2-10 Determine the information capacity in bps for a circuit with a 100-kHz band

Question

2-10
Determine the information capacity in bps for a circuit with a 100-kHz bandwidth and signal-to-noise ratio of 40 dB(10,000)

2-11
Determine the number of conditions possible for a binary code containing the following number of bits
a. 3
b. 5
c. 7
d. 12

2.12
Determine the highest bit rate possible for a circuit propagating a four-bit binary code with a banwitdh of 10,000Hz

2.16
Determine the minimum bandwidth, baud, and bandwidth efficiency for the following bit rates and modulation schemes-BPSK, QPSK, 8-PSK, and 16-PSK.
a. fb = 2400 bps
b. fb = 4800 bps
c. fb = 9600 bps

Explanation / Answer

2-10 using shannon-hartley theorem signal to noise ratio, SNR = 20 dB or 10000 that is , S/N = 10000 Bandwidth, B = 100 kHz C = B*log(1 + S/N) [log is taken in base 2] C = 100*log(1+10000) kbps C = 1328785.66 bps 2-11 If by number of conditions it means the number of different combinations then, for n bits, number of different conditions = 2^n for exaple for n = 2 , (00,01,10,11) that is total 2^2 = 4 using this a. = 2^3 = 8 b. = 2^5 = 32 c. = 2^7 = 128 d. = 2^12 = 4096 2-12 M - number of different levels in signal M = 2^no. of bits = 2^4 B - bandwidth B = 10000 Hz highest bit rate possible = 2*B*log(M) [log is in base 2] highest bit rate possible = 2*10000*log(2^4) bps highest bit rate possible = 80000 bps 2-16 don't know this one

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