TECHNICAL SI 2018 - Question 61

In superconductivity, the electrical resistance of material becomes

a) zero

b) infinite

c) finite

d) unity

Correct Answer: Option A 

Superconductivity - property of certain materials to conduct DC current without energy loss when they are cooled below certain temperature (critical temperature). 

No energy loss means, $I^2\;R\;t=0$. Here, $I$ and $t$ can't be zero. Hence $R=0$

TECHNICAL SI 2018 - Question 61

In superconductivity, the electrical resistance of material becomes

a) zero

b) infinite

c) finite

d) unity

Correct Answer: Option A 

Superconductivity - property of certain materials to conduct DC current without energy loss when they are cooled below certain temperature (critical temperature). 

No energy loss means, $I^2\;R\;t=0$. Here, $I$ and $t$ can't be zero. Hence $R=0$

The Number system

 ஒரு அழகிய கிராமத்தில்... அழகிய காலை வேளையில்... 

குவா... குவா...

என்ற அழுகையுடன் குழந்தை பிறக்கிறது...

அப்படியே கட் பண்ணி... ரெண்டு வருஷம் கழிச்சி பாத்தா... அம்மா ABCD சொல்லி கொடுத்துட்டு இருப்பாங்க... அப்பா one... two... three... சொல்லி கோடுத்துட்டு இருப்பாரு... 

அன்றைக்கு ஆரம்பித்து இன்று வரை நாம் கற்றுக் கொள்ளும் எண்கள் (numerals)

"0, 1, 2, 3, 4..." 

 "௧ ௨ ௩ ௪ ௫ ௬..." 

"I, II, III, IV, V, VI,.." 

இப்படி எத்தனை வகையான எண்களை படித்தாலும்... Addition... subtraction... multiplication.. division என அனைத்திலும் நமக்கு பயன்படுவது "$0, 1, 2, 3, 4, 5, 6, 7, 8, 9$" தான்... இந்த பத்து எண்கள் தான் எவ்வளவு பெரிய எண்களை உருவாக்குவதற்கும் அடிப்படை (base).  எனவே இதை நாம் base-$10$ number system எனக் குறிப்பிடலாம். Deca என்பது எண் $10$ஐ குறிப்பது.. எனவே Base $10$ number systemஐ Decimal number system என்று அழைக்கிறார்கள்.

Decimal numbers:

Numbers: $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$

Base: 10

Use: Humans can understand

$9876$

$-235$

$365.345$

இந்த மூன்றில் எது decimal number??

Decimal number system என்பது ஒரு பொதுப்பெயர் (generic term). இங்கே இந்த பத்து எண்களை $(0, 1, 2, 3, 4, 5, 6, 7, 8, 9)$ அடிப்படையாகக் (base) கொண்ட அனைத்து எண்களுமே decimal numbers தான். அது positive number ஆக இருக்கலாம், negative number ஆக இருக்கலாம் அல்லது decimal point number ஆகக் கூட இருக்கலாம். எனவே மேலே காட்டப்பட்டுள்ள மூன்றுமே decimal numbers தான். அதில் தெளிவு வேண்டும்.

Machineகளை பொறுத்த வரைக்கும், அதற்கு தெரிந்தது இரண்டு தான். 

Wireல current இருக்கா இல்லையா??

 Switch ON பண்ணா ஓடும், OFF பண்ணா நின்னுடும். 

ON or OFF

" $0$ and $1$"

இந்த $0$ and $1$ ஐ அடிப்படையாக வைத்து தான் machineல் பல complex ஆன வேலைகளை செய்ய முடியும். இரண்டு எண்கள் தான் அடிப்படை (base $2$) எனும் போது இதை நாம் Base $2$ number system  எனலாம். Bi என்பது எண் $2$ஐ குறிப்பது.. எனவே Base $2$ number systemஐ Binary number system என்று அழைக்கிறார்கள்.

Binary numbers:

Numbers: $0$ and $1$

Base: $2$

Use: Machines can understand

மனிதர்களாகிய நமக்கு புரிவது decimal number system, machineகளுக்கு புரிவது binary number system.  

The Number system

 ஒரு அழகிய கிராமத்தில்... அழகிய காலை வேளையில்... 

குவா... குவா...

என்ற அழுகையுடன் குழந்தை பிறக்கிறது...

