GATE EC 2015 Official Paper: Shift 1 (Previous Year Paper)

Here, the principal is presenting something to the chief guest as a token of appreciation, thus, the perfect fit for the blank will be some ‘souvenir’. ‘Moment’ is completely illogical as it refers to a very brief period of time, also, ‘momentum’ refers to the impetus gained by a moving object that is irrelevant to the context. ‘Momento’ is a misspelled word for ‘memento’ that is related to remember or ‘souvenir’, hence, option B is the correct one.

Here, the sounds of various species are given and we are asked the sound made by frog. ‘Roar’ is a loud cry made by lion or other such large animals whereas ‘hiss’ is the sound made by a snake. ‘Patter’ is a repeated light tapping sound thus, ‘croak’ is the perfect answer. Hence, option A is the correct one.

Educe means ‘to bring out or develop (something latent or potential)’. ‘Educate’ is completely irrelevant and ‘exert’ is to apply a force that is again not a suitable synonym. ‘Extend’ is making larger which is not synonymous to educe. Thus, the correct answer is ‘extract’ which means to take out something with effort. Hence, option C is the correct one.

Calculation:

We have, 66 ∎ 6 \(= \frac{{66 - 6}}{{66 + 6}} = \frac{{60}}{{72}}\)

And,\(\left( {66{\rm{;\Delta ;}}6} \right) = \frac{{66 + 6}}{{66 - 6}} = \frac{{72}}{{60}}\)

Now, (66 ∎ 6) → (66 Δ 6) \(= \frac{{60}}{{72}} \times \frac{{72}}{{60}} = 1\)

\(\begin{array}{l} {\log _{\rm{x}}}\left( {\frac{5}{7}} \right){\rm{;}} = {\rm{;}}-\frac{1}{3}\ \Rightarrow {\rm{;}}{{\rm{x}}^{ - \frac{1}{3}}}{\rm{;}} = {\rm{;}}5/7\ \Rightarrow {\rm{;}}{{\rm{x}}^{\frac{1}{3}}}{\rm{;;}} = {\rm{;}}7/5\ \Rightarrow {\rm{;}}{\left( {{{\rm{x}}^{\frac{1}{3}}}} \right)^3}{\rm{;}} = {\rm{;}}{\left( {\frac{7}{5}} \right)^3}\ \Rightarrow {\rm{;x;}} = \frac{{343}}{{125}} \end{array}\)

‘Tuberculosis’ is a singular noun as it is a name of a disease. Even though it is connected to ‘its effect’ using a connector ‘together with’, the verb will still follow the first subject which is singular. Thus, ‘are one the’ and ‘rank as one’ in option D and A respectively are omitted as they are plural verbs. Option C is incorrect as the word ‘rank’ is followed by ‘of’ and is grammatically incorrect. Thus, the correct option is A.

From the line 3 of the paragraph it is evident that human progress is contradictory to environmental security. Hence, option B is the correct one.

The relation between numbers is as follows,

Middle number = (Sum of numbers on left side of middle number + sum of numbers on right side of middle number) /2

So we can see,

5 = (6 + 4)/ 2

7 = (7 + 4 + 2 + 1)/ 2

8 = (1 + 9 + 2 + 1 + 2 + 1)/2

5 = (4 + 1 + 2 + 3)/2

Thus,

(3 + 3)/2 = 3

Hence the missing number is 3.

We know that, volume of a cube = (side)³

∴ Volume of larger cube = 3³ = 27 unit³

And, volume of one smaller cube = 1³ = 1 unit³

Number of smaller cubes used = Volume of larger cube/Volume of one smaller cube

∴ Number of smaller cubes used = 27/1 = 27

We know that, a cube has 6 faces.

∴ Total number of faces of all smaller cubes = 6 × 27 = 162

Number of faces of smaller cubes visible on one face of larger cuve = 9

∴ Total number of visible faces of smaller cubes = 9 × 6 = 54

∴ Total number of faces that are not visible = 162 – 54 = 108

∴ required ratio = 54 : 108 = 1 : 2

The only logically inferred is that Humpty Dumpty falls only sometimes while having lunch as it is given that the wall breaks only sometimes though he has his lunch everyday on the wall.

Since system has infinitely many solutions we have rank of A = rank of [A : B] < (n = 3)

\(\begin{array}{l} \left[ {\begin{array}{*{20}{c}} 1&{ - 2}&3&:&{ - 1}\ 1&{ - 3}&4&:&1\ { - 2}&4&{ - 6}&:&{\rm{k}} \end{array}} \right]\ \left. {\begin{array}{*{20}{c}} {{{\rm{R}}_2} \to {{\rm{R}}_2} - {{\rm{R}}_1}}\ {{{\rm{R}}_3} \to {{\rm{R}}_3} + 2{{\rm{R}}_1}} \end{array}} \right} \to \left[ {\begin{array}{*{20}{c}} 1&{ - 2}&3&:&{ - 1}\ 0&{ - 1}&1&:&2\ 0&0&0&:&{{\rm{k}} - 2} \end{array}} \right] \end{array}\)

∴ k = 2

By Lagrange’s mean value theorem we have

\(f'(c) = \frac{{f(b) - f(a)}}{{b - a}}\)

\(\begin{array}{l} {{\rm{f}}'}\left( {\rm{x}} \right) = \frac{{{\rm{f}}\left( 1 \right) - {\rm{f}}\left( { - 1} \right)}}{{1 - \left( { - 1} \right)}}\ \end{array}\)

f(1) = 1 - 1 + 1 = 1

f(-1) = 1 - 1 - 1 = -1

\(f^1(x) = \frac{2}{2} = 1{\rm{;}}\ \)

0 - 2x + 3x² = 1 ⇒ 3x² - 2x - 1 = 0

\(\therefore {\rm{;x;}} = {\rm{;}}1{\rm{;and;}} - \frac{1}{3}\ \therefore {\rm{;x}} = - \frac{1}{3}{\rm{;only;lies;in;}}\left( { - 1,{\rm{;}}1} \right) \)

Since A and B are independent

\(\begin{array}{l} {\rm{P;}}\left( {{\rm{A}} \cap {\rm{B}}} \right){\rm{;}} = {\rm{;P;}}\left( {\rm{A}} \right).{\rm{P;}}\left( {\rm{B}} \right)\ {\rm{P}}\left( {\frac{{\rm{A}}}{{\rm{B}}}} \right) = \frac{{{\rm{P}}\left( {{\rm{A}} \cap {\rm{B}}} \right)}}{{{\rm{P}}\left( {\rm{B}} \right)}} = \frac{{{\rm{P}}\left( {\rm{A}} \right).{\rm{P}}\left( {\rm{B}} \right)}}{{{\rm{P}}\left( {\rm{B}} \right)}} = {\rm{P}}\left( {\rm{A}} \right)\ {\rm{P}}\left( {{\rm{\bar A}} \cap {\rm{\bar B}}} \right) = {\rm{P}}\left( {{\rm{\bar A}}} \right).{\rm{P}}\left( {{\rm{\bar B}}} \right) = \left[ {1 - {\rm{P}}\left( {\rm{A}} \right)} \right]\left[ {1 - {\rm{P}}\left( {\rm{B}} \right)} \right]\ {\rm{P}}\left( {{\rm{A}} \cup {\rm{B}}} \right) = {\rm{P}}\left( {\rm{A}} \right) + {\rm{P}}\left( {\rm{B}} \right) - {\rm{P}}\left( {{\rm{A}} \cap {\rm{B}}} \right) \end{array}\)

Residue of \(\frac{{\rm{z}}}{{{{\rm{z}}^2} - 1}}{\rm{;at;z}} = 1{\rm{;is}} = \begin{array}{*{20}{c}} {{\rm{lim}}}\ {{\rm{z}} \to 1} \end{array}\left( {{\rm{z}} - 1} \right).\frac{{\rm{Z}}}{{{{\rm{z}}^2} - 1}} = \begin{array}{*{20}{c}} {{\rm{lim}}}\ {{\rm{z}} \to 1} \end{array}{\rm{;}}\frac{{\rm{z}}}{{{\rm{z}} + 1}} = \frac{1}{2}\)

\(\begin{array}{l} \mathop \oint \limits_{\rm{c}}^{\rm{;}} {{\rm{z}}^2}.{\rm{dz}} = 0\ \frac{1}{{2{\rm{\pi i}}}}{\rm{;}}\oint \frac{1}{{\rm{z}}}.{\rm{dz}} = \frac{1}{{2{\rm{\pi i}}}}.2{\rm{\pi i}}\left( 1 \right) = 1\ {\rm{z;}} = {\rm{;x;}} + {\rm{;iy;;and;\bar z}} = {\rm{x}} - {\rm{iy;i}}.{\rm{e}}.{\rm{;u}} = {\rm{x;and;v}} = - {\rm{y}}\ {{\rm{u}}_{\rm{x}}} = 1{\rm{;and;}}{{\rm{v}}_{\rm{y}}} = - 1 \end{array}\)

\({{\rm{u}}_{\rm{X}}} \ne {{\rm{v}}_{\rm{y}}} \to {\rm{\bar z}}\) is not Analytic Analytic

\(\begin{array}{l} {\rm{AX;}} = {\rm{;\lambda X}}\ \left[ {\begin{array}{*{20}{c}} 4&1&2\ {\rm{p}}&2&1\ {14}&{ - 4}&{10} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1\ 2\ 3 \end{array}} \right] = {\rm{\lambda }}\left[ {\begin{array}{*{20}{c}} 1\ 2\ 3 \end{array}} \right]\ \therefore {\rm{;}}4{\rm{;}} + {\rm{;}}2{\rm{;}} + {\rm{;}}6{\rm{;}} = {\rm{;\lambda ;;;}} \Rightarrow {\rm{;\lambda ;}} = {\rm{;}}12\ {\rm{p;}} + {\rm{;}}4{\rm{;}} + {\rm{;}}3{\rm{;}} = {\rm{;}}2{\rm{\lambda ;;;;}} \Rightarrow {\rm{;P;}} = {\rm{;}}24{\rm{;}}-{\rm{;}}7{\rm{;}} = {\rm{;}}17 \end{array}\)

Concept:

In the Series RLC circuit,

Under resonance condition, the voltage across the L & C components are Quality factor (Q) times the supply voltage. Hence, the series RLC circuit is considered as

the voltage amplification circuit.

VL = Vc = Q × VS

Calculation:

\(\begin{array}{l} {{\rm{V}}_{\rm{c}}} = {\rm{QV}}\angle - 90^\circ \ {\rm{Q}} = \frac{{{{\rm{\omega }}_0}{\rm{L}}}}{{\rm{R}}} = \frac{1}{{\rm{R}}}\sqrt {\frac{{\rm{L}}}{{\rm{C}}}} = \frac{{10}}{4} = 2.5\ {{\rm{V}}_{\rm{c}}} = 25\angle - 90^\circ \ \left| {{{\rm{V}}_{\rm{c}}}} \right| = 25{\rm{V}} \end{array}\)

Note:

The Q-factor is the short form for the Quality factor.

It is defined as the ratio of the amount of energy stored in the electrical elements and the amount of energy lost per cycle through the electrical elements.

The Q-factor for the parallel RLC circuit is high when (Q-factor ≥ 10).

