rm{r}}^2}}}(n+2){{}}\left( {{r^{\left( {n + 1} \right)}}} \right)$

⇒ For  $\vec \nabla .{\rm{\vec A}} = 0,~n+2 =0$

$n = - 2$

Concept:

For a geometric progression given as:

a, ar2, ar3, …, ∞

Where,

a > 0 and 0 < r < 1

The sum is given by:

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

Calculation:

The probability of appearing a head is 1/2. If the number of required tosses is odd, we have the following sequence of events.

H, TTH, TTTTH, …..

The required Probability P will be:

$P = \frac{1}{2} + {\left( {\frac{1}{2}} \right)^3} + {\left( {\frac{1}{2}} \right)^5} + \ldots$

$= \frac{{\frac{1}{2}}}{{1 - \frac{1}{4}}} = \frac{2}{3}$

M1 will be in linear region if V_out>V_in + V_T

At the edge of the linear region V_out= V_in + V_T

$\begin{array}{l} \frac{1}{2}{K_n}{\left( {{V_{in}} - {V_{thn}}} \right)^2} = {K_p}\left[ {\left( {{V_{DD}} - {V_{in}} - {V_{thp}}} \right)\left( {{V_{DD}} - {V_{out}}} \right) - \frac{1}{2}{{\left( {{V_{DD}} - {V_{out}}} \right)}^2}} \right]\ Or,\ \frac{1}{2}{K_n}{\left[ {{V_{in}} - {V_T}} \right]^2} = {K_p}\left[ {\left( {{V_{DD}} - {V_{in}} - {V_T}} \right)\left( {{V_{DD}} - {V_{in}} - {V_T}} \right) - \frac{1}{2}{{\left( {{V_{DD}} - {V_{in}} - {V_T}} \right)}^2}} \right]\ Or,\ \ \ {\left[ {{V_{in}} - {V_T}} \right]^2} = {\left( {{V_{DD}} - {V_{in}} - {V_T}} \right)^2}\ Or,\ \ \ {\left( {{V_{in}} - 1} \right)^2} = {\left( {4 - {V_{in}}} \right)^2}\ Or,\ \ \ {V_{in}} - 1 = \left( {4 - {V_{in}}} \right)\ Or,\ \ \ {V_{in}} = 2.5 \end{array}$

So, for Vin< 2.5 the PMOS will be in a linear region so the correct option is (1)

Concept:

The decision making algorithm used in optimum receiver is as follows:

If r₀ is received, m₀ is chosen only if:

p(r₀ / m₀) p(m₀) > p(r₀ / m₁) p(m₁)

And if r₁ is received, m₁ is chosen only if:

p(r₁ / m₁) p(m₁) > p(r₁ / m₀) p(m₀)

Analysis:

Given:

p(m₀) = 0.9

p(m₁) = 1 – 0.9 = 0.1

p(r₀ / m₀) = 0.9

p(r₀ / m₁) = 0.4

p(r₁ / m₀) = 0.1

p(r₁ / m₁) = 0.6

Now, test p(r₀ / m₀) p(m₀) > p(r₀ / m₁) p(m₁)

0.9 × 0.9 > 0.4 × 0.1 ; (true)

Decision: Choose m₀ for received r₀.

Test p(r₁ / m₁) p(m₁) > p(r₁ / m₀) p(m₀)

0.6 × 0.1 > 0.1 × 0.9 ; (not true)

Decision: Choose m₀ for received r₁.

In fact, in the given situation, the optimum receiver always decides in favor of m₀ irrespective of whether r₀ or r₁ is received.

Now, the Probability of correct reception will be:

P_c = p(r₀ / m₀) p(m­₀) + p(r₁ / m₀) p(m₀)

= 0.9 × 0.9 + 0.1 × 0.9 = 0.9

BER = p(ε) = 1 - P_c

BER = 1 – 0.9 = 0.1