diff options
Diffstat (limited to '3161')
68 files changed, 2483 insertions, 0 deletions
diff --git a/3161/CH1/EX1.1/Ex1_1.sce b/3161/CH1/EX1.1/Ex1_1.sce new file mode 100755 index 000000000..27e052bfd --- /dev/null +++ b/3161/CH1/EX1.1/Ex1_1.sce @@ -0,0 +1,24 @@ +clc;
+//page 12
+//problem 1.1
+
+//Given signal u = 2*exp(-3*t)
+
+//Since the function integral does not accept %inf as limit we need to use approximation by changing variables.
+
+//First the signal is to be expressed in terms of 'x'.
+
+function y=Signal(x);
+y=2*exp(-3*x);
+endfunction;
+
+//We then substitute x = tan(z), and then express the given signal wrt 'z' and not 'x'.
+
+function y=Gmodified(z);
+x=tan(z);
+y=(Signal(x))^2/(cos(z))^2;
+endfunction;
+
+E = intg(0,atan(10),Gmodified)
+
+disp(E,'The energy of this signal is ');
diff --git a/3161/CH1/EX1.2/Ex1_2.sce b/3161/CH1/EX1.2/Ex1_2.sce new file mode 100755 index 000000000..076106ac0 --- /dev/null +++ b/3161/CH1/EX1.2/Ex1_2.sce @@ -0,0 +1,13 @@ +clc;
+//page 12
+//problem 1.2
+
+//Given signal u = 2*sin(0.5*%pi*t)
+
+//Since u is periodic, averaging over -infinity to + infinity will give the same result as averaging over -2 to 2, where 4 is the time period.
+
+t0 = -2;
+t1 = 2;
+E = integrate('(2*sin(0.5*%pi*t))^2','t',t0,t1)/4;
+
+disp(E,'The power of the signal is ');
diff --git a/3161/CH1/EX1.3/Ex1_3.sce b/3161/CH1/EX1.3/Ex1_3.sce new file mode 100755 index 000000000..f7fec6d4d --- /dev/null +++ b/3161/CH1/EX1.3/Ex1_3.sce @@ -0,0 +1,38 @@ +clc;
+//page 18
+//problem 1.3
+
+//u1(T) vs T
+T = [-5:0.0082:5];
+u1(T<=0) = 0;
+u1(T>0) = 1;
+xlabel('T');
+ylabel('u(T)')
+
+subplot(131);
+plot2d(T,u1);
+
+//u2(T-t) vs T
+//Shifting the given signal by t units to the right, we get
+//Let us assume the amount of time to be shited is 3 units
+t = 3;
+
+T = [-5:0.0082:5];
+u2(T<=t) = 0;
+u2(T>t) = 1;
+xlabel('T');
+ylabel('u(T - t)')
+
+subplot(132);
+plot2d(T,u2);
+
+//u(t - T) = u(-(T - t))
+
+T = [-5:0.0082:5];
+u3(T>=t) = 0;
+u3(T<t) = 1;
+xlabel('T');
+ylabel('u(t - T)')
+
+subplot(133);
+plot2d(T,u3);
diff --git a/3161/CH1/EX1.4/Ex1_4.sce b/3161/CH1/EX1.4/Ex1_4.sce new file mode 100755 index 000000000..a840f0954 --- /dev/null +++ b/3161/CH1/EX1.4/Ex1_4.sce @@ -0,0 +1,45 @@ +clc;
+//page 18
+//problem 1.4
+
+//u1(t)
+t = [-5:0.0082:5];
+u1(t<=0) = 0;
+u1(t>0) = 1;
+
+xlabel('t');
+ylabel('u(t)')
+
+subplot(131);
+plot2d(t,u1);
+
+//u2(t-T)
+//Shifting the given signal by t units to the right, we get
+//Let us assume the amount of time to be shited is 3 units
+T = 3;
+
+t = [-5:0.0082:5];
+u2(t<=T) = 0;
+u2(t>T) = 1;
+
+xlabel('t');
+ylabel('u(t-T)')
+
+subplot(132);
+plot2d(t,u2);