\({\rm{Q}} = 2{\rm{\pi }}\frac{{{\rm{Amount;of;energy;store;}}}}{{{\rm{Amount;of;energy;lost}}}}\)

Or Q-factor is also given by

\({\rm{Q}} = \frac{{{\rm{The;resonant;frequency;;of;the;parallel;RLC;circuit}}}}{{{\rm{Bandwith;of;the;parallel;RLC;circuit;}}}}\)

The quality for the series RLC circuit is given by

\(Q = \dfrac{ω_0L}{R} = \dfrac{1}{ω_0 RC} = \dfrac{1}{R} \sqrt {\dfrac{L}{C}}\)

Concept:

Wheat-stone bridge:

The Wheatstone bridge circuit is a series-parallel arrangement of resistances connected between a voltage supply terminal and ground producing zero voltage difference between the two parallel branches when the bridge is balanced.

A Wheatstone bridge circuit has two input terminals and two output terminals consisting of four resistors configured in a diamond-like arrangement as shown. This is typical of how the Wheatstone bridge is drawn.

The bridge is said to be balanced when the current through the galvanometer is 0. (i.e. voltage difference between parallel branch nodes is 0)

Balanced condition,

\(\frac{{{R_1}}}{{{R_2}}} = \frac{{{R_3}}}{{{R_X}}}\)

Also called balanced bridge condition.

Calculation:

Redraw the given circuit we get,

By bridge condition

\(,{{\rm{R}}_{{\rm{eq}}}} = \frac{{\rm{R}}}{3} = 100{\rm{\Omega }}\)

Concept:

Nodal Analysis:

Nodal analysis is a method of analyzing network with the help of KCL equations.

For a network of N nodes, the number of simultaneous equations to be solved to get the unknowns

= Number of KCL equations

= N - 1

= one less than the number of nodes

Mesh Analysis:

Mesh analysis is a method of analyzing network with the help of KVL equations.

For a network having N nodes and B branches, the number of simultaneous equations to be solved to get the unknowns

= Number of KVL equations

= number of independent loop equations

= B - N + 1

Calculation:

By nodal analysis we get,

Find the value of I

-5 + I + I + 2I = 0

4I = 5

I = 5/4 A

The value of V1 is given by

V₁ = 4I

V1 = 5 Volts

By using mesh analysis we get V₂ as

V₂ = 5 × 4 +  V1

V₂ = 20 + 5

V₂ = 25 Volts

A negative differential resistance region is observed if both the P and N regions are heavily doped. One such example is the tunnel diode.

Concept:

The conductivity of the sample \({\rm{\sigma }} = {\rm{q}}\left( {{\rm{n}}{{\rm{\mu }}_{\rm{n}}} + {\rm{p}}{{\rm{\mu }}_{\rm{p}}}} \right)\) if, n ≫ p, we can approximate \({\rm{\sigma }} \approx {\rm{qn}}{{\rm{\mu }}_{\rm{n}}}\)

Application: 

The conductivity of the sample \({\rm{\sigma }} = {\rm{q}}\left( {{\rm{n}}{{\rm{\mu }}_{\rm{n}}} + {\rm{p}}{{\rm{\mu }}_{\rm{p}}}} \right)\) since, n ≫ p, we can approximate \({\rm{\sigma }} \approx {\rm{qn}}{{\rm{\mu }}_{\rm{n}}}\)

Using \({\rm{n}} = {{\rm{N}}_{\rm{D}}} = {10^{16}}{\rm{c}}{{\rm{m}}^{ - 3}}\), we have \({\rm{\sigma }} = 1.6 \times {10^{ - 19}} \times {10^{16}} \times 1200\)

\(\Rightarrow {\rm{\sigma }} = 1.92\frac{{{\rm{mho}}}}{{{\rm{cm}}}}\)

Now, resistivity \({\rm{\rho }} = \frac{1}{{\rm{\sigma }}} = \frac{1}{{1.92}} = 0.52{\rm{;\Omega cm}}\)

During the positive half cycle, both the diodes will be forward biased.

Applying KVL, we get:

V_out = - V_in

During the negative half-cycle, both the diodes will be reverse biased, therefore:

V_out = 0 V

First, we will check if the Zener diode is ON or OFF.

Assuming the Zener diode to be OFF, the voltage across the Zener diode is calculated as:

\(V_0=\frac{{10\left( 1 \right)}}{{1 + 1}} = \frac{{10}}{2} = 5{\rm{V}}\)

Since this is less than the breakdown voltage of the Zener diode, the Zener diode will not be regulating and the output voltage will simply be the fixed at 5 V.

∴ The output voltage V₀ = 5V

Concept:

RC circuit with the presence of source after the switching action is shown as:

The voltage across the capacitor for t > 0 is defined by

\(V_C(t) = V(∞ ) + [V(0) - V(∞)]e^{-\frac{t}{τ}}\)

V(0): Voltage across the capacitor before switch operation, it is previous steady-state

V(∞): Voltage across the capacitor after the switch operation, it is next steady-state

When the initial condition is given

let V(0)  = V₀

V_C(t) = V_S + [V₀ - V_S]e^-t/τ  

Here V_S is the source voltage.

The response of the voltage across the capacitor is

Calculation:

Here the capacitor is initially uncharged i.e.

\({{\rm{V}}_{\rm{C}}}\left( 0 \right){\rm{;}} = {\rm{;}}0\)

The capacitor will be charged to supply voltage 3V when the switch is in position B for ∞ time.

The voltage across the capacitor at t = ∞ is 3 V

So we need to find the capacitor voltage

\({{\rm{V}}_{\rm{C}}}\left( {\rm{t}} \right) = {{\rm{V}}_{\rm{c}}}\left( ∞ \right) + \left[ {{{\rm{V}}_{\rm{c}}}\left( 0 \right) - {{\rm{V}}_{\rm{c}}}\left( ∞ \right)} \right]{{\rm{e}}^{ - {\rm{t}}/{\rm{τ }}}} \)

\(= 3 - 3{{\rm{e}}^{ - \frac{{\rm{t}}}{{\rm{τ }}}}}\)

\(\ {\rm{τ }} = {\rm{RC}} \)

\(= 120 \times 0.1 \times {10^{ - 6}} \)

\(= 12{\rm{\mu sec}} \)

\({{\rm{i}}_{\rm{c}}} = {\rm{C}}\frac{{{\rm{d}}{{\rm{V}}_{\rm{c}}}}}{{{\rm{dt}}}}\)

\( = \left( {0.1 \times {{10}^{ - 6}}} \right)\frac{{\rm{d}}}{{{\rm{dt}}}}\left[ {3 - 3{{\rm{e}}^{ - \frac{{\rm{t}}}{{\rm{τ }}}}}} \right] \)

\(= \frac{{3 \times 0.1}}{{\rm{τ }}}{{\rm{e}}^{ - \frac{{\rm{t}}}{{\rm{τ }}}}} = \frac{{0.3}}{{12}} {{\rm{e}}^{ - \frac{{\rm{t}}}{{\rm{τ }}}}} \)

So, the instantaneous power of the source \(= {\rm{;V}}\left( {\rm{t}} \right){\rm{i}}\left( {\rm{t}} \right)\)

\(\begin{array}{l} {\rm{P}}\left( {\rm{t}} \right) = \left( 3 \right)\left[ {\frac{{0.3}}{{12}}\times{{\rm{e}}^{ - \frac{{\rm{t}}}{{\rm{τ }}}}}} \right] = \frac{{0.9}}{{12}}{{\rm{e}}^{ - \frac{{\rm{t}}}{{\rm{τ }}}}}\ \to {\rm{E}} = \mathop \smallint \limits_0^∞ {\rm{Pd}}\left( {\rm{t}} \right)\ = \mathop \smallint \limits_0^∞ \frac{{0.9}}{{12}}{{\rm{e}}^{ - \frac{{\rm{t}}}{{\rm{τ }}}}}{\rm{dt}}\ = \left( { - {\rm{τ }}} \right)\frac{{0.9}}{{12}}\left[ {{{\rm{e}}^{ - \frac{{\rm{t}}}{{\rm{τ }}}}}} \right]_0^∞ = {\rm{τ }}\left( {\frac{{0.9}}{{12}}} \right)\ = 12 \times {10^{ - 6}} \times \frac{{0.9}}{{12}}\ = {\rm{;}}0.9{\rm{;\mu J}} \end{array}\)

The result of the addition is stored in the accumulator A and any overflow bit (carry) that is generated gets updated in the flag register F.

The size of a memory register is given by \({\rm{M}} \times {\rm{N}}\) where \({\rm{M}}\) is the number of rows and \({\rm{N}}\) is the number of columns. Now, the given memory array is in the size of a square, i.e, the number of rows is equal to the number of columns. Thus, \({\rm{M}} = {\rm{N}}\). So, we have \({{\rm{M}}^2} = 16384\).

\(\Rightarrow {\rm{M}} = 128\)

Now, the rows are fetched by the address lines and all the registers of a particular row are handled by data lines. Now, for a register with 128 flip flops can be handled by \({\log _2}128 = 7\) data lines. Similarly, the number of address lines required by the row decoder is also 7.

Given,

The increment in the output voltage on changing just one bit is 0.625 V.

Hence, the resolution is 0.625 V.

Now the decimal equivalent of 1111 is 15.

Thus, the output voltage is 15 × 0.625 = 9.375 V.

Concept:

Unit impulse function:

It is defined as, \(\delta \left( t \right) = \left{ {\begin{array}{*{20}{c}} {\infty ,;;t = 0}\ {0,;;t \ne 0} \end{array}} \right.\)

Properties:

1. \(\mathop \smallint \limits_{ - \infty }^\infty \delta \left( t \right)dt = 1\) 2. \(\delta \left( {at} \right) = \frac{1}{{\left| a \right|}}\delta \left( t \right)\) 3. x(t) δ(t – t0) = x(t0) ⇒ x(t) δ(t) = x(0) 4. \(\mathop \smallint \limits_{ - \infty }^\infty x\left( t \right)\delta \left( {t - {t_o}} \right)dt = x\left( {{t_0}} \right)\)  5. \(\mathop \smallint \limits_{ - \infty }^T x\left( t \right)\delta \left( {t - {t_o}} \right)dt = 0\)  If T < t0 6. \(\mathop \smallint \limits_{ - \infty }^\infty f\left( t \right)\delta \left( {at + b} \right)dt = \mathop \smallint \limits_{ - \infty }^\infty f\left( t \right)\frac{1}{{\left| a \right|}}\delta \left( {t + \frac{b}{a}} \right)dt\) 7. \(\mathop \smallint \limits_{ - \infty }^\infty x\left( t \right){\delta ^n}\left( {t - {t_o}} \right)dt = {\left. {\frac{{{d^n}x(t)}}{{d{t^n}}}} \right|_{t = {t_0}}}\)

Calculation:

Using properties of impulse

\(\begin{array}{l} {\rm{x}}\left( { - {\rm{t}}} \right){\rm{*\delta }}\left( { - {\rm{t}} - {{\rm{t}}_0}} \right){\rm{;}} = {\rm{;x}}\left( { - {\rm{t}}} \right){\rm{*\delta }}\left( {{\rm{t}} + {{\rm{t}}_0}} \right)\ = {\rm{;x}}\left( { - \left( {{\rm{t}} + {{\rm{t}}_0}} \right)} \right)\ = {\rm{;x}}\left( { - {\rm{t}} - {{\rm{t}}_0}} \right) \end{array}\)

Concept:

The instantaneous power of a signal is calculated as:

\(p\left( t \right) = \smallint {s^2}\left( t \right)dt\)

The average power will be:

\({P_{avg}} = \frac{1}{T}\mathop \smallint \limits_0^T {|s^2}\left( t \right)|dt\)

Calculation:

\(x\left( {\frac{{t - 1}}{2}} \right) = x\left( {\frac{1}{2}\left( {t - 1} \right)} \right)\)

So, we first scale x(t) by 1/2.and then shift the result by 1 unit right. The resulting waveform is obtained as:

Now, the average power is given by:

\({P_{av}} = \frac{1}{T}\mathop \smallint \limits_T {\left| {g\left( t \right)} \right|^2}dt\)

We have the magnitude of waveform g(t) as shown below.

