A noisy channel which has a bandwidth of 400KHz and theoretical channel capacity of 4.01Mbps.
According to Shannon Capacity formula:
a. Calculate the Signal Power received at the receiver end if the Noise Power is .001 Watt.
b. Calculate the signal to noise ratio of the channel in dB (SNRdB).
C=Bandwidth*log2(1+S/N)
4.01e6=400e3log2(1+S/N)
10=log2(1+S/N)
1024=1+S/N
S/N=1024
S=1023*.001 watts= about 1 watt
db gain= 10 log (S/N)=30 db
checkthis, it has been 30 years since I used Shannon.
@bobpursley You did:
4.01e6=400e3log2(1+S/N)
10=log2(1+S/N)
Now question is that after solving C and B i.e., C/B we get following result:
=4010000/400000
=401/40
=10.025
But you wrote only 10 here. Question is why?
I reduced it to 10 because of significant events rules.
To calculate the Signal Power received at the receiver end, we need to use the Shannon Capacity formula and rearrange it to solve for Signal Power.
The Shannon Capacity formula is:
C = B * log2(1 + (S/N))
Where:
C = channel capacity in bits per second (bps)
B = bandwidth in hertz (Hz)
S = signal power
N = noise power
In this case, we know:
C = 4.01 Mbps = 4.01 x 10^6 bps
B = 400 KHz = 400 x 10^3 Hz
N = 0.001 Watt
To solve for the Signal Power (S), we can rearrange the formula as follows:
S = (2^(C/B) - 1) * N
a. Calculate the Signal Power received at the receiver end:
Plug the values into the formula:
S = (2^(4.01 x 10^6 / (400 x 10^3)) - 1) * 0.001
S ≈ 22.640 Watt
Therefore, the Signal Power received at the receiver end is approximately 22.640 Watt.
b. Calculate the signal-to-noise ratio of the channel in dB (SNRdB):
The signal-to-noise ratio (SNR) can be calculated using the formula:
SNR = S / N
To convert the SNR from a linear scale to decibels (dB), we can use the following formula:
SNRdB = 10 * log10(SNR)
Plug the values into the formulas:
SNR = 22.640 / 0.001
SNR ≈ 22640
SNRdB = 10 * log10(22640)
SNRdB ≈ 43.027 dB
Therefore, the signal-to-noise ratio (SNR) of the channel is approximately 43.027 dB.