+
+
+//u(t) - u(t - T)
+
+t = [-5:0.0082:5];
+u3 = u1 - u2;
+
+xlabel('t');
+ylabel('u(t) - u(t-T)')
+
+subplot(133);
+plot2d(t,u3);
+
+
+
+
diff --git a/3161/CH1/EX1.5/Ex1_5.sce b/3161/CH1/EX1.5/Ex1_5.sce new file mode 100755 index 000000000..435b55a3a --- /dev/null +++ b/3161/CH1/EX1.5/Ex1_5.sce @@ -0,0 +1,41 @@ +clc;
+//page 18
+//problem 1.5
+
+//V1(t) = u(t) - u(t - 5)
+t = [-5:0.1:5];
+V1(t<=0) = 0;
+V1(t>0) = 1;
+
+xlabel('t');
+ylabel('V1(t)')
+subplot(121);
+plot2d(t,V1);
+
+
+//V2(t) = 2*t*(u(t) - u(t - 3))
+t = [0:0.1:3];
+V2 = 2*t;
+
+xlabel('t');
+ylabel('V2(t)')
+subplot(122);
+plot2d(t,V2);
+
+//Autocorrelation R12(0) = R
+
+R = integrate('2*t','t',0,3);
+
+E1 = integrate('1','t',0,5);
+
+//In the textbook, E2 has been computed as 18 instead of 36
+E2 = integrate('4*t^2','t',0,3);
+
+c = R/(E1*E2)^0.5;
+
+disp(R,'The correlation term R12(0) is ');
+disp(E1,'The autocorrelation term R1(0) is ');
+disp(E2,'The autocorrelation term R2(0) is ');
+
+
+
diff --git a/3161/CH1/EX1.6/Ex1_6.sce b/3161/CH1/EX1.6/Ex1_6.sce new file mode 100755 index 000000000..c11c4246b --- /dev/null +++ b/3161/CH1/EX1.6/Ex1_6.sce @@ -0,0 +1,38 @@ +clc;
+//page 19
+//problem 1.6
+
+//V1(t) = u(t) - u(t - 5)
+t = [-5:0.1:5];
+V1(t<=0) = 0;
+V1(t>0) = 1;
+
+xlabel('t');
+ylabel('V1(t)')
+subplot(121);
+plot2d(t,V1);
+
+
+//V2(t) = 2*t*(u(t) - u(t - 3))
+t = [0:0.1:3];
+V2 = 2*t;
+
+xlabel('t');
+ylabel('V2(t)')
+subplot(122);
+plot2d(t,V2);
+
+//Autocorrelation R12(1) = Ra
+//The range is t = 0 to 2, as signal V2(t) has been shifted left by one unit, V2(t-1)
+
+Ra = integrate('2*(t+1)','t',0,2);
+
+disp(Ra,'The correlation term R12(1) is ');
+
+//Autocorrelation R12(-1) = Rb
+//The range is t = 1 to 4, as signal V2(t) has been shifted right by one unit, V2(t+1)
+
+Rb = integrate('2*(t-1)','t',1,4);
+
+disp(Rb,'The correlation term R12(-1) is ');
+
diff --git a/3161/CH10/EX10.3/Ex10_3.sce b/3161/CH10/EX10.3/Ex10_3.sce new file mode 100755 index 000000000..998794f6c --- /dev/null +++ b/3161/CH10/EX10.3/Ex10_3.sce @@ -0,0 +1,26 @@ +clc;
+//page 520
+//problem 10.3
+
+// Part (a)
+
+//Input SNR SNR_ip
+SNR_ip = 1000;
+
+//Beta B
+B = 10;
+
+//Output SNR SNR_op
+SNR_op = (1.5*(B^2)*SNR_ip)/(1 + (12*B/%pi)*(SNR_ip)*exp(-0.5*(1/(B+1))*(SNR_ip)));
+
+disp('Output SNR is '+string(10*log10(SNR_op))+' dB');
+
+// Part (b)
+
+//Input SNR SNR_ip
+SNR_ip = 10;
+
+//Output SNR SNR_op
+SNR_op = (1.5*(B^2)*SNR_ip)/(1 + (12*B/%pi)*(SNR_ip)*exp(-0.5*(1/(B+1))*(SNR_ip)));
+
+disp('Output SNR is '+string(10*log10(SNR_op))+' dB');
diff --git a/3161/CH10/EX10.5/Ex10_5.sce b/3161/CH10/EX10.5/Ex10_5.sce new file mode 100755 index 000000000..fd4570349 --- /dev/null +++ b/3161/CH10/EX10.5/Ex10_5.sce @@ -0,0 +1,32 @@ +clc;