அப்படியே கட் பண்ணி... ரெண்டு வருஷம் கழிச்சி பாத்தா... அம்மா ABCD சொல்லி கொடுத்துட்டு இருப்பாங்க... அப்பா one... two... three... சொல்லி கோடுத்துட்டு இருப்பாரு... 

அன்றைக்கு ஆரம்பித்து இன்று வரை நாம் கற்றுக் கொள்ளும் எண்கள் (numerals)

"0, 1, 2, 3, 4..." 

 "௧ ௨ ௩ ௪ ௫ ௬..." 

"I, II, III, IV, V, VI,.." 

இப்படி எத்தனை வகையான எண்களை படித்தாலும்... Addition... subtraction... multiplication.. division என அனைத்திலும் நமக்கு பயன்படுவது "$0, 1, 2, 3, 4, 5, 6, 7, 8, 9$" தான்... இந்த பத்து எண்கள் தான் எவ்வளவு பெரிய எண்களை உருவாக்குவதற்கும் அடிப்படை (base).  எனவே இதை நாம் base-$10$ number system எனக் குறிப்பிடலாம். Deca என்பது எண் $10$ஐ குறிப்பது.. எனவே Base $10$ number systemஐ Decimal number system என்று அழைக்கிறார்கள்.

Decimal numbers:

Numbers: $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$

Base: 10

Use: Humans can understand

$9876$

$-235$

$365.345$

இந்த மூன்றில் எது decimal number??

Decimal number system என்பது ஒரு பொதுப்பெயர் (generic term). இங்கே இந்த பத்து எண்களை $(0, 1, 2, 3, 4, 5, 6, 7, 8, 9)$ அடிப்படையாகக் (base) கொண்ட அனைத்து எண்களுமே decimal numbers தான். அது positive number ஆக இருக்கலாம், negative number ஆக இருக்கலாம் அல்லது decimal point number ஆகக் கூட இருக்கலாம். எனவே மேலே காட்டப்பட்டுள்ள மூன்றுமே decimal numbers தான். அதில் தெளிவு வேண்டும்.

Machineகளை பொறுத்த வரைக்கும், அதற்கு தெரிந்தது இரண்டு தான். 

Wireல current இருக்கா இல்லையா??

 Switch ON பண்ணா ஓடும், OFF பண்ணா நின்னுடும். 

ON or OFF

" $0$ and $1$"

இந்த $0$ and $1$ ஐ அடிப்படையாக வைத்து தான் machineல் பல complex ஆன வேலைகளை செய்ய முடியும். இரண்டு எண்கள் தான் அடிப்படை (base $2$) எனும் போது இதை நாம் Base $2$ number system  எனலாம். Bi என்பது எண் $2$ஐ குறிப்பது.. எனவே Base $2$ number systemஐ Binary number system என்று அழைக்கிறார்கள்.

Binary numbers:

Numbers: $0$ and $1$

Base: $2$

Use: Machines can understand

மனிதர்களாகிய நமக்கு புரிவது decimal number system, machineகளுக்கு புரிவது binary number system.  

TRB 2006 - Question 1

An amplifier A has $6$ dB gain and $50$ ohm input and output impedance. The noise figure of this amplifier is $3$ dB

A cascade of two such amplifiers will have noise figure of 

a) $6$ dB 

b) $8$ dB

c) $12$ dB

d) none of these

Correct Answer: Option D

Solution:

Gain of A = $6\;dB$ 

$10\;log_{10}(G)=6\;dB$ 

$G=10^{6/10}=10^{0.6}$ 

$G=3.98$ 

Noise figure = $3\;dB$ 

$10\;log_{10}(G)=3\;dB$ 

$F=10^{3/10}=10^{0.3}$ 

$F=1.995$ 

Equivalent noise figure of cascaded amplifiers, $$F_{cas}=F_1+\frac{F_2-1}{G_1}+\frac{F_3-1}{G_1\;G_2}+...$$ $$F_{cas}=1.995+\frac{1.995-1}{3.98}$$ $$F_{cas}=1.995+\frac{0.995}{3.98}$$ $$F_{cas}=1.995+0.25$$ $$F_{cas}=2.245$$ $$F_{cas}(dB)=10\;log_{10}(2.245)$$ $$F_{cas}(dB)=3.512\;dB$$

TRB 2006 - Question 1

An amplifier A has $6$ dB gain and $50$ ohm input and output impedance. The noise figure of this amplifier is $3$ dB

