QPSK and PCM-TDM Systems

Topics: Decibel, Signal processing, Baseband Pages: 1 (323 words) Published: October 23, 2014

Homework 7_1
ECET310
For a QPSK system and the given parameters, determine
Carrier power in dBm
Noise power in dBm
Noise power density in dBm
Energy per bit in dBJ
Carrier-to-noise power ratio
Eb/No ratio
PC=10-12 W,fb=20Kbps
PN=10-16 W,B=60 KHz
10 log (10^-12/.001)=-90dBm
10 log(10^-16/.001)= -130dBm
10 log(pn/.001)- 10logB= NdBm- 10logB = -130dBm-10log(60khz)= -177.8dBm 10 log(pc/fb)= 10 log(10^-12/20kbps)= -163dBj
Pc/pn= 10 log (pc/pn)= 10 log (10^-12/10^-16)= 40dB, 41dB
Eb/No (dB) = 10log(C/N) + 10log(B/fb) = 40+ 10 log(60khz/20kbps)=44.77dB Which system requires the highest Eb/No ratio for a probability of error of 10-7, four-level QAM or 8-PSK system? The level 4 QAM system.

Determine the minimum bandwidth required to achieve a P(e) of 10-4 for a 16-PSK system operating at 25 Mbps with a carrier-to-noise ratio of 10 dB.
B / fb = Eb / No – C / N = 18.3 dB – 10 dB = 8.3 dB Log^-1(8.3/10)=6.76
B= 6.76*25mbps= 169Mbps

A PCM-TDM system multiplexes 32 voice channels each with a bandwidth of 0 Khz to 4 KHz. Each sample is encoded with an 8-bit PCM code. UPNRZ encoding is used. Determine: The line data rate
(32)(8000)(8)=2.04mbps
Minimum sampling rate
=2(4khz)=8khz
Minimum Nyquist bandwidth
UPnRZ= Fb= 2.04mbps
Consider a PCM-TDM system in which 24 signals are to be processed. Each signal has a baseband bandwidth of 3 KHz. The sampling rate has to be 33.3% higher than the theoretical minimum, and 8 bits are to be used. Conventional NRZ-L encoding format is used. An extra bit is added to each frame for sync. Determine: Required sampling rate

2(3khz)=6khz
Line speed
(24)(8000)(8)=1.53mbps
Minimum required bandwidth
1.53mbps/2=768kbps
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