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

Let us look at the meanings of the given words:

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

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

The complete sentence should be:

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

Hence, option 3 is correct.

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

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

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

For example: He is taller than his brother.

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

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

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

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

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

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

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

The path of Fatima is shown

From the figure

x² + 6² = 10²

x² = 64

x = 8 km

Hence, Fatima should move 8 km to East.

n(C) = 329

n(P) = 186

n(M) = 295

n(C⋂P) = 83

n(C⋂M) = 217

n(P⋂M) = 63

n(P⋃C⋃M) = 500

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

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

n(P⋂C⋂M) = 53

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

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

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

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

Hence, the correct answer is option 1.

The given information can be represented as shown below.

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

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

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

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

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

According to questions

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

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

⇒ x = 3600

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

Total number of 3 digits numbers

= 999 - 99 = 900

Case I:

21 are together as shown

Total cases = 10

Case 2:

21 are together as shown

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

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

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

P to Q → 4 contour lines

P to R → 7 contour lines

P to S → 3 contour lines

P to T → 3 contour lines

Hence, steepest path is from P to R.

Given Differential equation:

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

Auxillary equation = D² + 2D – 5 = 0

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

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

Therefore the general solution is;

y = k₁e^(-1+√6)x + k₂e^(-1-√6)x

Concept:

For simple poles at z = a, b, c…

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

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

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

Calculation:

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

For a simple pole at z = 4

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

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

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

Residue will be  

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

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

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

Bandpass Filter:

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

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

It attenuates signals with frequencies below fc1 and above fc2.

Calculation:

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

Since  δ(n) ↔ 1

δ(n - to) ↔ e-jωto

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

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

= 5(1 – e^-4jω) + 7(e^-j3ω – e^-jω)

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

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

= e^-j2ω [10j sin 2ω – 14j sin ω]

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

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

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

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

Concept:

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

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

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

Calculation:

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

Delaying the input first;

x(u) → x(u – t₀)

let,  u – t₀ = λ

du = dλ

When, u = t – T

λ = t – t₀ – T

and when u = t,

λ = t – t₀

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

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

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

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

So the system is Time-Invariant.

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

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

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

The output characteristics will now become:

now is no longer infinite.

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

Concept:

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

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

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

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

Calculation:

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

So, the output voltage for the positive half-cycle

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

⇒ V_out = 14 – 6.5 V = 7.5 V

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

V₀⁺(C) = 6.5 V only

The circuit for negative half-cycle in as shown:

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

V_out + 6.5 + 14 = V_out (Applying KVL)

V­_out = -14 – 6.5 = - 20.5 V

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

So, the maximum V_out will be 7.5 V

And the minimum V_out will be -20.5 V

⇒ If P = high:

PMOS is OFF

NMOS is ON 

then y = Q̅

⇒If P = low

PMOS is ON

NMOS is OFF 

then  y = Q

The truth table of the above will be as shown:

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

= P ⊕ Q

So, Ex – OR Operation is being performed.

F' = X̅.Y + X.0 = X̅.Y

F = Z̅.F' + Z.F̅'

= Z̅.(X̅.Y) + Z.(X̅.Y)'   ---(1)

$\overline{XY}=\overline{X}+\overline{Y}$

Using the above in Equation (1):

= Z̅.X̅.Y + Z.(X + Y̅)

= X̅YZ̅ + XZ + Y̅Z

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

In a DRAM:

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

Concept:

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

 

Explanation:

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

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

The system is still unstable.

Concept:

Binary symmetric channel:

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

C_m = max [I(x, y)]

When I(x, y) = Mutual information

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

Calculation:

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

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

For p = 0;

From Equation - (1),

C_m = 1;

For p $= \frac{1}{2}$

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

= 0

And for p = 1;

C_m = 1

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

Option (C) →

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

V_a = V_b

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

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

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

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

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

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

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

Using equation (1)

$\Rightarrow \frac{{{E_a}}}{{{E_b}}} = \frac{b}{a}$ 

⇒ Given, b > a

So, E_a > E_b

Solving

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

Solving the inner integral first,

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

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

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

Equation (1) now becomes,

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

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

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

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

Solving for the inner integral first,

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

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

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

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

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

Equation (2) now becomes,

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

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

$\Rightarrow \frac{1}{2} - 1 \Rightarrow - \frac{1}{2} = - 0.5$

Concept:

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

Residue theorem as;

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

Calculation:

Given C, |Z|= 3

|x + iy| = 3

$\sqrt {{x^2} + {y^2}} = 3$

x² + y² = 9

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

Poles are defined by z² + 1 = 0

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

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

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

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

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

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

Now, using Cauchy’s Residue Theorem,

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

= 2πi (ieⁱ – ie^-i)

= (2πi) (i) (eⁱ – e^-i)

= 2πi² (2i sin (1))

⇒ -4πi sin (1)

Concept:

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

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

Calculation:

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

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

This has to happen for F̅ to be irrotational.

