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Question 2.4: For a BPSK modulator with a carrier frequency of 70 MHz and ......

For a BPSK modulator with a carrier frequency of 70 MHz and an input bit rate of 10 Mbps, determine the maximum and minimum upper and lower side frequencies, draw the output spectrum, determine the minimum Nyquist bandwidth, and calculate the baud.

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Substituting into Equation 20 yields

BPSK output = [sin(2πf_{a}t)] × [sin(2πf_{c}t)]           (20)

output = (sin ω_{a}t)(sin ω_{c}t)

= [sin 2π(5 MHz)t][sin 2π(70 MHz)t]

\underbrace{=\frac{1}{2}\cos\,2π (70 \,MHz\, – \,5 \,MHz)t\ }_\text{lower side frequency}\underbrace{-\, \frac{1}{2}\cos\,2π (70 \,MHz + 5 \,MHz)t}_\text{upper side frequency}

Minimum lower side frequency (LSF):

LSF = 70 MHz  – 5 MHz  = 65 MHz

Maximum upper side frequency (USF):

USF = 70 MHz + 5 MHz = 75 MHz

Therefore, the output spectrum for the worst-case binary input conditions is as follows:
The minimum Nyquist bandwidth (B) is

B = 75 MHz – 65 MHz = 10 MHz

and the baud = f_{b} or 10 megabaud.

fig_ch2_ex4

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