GATE EC 2017 Official Paper: Shift 1 (Previous Year Paper)

The correct answer is 'hurtful'.

Key Points

• 'Sharp tongue' is an idiomatic phrase which means a tendency to speak in a malicious or in highly critical manner. 
• Example: The mean old lady who lived on the corner made a reputation among us kids with her sharp tongue.
• Thus, out of all the options, only 'hurtful' fits the blank.
• Therefore, the correct answer is option 1.

The correct sentence is: She has a sharp tongue and it can occasionally turn hurtful.

The correct answer is 'could have, been'.

Key Points

• The given sentence is an example of the third conditional sentence.
• Third conditional sentence: It is used to talk about something in the past that did not happen.
• Structure:
• if + past perfect, would + have + past participle.
• Example:
• If I’d known, I would have worn something nicer.
• I definitely would’ve remembered if you’d told me!
• Note: Would have can be replaced with 'could have', 'might have', etc.
• Thus, 'Could have' will be used in the first blank, and 'been' will be used in the second blank.

Therefore, the correct sentence is: 'I could have made arrangements had I been informed earlier.'

Additional Information

• Conditional sentence: As the name suggests it is used to tell conditions. This type of sentence has two clauses one is a conditional clause which is used to tell conditions and another is the main clause.
• Generally, the Conditional clause is succeeded by the main clause but it can be reversed.

Let Initial household consumption be “H” and Other consumption be “O”

During Summer

Household consumption = H – 20% of H = 0.8 H

Other consumption = O + 70% of O = 1.7 O

Overall Consumption:

0.8 H + 1.7 O ………(1)

According to the question, the overall consumption reduces by 25 %, i.e.

Overall Consumption will now be:

= (O + H) – 25% of (O + H) = 0.75 (O + H) ..(2)

From Equation (1) and (2), we get:

0.8 H + 1.7 O = 0.75 (O+H)

0.05 H = -0.95 O

\(\frac{H}{O} = - \frac{{0.95}}{{0.05}}\)

Since the ratio cannot be negative, there are errors in the Official's statement.

In a pie chart

360° represents 100%

1% represents 3.6°

40% represents 144°

Explanation:

The least possible Venn diagram is:

Conclusion i → False (as it is possible but not definite)

Conclusion ii → True (as some shelves are chairs and all chairs are benches)

Conclusion iii → False (as it is possible but not definite)

Conclusion iv → False (as it is possible but not definite)

Hence, only conclusion ii follows.

The correct answer is 'Hostile'.

Key Points

• The passage lists the various events that happened in the contemporary history of the Indian subcontinent. The word 'antagonistic' is used to describe the relationship between India and Pakistan after the Partition by the Britishers.
• Let's explore the meaning of the given word and the marked word
• Antagonistic: showing or feeling active opposition or hostility toward someone or something.
• Example: My step-brother has always been very antagonistic towards me, never sharing his things or spending any time with me if he can help it.
• Hostile: unfriendly; antagonistic.
• Example:  Their hostile looks showed that he was unwelcome.
• Thus, from the above-given explanation, we can say that 'Hostile' is closest in meaning to the word 'Antagonistic'.
• Therefore, the correct answer is option 4.

Additional Information

• Let's explore the meaning of other options:
• Impartial: not partial or biased
• Argumentative: having or showing a tendency to disagree or argue with other people in an angry way
• Separated: divided into constituent or distinct elements.

The seating arrangement considering all given conditions is shown:

From the seating arrangement, person third to the left of V is X

Length of truck + gap required = 10 + 20 = 30 m

Length of car + gap required = 5 + 15 = 20 m

Total distance is need for truck and car for passing alternatively = 30 + 20 = 50 m

Lets us call this 50 m as 1 unit

Speed = 36 km/h

So, for 1 hour the distance is 36 km = 36000 meters

Total number of units that can pass are:

\(\frac{{36000}}{{50}} = 720;units\)

Total number of vehicles that can pass = 720 × 2 = 1440

Case 1 ( 1 Indian)

The possible subgroups can be

1 Indian with 0 Chinese

1 Indian with 1 Chinese

1 Indian with 2 Chinese

1 Indian with 3 Chinese

Sum of case 1 Possible cases :

Case 2 ( 2 Indians)

The possible subgroups can be

2 Indian with 0 Chinese

2 Indian with 1 Chinese

2 Indian with 2 Chinese

2 Indian with 3 Chinese

Sum of case 2 Possible cases :

Case 3 ( 3 Indians)

The possible subgroups can be

3 Indian with 0 Chinese

3 Indian with 1 Chinese

3 Indian with 2 Chinese

3 Indian with 3 Chinese

Sum of case 2 Possible cases :

Total cases :

24 + 24 + 8 = 56

During the first path

550 → 525→500→ 475

The movement is downward

During the second path

475 → 500 → 525 → 550

The movement is upwards

The last path

575 → 550

Path is down

Total path movement from P to Q is

Down → Up → Down

\(A = \left[ {\begin{array}{*{20}{c}} 1&2&3&4&5\ 5&1&2&3&4\ 4&5&1&2&3\ 3&4&5&1&2\ 2&3&4&5&1 \end{array}} \right]\)

The Eigenvalues of a Matrix are given by:

|A – λ I| = 0

\(\left[ {\begin{array}{*{20}{c}} {1 - \lambda }\ {\begin{array}{*{20}{c}} 5\ {\begin{array}{*{20}{c}} 4\ {\begin{array}{*{20}{c}} 3\ 2 \end{array}} \end{array}} \end{array}} \end{array}\begin{array}{*{20}{c}} 2\ {\begin{array}{*{20}{c}} {1 - \lambda }\ {\begin{array}{*{20}{c}} 5\ 4\ 3 \end{array}} \end{array}} \end{array}\begin{array}{*{20}{c}} 3\ {\begin{array}{*{20}{c}} 2\ {\begin{array}{*{20}{c}} {1 - \lambda }\ 5\ 4 \end{array}} \end{array}} \end{array}\begin{array}{*{20}{c}} 4\ {\begin{array}{*{20}{c}} 3\ {\begin{array}{*{20}{c}} 2\ {1 - \lambda }\ 5 \end{array}} \end{array}} \end{array}\begin{array}{*{20}{c}} 5\ {\begin{array}{*{20}{c}} 4\ {\begin{array}{*{20}{c}} 3\ 2\ {1 - \lambda } \end{array}} \end{array}} \end{array}} \right] = 0\)

