16 clear; Fs 32, Sample rate (kHz) Rs 1.6; Symbol rate (kHz) Fs /Rs; No. of samp

Question

16 clear; Fs 32, Sample rate (kHz) Rs 1.6; Symbol rate (kHz) Fs /Rs; No. of samples per symbol 30; No. of bits to simulate Nb Discrete B (z) /A(z) model of telephone channel A 11.00, 2.838, 3.143 1.709, 0.458 0.0491; B 11 Pulse Shape the data pulse lones (1,Ns) bipolar NRZ data sign (randn (1,Nb)) t random data Sig pulse data; Sig Sig (t); Pass Signal Through the Channel RxSig filter (B,A,Sig): Plotting plot (real (RDSig) hold on; plot (Sig, r'); legend ("Received signal Transmitted signal hold off; xlabel ("Time Samples ylabel Amplitude');

Explanation / Answer

a) Manchester line coding has a [1 0] corresponding to the [1] of NRZ polar coding. Hence, while defining pulse as an array of ones, we need to change it and implement the following code instead of the pulse state at line 14.

pulse = [ones(1, Ns);-1.*ones(1, Ns)];
data_rep = sign(randn(1, Nb));
data = [data_rep; data_rep];

When we increase the data rate, we are able to send the same data as the NRZ coding at 1.6 kbps because now we are sending two bits in Manchester coding style instead of th eone bit in NRZ coding style.

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