Taking the sample period -3 ≤ t ≤ 3, we obtain:

\({P_{av.}} = \frac{1}{6}\mathop \smallint \limits_{ - 3}^3 {\left| {g\left( t \right)} \right|^2}dt\)

\({P_{av.}} = \frac{1}{6}\mathop \smallint \limits_{ - 3}^3 {\left| {g\left( t \right)} \right|^2}dt\)

\(= \frac{1}{6}\left[ {\mathop \smallint \limits_{ - 1}^1 {{\left( {\frac{{ - 3}}{2}t + \frac{3}{2}} \right)}^2}dt + \mathop \smallint \limits_1^3 {{\left( {\frac{3}{2}t - \frac{3}{2}} \right)}^2}dt} \right]\)

\( = \frac{3}{8}\left[ {\mathop \smallint \limits_{ - 1}^1 {{\left( {1 - t} \right)}^2}dt + \mathop \smallint \limits_1^3 {{\left( {t - 1} \right)}^2}dt} \right]\)

\( = \frac{3}{8}\left{ {\left[ { - \frac{{{{\left( {1 - t} \right)}^3}}}{3}} \right]_{ - 1}^1 + \left[ {\frac{{{{\left( {t - 1} \right)}^3}}}{3}} \right]_1^3} \right}\)

\( = \frac{3}{8}\left{ {\frac{8}{3} + \frac{8}{3}} \right}\)

= 2

Concept:

Negative feedback in a closed-loop reduces gain improves the bandwidth and disturbance rejection. Also, it reduces the sensitivity to parameter variation.

• Negative feedback in a control system reduces the overall gain.
• It reduces the sensitivity of output to input variation, distortion, and noise reduction.
• It improves bandwidth and input and output impedances (reducing impedance and increasing bandwidth which are desired in most cases)

Important Points

Positive feedback:

In open-loop system, we do not have any feedback path but in closed-loop system, we have a feedback path.

The transfer function of closed-loop transfer function, with positive feedback, is given by \(\frac{{G\left( s \right)H\left( s \right)}}{{1 - G\left( s \right)H\left( s \right)}}\)

The overall gain of the transfer function increases with positive feedback. 

Negative feedback:

The transfer function of closed loop transfer function, with negative feedback, is given \(\frac{{G\left( s \right)H\left( s \right)}}{{1 + G\left( s \right)H\left( s \right)}}\)

The overall gain of the transfer function decreases with negative feedback.

The characteristic equation is \(1 + {\rm{G}}\left( {\rm{s}} \right){\rm{H}}\left( {\rm{s}} \right) = 0\)

\(\begin{array}{l} \Rightarrow 1 + \frac{{\rm{k}}}{{{\rm{s}}\left( {{\rm{s}} + 1} \right)\left( {{\rm{s}} + 3} \right)}} = 0\ \Rightarrow {{\rm{s}}^3} + 4{{\rm{s}}^2} + 3{\rm{s}} + {\rm{k}} = 0 \end{array}\)

Solving the Routh’s table,

For finding imaginary axis poles, we set \(\frac{{12 - {\rm{k}}}}{4} = 0\)

Thus, \({\rm{k}} = 12\)

Given transfer function

\({\rm{G}}\left( {{\rm{jω }}} \right) = \frac{{10\left( {{\rm{jω }} + 1} \right)}}{{\left( {{\rm{jω }} + 10} \right)}}\) (it is Lead compensator)

Magnitude of the transfer function will be

\( {\rm{M}} = \frac{{10\sqrt {{{\rm{ω }}^2} + 1} }}{{\sqrt {{{\rm{ω }}^2} + 100} }}\)

Phase of the transfer function will be

\({\rm{;;}}ϕ = {\tan ^{ - 1}}\left( {\rm{ω }} \right) - {\tan ^{ - 1}}\left( {\frac{{\rm{ω }}}{{10}}} \right)\)

At ω = 0 

Magnitude M₁ = 1 and phase ϕ₁ = 0° 

At ω = ∞ 

Magnitude M₂ = 10 and phase ϕ₂ = 0° 

At ω = 2

Magnitude M₃ = 2.19 and phase ϕ₃ = 52.12° 

The polar plot considering the above magnitude and phase values will be

Zero is nearer to the imaginary axis than the pole, hence the plot will move in a clockwise direction.

Concept: In Delta modulation Slope Overload error occurs if the step size is small or Step size × sampling frequency  ≤ slope of m(t).

                Therefore optimum step size is required to avoid the Slope overload error  i.e The slope of m(t) ≤ Step size × sampling frequency

Application: For preventing slope overload, we should have:

The slope of m(t) ≤ Step size × sampling frequency

For sinusoidal signal, if \({{\rm{V}}_{{\rm{input}}}} = {{\rm{A}}_{\rm{m}}}\sin \left( {{{\rm{\omega }}_{\rm{m}}}{\rm{t}}} \right)\) 

\({\rm{slop}}{{\rm{e}}_{{\rm{input}}}} = {\left( {\frac{{{\rm{d}}{{\rm{V}}_{{\rm{input}}}}}}{{{\rm{dt}}}}} \right)_{{\rm{max}}}} = {{\rm{A}}_{\rm{m}}}{{\rm{\omega }}_{\rm{m}}}\)

\(\ {{\rm{A}}_{\rm{m}}}{{\rm{\omega }}_{\rm{m}}} \le \frac{{\rm{\Delta }}}{{{{\rm{T}}_{\rm{s}}}}} = \frac{{{\rm{step;size}}}}{{{\rm{sampling;interval}}}} = {\rm{\Delta ;}}{{\rm{f}}_{\rm{s}}}\)

\({{\rm{A}}_{\rm{m}}}2{\rm{\pi }}\left( {2 × {{10}^3}} \right) \le 2 × {10^4} × 0.1\)

\(\therefore {{\rm{A}}_{\rm{m}}} \le \frac{1}{{2{\rm{\pi }}}} \)

Concept: AM, DSBSC, SSBSC, and VSB modulated signals are Bandpass signals.

Application: We have the signal.

\({\rm{s}}\left( {\rm{t}} \right) = {\rm{m}}\left( {\rm{t}} \right)\cos \left( {2{\rm{\pi }}{{\rm{f}}_{\rm{c}}}{\rm{t}}} \right) + {\rm{\hat m}}\left( {\rm{t}} \right)\sin \left( {2{\rm{\pi }}{{\rm{f}}_{\rm{c}}}{\rm{t}}} \right)\)

Here, \({\rm{s}}\left( {\rm{t}} \right)\) represents SSB – Lower sideband, and thus it is a bandpass signal.

Concept: For the infinitely long current-carrying conductor, the magnetic field is given by

\(\begin{array}{l} {\rm{\vec H}} = \frac{{\rm{I}}}{{2{\rm{\pi r}}}}{{{\rm{\hat a}}}_\phi }\ \Rightarrow \left| {{{{\rm{\vec H}}}_\phi }} \right| \propto \frac{1}{{\rm{r}}} \end{array}\)

Application: As

 \(\begin{array}{l} {\rm{\vec H}} = \frac{{\rm{I}}}{{2{\rm{\pi r}}}}{{{\rm{\hat a}}}_\phi }\ \Rightarrow \left| {{{{\rm{\vec H}}}_\phi }} \right| \propto \frac{1}{{\rm{r}}} \end{array}\)

Hence, the correct plot will be

Concept: General form of electrical field directed in Positive y-direction is given by \({\rm{\vec E}}\left( {{\rm{z}},{\rm{t}}} \right) = A\cos \left( {{\rm{\omega t}} - {\rm{β z}}} \right){{\rm{\hat a}}_{\rm{y}}}\frac{{\rm{V}}}{{\rm{m}}}\) where \({\rm{β }} = \frac{{2{\rm{\pi }}}}{{\rm{\lambda }}} \)

Application:  Given The electric field

\({\rm{\vec E}} = \left( {{\rm{z}},{\rm{t}}} \right) = 2\cos \left( {{{10}^8}{\rm{t}} - \frac{{\rm{z}}}{{\sqrt 2 }}} \right){{\rm{\hat a}}_{\rm{y}}}\frac{{\rm{V}}}{{\rm{m}}}\)

Comparing with

\({\rm{\vec E}}\left( {{\rm{z}},{\rm{t}}} \right) = 2\cos \left( {{\rm{\omega t}} - {\rm{β z}}} \right){{\rm{\hat a}}_{\rm{y}}}\frac{{\rm{V}}}{{\rm{m}}}\)

We have, phase constant, \({\rm{β }} = \frac{{2{\rm{\pi }}}}{{\rm{\lambda }}} = \frac{1}{{\sqrt 2 }}\)

\(\Rightarrow {\rm{\lambda }} = 2\sqrt 2 {\rm{\pi }} = 8.88{\rm{m}}\)

\(\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{t}}^2}}} + 2.\frac{{{\rm{dy}}}}{{{\rm{dt}}}} + {\rm{y}} = 0\)

Characteristics solution \({{\rm{Y}}_{\rm{c}}}{\rm{;}} = {\rm{;}}\left( {{{\rm{C}}_1}{\rm{;}} + {\rm{;}}{{\rm{C}}_2}{\rm{;t}}} \right){{\rm{e}}^{ - {\rm{t}}}}{\rm{;}}\)

\(\begin{array}{l} {\rm{Y;}}\left( 0 \right){\rm{;}} = {\rm{;}}{{\rm{C}}_1}{\rm{;}} = {\rm{;}}1\ {{\rm{Y}}_1}\left( {\rm{x}} \right){\rm{;}} = {\rm{;}} - {\rm{;}}\left( {{{\rm{C}}_1}{\rm{;}} + {\rm{;}}{{\rm{C}}_2}{\rm{;t}}} \right){{\rm{e}}^{ - {\rm{t}}}}{\rm{;}} + {\rm{;}}{{\rm{e}}^{ - {\rm{t}}}}.{\rm{;}}{{\rm{C}}_2}\ {{\rm{Y}}_1}{\rm{;}}\left( 0 \right){\rm{;}} = {\rm{;}} - {\rm{;}}{{\rm{C}}_1}{\rm{;}} + {\rm{;}}{{\rm{C}}_2}{\rm{;}} = {\rm{;}}1{\rm{;;;}} \Rightarrow {\rm{;}}{{\rm{C}}_2}{\rm{;}} = {\rm{;}}2\ \therefore {\rm{;y;}} = {\rm{;}}\left( {1{\rm{;}} + {\rm{;}}2{\rm{t}}} \right){{\rm{e}}^{ - {\rm{t}}}} \end{array}\)