Apply Row Transformation

R₁ → R₁ + R₂ + R₃ + R₄ + R₅

\(\left[ {\begin{array}{*{20}{c}} {15 - \lambda }\ {\begin{array}{*{20}{c}} 5\ {\begin{array}{*{20}{c}} 4\ {\begin{array}{*{20}{c}} 3\ 2 \end{array}} \end{array}} \end{array}} \end{array}\begin{array}{*{20}{c}} {;15 - \lambda }\ {\begin{array}{*{20}{c}} {1 - \lambda }\ {\begin{array}{*{20}{c}} 5\ 4\ 3 \end{array}} \end{array}} \end{array}\begin{array}{*{20}{c}} {;15 - \lambda ;}\ {\begin{array}{*{20}{c}} 2\ {\begin{array}{*{20}{c}} {1 - \lambda }\ 5\ 4 \end{array}} \end{array}} \end{array}\begin{array}{*{20}{c}} {15 - \lambda ;}\ {\begin{array}{*{20}{c}} 3\ {\begin{array}{*{20}{c}} 2\ {1 - \lambda }\ 5 \end{array}} \end{array}} \end{array}\begin{array}{*{20}{c}} {15 - \lambda }\ {\begin{array}{*{20}{c}} 4\ {\begin{array}{*{20}{c}} 3\ 2\ {1 - \lambda } \end{array}} \end{array}} \end{array}} \right]\)

\(\left( {15 - \lambda } \right)\left( {\begin{array}{*{20}{c}} 1\ {\begin{array}{*{20}{c}} 5\ {\begin{array}{*{20}{c}} 4\ {\begin{array}{*{20}{c}} 3\ 2 \end{array}} \end{array}} \end{array}} \end{array}\begin{array}{*{20}{c}} 1\ {\begin{array}{*{20}{c}} {;1 - \lambda }\ {\begin{array}{*{20}{c}} 5\ 4\ 3 \end{array}} \end{array};} \end{array}\begin{array}{*{20}{c}} 1\ {\begin{array}{*{20}{c}} 2\ {\begin{array}{*{20}{c}} {1 - \lambda }\ 5\ 4 \end{array};} \end{array}} \end{array}\begin{array}{*{20}{c}} 1\ {\begin{array}{*{20}{c}} 3\ {\begin{array}{*{20}{c}} {;2}\ {1 - \lambda }\ 5 \end{array}} \end{array}} \end{array}\begin{array}{*{20}{c}} 1\ {\begin{array}{*{20}{c}} 4\ {\begin{array}{*{20}{c}} 3\ 2\ {;1 - \lambda } \end{array}} \end{array}} \end{array}} \right)\)

It is given that the matrix A has only one real eigenvalue, i.e.

⇒ 15 - λ = 0

⇒ λ = 15

The other 4 eigenvalues are complex conjugates .

\(\left[ {\begin{array}{*{20}{c}} 5&{10}&{10}\ 1&0&2\ 3&6&6 \end{array}} \right];\)

Converting it into Echelon form, we carry out the following transformations:

R₂ ↔ R₁

\(\left[ {\begin{array}{*{20}{c}} 1&0&2\ 5&{10}&{10}\ 3&6&6 \end{array}} \right]\)

R₂ → R₂ – 5R₁

R₃ → R₃ – 3R₁

\(\left[ {\begin{array}{*{20}{c}} 1&0&2\ 0&{10}&0\ 0&6&0 \end{array}} \right]\)

\({{\rm{R}}_3}{\rm{;}} \to {\rm{;}}{{\rm{R}}_3}{\rm{}} - \frac{6}{{10}}{R_2}\)

\(\left[ {\begin{array}{*{20}{c}} 1&0&2\ 0&{10}&0\ 0&0&0 \end{array}} \right]\)

Since the number of non-zeros Rows in the above Matrix is 2, the rank of the matrix will be 2.

Concept:

The linear dependency or Independency of function fi can be found using Wronskion Matrix.

An analytic function is linearly dependent if:

\(\left| {\begin{array}{*{20}{c}} {{f_1}\left( x \right)}&{{f_2}\left( x \right)}&{{f_3}\left( x \right)}\ {f_1'\left( x \right)}&{f_2'\left( x \right)}&{f_3'\left( x \right)}\ {f_1{''}\left( x \right)}&{f_2{''}\left( x \right)}&{f_3{''}\left( x \right)} \end{array}} \right| = 0\)

Application:

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

Determinant ≠ 0

Function y₁, y₂, y­₃ are linearly independent.

Concept:

If an event ‘A’ with ‘n’ equal possible outcomes is repeated ‘k’ times then the total number of possible outcome is n^k

Calculation:

Fair dice has 6 outcomes

For 3 dices, the total number of outcomes will be:

6³ = 216

Out of 216 outcomes, the favorable (all three dice have the same number of dots) are:

⇒ ((1, 1,1) (2, 2,2) (3, 3,3), (4,4,4) (5, 5,5) (6, 6,6))

∴ The required probability will be:

\(Probability = \frac{6}{{216}} = \frac{1}{{36}} = 0.027\)

Concept:

An LTI system is stable if and only if the ROC of the impulse function H(s) includes the jω axis.

For Causal System → ROC is to the right side of the rightmost pole.

For Anti Causal System → ROC is to the left side of the left-most pole.

Observations:

• For a causal system to be stable, the poles must lie on the left half of the complex plane (to include the jω axis)
• A causal system with a pole on the right side cannot be BIBO stable because it's ROC can never include the jω axis. (Statement (II) is therefore correct)
• A BIBO system with a pole in the right half of the complex plane is stable if the system is anti-causal, as this will include the jω axis. (Statement (I) is therefore incorrect)

Concept:

For a system to be causal, the Present output should depend on the present or past input only.

For a system to be Stable, a Bounded input should produce a Bounded Output.

Analysis:

\(y\left( n \right) = \left{ {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} {n|x\left( n \right)|}&{0 \le n \le 10} \end{array}}\ {\begin{array}{*{20}{c}} {x\left( n \right) - x\left( {n - 1} \right)}&{otherwise} \end{array}} \end{array}} \right.\)

For 0 < n < 10,

y(n) = n|x(n)|

Let x(n) is bounded, i.e. for -∞ ≤ n ≤ ∞, x(n) ≤ M, where M is finite.

So, for -∞ ≤ n ≤ ∞, |n.x(n)| will also approach a finite value. Hence in this interval y(n) is bounded. i.e. the system is stable.

Since the output y(n) is depending on the present value of the input only, the system in this interval is also causal.

For n < 0 and n > 10:

y(n) = x(n) - x(n-1).

Let input x(n) is bounded. ∴ y(n) = x(n) - x(n-1) will also be bounded, i.e. it will go to a finite value.

Hence in this interval, the system is said to be stable.

Also, because the output y(n) depends on the present and the past input values only, the system is causal as well in this interval.