A cascade of two such amplifiers will have noise figure of 

a) $6$ dB 

b) $8$ dB

c) $12$ dB

d) none of these

Correct Answer: Option D

Solution:

Gain of A = $6\;dB$ 

$10\;log_{10}(G)=6\;dB$ 

$G=10^{6/10}=10^{0.6}$ 

$G=3.98$ 

Noise figure = $3\;dB$ 

$10\;log_{10}(G)=3\;dB$ 

$F=10^{3/10}=10^{0.3}$ 

$F=1.995$ 

Equivalent noise figure of cascaded amplifiers, $$F_{cas}=F_1+\frac{F_2-1}{G_1}+\frac{F_3-1}{G_1\;G_2}+...$$ $$F_{cas}=1.995+\frac{1.995-1}{3.98}$$ $$F_{cas}=1.995+\frac{0.995}{3.98}$$ $$F_{cas}=1.995+0.25$$ $$F_{cas}=2.245$$ $$F_{cas}(dB)=10\;log_{10}(2.245)$$ $$F_{cas}(dB)=3.512\;dB$$

TRB 2006 - Question 2

A Very lossy $\lambda/4$ long $50\;\Omega$ transmission line is open circuited at the load end. The input impedance measured at the other end of the line is approximately.

a) $0$

b) $50\;\Omega$

c) $\infty$

d) None of these

Correct Answer: Option B

Solution:
As soon as we saw $Z_{in}$ for $\lambda/4$ transmission line, we tempt to use the formula,
$$Z_{in}=Z_0^2/Z_L$$ $$Z_{in}=50^2/\infty$$ $$Z_{in}=0\;\Omega$$

Hence, we choose the option A.

But the above formula is for lossless $\lambda/4$ transmission line. 

In our question, they have asked for very lossy transmission line.

How to approach this question?

Case 1: Lossless transmission line matched with $Z_0$

Since the transmission line is matched, there will be no standing wave and hence $Z_{in}$ will be equal to $Z_0$.

Case 2: Lossless transmission line not matched with $Z_0$

Since the transmission line is not matched, there will be reflections at the load and hence standing wave exists. Whenever standing wave is there, the impedance of the transmission line will vary in accordance voltage and current at that point.

Case 3: Very lossy transmission line not matched with $Z_0$

Since the transmission line is not matched, there will be reflections at the load and hence standing wave exists. But here, standing wave gets attenuated since the transmission line is very lossy. There will be no standing wave at input side. Hence input impedance is equal to characteristic impedance $Z_0$.

Hence the answer is Option B

TRB 2006 - Question 2

A Very lossy $\lambda/4$ long $50\;\Omega$ transmission line is open circuited at the load end. The input impedance measured at the other end of the line is approximately.

a) $0$

b) $50\;\Omega$

c) $\infty$

d) None of these

Correct Answer: Option B

Solution:
As soon as we saw $Z_{in}$ for $\lambda/4$ transmission line, we tempt to use the formula,
$$Z_{in}=Z_0^2/Z_L$$ $$Z_{in}=50^2/\infty$$ $$Z_{in}=0\;\Omega$$

Hence, we choose the option A.

But the above formula is for lossless $\lambda/4$ transmission line. 

In our question, they have asked for very lossy transmission line.

How to approach this question?

Case 1: Lossless transmission line matched with $Z_0$
Since the transmission line is matched, there will be no standing wave and hence $Z_{in}$ will be equal to $Z_0$.

Case 2: Lossless transmission line not matched with $Z_0$
Since the transmission line is not matched, there will be reflections at the load and hence standing wave exists. Whenever standing wave is there, the impedance of the transmission line will vary in accordance voltage and current at that point.

Case 3: Very lossy transmission line not matched with $Z_0$
Since the transmission line is not matched, there will be reflections at the load and hence standing wave exists. But here, standing wave gets attenuated since the transmission line is very lossy. There will be no standing wave at input side. Hence input impedance is equal to characteristic impedance $Z_0$.

