Examination of digital link and ISI
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A digital communication system transmits binary data using polar NRZ pulses of current over a baseband link with a bandwidth of 20 kHz. The transmitter and receiver have ideal "Root Raised Cosine" (RRC) filters with roll off factor = 0.5.
a. What is the minimum bandwidth that the link must provide to avoid generating ISI (inter symbol interference) at the receiver?
b. What transfer function must the link have over this bandwidth to avoid the generation of ISI?
c. A radio link has a bandwidth of 15 kHz. What is the maximum bit rate at which binary data can be transmitted without ISI if the transmitter and receiver have ideal RRC filters with roll off factor = 0.5 using BPSK modulation?
d. What is the noise bandwidth of the receiver?
e. What is the requirement for the phase response (transfer function) of each of these links to avoid distortion of the transmitted waveform?
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Solution Summary
A digital communication system transmits binary data using polar NRZ pulses of current over a baseband link with a bandwidth of 20 kHz. The transmitter and receiver have ideal "Root Raised Cosine" (RRC) filters with roll off factor = 0.5.
a. the minimum bandwidth that the link must provide to avoid generating ISI (inter symbol interference) at the receiver is deduced
b. The transfer function for link to have over this bandwidth to avoid the generation of ISI is deduced
c. A radio link has a bandwidth of 15 kHz. The maximum bit rate at which binary data can be transmitted without ISI if the transmitter and receiver have ideal RRC filters with roll off factor = 0.5 using BPSK modulation is deduced
d. Te noise bandwidth of the receiver is then deduced
e. The requirement for the phase response (transfer function) of each of these links to avoid distortion of the transmitted waveform is deduced
Solution Preview
a. Symbol Rate SR is given in terms of the real (occupied) bandwidth B and the bandwidth "rolloff" factor r; as defined by (1).
SR = B/(1+r) (1)
So the symbol rate of the link is
SR = 20/(1+0.5) = 20/1.5 kSps
SR = 13.33 kSps
Minimum bandwidth required so that no ISI is produced (BMIN) is defined by Nyquist ISI criteria [1] and equation (2) as half the Symbol Rate.
BMIN = SR/2 (2)
BMIN = 6.67 kHz
...
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