∴ After considering both the intervals we can conclude that the system is both stable and causal.

AC Equivalent circuit:

Form voltage division

\({V_1} = \frac{{100\angle 0\left( 1 \right)}}{{1 + \left( {j{\rm{\omega }} + 1} \right)}}\)

\({V_1} = \frac{{100\angle 0}}{{2 + j{\rm{\omega }}}}\)

\(\angle {V_1} = 0 - {\tan ^{ - 1}}\left( {{\rm{\omega }}/2} \right)\)

\({V_2} = \frac{{100\angle 0\left( {1 + j{\rm{\omega }}} \right)}}{{2 + j{\rm{\omega }}}}\)

\(\angle {V_2} = {\tan ^{ - 1}}\left( {\frac{{\rm{\omega }}}{1}} \right) - {\tan ^{ - 1}}\left( {\frac{{\rm{\omega }}}{2}} \right)\)

∠V₁ - ∠V₂ = π/4

\(-{\tan ^{ - 1}}\left( {{\rm{\omega }}/2} \right)-tan^{-1}\omega +{\tan ^{ - 1}}\left( {{\rm{\omega }}/2} \right)=\frac{\pi}{4}\)

⇒ |tan^-1 ω| = π/4

ω = 1 rad/sec

Concept:

For an odd signal x(t) = -x(-t).

The average value of an odd periodic signal is 0.

For half-wave symmetric signal:

x (t ± T/2) = - x(t)

A half-wave symmetric signal only has odd harmonics in its Fourier series representation.

Calculation:

The given function is odd, hence

a_o (Average Value) = 0 and a_n = 0

Also, since x(t) = -x (t – π/ ω0)

It is also half-wave symmetric.

Hence the signal will contain only odd harmonics.

Concept:

If the substituting atom gives one extra electron in the outermost shell then n-type semiconductor is formed. If substituting atom has one electron less in the outermost shell then p-type semiconductor is formed.

Si → 4e^- in the outermost shell

Ga → 3e^- in the outermost shell

As → 5e^- in the outermost shell

Application:

Note:  Here, GaAs is doped with Silicon. So, Silicon is the substituting atom.

For Arsenic sites, Silicon will act as a p-type dopant as it has 1 electron less than Arsenic in its outermost shell.

And for Gallium sites, Silicon will act a n-type dopant as it has 1 electron more than Gallium in its outermost shell.

Si → Ga → 1e^- extra → n-type

Si → As → 1e^- less → p-type

Concept:

The Built-in Potential of a p-n junction diode is given by:

\({V_o} = \frac{{kT}}{q}\ln \left( {\frac{{{N_A}{N_D}}}{{n_i^2}}} \right)\)

Where NA = doping density of holes

ND = doping density of electrons

Also, the law of mass-Action states that:

\(n_i^2 = {N_A}{N_D}\)

Calculation:

 

n⁺ = 10¹⁸/cm³

n = 10¹⁵/cm³

n_i = 10¹⁰/cm³

The n⁺ Junction here has more electrons than the n Junction [Analogous to p-n Junction]

Hole concertation in n – region using mass action law will be:

 \(p= \frac{{n_i^2}}{N} = \frac{{{{10}^{20}}}}{{{{10}^{15}}}}\) = N_A

Putting on the respective values, we get:

\({V_o} = \frac{{KT}}{2}{l_n}\left( {\frac{{N_D^ + + {N_A}}}{{n_i^2}}} \right) = 0.173V\)

Observation 1:

Consider the hole concertation in the base, near the collector-base junction.

Minority carrier concertation (∆p_b) is high. This is because excess holes are supplied by the collector (P).

This means C.B. Junction is forward Biased.

Observation 2:

The minority hole carrier concertation in a base region near the Emitter – base Junction is low, i.e. holes are extracted from the base.

This indicates that the Emitter Base junction is Reversed Biased.

∴ The mode of operation is Reverse Active.

Since the given Op-Amp has a positive-feedback connection, the output will be saturated to either + 15V or -15 V.

Apply KCL at node V, we get:

\( \frac{{V - {V_{out}}}}{{10}} + \frac{{V - 3}}{5} + 0 = 0\)

Due to the high input impedance of the Op-Amp, the input current to the Op-amp will be 0.

\(V = \frac{{{V_{out}} + 6}}{3}\)

For an upper saturated voltage output of V_out = +15 V, we get the threshold voltage V of:

\(V =V_{in}= \frac{{15 + 6}}{3} \)

\(V= \frac{{21}}{3}=7~V\)

For a lower saturated voltage of V_{out =}  -15 V, we get the threshold voltage V of:

\(V = V_{in}=\frac{{ - 15 + 6}}{3}\)

\(V== \frac{{ - 9}}{3}=-3~V\)

Transconductance Amplifier gain \( = \frac{{{I_{out}}}}{{{V_{in}}}}\) 

Thus, the input to the transconductance Amplifier is a voltage, and the output is current.

To effectively couple voltage from the source, R_in should be high.

To effectively deliver current to load, R_out should be High.

The miller effect causes increase in the Input capacitance of Common – Emitter Amplifier

Upper cutoff frequency \(\left( {{f_h}} \right)\propto \frac{1}{{{C_{input}}}}\) 

Therefore, there is decrease in upper-cut off frequency & Bandwidth of CE Amplifier

Let as assume t_pd 1 < t_pd 2

x changes state first then y changes

1^st output of X (P = 1, y = 0) ⇒ X₁ = 1

Next output of Y (Q = 1, X₁ = 1) ⇒ Y₁ = 0

2^nd output of X (P = 1, y₁ = 0) ⇒ 1

Hence output x = 1 y = 0 (if t_pd1 < t_pd2)

& Output X = 0 Y = 1 (if t_pd2 < t_pd1)

Concept:

The time required to execute an instruction is given by:

(No. of T - states) × T_clk

T_clk = Time period of the clock.

Calculation:

Given, clock frequency f_clk = 5 MHz

∴ The time period of the clock will be:

\({T_{clk}} = \frac{1}{5} = 0.2~\mu ;sec\)

We can write:

1.4 = n × 0.2

n = 7

Concept:

The duty cycle of a waveform is defined as:

Duty Cycle \(=\frac{T_{on}}{T}\)

where T = Time period of the waveform

Calculation:

The output of a D-latch is high when both the input (CLK 1) and clock (CLK 2) are high i.e. the output wave-form is AND condition of the 2 Clock input.

This is explained with the help of the following diagrams:

 

Now, the ON time will be:

\({T_{on}} = \frac{{{T_{clk}}}}{2} - \frac{{{T_{clk}}}}{5}\)

\(= \frac{{3{T_{clk}}}}{{10}}\)

The duty cycle will the ratio of the on-time with the time period, i.e.