\(\begin{array}{l} {\rm{P}} = {{\rm{x}}^3}{\rm{y;}}{{{\rm{\hat a}}}_{\rm{x}}} - {{\rm{x}}^2}{{\rm{y}}^2}\widehat {{\rm{ay}}} - {{\rm{x}}^2}{\rm{yz\hat az}}\ \end{array}\)

\(\nabla .{\rm{P}} = 3{{\rm{x}}^2}{\rm{y}} - {{\rm{x}}^2}2{\rm{y}} - {{\rm{x}}^2}{\rm{y}} = 0{\rm{;}} \Rightarrow {\rm{solenoidal;}}\)

\( \nabla \times {\rm{P}} = \left| {\begin{array}{*{20}{c}} {\rm{i}}&{\rm{j}}&{\rm{k}}\ {\frac{\partial }{{\partial {\rm{x}}}}}&{\frac{\partial }{{\partial {\rm{y}}}}}&{\frac{\partial }{{\partial {\rm{z}}}}}\ {{{\rm{x}}^3}{\rm{y}}}&{ - {{\rm{x}}^2}{{\rm{y}}^2}}&{ - {{\rm{x}}^2}{\rm{yz}}} \end{array}} \right|\)

\(\Rightarrow {\rm{i}}\left( { - {{\rm{x}}^2}{\rm{z}} - 0} \right) - {\rm{j}}\left( { - 2{\rm{xyz}}} \right) + {\rm{k}}\left( { - 2{\rm{x}}{{\rm{y}}^2} - {{\rm{x}}^3}} \right) \ne 0\)

∴ ∇ × P ≠ 0   ⇒ not irrotational

f(x) = e-x(x2 + x + 1)

At x = 0, f(0) = 1

And  \({\rm{x}} = \infty ,{\rm{f}}\left( \infty \right) = 0\)

For maxima and minima we put \({\rm{f'}}\left( {\rm{x}} \right) = 0\)

\(\begin{array}{l} \Rightarrow {\rm{f'}}\left( {\rm{x}} \right) = {{\rm{e}}^{ - {\rm{x}}}}\left( {2{\rm{x}} + 1} \right) - {{\rm{e}}^{ - {\rm{x}}}}\left( {{{\rm{x}}^2} + {\rm{x}} + 1} \right) = 0\ \Rightarrow {{\rm{e}}^{ - {\rm{x}}}}\left( {2{\rm{x}} + 1 - {{\rm{x}}^2} - {\rm{x}} - 1} \right) = 0\ \Rightarrow - {{\rm{e}}^{ - {\rm{x}}}}\left( {{{\rm{x}}^2} - {\rm{x}}} \right) = 0\ \Rightarrow {\rm{x}} = 0,1 \end{array}\)

Also, \({\rm{f''}}\left( {\rm{x}} \right) = - {{\rm{e}}^{ - {\rm{x}}}}\left( {2{\rm{x}} - 1} \right) + {{\rm{e}}^{ - {\rm{x}}}}\left( {{{\rm{x}}^2} - {\rm{x}}} \right) = 0\)

\(\Rightarrow {{\rm{e}}^{ - {\rm{x}}}}\left( {{{\rm{x}}^2} - 3{\rm{x}} + 1} \right) = 0\)

Now, at \({\rm{x}} = 0,{\rm{;f''}}\left( {\rm{x}} \right) > 0\)

Thus, \({\rm{x}} = 0\) is a minima.

At \({\rm{x}} = 1,{\rm{;f''}}\left( 1 \right) = {{\rm{e}}^{ - 1}}\left( {1 - 3 + 1} \right) = - {{\rm{e}}^{ - 1}} < 0\)

Thus, \({\rm{x}} = 1\) point is a maxima.

Only second option qualifies for this.

Let 2x, 2y be the length, breadth of a rectangle

∴ Area of rectangle = 4xy I.e. 2x.2y

Given equation of ellipse is  x2 + 4y2 = 1

\({\rm{Let;f;}} = {\rm{;}}{\left( {{\rm{area}}} \right)^2}{\rm{;}} = {\rm{;}}16{{\rm{x}}^2}{{\rm{y}}^2} = 16\left( {1{\rm{;}}-{\rm{;}}4{{\rm{y}}^2}} \right){\rm{;}}\left( {{{\rm{y}}^2}} \right)\)

For maxima (or) minima \({{\rm{f}}^1}\left( {\rm{y}} \right){\rm{;}} = {\rm{;}}0\)

\(\begin{array}{l} {{\rm{f}}^1}\left( {\rm{y}} \right) = 16\left( {1 - 4{{\rm{y}}^2}} \right)\left( {2{\rm{y}}} \right) + 16{{\rm{y}}^2}\left( { - 8{\rm{y}}} \right) = 0\ = 32{\rm{y}} - 128{{\rm{y}}^3} - 128{{\rm{y}}^3} = 0\ {\rm{y}}\left( {32 - 256{{\rm{y}}^2}} \right) = 0 \end{array}\)

y = 0 and \({\rm{y}} = \sqrt {\frac{{32}}{{256}}} = \frac{1}{{\sqrt {8{\rm{;}}} }}\)

For maximum area \({\rm{y}} = \frac{1}{{\sqrt 8 }}\) and \({{\rm{x}}^2} = 1 - 4{{\rm{y}}^2} = 1 - 4\left( {\frac{1}{8}} \right) = \frac{1}{2}\)

\(\therefore {\rm{x}} = \frac{1}{{\sqrt 2 }}{\rm{;and;y}} = \frac{1}{{\sqrt 8 }}\)

Maximum area = \(4{\rm{xy}} = \frac{4}{{\sqrt {16} }} = 1\)

The characteristic equation of a series RLC circuit is given by,

\({\rm{C}}.{\rm{E}} = {{\rm{s}}^2} + \frac{{\rm{R}}}{{\rm{L}}}{\rm{s}} + \frac{1}{{{\rm{LC}}}}\)

We have \({{\rm{\omega }}_{\rm{n}}} = \frac{1}{{\sqrt {{\rm{LC}}} }}\)

And \(2{\rm{\zeta }}{{\rm{\omega }}_{\rm{n}}} = \frac{{\rm{R}}}{{\rm{L}}}\)

\(\begin{array}{l} \Rightarrow {{\rm{\omega }}_{\rm{n}}} = \frac{{\rm{R}}}{{\rm{L}}} \times \frac{1}{2} \times \sqrt {{\rm{LC}}} \ \Rightarrow {{\rm{\omega }}_{\rm{n}}} = \frac{{\rm{R}}}{2}\sqrt {\frac{{\rm{C}}}{{\rm{L}}}} \end{array}\)

Concept:

At steady-state conditions, the capacitor acts as an open circuit, and the inductor acts as a short circuit.

Calculation:

\({{\rm{V}}_{\rm{c}}}\left( 0 \right) = 0{\rm{V}}\)

At t = ∞, C is open circuit

\(\begin{array}{l} {{\rm{V}}_{\rm{c}}}\left( \infty \right) = 4{\rm{V}}\ {\rm{\tau }} = {{\rm{R}}_{{\rm{th}}}}{\rm{C}} = \left( {3||2} \right) \times \frac{5}{6} = \frac{6}{5} \times \frac{5}{6} = 1{\rm{sec}}\ {{\rm{V}}_{\rm{c}}}\left( {\rm{t}} \right) = {{\rm{V}}_{\rm{c}}}\left( \infty \right) + \left[ {{{\rm{V}}_{\rm{c}}}\left( {{0^ - }} \right) - {{\rm{V}}_{\rm{c}}}\left( \infty \right)} \right]{{\rm{e}}^{ - \frac{{\rm{t}}}{{\rm{\tau }}}}}\ = 4 - 4{{\rm{e}}^{ - {\rm{t}}}}\ {{\rm{V}}_{\rm{c}}}\left( 1 \right) = 4 - 4{{\rm{e}}^{ - 1}} = 2.528{\rm{V}} \end{array}\)

Concept:

Maximum Power Transfer Theorem:

The maximum power transfer theorem states that the maximum power transfer from the AC network to the load when the load impedance is equal to the complex conjugate of Thevenin’s equivalent impedance.

ZL = RL + jXL = Rth – jXth = Zth*

If the load is purely resistive, the load resistance is equal to the magnitude of Thevenin impedance.

\({R_L} = \left| {{Z_{th}}} \right| = \sqrt {R_{th}^2 + X_{th}^2} \)

Calculation:

We have \(\rm V_{Th(RMS)}=4\times \frac{j2}{2+j2}=2\sqrt{2}\angle45^\circ\)

And, \(\rm R_{Th}=2||j2=\frac{2\times j2}{2+j2}=\sqrt{2}\angle45^\circ\)

Now, \(\rm R_L=|R_{Th}|=\sqrt{2}\ \Omega\)

Thus, maximum power transferred, 

\(\rm P_{max}=\left|\frac{2\sqrt{2}\angle45^\circ}{\sqrt{2}+1+j}\right|^2\times \sqrt{2}=1.66\ W\)

Concept:

 junction capacitance

\({{\rm{C}}_{\rm{j}}} \propto \frac{1}{{\sqrt {1 - \frac{{{{\rm{V}}_{\rm{a}}}}}{{{{\rm{V}}_{{\rm{bi}}}}}}} }}\)

Where V_a is applied bias, Thus,

\(\frac{{{{\rm{C}}_{{\rm{j}}1}}}}{{{{\rm{C}}_{{\rm{j}}2}}}} = \frac{{\sqrt {1 - \frac{{{{\rm{V}}_{{\rm{a}}2}}}}{{{{\rm{V}}_{{\rm{bi}}}}}}} }}{{\sqrt {1 - \frac{{{{\rm{V}}_{{\rm{a}}1}}}}{{{{\rm{V}}_{{\rm{bi}}}}}}} }}\)

Application:

We have \({{\rm{V}}_{{\rm{bi}}}} = 0.75{\rm{V}},{\rm{;}}{{\rm{C}}_{{\rm{j}}1}} = 5{\rm{pF;and;}}{{\rm{V}}_{{\rm{a}}1}} = - 1.25{\rm{V}}\)

Thus,

\(\begin{array}{l} \frac{5}{{{{\rm{C}}_{{\rm{j}}2}}}} = \frac{{\sqrt {1 - \frac{{\left( { - 7.25} \right)}}{{0.75}}} }}{{\sqrt {1 - \frac{{\left( { - 1.25} \right)}}{{0.75}}} }}\ {{\rm{C}}_{{\rm{j}}2}} = 5 \times \frac{{\sqrt {1 + \frac{5}{3}} }}{{\sqrt {1 + \frac{{29}}{3}} }}\ \Rightarrow {{\rm{C}}_{{\rm{j}}2}} = 5 \times \frac{{1.633}}{{3.266}} = 2.5{\rm{pF}} \end{array}\)

Concept:

The drain current when the MOSFET is in saturation is given by:

\({I_D} = \frac{1}{2}{\mu _n}{C_{ox}}\left( {\frac{W}{L}} \right) \times {\left( {{V_{GS}} - {V_T}} \right)^2}\)

VT  = Threshold

The transconductance (gm) is defined as the change in the output current with a change in the Gate to source voltage, i.e.