Hence the answer is Option B

TRB 2006 - Question 3

In a JK flipflop, $J=\bar Q$ and $K=1$. Assuming the flip flop was initially cleared and clocked for $6$ pulses, the sequence at the $Q$ will be

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a) $010000$ 
b) $011001$
c) $010010$
d) $010101$

Correct Answer: Option D

The characteristic equation of JK flip-flop is $$Q^+=J\bar{Q}+\bar{K}Q$$ Subs the given inputs, $J=\bar{Q}$ and $K=1$ $$Q^+=\bar{Q}\bar{Q}+0Q$$ $$Q^+=\bar{Q}+0$$ $$Q^+=\bar{Q}$$ From this we can say, for every clock pulse, the output of the JK flip-flop gets toggled. Hence $010101$ is the answer

TRB 2006 - Question 3

In a JK flipflop, $J=\bar Q$ and $K=1$. Assuming the flip flop was initially cleared and clocked for $6$ pulses, the sequence at the $Q$ will be

a) $010000$ 
b) $011001$
c) $010010$
d) $010101$

Correct Answer: Option D

The characteristic equation of JK flip-flop is $$Q^+=J\bar{Q}+\bar{K}Q$$ Subs the given inputs, $J=\bar{Q}$ and $K=1$ $$Q^+=\bar{Q}\bar{Q}+0Q$$ $$Q^+=\bar{Q}+0$$ $$Q^+=\bar{Q}$$ From this we can say, for every clock pulse, the output of the JK flip-flop gets toggled. Hence $010101$ is the answer

TRB 2006 - Question 4

The Gate delay of an NMOS inverter is dominated by charge time rather than discharge time because

a)the driver transistor has larger threshold voltage than load transistor 

b) the driver transistor has larger leakage currents compared to load transistor

c) the load transistor has a smaller W/L ratio compared to the driver transistor

d)none of these

Correct Answer: Option C

TRB 2006 - Question 4

The Gate delay of an NMOS inverter is dominated by charge time rather than discharge time because

a)the driver transistor has larger threshold voltage than load transistor 

b) the driver transistor has larger leakage currents compared to load transistor

c) the load transistor has a smaller W/L ratio compared to the driver transistor

d)none of these

Correct Answer: Option C

TRB 2006 - Question 5

Boolean function of $A+BC$ is reduced form of

a) $AB + BC$
b) $(A+B) (A+C)$
c) $A'B+AB'C$
d) $(A+C) B$

Correct Answer: Option B

Option A: $AB + BC$ can't be reduced further
Option B: $$=(A+B) (A+C)$$ $$=AA+AC+AB+BC$$ $$=A+AC+AB+BC$$ $$=A(1+C)+AB+BC$$ since $1+C=1$ $$=A+AB+BC$$ $$=A(1+B)+BC$$ $$=A+BC$$ Option C: $A'B+AB'C$ can't be reduced further
Option D: $(A+C) B$ can't be reduced further

TRB 2006 - Question 5

Boolean function of $A+BC$ is reduced form of

a) $AB + BC$
b) $(A+B) (A+C)$
c) $A'B+AB'C$
d) $(A+C) B$

Correct Answer: Option B

Option A: $AB + BC$ can't be reduced further
Option B: $$=(A+B) (A+C)$$ $$=AA+AC+AB+BC$$ $$=A+AC+AB+BC$$ $$=A(1+C)+AB+BC$$ since $1+C=1$ $$=A+AB+BC$$ $$=A(1+B)+BC$$ $$=A+BC$$ Option C: $A'B+AB'C$ can't be reduced further
Option D: $(A+C) B$ can't be reduced further

TRB 2006 - Question 6

One year in Earth and Mars

a) differs

b) remains the same 

c) differs and in Mars it is more than 365 days

d) decreases in Mars compared to Earth

Correct Answer: Option C 

Earth: 365 days

Mars: 687 days

TRB 2006 - Question 6

One year in Earth and Mars

a) differs

b) remains the same 

c) differs and in Mars it is more than 365 days

d) decreases in Mars compared to Earth

Correct Answer: Option C 

Earth: 365 days

Mars: 687 days

TRB 2006 - Question 7

CDMA is newest and latest used access method using satellites and popular in _____ applications and not much in ______ applications

a) Commercial, military 

b) military, commercial

c) military, defence

d) research, commercial

Correct Answer: Option B

From the book, "Data Communications and Networks" authored by Achyut S Godbole, 

CDMA is newest and latest used access method using satellites. It is popular in military installations but not so much in the case of commercial applications.