\(Duty;cycle = \frac{{\frac{{3;{T_{CLK}}}}{{10}}}}{{{T_{CLK}}}} \times 100;\)

= 30%

Concept: 

 

Calculation:

The integer 'p' represents the type of the system.

It is given that for a step input e_ss = 0

For a Ramp input e_ss = 6, i.e. [Non Zero Finite]

∴ We conclude that it is a Type = 1 system.

Concept:

The slope of Bode plot increase of + 20 dB/decade for every zero and decrease by -20 dB/decade for every pole.

Calculation:

Since the final slope of the Bode Plot is – 60 dB/decade:

Poles (q) > Zeros (p) by 3

p – q = 3

Option (A) satisfies this condition.

Let H(f) be TF of phase-lag controller

\(H\left( f \right) = \frac{{s + a}}{{s + b}}\)

\(\angle H\left( b \right) = {\tan^{-1}}\left( {\frac{{\rm{\omega }}}{a}} \right){ - ;{tan^{ - 1}}}\left( {\frac{{\rm{\omega }}}{b}} \right)\)

For lag Network ∠H(b) = -ve

\({\tan ^{ - 1}}\left( {\frac{{\rm{\omega }}}{a}} \right) < {\tan ^{ - 1}}\left( {\frac{{\rm{\omega }}}{b}} \right)\)

\(\left( {\frac{{\rm{\omega }}}{a}} \right) < \left( {\frac{{\rm{\omega }}}{b}} \right)\)

b < a

Pole is near to j ω axis than zero

Concept:

Mutual information of two random variables is a measure to tell how much one random variable tells about the other.

It is mathematically defined as:

I(X₁, X₂) = H(X₁) – H(X₁/X₂)

Application:

Since X₁ and X₂ are independent, we can write:

H(X₁/X₂) = H(X₁)

I(X₁,X₂ ) = H(X₁) – H(X₁)

= 0

DPCM transmitter is as shown:

We observe that the input to quantizer is the difference between the sampled message signal and its predicted value

Concept:

Nyquist ISI criterium states that for ISI – free response:

\(h\left( {n{T_s}} \right) = \left{ {\begin{array}{*{20}{c}} {1;}&{n = 0}\ {0;}&{n \ne 0} \end{array}} \right.\)

i.e.

\(\frac{1}{{{T_s}}}\mathop \sum \limits_{T = - \infty }^{{T_\infty }} H\left( {b - \frac{k}{{{T_S}}}} \right) = 1\forall ;f\)

\(\mathop \sum \limits_{T = - \infty }^{{T_\infty }} H\left( {b - \frac{k}{{{T_S}}}} \right) = T_s = Constant;\forall ;f\)

Calculation:

\(f_s = \frac{1}{{{T_s}}} = 2k;symbols/sec\)

\(\mathop \sum \limits_{n = - \infty }^{ + \infty } p\left( {f - n;{f_s}} \right) = {T_s}\)

\(\mathop \sum \limits_{n = - \infty }^{ + \infty } P\left( {f - n;2k} \right) = 2k\)

We observe that only option (B) satisfies this condition, i.e.

\(\mathop \sum \limits_{n = - \infty }^{ + \infty } P\left( {b - n2k} \right) = Constant\)

Concept:

The propagation constant for a transmission line is given as:

γ = α + j β

α is the attenuation constant (usually expressed in Neper/sec)

where 1 Np = 8.686 dB

β is phase constant

Calculation:

γ = (0.1 + j 40) m-1 

α = 0.1 Np/m

Since 1 Np = 8.686 dB, the absolute value of α in dB/m will be:

α = 0.1 × 8.686

α = 0.868 dB/m

Concept:

The capacity of a wireless link is given by Shanon Hartley theorem as:

\(C = B~lo{g_2}\left( {1 + \frac{S}{N}} \right)\)

Channel capacity depends on \(\left( {\frac{S}{N}} \right)\), and hence on S.

S = Received power at the receiver = P_r

\(S = {P_r} = \frac{{{P_T}{G_t}{G_r}}}{{{{\left( {\frac{{4;{\bf{\pi }}R}}{\lambda }} \right)}^2}}}\)

\(= \frac{{{P_t}\left( {\frac{{{G_\pi }}}{{{\lambda ^2}}}{A_{te}}} \right)\left( {\frac{{4\pi }}{{{\lambda ^2}}}{A_{re}}} \right)}}{{{{\left( {\frac{{4\pi R}}{\lambda }} \right)}^2}}}\)

Calculation:

It is given,

A_te' = 2A_te

A_re' = 2 A_re

R' = 2 R

\({P_{r'}} = \frac{{{P_t}\left( {\frac{{4\pi ;{A_{te}}}}{{{\lambda ^2}}}} \right)\left( {\frac{{4\pi }}{{{\lambda ^2}}}{A_{rc}}} \right)2 \times 2}}{{{{\left( {\frac{{4\pi R}}{\lambda }} \right)}^2}{{\left( 2 \right)}^2}}}\)

P_r' = P_r

S' = S

\(\frac{{S'}}{N} = \frac{S}{N}\)

⇒ C' = C, i.e.

The maximum capacity of the channel remains unchanged.

Concept:

Series expansion of \({e^x} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \ldots \)

Calculation:

Series expansion of \({e^{x + {x^2}}}\)

\({e^{x + {x^2}}} = 1 + \left( {x + {x^2}} \right) + \frac{{{{\left( {x + {x^2}} \right)}^2}}}{{2!}} + \frac{{{{\left( {x + {x^2}} \right)}^3}}}{{3!}} + \ldots \)

\({e^{x + {x^2}}} = 1 + x + {x^2} + \left( {\frac{{{x^2} + {x^4} + 2{x^3}}}{2}} \right) + \left( {\frac{{{x^3} + {x^6} \ldots }}{6}} \right)\)

Neglecting higher powers of x, we can write:

\(= 1 + x + \frac{{3{x^2}}}{2} + \frac{7}{6}{x^3}\)

Concept:

The volume of the solid is calculated as:

\(V = \iiint_R;dx;dy;dz = \iiint r;dr;d\theta ;dz;\)

Calculation:

The given region is:

x² + y² ≤ z³

0 ≤ z ≤ 1

We can parametrically define the region as:

x = r cos θ

y = r sin θ

Squaring and adding

\(x^2+y^2=r^2\)

Let PQ be the radius at any location z from the axis:

r = PQ = \(\sqrt{x^2+y^2}\).