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

\({g_m} = {\rm{;}}{\mu _n}{C_{ox}}\left( {\frac{W}{L}} \right);\left( {{V_{GS}} - {V_T}} \right)\)

The output resistance (r0) is given by:

\({r_o} = \frac{{{V_A}}}{{{I_D}}}\)

\({V_A} = \frac{1}{\lambda }\)

λ = Channel length parameter

Calculation:

Given: ID = 1 mA

Channel length modulation coefficient = 0.05 V-1

\({V_A} = \frac{1}{\lambda } = 20\)               

\({r_o} = \frac{{{V_A}}}{{{I_D}}} = \frac{{20}}{{1;mA}};=20;k\Omega\)

Hence the value output impedance is 20 kohm.

Concept:

The reverse saturation hole current is given by,

\({{\rm{J}}_{{\rm{po}}}} = {\rm{q\eta }}_{\rm{i}}^2 \cdot \left( {\frac{{{{\rm{D}}_{\rm{p}}}}}{{{{\rm{L}}_{\rm{p}}}{{\rm{N}}_{\rm{D}}}}}} \right)\). 

Application:

We have \({{\rm{L}}_{\rm{p}}} = \sqrt {{{\rm{D}}_{\rm{p}}}{{\rm{\tau }}_{\rm{p}}} = \sqrt {36} \times 100 \times {{10}^{ - 6}}}\)

\(\begin{array}{l} \Rightarrow {{\rm{L}}_{\rm{p}}} = 60 \times {10^{ - 3}}{\rm{cm}}\ {{\rm{J}}_{{\rm{po}}}} = 1.6 \times {10^{ - 19}} \times {\left( {{{10}^{10}}} \right)^2} \times \left[ {\frac{{36}}{{60 \times {{10}^{ - 3}} \times {{10}^{15}}}}} \right]\ \Rightarrow {{\rm{J}}_{{\rm{po}}}} = 9.6\frac{{{\rm{pA}}}}{{{\rm{c}}{{\rm{m}}^2}}} \end{array}\)

Now, current density at V = 208 mV

\(\begin{array}{l} {{\rm{J}}_{\rm{p}}} = {{\rm{J}}_{{\rm{po}}}}\left( {{{\rm{e}}^{\frac{{\rm{V}}}{{{{\rm{V}}_{\rm{T}}}}}}} - 1} \right)\ = 9.6 \times {10^{ - 12}}\left( {{{\rm{e}}^{\left( {\frac{{208}}{{26}}} \right)}} - 1} \right)\ \Rightarrow {{\rm{J}}_{\rm{p}}} = 28.61\frac{{{\rm{nA}}}}{{{\rm{c}}{{\rm{m}}^2}}} \end{array}\)

Concept: 

Canonical form:  Any Boolean function that is expressed as a sum of minterms or as a product of max terms is said to be in its canonical form.

There are two types of canonical forms:

SOP:

Sum of products or sum of minterms

In the SOP (sum of product) form, a minterm is represented by 1.

Example of SOP: XY + X’Y’

POS:

Product of sums or product of max terms

In POS (product of sum) form, a maxterm is represented by 0.

Example of POS: (X+Y) (X’+Y’)

Calculation:

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

The above expression can be represented in k-map as follows

\(\begin{array}{l} {\rm{F}}\left( {{\rm{x}},{\rm{y}},{\rm{z}}} \right) = {\rm{;}}\sum {\rm{m;}}\left( {2,4,6,7} \right) = \prod {\rm{M;}}\left( {0,1,3,5} \right)\ = {\rm{;}}\left( {{\rm{x;}} + {\rm{;y;}} + {\rm{;z}}} \right){\rm{;}}\left( {{\rm{x;}} + {\rm{;y;}} + {\rm{;\bar z}}} \right)\left( {{\rm{x;}} + {\rm{;\bar y}} + {\rm{;\bar z}}} \right)\left( {{\rm{\bar x}} + {\rm{;y;}} + {\rm{;\bar z}}} \right) \end{array}\)

When, \({\rm{A}} = 0,{\rm{;B}} = 1,{\rm{X}} = 0,{\rm{Y}} = 0,{\rm{;C}} = 0 \Rightarrow {\rm{Z}} = 0\)

At, \({\rm{t}} = {0^ + },{\rm{;A}} = 1,{\rm{;B}} = 0,{\rm{X}} = 0,{\rm{;Y}} = 0,{\rm{;C}} = 1 \Rightarrow {\rm{Z}} = 0\)

At, \({\rm{t}} = {20^ + }{\rm{;ns}},{\rm{A}} = 1,{\rm{;B}} = 0,{\rm{X}} = 1,{\rm{;Y}} = 0,{\rm{;C}} = 1 \Rightarrow {\rm{Z}} = 1\)

At, \({\rm{t}} = {40^ + }{\rm{;ns}},{\rm{A}} = 1,{\rm{;B}} = 0,{\rm{X}} = 1,{\rm{;Y}} = 1,{\rm{;C}} = 1 \Rightarrow {\rm{Z}} = 1\)

At, \({\rm{t}} = {60^ + }{\rm{;ns}},{\rm{A}} = 1,{\rm{;B}} = 0,{\rm{X}} = 1,{\rm{;Y}} = 1,{\rm{;C}} = 1 \Rightarrow {\rm{Z}} = 0\)

Thus, \({\rm{Z}} = 1\) exists from \({\rm{t}} = {20^ + }\) to \({\rm{t}} = 60\) nsec.

\(\Rightarrow {\rm{Z}} = 1\) for 40 nsec.

\({\rm{M}}\left( {{\rm{a}},{\rm{b}},{\rm{c}}} \right){\rm{;}} = {\rm{;ab;}} + {\rm{;bc;}} + {\rm{;ac}} = {\rm{;\Sigma m;}}\left( {3,5,6,7} \right)\)

\(\begin{array}{l} \overline {{\rm{M}}\left( {{\rm{a}},{\rm{b}},{\rm{c}}} \right)} = \overline {{\rm{\Sigma m}}\left( {3,5,6,7} \right)} = {\rm{\Sigma m}}\left( {0,1,2,4} \right)\ {\rm{M}}\left( {{\rm{a}},{\rm{b}},{\rm{\bar c}}} \right) = {\rm{ab}} + {\rm{b\bar c}} + {\rm{a\bar c}} = {\rm{\Sigma m}}\left( {2,4,6,7} \right) \end{array}\)

K-map of \({\rm{M}}\left( {{\rm{a}},{\rm{b}},{\rm{\bar c}}} \right)\)

\({\rm{c;}} = {\rm{;\Sigma ;m}}\left( {1,3,5,7} \right)\)

K-map of \({\rm{f}}\left( {{\rm{a}},{\rm{b}},{\rm{c}}} \right) = {\rm{c}}\)

Now, since \({\rm{;M}}\left( {{\rm{a}},{\rm{b}},{\rm{c}}} \right){\rm{;}} = {\rm{;ab;}} + {\rm{;bc;}} + {\rm{;ac}}\)

∴ Required function,

\(\begin{array}{l} \Rightarrow {\rm{M}}\left( {\overline {{\rm{M}}\left( {{\rm{a}},{\rm{b}},{\rm{c}}} \right)} ,{\rm{M}}\left( {{\rm{a}},{\rm{b}},{\rm{\bar c}}} \right),{\rm{c}}} \right) = \overline {{\rm{M}}\left( {{\rm{a}},{\rm{b}},{\rm{c}}} \right)} .{\rm{M}}\left( {{\rm{a}},{\rm{b}},{\rm{\bar c}}} \right) + {\rm{M}}\left( {{\rm{a}},{\rm{b}},{\rm{\bar c}}} \right).{\rm{c}} + \overline {{\rm{M}}\left( {{\rm{a}},{\rm{b}},{\rm{c}}} \right)} .{\rm{c}}\ = {\rm{\Sigma m;}}\left( {2,{\rm{;}}4,{\rm{;}}6,{\rm{;}}7} \right).{\rm{\Sigma m;}}\left( {0,1,2,4} \right){\rm{;}} + {\rm{;\Sigma m}}\left( {2,4,6,7} \right){\rm{;}}.{\rm{;\Sigma m}}\left( {1,3,5,7} \right){\rm{;}} + {\rm{;\Sigma m}}\left( {0,1,2,4} \right){\rm{;}}.{\rm{;\Sigma m}}\left( {1,3,5,7} \right){\rm{;;;}}\ = {\rm{;\Sigma ;m}}\left( {2,4} \right){\rm{;}} + {\rm{;\Sigma m}}\left( 7 \right){\rm{;}} + {\rm{;\Sigma m;}}\left( 1 \right)\ = {\rm{;\Sigma m;}}\left( {1,2,4,7} \right){\rm{;}} = {\rm{;a;}} \odot {\rm{;b;}} \odot {\rm{;c;}} = {\rm{;a;}} \oplus {\rm{;b;}} \oplus {\rm{;C;;;;}} \end{array}\)

Note:

For odd number of inputs, EX-OR output = EX – NOR output and for even number of inputs, \({\rm{EX}} - {\rm{OR;output}} = \overline {{\rm{EX}} - {\rm{NOR}}} {\rm{;output}}\).

Since, gate is shorted to drain.

Thus, the MOSFET is in saturation.

Current is saturation is given by

\({{\rm{I}}_{\rm{D}}} = \frac{1}{2}{{\rm{\mu }}_{\rm{n}}}{{\rm{C}}_{{\rm{ox}}}}\frac{{\rm{W}}}{{\rm{L}}}{\left( {{{\rm{V}}_{{\rm{GS}}}} - {{\rm{V}}_{{\rm{TH}}}}} \right)^2}\)

We have \({{\rm{V}}_{{\rm{GS}}}} = {{\rm{V}}_{{\rm{DD}}}} - {{\rm{V}}_{{\rm{SS}}}}\)

And let \(\frac{1}{2}{{\rm{\mu }}_{\rm{n}}}{{\rm{C}}_{{\rm{ox}}}}\frac{{\rm{W}}}{{\rm{L}}} = {{\rm{K}}_{\rm{n}}},\) then

\({{\rm{I}}_{\rm{D}}} = {{\rm{K}}_{\rm{n}}}{\left( {{{\rm{V}}_{{\rm{DD}}}} - {{\rm{V}}_{{\rm{TH}}}} - {{\rm{V}}_{{\rm{SS}}}}} \right)^2}\)

Let \({{\rm{V}}_{{\rm{DD}}}} - {{\rm{V}}_{{\rm{TH}}}} = {\rm{A}} = {\rm{constant}}\)

\(\Rightarrow {{\rm{I}}_{\rm{D}}} = {{\rm{K}}_{\rm{n}}}{\left( {{\rm{A}} - {{\rm{V}}_{{\rm{SS}}}}} \right)^2}\)

This is equation of parabola

For \({{\rm{V}}_{{\rm{SS}}}} = 0,{{\rm{I}}_{\rm{D}}} = {{\rm{K}}_{{\rm{nA}}}}({\rm{maximum}})\)

And \({{\rm{V}}_{{\rm{SS}}}} = {{\rm{V}}_{{\rm{DD}}}} - {{\rm{V}}_{{\rm{TH}}}},{{\rm{I}}_{\rm{D}}} = 0\)

So, we have, the plot given by,

Current through 50Ω resistor \({{\rm{I}}_{50{\rm{\Omega }}}} = \frac{1}{{50}} = 20{\rm{mA}} = {{\rm{I}}_{100{\rm{\Omega }}}}\)

