need complete step by step solutions please. will rate 5 stars The message signa

Question

need complete step by step solutions please. will rate 5 stars

The message signal x(t), in the following system has a bandwidth of 10 kHz. The center frequency of the NBFM signal varphi1(t) is 10.1 MHz, the peak frequency deviation Deltaf1 = 2 kHz and the amplitude of the NBFM signal is 1 V. The broadcast signal is to have a center frequency of 104.5 MHz and a peak frequency deviation of no less than 50 kHz. Select the power-of-two frequency mulitiplication factor n and the frequency of the local oscillator f2. Select the center frequency of the BPF filter and the bandwidth of the BPF. Select the gain of the power amplifier A to produce an average power of 100 W.

Explanation / Answer

according to data given:

fc = 104.5 MHz

center frequency fc1 = 10.1 MHz

delta f = frequency deviation in broadcasted message = 50 khz = 50000 hz

delta f0 =frequency deviation in NBFM signal = 2 khz = 2000 hz

N = delta f / delta f0 = 50000/2000 = 25 = N answer

let frequency of local oscillator be fL

so, |f1-fL| = fc

f1 = N f0 = 25 X 10.1 MHz = 252.5 Mhz

hence, fL = f1-fc = 252.5-104.5 = 148 MHz ANSWER frequency of oscillator

BPF:

center frequency of BPF = fc = 104.5 MHz

BW = 2{delta f + fm} = 2 {50 KHz + 10.1 MHz} = 20.3 MHz ANSWER

Amplitude square / 2 = power => A^2 / 2 = 100 => A = sqrt 200 = 14.14 V

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