\(r^2={x^2+y^2}\)

The volume of the region revolved around z-axis is given by:

\(\int_0^1\pi(PQ)^2dz=\int_0^1\pi z^3dz=\pi\left[\frac{z^4}{4}\right]_0^1=\frac{\pi}{4}=0.7853;\)

\(I = \mathop \smallint \limits_{\left( {0,2,1} \right)}^{\left( {4,1, - 1} \right)} \left( {2z;dx + 2y;dy + 2x;dz} \right)\)

\(I = \mathop \smallint \limits_{\left( {0,2,1} \right)}^{\left( {4,1, - 1} \right)} \left( {2z;dx + 2y;dy + 2x;dz} \right)\)

\(I = \mathop \smallint \limits_{\left( {0,2,1} \right)}^{\left( {4,;1, - 1} \right)} \left( {2d\left( {xz} \right) + 2y;dy} \right)\)

\( = 2\left( {xz} \right)_{\left( {0,2,1} \right)}^{\left( {4,1, - 1} \right)} + 2\left( {\frac{{{y^2}}}{2}} \right)_{\left( {0,2,1} \right)}^{\left( {4,1, - 1} \right)}\)

\( = \left[ {2xz + {y^2}} \right]_{\left( {0,2,1;} \right)}^{\left( {4,1, - 1} \right)}\)

\(\left[ {2\left( 4 \right)\left( { - 1} \right) + {{\left( 1 \right)}^2}} \right] - {\left[ 2 \right]^2}\)

= [-8 + 1] – 4

= -7 - 4

= -11

Concept:

Variable separable

Equation of the form \(\frac{dy}{dx}=f(ax;+;by;+;c)\) can be reduced to variable separable form by putting ax + by + c = t

Calculation:

Given:

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

Substitute x + y – 1 = t

\(1 + \frac{{dy}}{{dx}} = \frac{{dt}}{{dx}}\)

\(\frac{{dy}}{{dx}} = \frac{{dt}}{{dx}} - 1\)

\(\frac{{dt}}{{dx}} - 1 = {t^2}\)

\(\frac{{dt}}{{dx}} = 1 + {t^2}\)

\(\smallint \frac{{dt}}{{1 + {t^2}}} = \smallint dx\)

tan^-1 t = x + c

t = x + y – 1

tan^-1 (x + y – 1) = x + c

x + y – 1 = tan (x + c)

y = 1 – x + tan (x + c)

Concept:

The (n + 1)^th iteration of x in the newton Raphson is given by:

\({x_{x + 1}} = {x_n} - \left( {\frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}} \right)\)

Calculation:

Let, f(x) = x³ + x – 1

f'(x) = 3x² + 1

x₀ = 1

\({x_1} = {x_0} - \left( {\frac{{f'\left( {{x_0}} \right)}}{{f'\left( {{x_0}} \right)}}} \right)\)

\(= 1 - \frac{1}{4}=\frac{3}{4}\)

x1 = 0.75

Similarly, 

\({x_2} = {x_1} - \frac{{f\left( {{x_1}} \right)}}{{f'\left( {{x_1}} \right)}}\)

\(= 0.75 - \frac{{0.1718}}{{2.6875}}\)

= 0.75 – 0.0639

x₂ = 0.6861

Concept:

If x(t) → {a_k}, with period = T, then

x(at) → {a_k}, with period = T/a

We observe that the Fourier series coefficient are unaffected by time scaling.

Application:

Given, x(t) → {a_k}, with period = 1 sec

x(3t) → {a_k}, with period = 1/3 sec

∴ The Angular frequency will be:

\(ω = \frac{{2\pi }}{T}\)

\(ω= \frac{{2\pi }}{{\frac{1}{3}}}\)

ω = 6π rad/sec

Concept:

• For linearity and time invariance output must be weighted superposition of time-shifted impulses.
• This weighted superposition is termed as convolution sum for discrete-time systems and convolution integral for continuous-time. And it is determined by the symbol (∗ )
• If two systems are cascaded then the resultant signal is convolution in the time domain and multiplication in the frequency domain, below diagrams, shows that.

 

Symbolically resultant is represented as:

y(t) = x(t) ∗ h(t)

symbolically this is represented as:

y[n] = x[n] ∗ h[n]

Calculation:

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

h(n) = {1, a, b}

x(n)*h(n) =

y(n) = {1, a + 2, 1 + 2a + b, a + 2b, b}

y(1) = a + 2 = 3

a = 1

y(2) = 1 + 2a + b = 4

1 + 2 + b = 4

b = 1

y(n) = {1, 3, 4, 3, 1}

10 y(3) + y(4) = 10 (3) + 1 = 31

Concept:

The DTFT of an aperiodic signal x(n) is defined as:

\(X\left( {{e^{j\omega }}} \right) = \mathop \sum \limits_{n = - \infty }^{ + \infty } x\left( n \right){e^{ - j{\rm{\omega }}n}}\)

Application:

\(h\left( n \right) = \left{ {\frac{1}{3},\frac{1}{3},\frac{1}{3}} \right}\)

\(H\left( {{e^{j\omega }}} \right) = \frac{1}{3} + \frac{1}{3}{e^{ - j\omega }} + \frac{1}{3}{e^{ - j2\omega }}\)

\(= \frac{1}{3}{e^{ - j\omega }}\left[ {{e^{j\omega }} + {e^{ - j\omega }}} \right] + \frac{1}{3}{e^{ - j\omega }}\)

\(= \frac{2}{3}{e^{ - j\omega }}\cos \omega + \frac{1}{3}{e^{ - j\omega }}\)

\(= \frac{1}{3}{e^{ - j\omega }}\left[ {1 + 2\cos \omega } \right]\)

At ω = ω₀

\(H\left( {{e^{j{\omega _0}}}} \right) = 0\)

We can, therefore, write:

1 + 2 cos ω₀ = 0

cos ω₀ = -1/2

\(\cos {\omega _0} = \cos \frac{{2\pi }}{3}\)

\(\omega_o = \frac{{2\pi }}{3} = 2.09;radians\)

Using voltage division rule

V₂ = V_in (\(\frac{{{Z_2}}}{{{Z_2} + ;{Z_1}}}\))

V₁ = V_in (\(\frac{{{Z_1}}}{{{Z_2} + ;{Z_1}}}\))

\(\frac{{{V_2}}}{{{V_1}}} = \frac{{{Z_2}}}{{{Z_1}}} = \frac{{5 + \frac{1}{{j\omega C}}}}{{4 + j\omega L}}\)

\(= \frac{{5 - \frac{j}{{3~\times{{\frac{1}{36}}}}}}}{{4 + j\left( 3 \right)\left( 1 \right)}}\)