∴ Current through the branches of bridge \(= \frac{{20{\rm{mA}}}}{2} = 10{\rm{mA}}\)

\(\begin{array}{l} \therefore{{\rm{V}}_0} = 250\left( {1 + {\rm{\delta }}} \right){\rm{.}}\frac{{10}}{{100}} - 250\left( {1 - {\rm{\delta }}} \right){\rm{.}}\frac{{10}}{{1000}}\ = 2.5\left[ {1 + {\rm{\delta }} - 1 + {\rm{\delta }}} \right] = 2.5 \times 2 \times 0.05 = 0.25\ = 250{\rm{mV}} \end{array}\)

The feedback circuit can be drawn as:

∴ \(\frac{{{{\rm{V}}_{\rm{f}}}}}{{{{\rm{V}}_{{\rm{out}}}}}} = \frac{{{\rm{Z}}_1}}{{{{\rm{Z}}_1} + {{\rm{Z}}_2}}}\)

Where:

\({{\rm{Z}}_1} = \frac{{\frac{1}{{\rm{R}}} \times \frac{1}{{2{\rm{sC}}}}}}{{\frac{1}{{\rm{R}}} + \frac{1}{{2{\rm{sC}}}}}} = \frac{1}{{{\rm{R}} + 2{\rm{sC}}}}\)

\({{\rm{Z}}_2} = 2{\rm{R}} + \frac{1}{{{\rm{sC}}}} = \frac{{2{\rm{sCR}} + 1}}{{{\rm{sC}}}}\)

It is a Wein bridge oscillator with a frequency of:

\({\rm{\omega }} = \frac{1}{{{{\rm{R}}_1}{{\rm{C}}_1}}} = \frac{1}{{2{\rm{RC}}}}\)

\(\therefore{{\rm{Z}}_1} = \frac{1}{{{\rm{R}} + 2{\rm{j\omega c}}}} \)

\(Z_1= \frac{1}{{{\rm{R}} + \frac{{\rm{j}}}{{\rm{R}}}}} = \frac{{\rm{R}}}{{{{\rm{R}}^2} - {\rm{j}}}}\)

\(\ {{\rm{Z}}_2} = \frac{{2{\rm{sCR}} + 1}}{{{\rm{sC}}}} = 2\left( {{\rm{R}} - {\rm{jR}}} \right)\)

From Bark Hausen Criterion Aβ = 1, we get:

\(\therefore{\rm{\beta }} = \frac{{{{\rm{Z}}_2}}}{{{{\rm{Z}}_1} + {{\rm{Z}}_2}}} = \frac{1}{5}\)   

\(1 + \frac{{{{\rm{R}}_1}}}{{{{\rm{R}}_2}}} = 5 \Rightarrow {{\rm{R}}_1} = 4{{\rm{R}}_2}\)

Apply KVL we get,

 \(-{{\rm{V}}_{{\rm{B}}{{\rm{E}}_1}}} + {{\rm{V}}_{12}} + {{\rm{V}}_{{\rm{B}}{{\rm{E}}_2}}} = 0\)

\(\therefore{{\rm{V}}_{12}} = {{\rm{V}}_{{\rm{B}}{{\rm{E}}_1}}}-{{\rm{V}}_{{\rm{B}}{{\rm{E}}_2}}}\)

But \({{\rm{I}}_1} = {\rm{e}}\frac{{{{\rm{V}}_{{\rm{BE}}1}}}}{{{\rm{VT}}}}{\rm{and}}{{\rm{I}}_2} = {\rm{e}}\frac{{{{\rm{V}}_{{\rm{BE}}2}}}}{{{\rm{VT}}}}\)

\(\begin{array}{l} \frac{{{{\rm{I}}_1}}}{{{{\rm{I}}_2}}} = {\rm{e}}\left( {\frac{{{{\rm{V}}_{{\rm{BE}}1}} - {{\rm{V}}_{{\rm{BE}}2}}}}{{{\rm{VT}}}}} \right) = 20\ \therefore{{\rm{V}}_{{\rm{BE}}1}} - {{\rm{V}}_{{\rm{BE}}2}} = {{\rm{V}}_{\rm{T}}}\ln 20 = \frac{{\left( {323{\rm{^\circ k}}} \right)}}{{11600}}\ln 20\ \therefore{{\rm{V}}_{{\rm{BE}}1}} - {{\rm{V}}_{{\rm{BE}}2}} = {{\rm{V}}_{\rm{T}}}\ln 20 = 83.42{\rm{mV}} \end{array}\)

\(\begin{array}{l} \left[ {\begin{array}{*{20}{c}} {\rm{p}}\ {\rm{q}}\ {\rm{r}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&1&1\ 1&{{\rm{w}}_3^1}&{{\rm{w}}_3^2}\ 1&{{\rm{w}}_3^2}&{{\rm{w}}_3^1} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&0&0\ 0&{{\rm{w}}_3^2}&0\ 0&0&{{\rm{w}}_3^4} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\frac{{\rm{A}}}{3}}\ {\frac{{\rm{B}}}{3}}\ {\frac{{\rm{C}}}{3}} \end{array}} \right]\ \left[ {\begin{array}{*{20}{c}} {\rm{p}}\ {\rm{q}}\ {\rm{r}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&{{\rm{w}}_3^2}&{{\rm{w}}_3^4}\ 1&{{\rm{w}}_3^3}&{{\rm{w}}_3^6}\ 1&{{\rm{w}}_3^4}&{{\rm{w}}_3^5} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\frac{{\rm{A}}}{3}}\ {\frac{{\rm{B}}}{3}}\ {\frac{{\rm{C}}}{3}} \end{array}} \right] \end{array}\)

We know that \({{\rm{w}}_3} = {{\rm{e}}^{{\rm{j}}\frac{{2{\rm{\pi }}}}{3}}}\) which is cube root of unity.so \(1 + {\rm{w}} + {{\rm{w}}^2} = 0{\rm{;and;}}{{\rm{w}}^3} = 1\)

\(\begin{array}{l} \therefore \left[ {\begin{array}{*{20}{c}} {\rm{p}}\ {\rm{q}}\ {\rm{r}} \end{array}} \right] = \frac{1}{3}\left[ {\begin{array}{*{20}{c}} 1&{{\rm{w}}_3^2}&{{\rm{w}}_3^1}\ 1&1&1\ 1&{{\rm{w}}_3^1}&{{\rm{w}}_3^2} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\rm{A}}\ {\rm{B}}\ {\rm{C}} \end{array}} \right]\ \therefore \left[ {\begin{array}{*{20}{c}} {\rm{p}}\ {\rm{q}}\ {\rm{r}} \end{array}} \right] = \frac{1}{3}\left[ {\begin{array}{*{20}{c}} 1&{{\rm{w}}_3^2}&{{\rm{w}}_3^1}\ 1&1&1\ 1&{{\rm{w}}_3^1}&{{\rm{w}}_3^2} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&1&1\ 1&{{\rm{w}}_3^{ - 1}}&{{\rm{w}}_3^{ - 2}}\ 1&{{\rm{w}}_3^{ - 2}}&{{\rm{w}}_3^{ - 1}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\rm{a}}\ {\rm{b}}\ {\rm{c}} \end{array}} \right]\ \therefore \left[ {\begin{array}{*{20}{c}} {\rm{p}}\ {\rm{q}}\ {\rm{r}} \end{array}} \right] = \frac{1}{3}\left[ {\begin{array}{*{20}{c}} 0&0&3\ 3&0&0\ 0&3&0 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\rm{a}}\ {\rm{b}}\ {\rm{c}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\rm{c}}\ {\rm{a}}\ {\rm{b}} \end{array}} \right] \end{array}\)

Concept:

Poles and Zeros of a transfer function are the frequencies for which the value of the denominator and numerator of transfer function becomes zero respectively.

The values of the poles and the zeros of a system determine whether the system is stable, and how well the system performs.

Calculation:

\(\begin{array}{l} \frac{{{\rm{Y}}\left( {\rm{z}} \right)}}{{{\rm{X}}\left( {\rm{z}} \right)}} = {\rm{H}}\left( {\rm{z}} \right) = \frac{1}{{1 - \frac{5}{6}{{\rm{z}}^{ - 1}} + \frac{{{{\rm{z}}^{ - 2}}}}{6}}} = \frac{{{{\rm{z}}^2}}}{{{{\rm{z}}^2} - \frac{5}{6}{\rm{z}} + \frac{1}{6}}}\ = \frac{{{{\rm{z}}^2}}}{{\left( {{\rm{z}} - \frac{1}{2}} \right)\left( {{\rm{z}} - \frac{1}{3}} \right)}} \end{array}\)

So, poles are \({\rm{z}} = \frac{1}{2},{\rm{z}} = \frac{1}{3}\)

Concept:

Causality and Stability:

For a system with rational transfer function H(z) to be causal, the ROC should lie outside the outermost pole, and for BIBO stability, the ROC should include the unit circle |z| = 1. 

∴ An LTI discrete-time causal system with the rational system function H(z) is said to be stable if all the poles of H(z) lie inside the unit circle.

Calculation:

The transfer function can be written as:

\({\rm{H}}\left( {\rm{z}} \right) = \frac{{{\rm{k}}{{\rm{z}}^4}}}{{\left( {{\rm{z}} - 0.5 - {\rm{j}}0.5} \right)\left( {{\rm{z}} - 0.5 + {\rm{j}}0.5} \right)\left( {{\rm{z}} + 0.5 + {\rm{j}}0.5} \right)\left( {{\rm{z}} + 0.5 - {\rm{j}}0.5} \right)}}\)

\({\rm{H}}\left( {\rm{z}} \right) = \frac{{{\rm{k}}{{\rm{z}}^4}}}{{\left( {{{\rm{z}}^2} - {\rm{z}} + \frac{1}{2}} \right)\left( {{{\rm{z}}^2} + {\rm{z}} + \frac{1}{2}} \right)}}\)

Now, since the system is causal and stable we have,

\({\rm{h}}\left[ 0 \right] = \mathop {\lim }\limits_{{\rm{z}} \to \infty } {\rm{H}}\left( {\rm{z}} \right) = {\rm{k}} = 1{\rm{;}}\)

Thus, the transfer function will be:

\({\rm{H}}\left( {\rm{z}} \right) = \frac{{{{\rm{z}}^4}}}{{\left( {{{\rm{z}}^2} - {\rm{z}} + \frac{1}{2}} \right)\left( {{{\rm{z}}^2} + {\rm{z}} + \frac{1}{2}} \right)}}\)

H(z) = 1 - 0.25z^-4.......

h[n] = [1,0,0,0,-0.25,.....]

Now,

h[0] = 1

so,h[n] is real for all n

Note:

Causality:

A linear time-invariant discrete-time system is said to be causal if the impulse response h[n] = 0, for n < 0 and it is therefore right-sided.

The ROC of such a system H(z) is the exterior of a circle. If H(z) is rational, then the system is said to be causal if:

1. The ROC is the exterior of a circle outside the outermost pole; and

2. The degree of the numerator polynomial of H(z) should be less than or equal to the degree of the denominator polynomial.

Stability:

A discrete-time LTI system is said to be BIBO stable if the impulse response h[n] is summable, i.e.

