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authorpriyanka2015-06-24 15:03:17 +0530
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+clc;
+//page 436
+//problem 8.2
+
+//Given frequency range fc - fm = 0.995MHz to fc + fm = 1.005Mhz
+//Double side message bandwidth is fM
+fM= (1.005 - 0.995)*10^6 / 2;
+disp('Message bandwidth is '+string(fM)+' Hz');
+//The textbook contains a calculation error here.
+//The calculated fM in the textbook is 500kHz instead of 5kHz,
+//Following which all the solutions obtained here are erroneous.
+
+//Given input signal strength Si= 1mW
+//Let output signal strength be So
+//So=Si/2
+Si= 10^(-3);
+So= Si/2;
+disp('Signal output strength is '+string(So)+' dB');
+
+//Given Power Spectral Density n = 10^-9 W/Hz
+//Let output noise strength be No
+n= 10^-9;
+No= (n*fM)/2;
+disp('Output Noise Strength is '+string(No)+' dB');
+
+//Let SNR at filter output be SNR
+SNR= So / No;
+disp('Output SNR of the DSB-SC wave is '+string(SNR)+' dB');
+
+//By reduction of message signal Bandwidth the Output Noise strength changes
+//Let the new output noise strength, bandwidth and SNR be be No_new, fM_new and SNR_new respectively
+fM_new = 75/100*fM;
+No_new = n*fM_new/4;
+SNR_new = So / No_new;
+disp('Changed SNR is '+string(SNR_new)+' dB');
+
+