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

TRB 2006 - Question 10

A voice graded channel of the telephone network has a Bandwidth of $3.43\; kHz$. The channel capacity of the channel for a SNR of $30dB$ is 

a) $33888.56928 \;bits/sec$

b) $1000\; bits/sec$ 
c) $10\; bits/sec $
d) $10^{-6}\; bps$

Correct Answer: Option A

SNR(dB) = $30$ dB

$10\;log_{10}(SNR) = 30$ 

SNR$=10^{3}=1000$ 

Channel capacity, $$C=B\;log_2(1+SNR)$$ $$=3.4×10^3\; log_2(1+1000)$$$$=3.4×10^3\; log_2(1001)$$ $$=3.4×10^3\; log_2(1001)$$ $$=3.4×10^3\; log_2(1001)$$ $$=3.4×10^3×9.96722625$$ $$=33888.56928\; bits/sec$$ 

If calc is not allowed, round off $1001$ to nearest $2^x$ value. Hence $$C\lt 3.4×10^3\; log_2(1024)$$ $$C\lt 3.4×10^3\; log_2(2^{10})$$ $$C\lt 34000\;bps$$