\(\mathop \sum \nolimits_{n = - \infty }^\infty \left| {h\left[ n \right]} \right| < \infty \)

z-transform of h[h] is given as:

\(H\left( z \right) = \mathop \sum \nolimits_{n = - \infty }^\infty h\left[ h \right]{z^{ - n}}\)

Let z = ejΩ (which describes a unit circle in the z-plane), then

\(\left| {H\left[ {{e^{j{\rm{\Omega }}}}} \right]} \right| = \left| {\mathop \sum \nolimits_{n = - \infty }^\infty h\left[ n \right]{e^{ - j{\rm{\Omega }}n}}} \right|\)

The above equality can be written as:

\( \le ;\mathop \sum \nolimits_{n = - \infty }^\infty \left| {h\left[ n \right]{e^{ - j{\rm{\Omega }}n}}} \right|\)

\(= \mathop \sum \nolimits_{n = - \infty }^\infty \left| {h\left[ n \right]} \right| < \infty \)

This is the condition for stability. Thus we can conclude that an LTI system is stable if the ROC of the system function H(z) contains the unit circle |z| = 1

Form magnitude condition:

\(\left| {{\rm{G}}\left( {\rm{s}} \right){\rm{H}}\left( {\rm{s}} \right)} \right| = 1\)

Given,

As the plant has unity feedback, therefore H(s) = 1

\({\rm{G}}\left( {\rm{s}} \right) {\rm{H}}\left( {\rm{s}} \right)= \frac{{{\rm{K}}\left( {{\rm{s}} + 4} \right)}}{{\left( {{\rm{s}} + 8} \right)\left( {{{\rm{s}}^2} - 9} \right)}}\) 

The gain of the system will be

\(\left| {{\rm{G}}\left( {\rm{s}} \right){\rm{H}}\left( {\rm{s}} \right)} \right| = 1\)

\(\left| {\frac{{{\rm{K}}\left( { {\rm{s}} + 4} \right)}}{{\left( { {\rm{s}} + 8} \right)\left( {{{\left ({ {\rm{s}}} \right)}^2} - 9} \right)}}} \right| = 1\)

\(\therefore \left| {\frac{{{\rm{K}}\left( { - 1 + 2{\rm{j}} + 4} \right)}}{{\left( { - 1 + 2{\rm{j}} + 8} \right)\left( {{{\left( { - 1 + {\rm{j}}2} \right)}^2} - 9} \right)}}} \right| = 1\)

\(\Rightarrow\left| {\frac{{{\rm{K}}\left( {3 + 2{\rm{j}}} \right)}}{{\left( {7 + 2{\rm{j}}} \right)\left( { - 12 - 4{\rm{j}}} \right)}}} \right| = 1\)

\(\Rightarrow\frac{{{\rm{K}}\sqrt {9 + 4} }}{{\sqrt {49 + 4} \sqrt {144 + 16} }} = 1\)

\( {\rm{K}} = \frac{{\sqrt {53} .\sqrt {160} }}{{\sqrt {13} }} = 25.54 \)

Concept:

Lead compensator:

1. When sinusoidal input applied to this it produces sinusoidal output with the phase lead input.

2. It speeds up the Transient response and increases the margin for stability.

A circuit diagram is as shown:

Response is:

\(\frac{{{V_0}\left( s \right)}}{{{V_i}\left( s \right)}} = \frac{{{R_2}\left( {1 + s{C_1}{R_1}} \right)}}{{{R_1} + {R_2} + s{C_1}{R_1}{R_2}}}\)

\(\frac{{{V_0}\left( s \right)}}{{{V_i}\left( s \right)}} = \frac{{1 + sτ }}{{1 + \alpha sτ }}\)

Lead constant \(\alpha = \frac{{{R_2}}}{{{R_1} + {R_2}}}\) < 1

Lag compensator:

1. If the steady-state output has phase lag then the network is called lag compensator.

2. It improves steady-state behavior without affecting the transient response.

A circuit diagram is shown

Response is

\(\frac{{{V_0}\left( s \right)}}{{{V_i}\left( s \right)}} = \frac{{\frac{1}{{s{C_2}}} + {R_2}}}{{{R_2} + {R_1} + \frac{1}{{s{C_2}}}}}\)

\(\frac{{{V_0}\left( s \right)}}{{{V_i}\left( s \right)}} = \frac{{1 + sτ }}{{1 + \alpha sτ }}\)  where α > 1

Lag constant  \(\alpha = \frac{{{R_1} + {R_2}}}{{{R_2}}}\) 

Calculation:

Given the transfer function of the compensator 

\({{\rm{G}}_{\rm{e}}}\left( {\rm{s}} \right) = \frac{{{\rm{s}} + 2}}{{{\rm{s}} + 4}}{\rm{;}}\)

The compensator can be rearranged as follows

\({{\rm{G}}_{\rm{e}}}\left( {\rm{s}} \right) = \frac{{2\left( {1 + \frac{{\rm{s}}}{2}} \right)}}{{4\left( {1 + \frac{{\rm{s}}}{4}} \right)}} = \frac{1}{2}\left( {\frac{{1 + \frac{{\rm{s}}}{2}}}{{1 + \frac{{\rm{s}}}{4}}}} \right)\)

Compare this with the standard equation, \({\rm{K}}\left( {\frac{{1 + {\rm{sT}}}}{{1 + {\rm{\alpha sT}}}}} \right)\). 

\(\therefore {\rm{\alpha T}} = \frac{1}{4},{\rm{T}} = \frac{1}{2}{\rm{;and;\alpha }} = \frac{1}{2}\)

It is a lead compensator and time constant RC = τ = 1/2 = 0.5

Concept:

According to RH stability criteria there are two necessary conditions:

First is there should not be any sign changes in the first column of the tabulation and Second Is there should not be any missing coefficients in the equation 1 + G(s) H(s).

The RH criteria can be solved as follows:

\({s^3} - - - {\rm{;;;}}{{\rm{a}}_0} \searrow {\rm{;;;;;}} \swarrow {{\rm{a}}_2}{\rm{;;;;;;}}{{\rm{a}}_4}\)

\({s^2} - - - {\rm{;;;}}{{\rm{a}}_1}{\rm{;}} \nearrow {\rm{;;;}} \nwarrow {{\rm{a}}_3}{\rm{;;;;;;;}}0\)

\({s^1} - - - {\rm{;;;}}\frac{{{{\rm{a}}_1}{{\rm{a}}_2} - {{\rm{a}}_0}{{\rm{a}}_3}}}{{{{\rm{a}}_1}}}{\rm{;}} = {\rm{A;;;;;;;;;;}}0{\rm{;}}\)

\({s^0} - - - {\rm{;;;}}\frac{{{\rm{A}}{{\rm{a}}_3}}}{{\rm{A}}} = {{\rm{a}}_3}\)

Calculation:

Given the open-loop transfer function

\({\rm{G}}\left( {\rm{s}} \right) = \frac{{\left( {{\rm{s}}{{\rm{K}}_{\rm{P}}} + {{\rm{K}}_{\rm{I}}}} \right)}}{{{{\rm{s}}^2}{\rm{;}}\left( {{\rm{s}} + 2} \right)}}\)

The characteristic equation of the above transfer function will be

⇒ q(s) = s³ + 2s² + sK_P + K_I = 0

From RH-criteria

\(\begin{array}{*{20}{c}} {{s^3}}\ {{s^2}}\ {{s^1}}\ {{s^0}} \end{array}\left| {\begin{array}{*{20}{c}} 1&2\ K_P&{K_I}\ {\frac{2K_P-K_I}{K_P}}&0\ K_I&{} \end{array}} \right.\)

From the Routh criterion, the condition for stability is

⇒ 2K_P > K_I and K_I > 0

∴ \({{\rm{K}}_{\rm{P}}} > \frac{{{{\rm{K}}_{\rm{I}}}}}{2} > 0\)

Concept:

For a binary symmetric channel with crossover probability ‘p’. The channel capacity is given by:

Cs = 1 + (1 - p) log2 (1 - p) + p log2p

Calculation:

We have the Binary symmetric channel as shown below.

Here,

P [X = 0] = 0.2

P [X = 1] =0.8

According to Bayes’ theorem,

\(\begin{array}{l} {\rm{P}}\left( {\frac{{{\rm{X}} = 1}}{{{\rm{Y}} = 0}}} \right) = \frac{{{\rm{P}}\left( {\frac{{{\rm{Y}} = 0}}{{{\rm{X}} = 1}}} \right){\rm{P}}\left( {{\rm{X}} = 1} \right)}}{{{\rm{P}}\left( {\frac{{{\rm{Y}} = 0}}{{{\rm{X}} = 1}}} \right){\rm{P}}\left( {{\rm{X}} = 1} \right) + {\rm{P}}\left( {\frac{{{\rm{Y}} = 0}}{{{\rm{X}} = 0}}} \right){\rm{P}}\left( {{\rm{X}} = 0} \right)}}\ \end{array}\)

\( \frac{{\left( {\frac{1}{7}} \right)\left( {.8} \right)}}{{\left( {\frac{1}{7}} \right)\left( {.8} \right) + \left( {\frac{6}{7}} \right)\left( {.2} \right)}} = 0.4\)

Hence answer is 0.4

Concept: 

In the TDM system users use the channel on a time slot basis. If the number of users increases more than allowed then more such channels are required.

Application:

Given the bandwidth of a channel in the GSM system:

BW_channel = 200 kHz 

8 users share the channel of Bandwidth 200 kHz.

For 12 users talking in a cell, two such channels will be used.

So the Bandwidth will be:

=2 × 200 = 400 kHz

Concept: If a signal x(t) is multiplied by a sinusoidal signal then the Fourier transform of x(t) gets shifted by the frequency of the sinusoid.

Application: We have the input signal \({\rm{x}}\left( {\rm{t}} \right) = {\rm{m}}\left( {\rm{t}} \right)\cos \left( {2400{\rm{\pi t}}} \right){\rm{}}\) 

\(= {\rm{m}}\left( {\rm{t}} \right)\cos \left( {{\rm{\omega t}}} \right)\)

And \({\rm{\omega }} = 2400{\rm{;\pi rad}} \to {\rm{f}} = 1200{\rm{Hz}}\)

So, \({\rm{y}}\left( {\rm{t}} \right) = 10{\rm{x}}\left( {\rm{t}} \right) + {{\rm{x}}^2}\left( {\rm{t}} \right)\)

\(\begin{array}{l} = 10{\rm{m}}\left( {\rm{t}} \right)\cos \left( {2400{\rm{\pi t}}} \right) + {{\rm{m}}^2}\left( {\rm{t}} \right){\cos ^2}\left( {2400{\rm{\pi t}}} \right)\ = 10{\rm{m}}\left( {\rm{t}} \right)\cos \left( {{\rm{\omega t}}} \right) + {{\rm{m}}^2}\left( {\rm{t}} \right)\left[ {\frac{{\cos \left( {2{\rm{\omega t}}} \right) + 1{\rm{;}}}}{2}} \right] \end{array}\)

\(= \underbrace {\frac{{{{\rm{m}}^2}\left( {\rm{t}} \right)}}{2}}_{ + {\rm{we;frequency}}} + \underbrace {10{\rm{m}}\left( {\rm{t}} \right)\cos \left( {{\rm{\omega t}}} \right)}_{\left[ {{\rm{\omega }} - {\rm{W}},{\rm{\omega }} + {\rm{W}}} \right]} + \underbrace {\frac{{{{\rm{m}}^2}\left( {\rm{t}} \right)\cos \left( {2{\rm{\omega t}}} \right)}}{2}}_{\left[ {2{\rm{\omega }} - 2{\rm{W}},2{\rm{\omega }} + 2{\rm{W}}} \right]}\)

From the frequency plot, we conclude the following results.