\(\frac{{{V_2}}}{{{V_1}}} = \frac{{5 - j;12}}{{4 + j;3}}\)

\(\left| {\frac{{{V_2}}}{{{V_1}}}} \right| = \frac{{\sqrt {{5^2} + {{12}^2}} }}{{\sqrt {{3^2} + {4^2}} }}\)

\(= \frac{{13}}{5}\)

\(\left| {\frac{{{V_2}}}{{{V_1}}}} \right| = 2.6\)

Concept:

Calculation:

Taking Laplace to transform with given conditions

Applying KCL at V(s) node:

\(\frac{{V\left( s \right) - \frac{{15}}{s}}}{1} + \frac{{V\left( s \right)}}{s} + \frac{{V\left( s \right)}}{{2s}} = 0\)

\(V\left( s \right) = \frac{{30}}{{2s + 3}}\)

\(I\left( s \right) = \frac{{V\left( s \right)}}{{2s}} = \frac{{30}}{{2s\left( {2s + 3} \right)}}\)

\(I\left( s \right) = 5\left[ {\frac{1}{s} - \frac{1}{{s + \frac{3}{2}}}} \right]\)

Taking the inverse Laplace transform, we get:

i(t) = 5(1 – e^-3t/2)

Given, at t = t₀

i = 2 A

2 = 5 (1 – e^-3t/2)

\(\frac{2}{5} = 1 - {e^{ - \frac{{3t}}{2}}}\)

\({e^{ - \frac{{3t}}{2}}} = 1 - \frac{2}{5} = \frac{3}{5}\)

\(t = \frac{2}{3}\ln \left( {\frac{3}{5}} \right)\)

t = 0.3405 sec

Concept:

The current for the material is defined as:

J = σE

E = Electric field applied

σ = Conductivity of the material, and is given by:

σ = nµne

J = nµne E

Also, the drift velocity is given by:

vd = µn E

The current density now becomes:

J = n e vd

Calculation:

The electric field corresponding to the given bias voltage will be:

\(E = \frac{V}{l} = \frac{5}{{1 × {{10}^{ - 6}}}}\)

E = 5 × 10⁶ V/m, or

E = 5 × 10⁴ V/cm

From the given figure, the slope of the curve will be:

\( m= \frac{{{{10}^7}}}{{5 × {{10}^5}}}=20\)

The equation of the line is:

v_d = 20 × E (y = mx)

For E = 5 × 104 V/cm, the drift velocity will be:

v_d = 20 × 5 × 104

v_d = 10⁶ cm/s

The current density will be:

J = 10¹⁶ × 1.6 × 10^-19 × 10⁶

J = 1.6 kA/cm²

Concept:

Diffusion current density due to holes is defined as:

\({J_{diff}} = - q;{D_p}\frac{{dn}}{{dx}}\)

\(\frac{{dn}}{{dx}}\) is the concentration gradient

Calculation:

At any position in semiconductor, the hole concentration is:

Hole concentration due to doping + Excess hole concentration, i.e.

\(p\left( x \right) = {G_{Lo}}\left( {1 - \frac{x}{L}} \right){\tau _{po}} + {P_{no}}\)

\(\frac{{dp;\left( x \right)}}{{dx}} = - \frac{{{G_{Lo}}}}{L}{\tau _{Po}} = \frac{{{{10}^{17}}\left( {{{10}^{ - 4}}} \right)}}{{{{10}^{ - 5}}}}\)

L = 10^-5 cm

\(J = \frac{{ + {{10}^2} \times 1.6 \times {{10}^{ - 19}} \times {{10}^{13}}}}{{{{10}^{ - 5}}}}\)

J = 16 A/cm²

Concept:

The depletion width for a PN junction is given by:

\(W = \sqrt {\frac{{2\epsilon {V_i}}}{{qN}}} \)

where \(N = \frac{{{N_A}{N_D};}}{{{N_A} + {N_D}}}\)

Also, the capacitance is defined as:

\({C_D} = \frac{{\epsilon A}}{W}\)

Calculation:

\(\begin{array}{l} {C_2} = \frac{{\epsilon A}}{{{W_2}}}\ {C_1} = \frac{{\epsilon A}}{{{W_1}}} \end{array}\)

\(\frac{{{C_2}}}{{{C_1}}} = \frac{{{W_1}}}{{{W_2}}}\)

\(\frac{{{C_2}}}{{{C_1}}} = \frac{{\sqrt {\frac{{2\epsilon {V_i}}}{{q{N_1}}}} }}{{\sqrt {\frac{{2\epsilon{V_i}}}{{q{N_2}}}} }}\)

\(\frac{{{C_2}}}{{{C_1}}} = \frac{{\sqrt {\frac{1}{{{N_1}}}} }}{{\sqrt {\frac{1}{{{N_2}}}} }} = \sqrt {\frac{{{N_2}}}{{{N_1}}}} \)

Putting on the respective values, we get:

\({N_1} = \frac{{{N_{{A_1}}}{N_{{D_1}}}}}{{{N_{{A_1}}} + {N_{{D_1}}}}} = \frac{{{{10}^{14}} \times {{10}^{14}}}}{{2 \times {{10}^{14}}}}\)

\({N_2} = \frac{{{N_{{A_2}}}{N_{{D_2}}}}}{{{N_{{A_L}}} + {N_{{D_2}}}}} = \frac{{{{10}^{16}} \times {{10}^{16}}}}{{2 \times {{10}^{16}}}}\)

\(\frac{{{C_2}}}{{{C_1}}} = \sqrt {\frac{{{N_2}}}{{{N_1}}}} = \sqrt {\frac{{{{10}^{16}}}}{{{{10}^{14}}}}} = 10\)

Applying KVL at the emitter-base loop, we get:

V_in – 12 i_B - V_BE = 0

2 – 12 i_B – 0.7 = 0

\({i_B}=\frac{{1.3}}{{12}} = 0.108;mA\)

At saturation V_CE = 0.2 V

\({I_{c;sat}} = \frac{{5 - 0.2}}{{4.8;k{\rm{\Omega }}}} = 1;mA\)

The total emitter current will be:

I_E = I_C + I_B

I_E = 1 + 0.108

IE = 1.108 mA

For saturation

\({I_{c;sat}} < \alpha {I_E}\)

\(\alpha > \frac{{{I_{csat}}}}{{{I_E}}}\)

\(\alpha> \frac{{1mA}}{{1.108;mA}};\)

α > 0.902

Concept:

The current in a MOSFET working in saturation is given by:

\({I_D} = \frac{1}{2}{\mu_n}{C_{ox}} \times \frac{W}{L}{\left( {{V_{gs}} - {V_{tn}}} \right)^2}\)

V_gs = Gate to source voltage

V_tn = Threshold voltage of the MOSFET.

