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 $$C\lt 3.4×10^3\; log_2(1024)$$ $$C\lt 3.4×10^3\; log_2(2^{10})$$ $$C\lt 34000\;bps$$

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 $$C\lt 3.4×10^3\; log_2(1024)$$ $$C\lt 3.4×10^3\; log_2(2^{10})$$ $$C\lt 34000\;bps$$

TRB 2006 - Question 11

 A memory system of size $16$ Kbytes is required to be designed using memory chips which have $12$ address lines and $4$ data lines each. 

The number of such chips required to design the memory system is 

a) $2$

b) $4$   

c) $8$     

d) $16$

Correct Answer: Option C

Solution:

👉Required memory size = $16$ kilobytes

As we know $1$ byte = $8$ bits and $1$ kilobytes = $1024$ bytes

Hence required memory size = $16×1024×8$ bits

👉Available memory chip has $12$ address lines and $4$ data lines, it's memory size is $2^{12}×4$ bits.

👉Required number of memory chips, $$=\frac{Required\; memory\; size}{ Available\; memory \;size}$$ $$=\frac{16×1024×8}{2^{12}×4}$$$$=\frac{2^4×2^{10}×2^3}{2^{12}×2^2}$$$$=2^{4+10+3-12-2}$$$$=2^3$$$$=8$$

We need $8$ memory chips.

TRB 2006 - Question 11

 A memory system of size $16$ Kbytes is required to be designed using memory chips which have $12$ address lines and $4$ data lines each. 

The number of such chips required to design the memory system is 

a) $2$

b) $4$   

c) $8$     

d) $16$

Correct Answer: Option C

Solution:

👉Required memory size = $16$ kilobytes

As we know $1$ byte = $8$ bits and $1$ kilobytes = $1024$ bytes

Hence required memory size = $16×1024×8$ bits

👉Available memory chip has $12$ address lines and $4$ data lines, it's memory size is $2^{12}×4$ bits.

👉Required number of memory chips, $$=\frac{Required\; memory\; size}{ Available\; memory \;size}$$ $$=\frac{16×1024×8}{2^{12}×4}$$$$=\frac{2^4×2^{10}×2^3}{2^{12}×2^2}$$$$=2^{4+10+3-12-2}$$$$=2^3$$$$=8$$

We need $8$ memory chips.

Circuit theory

Unit of a charge is

a) ampere

b) volt

c) coloumb

d) watt

Correct Answer: Option C

Solution:

Unit of charge is coloumb

Circuit theory

Unit of a charge is

a) ampere

b) volt

c) coloumb

d) watt

Correct Answer: Option C

Solution:

Unit of charge is coloumb

Circuit Theory

Which of the following amount of electrons is equivalent to $-3.941$ C of charge?

a) $1.628×10^{20}$

b) $1.24×10^{18}$

c) $6.482×10^{17}$

d) $2.46×10^{19}$

Correct Answer: Option D

Solution:

As we know, $6.24×10^{18}$ electrons consitute $-1$C of charge. Hence $-3.941$C of charge has $3.941×6.24×10^{18}$ no of electrons. $$=2.457×10^{19}$$