Results 1        \({\rm{\omega }} - {\rm{W}} > 700\)

                        \(1200 - {\rm{W}} > 700\)

                        \({\rm{W}} < 500\)

Results 2        \({\rm{\omega }} + {\rm{W}} < 1700\)

                        \(1200 + {\rm{W}} < 1700\)

                         \({\rm{W}} < 500\)

Results 3        \({\rm{\omega }} - {\rm{W}} > 2{\rm{W}}\)

                        \(1200 > 3{\rm{W}}\)

                        \({\rm{W}} < 400\)

Results 4        \(2{\rm{\omega }} - 2{\rm{W}} > 1700\)

                        \(2400 - 1700 > 2{\rm{W}}\)

                        \(2{\rm{W}} < 700\)

                        \({\rm{W}} < 350\)

Thus, the above conclusions result in \({\rm{W}} < 350\)

Concept: An error is said to occur if,  the receiver decides 1 for a transmitted bit 0  or receiver decides 0 for a transmitted bit 1 

Application: Given the conditional density function of R as

\(\begin{array}{l} {{\rm{f}}_{\left( {{\rm{R}}/0} \right)}}\left( {\rm{x}} \right) = \left{ {\frac{1}{{\begin{array}{*{20}{c}} 4\ 0 \end{array}}}:\begin{array}{*{20}{c}} { - 3 \le {\rm{x}} \le 1}\ {{\rm{otherwise}}} \end{array}} \right.\ {{\rm{f}}_{\left( {{\rm{R}}/1} \right)}}\left( {\rm{x}} \right) = \left{ {\frac{1}{{\begin{array}{*{20}{c}} 6\ {0,} \end{array}}},} \right.\begin{array}{*{20}{c}} { - 1 \le {\rm{x}} \le 5}\ {{\rm{otherwise}}} \end{array} \end{array}\)

Decision error probability that receiver decides 0 for a transmitted bit 1 is

\({{\rm{f}}_{\left( {{\rm{R}}/1} \right)}}\left( {{\rm{x}} = 0} \right) = \frac{1}{6}\)

Again, the decision error probability that receiver decides 1 for a transmitted bit 0 is

\({{\rm{f}}_{\left( {{\rm{R}}/0} \right)}}\left( {{\rm{x}} = 1} \right) = \frac{1}{4}\)

Hence, the minimum decision error probability is \({{\rm{f}}_{{\rm{R}}/1}}\left( {{\rm{x}} = 0} \right) = \frac{1}{6}\)

Concept: The general form of the magnetic field in TE mode is given by  \({{\rm{H}}_{\rm{z}}}\left( {{\rm{x}},{\rm{y}},{\rm{z}},{\rm{t}}} \right) = {{\rm{H}}_0}\cos \left( {\frac{{{\rm{m\pi }}}}{{\rm{a}}}{\rm{x}}} \right)\cos \left( {\frac{{{\rm{n\pi }}}}{{\rm{b}}}{\rm{y}}} \right)\cos \left( {{\rm{\omega t}} - {\rm{\beta z}}} \right)\frac{{\rm{A}}}{{\rm{m}}}\)

Application: Since the magnetic field has a z-component the mode must be transverse electric.

Comparing, with the z-component of the magnetic field in TE mode

\({{\rm{H}}_{\rm{z}}}\left( {{\rm{x}},{\rm{y}},{\rm{z}},{\rm{t}}} \right) = {{\rm{H}}_0}\cos \left( {\frac{{{\rm{m\pi }}}}{{\rm{a}}}{\rm{x}}} \right)\cos \left( {\frac{{{\rm{n\pi }}}}{{\rm{b}}}{\rm{y}}} \right)\cos \left( {{\rm{\omega t}} - {\rm{\beta z}}} \right)\frac{{\rm{A}}}{{\rm{m}}}\)

We have

\(\begin{array}{l} \frac{{{\rm{m\pi }}}}{{\rm{a}}} = 25{\rm{\pi }}\ \Rightarrow {\rm{m}} = 25{\rm{a}} = 25 \times 0.08 = 2 \end{array}\)

and \(\frac{{{\rm{n\pi }}}}{{\rm{b}}} = 30.3{\rm{\pi }}\) 

\(\Rightarrow {\rm{n}} = 30.3{\rm{b}} = 30.3 \times 0.033 = 1\)

Thus, the mode of propagation is \({\rm{T}}{{\rm{E}}_{21}}\)

Concept: Total time-averaged power can be obtained from Power density by integrating it over the given surface.

Application:

We have an electric field given by:

\({\rm{E}}\left( {{\rm{x}},{\rm{t}}} \right) = 24{\rm{\pi }}\cos \left( {{\rm{\omega t}} - {{\rm{k}}_0}{\rm{x}}} \right){{\rm{\hat a}}_{\rm{y}}}\)

Average Poynting vector:

\({\rm{\vec P}} = \frac{1}{2}\frac{{{\rm{E}}_{\rm{o}}^2}}{{{{\rm{\eta }}_{\rm{o}}}}}{{\rm{\hat a}}_{\rm{x}}} \)

\(\vec P = \frac{1}{2}\frac{{{{\left( {24{\rm{\pi }}} \right)}^2}}}{{120{\rm{\pi }}}}{{\rm{\hat a}}_{\rm{x}}} = 7.53{{\rm{\hat a}}_{\rm{x}}}\)

We have unit normal to plane \({\rm{f}} = {\rm{x}} + {\rm{y}}-1\) given by

\({{\rm{\hat a}}_{\rm{u}}} = \frac{{\vec \nabla {\rm{f}}}}{{\left| {\vec \nabla {\rm{f}}} \right|}} = \frac{{{{{\rm{\hat a}}}_{\rm{x}}} + {{{\rm{\hat a}}}_{\rm{y}}}}}{{\sqrt 2 }}\)

Now, the component of \({\rm{\vec P}} = 7.53{{\rm{\hat a}}_{\rm{x}}}\) along \({{\rm{\hat a}}_{\rm{u}}}\) is

\({{\rm{P}}_{\rm{n}}} = {\rm{\vec P}} \cdot {{{\rm{\hat a}}}_{\rm{u}}} = 7.53{{{\rm{\hat a}}}_{\rm{x}}} \cdot \left( {\frac{{{{{\rm{\hat a}}}_{\rm{x}}} + {{{\rm{\hat a}}}_{\rm{y}}}}}{{\sqrt 2 }}} \right)\)

\( {{\rm{P}}_{\rm{n}}} = \frac{{7.53}}{{\sqrt 2 }} = 5.32\)

Now, power passing through \(10{\rm{;cm}} \times {\rm{;}}10{\rm{;cm}}\) area plate is

\(\begin{array}{l} {{\rm{w}}_{\rm{E}}} = 5.32{\rm{;}} \times 100 \times {10^{ - 4}}\ {{\rm{w}}_{\rm{E}}} = 53.2{\rm{;mW}} \end{array}\)

Concept: 1. Intrinsic Impedance of any medium is given by \({{\rm{\eta }}} = \sqrt {\frac{{{{\rm{\mu }}}}}{{{\epsilon}}}} = \sqrt {\frac{{{{\rm{\mu }}_{\rm{r}}}{{\rm{\mu }}_{\rm{o}}}}}{{{\epsilon_{\rm{r}}}{\epsilon_{\rm{o}}}}}} \)

                2. Transmission coefficient is given by \({\rm{\tau }} = \frac{{2{{\rm{\eta }}_2}}}{{{{\rm{\eta }}_1} + {{\rm{\eta }}_2}}} \) and propagation constant is given by \({\rm{\gamma }} = \sqrt {{\rm{j\omega \mu }}\left( {{\rm{\sigma }} + {\rm{j\omega }}\epsilon} \right)}\)

We have \({{\rm{\eta }}_2} = \sqrt {\frac{{{{\rm{\mu }}_2}}}{{{\epsilon_2}}}} = \sqrt {\frac{{{{\rm{\mu }}_{\rm{r}}}{{\rm{\mu }}_{\rm{o}}}}}{{{\epsilon_{\rm{r}}}{\epsilon_{\rm{o}}}}}} = \sqrt {\frac{{{{\rm{\mu }}_0}}}{{{\epsilon_0}}}} = 120{\rm{\pi }}\) 

Calculation:

As we know the transmission coefficient

\({\rm{\tau }} = \frac{{2{{\rm{\eta }}_2}}}{{{{\rm{\eta }}_1} + {{\rm{\eta }}_2}}} = 1\)

Thus, \({{\rm{E}}_{\rm{t}}} = {{\rm{E}}_{\rm{i}}} = 10{\rm{;V}}/{\rm{m}}\)

Now, propagation constant \({\rm{\gamma }} = \sqrt {{\rm{j\omega \mu }}\left( {{\rm{\sigma }} + {\rm{j\omega }}\epsilon} \right)}\)

Now, since \({\rm{\sigma }} = 0\), we have

\(\begin{array}{l} {\rm{\gamma }} = \sqrt {{{\rm{j}}^2}{{\rm{\omega }}^2}{\rm{\mu }}\epsilon} \ \Rightarrow {\rm{\gamma }} = {\rm{j\omega }}\sqrt {{\rm{\mu }}\epsilon} \ \Rightarrow {\rm{\gamma }} = {\rm{j\omega }}\sqrt {{{\rm{\mu }}_{\rm{o}}}{\epsilon_{\rm{o}}}} \sqrt {{{\rm{\mu }}_{\rm{r}}}{\epsilon_{\rm{r}}}} \ \Rightarrow {\rm{\gamma }} = {\rm{j\omega }}\sqrt {{{\rm{\mu }}_{\rm{o}}}{\epsilon_{\rm{o}}}} \left( {1 - {\rm{j}}2} \right)\ \Rightarrow {\rm{\gamma }} = {\rm{\alpha }} + {\rm{j\beta }} = 2{\rm{\omega }}\sqrt {{{\rm{\mu }}_{\rm{o}}}{\epsilon_{\rm{o}}}} + {\rm{j\omega }}\sqrt {{{\rm{\mu }}_{\rm{o}}}{\epsilon_{\rm{o}}}} \end{array}\)

Comparing we have, \({\rm{\alpha }} = 2{\rm{\omega }}\sqrt {{{\rm{\mu }}_{\rm{o}}}{\epsilon_{\rm{o}}}} = 2 \times 2{\rm{\pi }} \times 1.1 \times {10^9} \times \frac{1}{{3 \times {{10}^8}}} = 46.1{\rm{;Np}}/{\rm{m}}\)

The electric field at \({\rm{z}} = 10{\rm{;cm}}\) is given by

\(\begin{array}{l} {{\rm{E}}_2} = {{\rm{E}}_1} \times {{\rm{e}}^{ - {\rm{\alpha z}}}} = 10 \times {{\rm{e}}^{ - 46.1 \times 0.1}}\ \Rightarrow {{\rm{E}}_2} = 0.1{\rm{;V}}/{\rm{m}} \end{array}\)