We need to send 256 kbps over a noiseless channel with a bandwidth 20 KHz.How many signal levels do we need?

Question

1)We need to send 256 kbps over a noiseless channel with a bandwidth 20 KHz.How many signal levels do we need?

 

2) We have a channel with a 1 Mhz bandwidth. The SNR for this channel is 63.What are the appropriate bit rate and singal level?

Answer 1

1)

$N = 2 * B *log_{2} L$

where,
L = Signal Levels, which we have to find
B = Bandwidth, given as 20 KHz
N = Rate, given as 256 Kbps

On putting these values, we get

$256 * 10^3 = 2 * 20 * 10^3 *log_{2} L$
$256 = 2 * 20 *log_{2} L$
$2^8 = 2^2 * 10 *log_{2} L$

$2^6 = 10 *log_{2} L$

$6.4 = log_{2} L$

So, $L$ is approx $85$

2)

$SNR_{db} = 10 * log_{10}SNR$

Given $SNR_{db}$ as $63$, so keeping in above formula, we get $SNR$ as

$63 = 10 * log_{10}SNR$

$6.3 = log_{10}SNR$

$SNR = 2,000,000$

We have, 

$BitRate = B * log_{2} (1 + SNR)$

where,

B = bandwidth, given as $1$ Mhz

$SNR$ we found as $2,000,000$

So, putting these values, we get

$BitRate = 1 * 10^6 * log_{2} (1 + 2000000)$

$BitRate = 1 * 10^6 * log_{2} (2000001)$

$BitRate = 1 * 10^6 * 20.93$

$BitRate = 20.93$ Mbps

In order to find levels, we have

$N = 2 * B *log_{2} L$

where,
L = Signal Levels, which we have to find
B = Bandwidth, given as 1 MHz
N = Rate, given as 20.93 Mbps

On putting these values, we get

$20.93 * 10^6 = 2 * 1 * 10^6 *log_{2} L$

$20.93 = 2 * 1 *log_{2} L$

$10.46 = log_{2} L$

So, $L$ is approx $1090$