Calculation:

\({V_{gs}} = {V_g} - {V_s}\)     ---(i)

\({V_g} = 8 \times \frac{5}{{3 + 5}} = 5~V\)     ---(ii)

Applying KVL from the source terminal to ground, we get:

V_s = I_D R_D

V_s = I_D(1k)

V_gs = 5 – ID(1k)

V_gs - V_t = 5 - ID(1k) – 1

Vgs - Vt = 4 - ID(1k)

Putting on the respective values, we get:

\({I_D} = \frac{1}{2} \times {10^{ - 3}}{\left( {4 - {I_D};1k} \right)^2}\)

\({10^6}I_D^2 - {10^4}{I_D} + 16 = 0\)

I_D = 8 mA or 2 mA

For, I_D = 8 mA     V_S = 8 V (not possible)

For I_D = 2 mA V_S = 2 V ( possible)

Hence, I_D = 2 mA

Concept:

The voltage gain of a C.E amplifier is given by:

\(\left| {{A_v}} \right| = \left| {{g_m}\left( {{R_C}||{R_L}} \right)} \right|\)

\({g_m} = \frac{{{I_C}}}{{{V_T}}}\)

I_C = DC collector current obtained by doing the DC analysis of the circuit.

Calculation:

DC equivalent circuit is as shown:

Using voltage division, we get:

\({V_B} = 12 \times \left( {\frac{{47}}{{47 + 73}}} \right) = 4.7V\)

Applying KVL in the B-E loop, we get:

V_B – 0.7 – I_E (2) = 0

\(\frac{{4.7 - 0.7}}{2} = {I_E} = 2mA \approx {I_C}\)

\(\left| {{A_v}} \right| = \frac{{\left| 2 \right|}}{{\left| {25} \right|}}\left( {2k//8k} \right) = 128\)

Concept:

For non-inverting op-amp, the voltage gain is written as:

\(\frac{{{V_o}}}{{{V_{in}}}} = \frac{{{A_{OL}}}}{{1 + {A_{OL}}β}} = {A_f}\)

A_OL = Open loop gain of the Op-Amp

β = Feedback factor

Also, the gain-bandwidth product of an Op-Amp always remains the same, i.e.

Gain × bandwidth (OLTF) = Gain × Bandwidth (CLTF)

Calculation:

Given A_OL = 10⁵

The circuit is redrawn as:

The feedback factor β is calculated as:

\(β = \frac{{{V_f}}}{{{V_o}}} = \frac{{1k}}{{1k + 79k}}\)

\(β = \frac{1}{{80}}\)

\({A_f} = \frac{{{{10}^5}}}{{1 + {{10}^5}\left( {\frac{1}{{80}}} \right)}} \approx 80\)

Short trick:

For high gain in case of a non-inverting amplifier, we get:

\({A_f} = 1 + \frac{{{R_b}}}{{{R_1}}}\)

\(A_f = 1 + \frac{{79}}{1} = 80\)

Since the gain-bandwidth product will remain the same, we can write:

10⁵ × 8 = 80 × (Bandwidth) _(CLTF)

\(Bandwidt{h_{\left( {CLTF} \right)}} = \frac{{{{10}^5} \times 8}}{{80}} \)

BW = 10 kHz

The frequency response of closed-loop op-amp gives:

\({A_v} = \frac{{{A_{vo}}}}{{\sqrt {1 + {{\left( {\frac{f}{{{f_h}}}} \right)}^2}} }}\)

f_h = Cut-off frequency

Putting on the respective values, we get:

\({A_v} = \frac{{80}}{{\sqrt {1 + {{\left( {\frac{{15}}{{10}}} \right)}^2}} }}\)

A_v = 44.4

F = ∑m (0, 2, 3, 5)

\(F = \bar A\bar C + \bar AB + A\bar BC\) 

\(F = \bar A\bar C + \bar AB + A\bar BC\)

Input to A is Exor of Previous value of A and D

New Input to B , C, D are previous values of A, B, C respectively.

Total clock cycles required is 10

Concept:

MVI → Move immediate data to register

ADD → Contents of operand + content of accumulator & result is stored in the accumulator

CMA → Compliment the contents of the accumulator

ANI → Logically AND the contents of the accumulator with 8-bit data and store data in the accumulator

Calculation:

MVI A, 33_H           A ← 33_H

MVI B, 78_H           B ← 33_H

Calculation

 

If one input of NAND is 0 output is 1

If 1 input of XOR is 1 the other input is inverted to given output

Assume X_in = 0

Input of B flip-flop = 1

Q_B = 1 always

Q_B = 1 Input of XOR

Output = Q_A ⊕ Q_B

Output = Q̅_A

Q_A toggles between 0 & 1

Output states when Q_in = 1

Q_A Q_B

\(Toggling \leftarrow \left[ {\begin{array}{*{20}{c}} 0&{1 \to always}\ 1&1 \end{array}} \right.\)

Hence, no. of possible states = 2

When input X_in = 0

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

Zeroes = -1 + j; -1 -j

Poles = 1, 2

Each pole will terminate at each zero as system gain K is increased

For 0 < K < 1.5

The poles are in right half of s-plane [unstable system]

For K = 1.5

Poles are on imaginary axis [marginally stable system]

K > 1.5

Poles are in the left half of s-Plane [stable system]

CE: s⁴ + 0s³ + 1s² + 0s + 1

Routh array

We have row zero at s3 row

Solving the auxiliary equation, we get:

s⁴ + s² + 1 = 0

By differentiating, we get:

4s³ + 2s = 0

The Routh array is modified as shown above.

Observations:

The row of zero indicates symmetric roots about the origin.

2 sign changes below row of zero indicate 2 poles in the right half of the s-plane.

∴ Two poles are on the right side and 2 poles symmetrically lying on left-half.

Concept:

For Nyquist stability

N = P

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

N = Number of encirclement about the point (-1 + j0)

N = +ve for anticlockwise

N = -ve for clockwise enduement

Calculation:

The polar plot is drawn as:

Let two points A and B representing -1 + j0 for different values of K,

The Nyquist plot is not encircling -1 + j 0 at A, i.e, K = 10

at B, the Nyquist plot is encircling -1 + j 0 i. e, K = 100

Let us now find the stability for K = 10 and K = 100 using Nyquist

Case : 1

K = 10

N = P - Z

Z = P - N

\(GH(s)=\dfrac{K}{s^3+4s^2+6s+4}\)

The Routh table can be drawn as follows:

 

No sign change, hence P = 0 and also N = 0

∴ Z = 0 → CLTP is stable

Case : 2

Z = P - N

N = -2 (two clockwise encirclement)

∴ Z = 2.

hence, CLTP is unstable.

