Telephone channels have a bandwidth of about 3.1 kHz. Answer the following quest

Question

Telephone channels have a bandwidth of about 3.1 kHz. Answer the following questions. a) If a telephone channel's signal-to-noise ratio (SNR) is 2,000 (the signal strength is 2,000 times as much as the noise strength), use the Shannon Equation to calculate the maximum data rate of a telephone channel? b) How fast could a telephone channel transmit data if the SNR were increased massively, from 2,000 to 20,000? (Note: This would not be realistic in practice.) c) With an SNR of 2,000, how fast could a telephone channel transmit data if the bandwidth were increased to 4 kHz? Show your work or no credit. d) What can you conclude from these calculations?

Explanation / Answer

a) Shannon's Theorem gives an upper bound to the capacity of a link, in bits per second (bps), as a function of the available bandwidth and the signal-to-noise ratio of the link.

The Theorem can be stated as:

C = B * log2(1+ S/N)

Where B isBandwidth in Hz ;S/N is signal to nise ratio and C is available channel capacity in bits per second

Here in the question we have B=3100 Hz

S/N=2000

C= 3100*log 2 (2001)

log base 2 of 2001 is equals to 10.966505451905741.

so = 3100*10.966=33996.46 bits/second=34 kbps

b)in this case SNR increased to 20000

so same theorem as explained above:

C = B * log2(1+ S/N)

Where B isBandwidth in Hz ;S/N is signal to nise ratio and C is available channel capacity

B=3100Hz but SNR changed to 20000.

C=3100* log base 2 of 20001 --------- log base 2of 20001 equals to 14.2877

=3100 * 14.2877

= 44292.118 bits/second= nearly 44.2 kbps

C) in this case B=4000 Hz=4 KHz

Applies the same theorem:

C = B * log2(1+ S/N)

C= 4000* log base 2 of 2001

=4000* 10.966=43864 bits/sec= Nearly 43.9 kbps

D) From this what we can concludes is the effective Capacity(Available channel capacity) will get increased more if we have larger bandwidth relative to SNR.

Because in the second case if we see we have increases the SNR which is inside log and which increases less value only as log(10*a) which equals to log 10 + log a where log 10 value is around 3.32

in the next case we have increased the bandwidth to some larger value which is multiplicant operand of our equation which ultimately results larger value

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.