Solving the determinant,

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

For the above to be zero,

-K₃ + 2 = 0;

K₁ = 0 and

K₂ - 3 = 0

K₁ = 0

K₂ =  3

K₃ = 2

Concept:

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

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

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

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

Calculation:

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

For T₁ MOSFET,

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

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

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

Given W₂ = 2W₁,

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

Putting these in Equation (1), we get,

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

\(\Rightarrow \frac{{{I_{D1}}}}{{{I_{D2}}}} = \frac{{{W_1}\left( {{V_{GS}} - {V_{th}}} \right)1^2}}{{8{W_1}\left( {{V{GS}} - {V_{th}}} \right)1^2}} \Rightarrow {I{D2}} = 8{I_{D1}}\)

Similarly,

\(\frac{{{g_{m1}}}}{{{g_{m2}}}} = \frac{{{W_1}{{\left( {{V_{GS}} - {V_{th}}} \right)}1}}}{{{W_2}{{\left( {{V{GS}} - {V_{th}}} \right)}_2}}}\)

\(\frac{{{g_{m1}}}}{{{g_{m2}}}} = \frac{{{W_1}{{\left( {{V_{GS}} - {V_{th}}} \right)}1}}}{{2{W_1}\left( 2 \right){{\left( {{V{GS}} - {V_{th}}} \right)}_1}}}\)

g_m2 = 4g_m

Concept:

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

Application:

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

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

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

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

So, the electric field distribution is as shown:

For the given figure,

Let x be the voltage at node A.

Now, ${V_{D{S_2}}} = 3 - x$

${V_{GS2}} - {V_t} = 2.5 - x - 1$

= 1.5 - x

We observe that, ${V_{D{S_2}}}$ is always > ${V_{G{S_2}}} - {V_t}$

So, M₂ will always be in saturation.

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

(I_DSat)₁ = (I_DSat)₂

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

⇒ (2 - 1)² = (2.5 - x - 1)²

⇒ 1 = 1.5 - x

x = 0.5 V

If ‘x’ were to be 0.5 V then for M₁,

V_DS = x - 0 = 0.5 V

and V_GS - V_t = 1 V

So, V_DS < V_GS - V_t

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

So, M₁ is in cut off.

Fromthe given figure

A = P̅ Q̅ R

B = P̅ Q R

C = P Q̅ R

F = A + B + C

= P̅ Q̅ R + P̅ Q R + P Q̅ R

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

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

Calculation

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

x₁ – x₂ = 0

x₁ = x₂         --------(1)

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

x₂ (t) + 2x₁ (t) 3x₂ (t) = r (t)

x₂ = -2x, -3x₂ + r          ------(2)

From (1) and (2)

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

c(t) = x₁ (t) [output]

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

Transfer function:

TF = C (SI – A)^-1 B + D

D = 0

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

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

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

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

C. E = s² + 3s + 2

s² + 2 ζ ω_ns + ω_n²

$w_n^2 = Z$

${\omega _n} = \sqrt 2$

2ζ ω_ns = 35

$\zeta = \frac{3}{{2\sqrt 2 }} > 1$

The system is overdamped.

Concept

From Nyquist criteria we have

N = P

Where,

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

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

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

Calculations:

P = open loop RHS poles

C.E = s³ + 3s² + 0s +10

From RH Criteria

Number of sign changes = P = 2

For 0 < K < 1

N = 0

From Nyquist criteria

N = P – Z

= 2 – Z

Z = 2

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

Given signal x(t) = sin (1400 πt)

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

The output spectrum will be as follows:

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

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

The output will be then

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

With frequencies = 2, 7, 11.

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

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

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

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

                                y₁             y₂

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

Assuming P(x₁) = P(x₂) = 1/2

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

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

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

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

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

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

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

(ii) For the channel to be deterministic;

H(Y/X) = 0

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

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

(iii) For the channel to be noiseless,

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

But, H(x/y) ≠ H(y/x) ≠ 0

Say it is not noiseless as well.

So, the channel is useless.

Concept:  

 

Cut-off frequency of waveguide (rectangular)

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

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

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

Calculation:

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

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

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

= 8.199 Hz

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

f_c for TE₀₁ mode = 14.74 Hz

f_c for TE₁₁ mode = 16.4 Hz

and

f_c for TE₂₀ mode = 13.144 Hz

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

95% of 13.144 Hz

⇒ 12.484 GHz