Consider a binary data sequence of bit rate 4978 Hz modulated by a carrier of 39661 Hz to obtain BPSK.*i) Calculate the bandwidthanswer format example if answer is 3.5khz write as 3500 ( only number) don't write as 3.5khz
Question
Consider a binary data sequence of bit rate 4978 Hz modulated by a carrier of 39661 Hz to obtain BPSK.*i) Calculate the bandwidthanswer format example if answer is 3.5khz write as 3500 ( only number) don't write as 3.5khz
Solution 1
In Binary Phase Shift Keying (BPSK), the bandwidth required is twice the bit rate.
So, the bandwidth = 2 * bit rate = 2 * 4978 Hz = 9956 Hz
So, the answer is 9956.
Solution 2
In Binary Phase Shift Keying (BPSK), the bandwidth required is equal to twice the bit rate.
So, the bandwidth = 2 * bit rate = 2 * 4978 Hz = 9956 Hz
So, the answer is 9956.
Similar Questions
Consider a BPSK modulated signal with bandwidth 4978 Hz i) Calculate the input data rateanswer format example if answer is 3.5khz write as 3500 ( only number) don't write as 3.5khz
Consider a BPSK modulated signal with bandwidth 19371 Hz i) Calculate the input data rate
choose the program for BPSK which is executable and will display error rate and demodulated data Select one:a.M = 2; % Modulation orderk = log2(M); % Bits per symbolsnr=10;numSymPerFrame = 50;dataIn = randi([0 k],numSymPerFrame,k)txSig = pskmod(dataIn,M)rxSig = awgn(txSig,snr)rxSym = pskdemod(rxSig,M);round(rxSym)nErrors = biterr(dataIn, rxSym);[nErrors err_rate] = biterr(dataIn, rxSym)b.M = 2; % Modulation orderk = log2(M); % Bits per symbolsnr=10;numSymPerFrame = 50;dataIn = randi([0 k],numSymPerFrame,k);txSig = pskmod(dataIn,M);rxSig = awgn(txSig,snr);rxSym = pskdemod(rxSig,M);round(rxSym)nErrors = biterr(dataIn, rxSym);[nErrors err_rate] = biterr(dataIn, rxSym)c.M = 2; % Modulation orderk = log2(M); % Bits per symbolsnr=10;numSymPerFrame = 50;dataIn = randi([0 k],numSymPerFrame,k);txSig = pskmod(dataIn,M)rxSig = awgn(txSig,snr)rxSym = pskdemod(rxSig,M);round(rxSym)nErrors = biterr(dataIn, rxSym);[nErrors err_rate] = biterr(dataIn, rxSym)d.M = 2; % Modulation orderk = log2(M); % Bits per symbolsnr=10;numSymPerFrame = 50;dataIn = randi([0 k],numSymPerFrame,k)txSig = pskmod(dataIn,M)rxSym = pskdemod(rxSig,M);nErrors = biterr(dataIn, rxSym);[nErrors err_rate] = biterr(dataIn, rxSym)
For FSK, the bit rate is equal to the baud rate (symbols/seconds).Group of answer choicesTrueFalse
if data sequence is 0 1 0 0 0 1 01 0the phase of BPSK waveform will vary as Select one:a. pi 0 pi pi pi 0 pi 0 pib. 0 0 pi pi 0 0 pi 0 pic. 0 0 pi pi pi 0 pi 0 pi
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.