Concept:

If two signals are uncorrelated then:

\(\mathop \smallint \limits_0^T {u_0}\left( t \right){u_1}\left( t \right) = 0\)

Calculation:

\(\smallint 5\cos \left( {20,000;\pi t} \right).5\cos \left( {22,000;\pi t} \right)dt = 0\)

\(\frac{{25}}{2}\smallint \left[ {\cos \left( {42000;\pi t} \right) + \cos \left( {2000;\pi t} \right)} \right]dt = 0\)

\(\frac{{25}}{2 }\left[ {\frac{{\sin \left( {42000;\pi T} \right)}}{{42000;\pi }} + \frac{{\sin \left( {2000;\pi T} \right)}}{{2000;\pi }}} \right] = 0\)

Both terms should be individually zero, i.e.

sin 2000 πT = 0

\(\begin{array}{l} \Rightarrow 2000;\pi T = \pi \left[ {smallest} \right]\ T = \frac{1}{{2000}} \end{array}\)

T = 0.5 msec

So, at T = 0.5 msec both terms are zero.

Concept:

(1). \(E\left[ {y\left( t \right)} \right] = H\left( 0 \right)E\left[ {x\left( t \right)} \right]\) 

(2). \(E\left[ {{x^2}\left( t \right)} \right] = \mathop \smallint \limits_{ - \infty }^\infty {S_X}\left( f \right)df \)

Application: 

In the given question,

H(0) = 1

\(E\left[ {y\left( t \right)} \right] = E\left[ {x\left( t \right)} \right]\)

\(E\left[ {{x^2}\left( t \right)} \right] = \mathop \smallint \limits_{ - \infty }^\infty {S_X}\left( f \right)df = 2\)

\(E\left[ {{y^2}\left( t \right)} \right] = \mathop \smallint \limits_{ - \infty }^{ + \infty } {S_y}\left( f \right)df\)

\( = \mathop \smallint \limits_{ - \infty }^{ + \infty } {S_x}\left( f \right){\left| {H\left( f \right)} \right|^2}df\)

\( = \mathop \smallint \limits_{ - \frac{1}{2}}^{ + \frac{1}{2}} {S_x}\left( f \right)df\)

2 - 2 e^-0.5

Low pass filter does not allow total power to pass from input to output.

Hence,

E[X²(t)] ≠ E[Y²(t)]

Concept:

Analysis:

x(t) = 4 cos 200(πt) + 8 cos (400 πt)

\(h\left( t \right) = \left{ {\begin{array}{*{20}{c}} {\frac{{2{\rm{sin}}\left( {300π t} \right)}}{{π t}}}&{t \ne 0}\ {600,}&{t = 0} \end{array}} \right.\)

y(t) = x(t) * h(t)

Y(ω) = X(ω) H(ω)

 

When X(ω) and H(ω) are multiplied, the resultant spectrum of Y(ω) will have impulses at -200π and + 200π,

Hence, the output signal y(t) will be:

y(t) = 2 × 4 cos (200 πt) = 8 cos (200 πt)

[∵ the magnitude of H(ω) at 200π = 2]

Maximum value = 8

Concept:

In circularly polarised wave the phase difference between x and y components is π/2

In linearly polarised → phase difference = 0 or π

Calculation:

Given the wave is circularly polarised the initial phase difference between x & y components is π/2

Additional π/2 difference is required to make it linearly polarised

⇒ β₁z – β₂z = π/2

Where Z is the distance travelled to become linearly polarised

\(\frac{\omega }{{{\nu _{px}}}}z - \frac{\omega }{{{\nu _{py}}}}z = \frac{\pi }{2}\)

\(\omega \left( {\frac{1}{{{\nu _{px}}}} - \frac{1}{{{V_{py}}}}} \right)z = \frac{\pi }{2}\)

\({\nu _{px}} = \frac{c}{{{n_x}}},{\nu _{py}} = \frac{c}{{{n_y}}}\)

\(2\pi f\left( {\frac{{{n_x}}}{c} - \frac{{{n_y}}}{c}} \right)z = \frac{\pi }{2}\)

\(\frac{{2\pi f}}{c}\left( {{n_x} - {n_y}} \right)z = \frac{\pi }{2}\)

λ = c/f

\(\frac{{2\pi }}{\lambda }\left( {{n_x} - {n_y}} \right)z = \frac{\pi }{2}\)

\(z = \frac{\pi }{2} \times \frac{\lambda }{{2\pi \left( {{n_x} - {n_y}} \right)}}\)

\( = \frac{{1.5 \times {{10}^{ - 6}}}}{{4 \times 0.0001}} = 0.375cm\)

Concept:

For plane polarised wave

\({\hat a_p}.\hat E = 0\)

Where,

 propagation vector

For circular polarisation

E_⊥ = E_llal & ϕ = 90°

For elliptical polarization

E_⊥ ≠ E_llal & ϕ = 90°

Calculation:

\(Propogation\space vector\space {\hat a_p} = \frac{{\bar V\left( {kr} \right)}}{{\left| {\bar V\left( {kr} \right)} \right|}}\)

V̅ (kr) = k(-x̂ + ŷ)

\(\left| {\bar V\left( {kr} \right)} \right| = k\sqrt 2 \)

\({\hat a_\phi } = \frac{{\bar V\left( {{k_r}} \right)}}{{\left| {\bar V\left( {{k_r}} \right)} \right|}} = \frac{{ - \hat x + \hat y}}{{\sqrt 2 }}\)

For plane wave

\({\hat a_\rho }.\hat E = 0\)

\(\left[ { - \frac{{\hat x + \hat y}}{{\sqrt 2 }}} \right]{E_0}\left[ {\hat x + \hat y + j2\hat z} \right]\)

\( = \frac{{ - {E_0}}}{{\sqrt 2 }} + \frac{{{E_0}}}{{\sqrt 2 }} + j0\)

= 0

Thus, given wave is plane polarised

Given wave is incidence an xy-plane ε in xy plane is llal polarised and along z is llal polarized

\({E_{11}} = {\left| E \right|_{xy}} = \sqrt {1 + 1} = \sqrt 2 \)

\({E_ \bot } = {\left| E \right|_z} = \sqrt {{2^2}} = 2\)

E₁₁ ≠ E_⊥ and ϕ = 90°

Elliptically polarised plane wave

For λ/2 dipole antenna the field radiated is zero along its axis

The direction of null

Occurs at ϕ = 45°

Axis of dipole is in xy-plane

Maximum field is radiated perpendicular to axis of dipole

θ = 90°