TRB 2006 - Question 7

CDMA is newest and latest used access method using satellites and popular in _____ applications and not much in ______ applications

a) Commercial, military 

b) military, commercial

c) military, defence

d) research, commercial

Correct Answer: Option B

From the book, "Data Communications and Networks" authored by Achyut S Godbole, 

CDMA is newest and latest used access method using satellites. It is popular in military installations but not so much in the case of commercial applications.

TRB 2006 -Question 8

The order in which the following satellite communication is accessed is

a) FDMA   TDMA   CDMA

b) CDMA   FDMA   TDMA

c) FDMA   CDMA   TDMA

d) Cellular   Telephony  AMPS   FDMA

Correct Answer: Option A

We can answer this question using, uplink power requirements

FDMA:

Using,

$$\frac{C}{N_0}=\frac{E_b}{N_0}+R$$

We can say Power level of carrier (C) depends on modulation rate.

In FDMA, we can say modulation rate (R) will be less, as total frequency spectrum is divided among earth stations and hence the EIRP (Equivalent isotropic radiated power) is less for uplink.

TDMA:

We can say TDMA uses modulation rate higher than FDMA and hence higher EIRP is required for the earth stations to uplink. 

Hence FDMA is most preferable than TDMA.

Likewise CDMA also has high modulation rate because of generation of pseudo noise like sequence.

FDMA > TDMA > CDMA

TRB 2006 -Question 8

The order in which the following satellite communication is accessed is

a) FDMA   TDMA   CDMA

b) CDMA   FDMA   TDMA

c) FDMA   CDMA   TDMA

d) Cellular   Telephony  AMPS   FDMA

Correct Answer: Option A

We can answer this question using, uplink power requirements

FDMA:

Using,

$$\frac{C}{N_0}=\frac{E_b}{N_0}+R$$

We can say Power level of carrier (C) depends on modulation rate.

In FDMA, we can say modulation rate (R) will be less, as total frequency spectrum is divided among earth stations and hence the EIRP (Equivalent isotropic radiated power) is less for uplink.

TDMA:

We can say TDMA uses modulation rate higher than FDMA and hence higher EIRP is required for the earth stations to uplink. 

Hence FDMA is most preferable than TDMA.

Likewise CDMA also has high modulation rate because of generation of pseudo noise like sequence.

FDMA > TDMA > CDMA

TRB 2006 - Question 9

 Is $x (t) = cos \Big (\frac{1}{3} t \Big ) + sin \Big (\frac{1}{4} t \Big )$ periodic?

If so, its period is

 a) not periodic, can't find its period

b) periodic, $24 \pi$

c) periodic, $3/4 \pi$

d) periodic, $12 \pi $

Correct Answer: Option B

$x (t) = cos \Big (\frac{1}{3} t \Big ) + sin \Big (\frac{1}{4} t \Big )$

Frequency and time period be, 

$2\pi f_1=1/3$ and $2\pi f_2=1/4$

$f_1=1/6\pi$ and $f_2=1/8\pi$

$T_1=6\pi$ and $T_2=8\pi$

To find sum of sinusoids periodic or not, 

$$\frac{T_1}{T_2}=\frac{6\pi}{8\pi}=\frac{6}{8}$$

Since $6/8$ is a rational number, the given sum of sinusoids is periodic.

Fundamental period, $$T=lcm(6\pi, 8\pi)$$ $$T=24\pi$$

TRB 2006 - Question 9

 Is $x (t) = cos \Big (\frac{1}{3} t \Big ) + sin \Big (\frac{1}{4} t \Big )$ periodic?

If so, its period is

 a) not periodic, can't find its period

b) periodic, $24 \pi$

c) periodic, $3/4 \pi$

d) periodic, $12 \pi $

Correct Answer: Option B

$x (t) = cos \Big (\frac{1}{3} t \Big ) + sin \Big (\frac{1}{4} t \Big )$

Frequency and time period be, 

$2\pi f_1=1/3$ and $2\pi f_2=1/4$

$f_1=1/6\pi$ and $f_2=1/8\pi$

$T_1=6\pi$ and $T_2=8\pi$

To find sum of sinusoids periodic or not, 

$$\frac{T_1}{T_2}=\frac{6\pi}{8\pi}=\frac{6}{8}$$

Since $6/8$ is a rational number, the given sum of sinusoids is periodic.

Fundamental period, $$T=lcm(6\pi, 8\pi)$$ $$T=24\pi$$