# ECET 375 homework week 2

## ECET 375 homework week 2

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Chapter 2
10.  Determine the information capacity in bps for a circuit with 100 kHz bandwidth and a signal-to-noise ratio of 40 dB (10,000).
I = (1+SNR)
Where,
I = Information capacity (bps)
B = Bandwidth (Hz)
SNR = signal to noise ratio (unitless)
I = 100 kHz*log2 (1+10,000) =1.33 Mbps

11. Determine the number of conditions possible for a binary code containing the following number of bits:
a. 3, 23 = 8
b. 5, 25 = 32
c. 7, 27 = 128
d. 12, 212 = 4096

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

Where,
B = Bandwidth
M = number of discrete signal or voltage levels

16. Determine the minimum bandwidth, baud and bandwidth efficiency for the following bit rates and modulation schemes -8-QAM and 16-QAM:
Baud rate = fb/N, N = log2M,   M = 2N
Minimum bandwidth = B = fb/N just like baud rate
B (Bandwidth Efficiency) = bit rate/minimum bandwidth

a. fb = 2400bps
(8-QAM) baud rate= 2400bps/3 = 800, minimum bandwidth = 2400bps/3 = 800 Hz, bandwidth efficiency = 2400bps/800 Hz = 3 bps per cycle of bandwidth
(16 - QAM) baud rate= 2400bps/4 = 600, minimum bandwidth = 2400bps/4 = 600 Hz, bandwidth efficiency = 2400bps/600 Hz = 4 bps per cycle of bandwidth

b. fb = 4800 bps
(8-QAM) baud rate= 4800bps/3 = 1600, minimum bandwidth = 4800bps/3 = 1600 Hz, bandwidth efficiency = 4800bps/1600 Hz = 3 bps per cycle of bandwidth
(16 - QAM) baud rate= 4800bps/4 = 1200, minimum bandwidth = 4800bps/4 = 1200 Hz, bandwidth efficiency = 4800bps/1200 Hz = 4 bps per cycle of bandwidth
c. fb = 9600 bps
(8-QAM) baud rate= 9600bps/3 = 3200, minimum bandwidth = 9600bps/3 = 3200 Hz, bandwidth efficiency = 9600bps/3200 Hz = 3 bps per cycle of bandwidth
(16 - QAM) baud rate= 9600bps/4 = 2400, minimum bandwidth = 9600bps/4 = 2400 Hz, bandwidth efficiency = 9600bps/2400 Hz = 4 bps per cycle of bandwidth

17. A probability of error of 10-6 means it is probable that there will...