+//page 533
+//problem 10.5
+
+//Given reference frequency is fref = 10 MHz
+fref = 10 * 10^6;
+
+//Given step frequency is fstep = 100 KHz
+fstep = 100 * 10^3;
+
+//Division ratio M
+M = fref/fstep;
+
+//Required output frequency F = 100.6 MHz
+F = 100.6 * 10^6;
+
+N = F/fstep;
+
+//Given P = 64
+P = 64;
+
+//Truncating value B = 15
+B = 15;
+
+A = N - P*B;
+
+disp('The value of A is '+string(A));
+disp('The value of B is '+string(B));
+disp('The value of M is '+string(M));
+
+
+
diff --git a/3161/CH10/EX10.7/Ex10_7.sce b/3161/CH10/EX10.7/Ex10_7.sce new file mode 100644 index 000000000..3cd75f1a8 --- /dev/null +++ b/3161/CH10/EX10.7/Ex10_7.sce @@ -0,0 +1,25 @@ +clc;
+//page 534
+//problem 10.7
+
+//Given reference frequency for PLL is fref = 0.48 MHz
+fref = 0.48 * 10^6;
+
+//Frequency divider N = 2000
+N = 2000;
+
+//Output Frequency fout
+fout = fref*N;
+
+//Output Frequency favg
+favg = (2000*15 + 2001*1)*(0.48/16) * 10^6;
+
+disp('Output frequency is '+string(fout)+' Hz');
+
+//Reference frequency is not subdivided before going to comparator and it is an integer divider in the feedback path the frequency resolution fres = 0.48 * 10^6;
+fres = 0.48 * 10^6;
+
+disp('Frequency resolution is '+string(fres)+' Hz');
+
+disp('Output frequency resolution is '+string(favg - fout)+' Hz');
+
diff --git a/3161/CH11/EX11.3/Ex11_3.sce b/3161/CH11/EX11.3/Ex11_3.sce new file mode 100644 index 000000000..a3254b02f --- /dev/null +++ b/3161/CH11/EX11.3/Ex11_3.sce @@ -0,0 +1,47 @@ +clc;
+//page 558
+//problem 11.3
+
+//Prior probability of s1 P_s1 = 0.4
+P_s1 = 0.4;
+
+//Prior probability of s2 P_s2 = 1 - P_s1
+P_s2 = 1 - P_s1;
+
+//Voltage level V1 = 1
+V1 = 1;
+
+//Voltage level V2 = -1
+V2 = -1;
+
+//Part a
+
+//Noise Variance sigma1 = 10^-3
+sigma1 = 10^-3;
+
+//Descision Threshold lambda1
+lambda1 = (V1+V2)/2 + (sigma1)*log(P_s2/P_s1)/(V1-V2);
+
+//Probability of error Pe
+Pe = 0.5*(2*P_s1 - P_s1*erfc(((V2-V1)/(2*sigma1*2^0.5)) + (sigma1)*log(P_s2/P_s1)/((V1-V2)*2^0.5)));
+
+disp('The decision threshold is '+string(lambda1)+' V');
+disp('The probability of error is approximately '+string(Pe));
+
+//Part b
+
+//Noise Variance sigma2 = 10^-1
+sigma2 = 10^-1;
+
+//Descision Threshold lambda2
+lambda2 = (V1+V2)/2 + (sigma2)*log(P_s2/P_s1)/(V1-V2);
+
+//Probability of error Pe
+Pe1 = 0.5*(2*P_s1 - P_s1*erfc(((V2-V1)/(2*sigma2*2^0.5)) + (sigma2)*log(P_s2/P_s1)/((V1-V2)*2^0.5)));
+
+//In the textbook Pe has been calulated to be 0.0021 because of the use of a very high precision calculator, unfortunately in scilab the function erfc approximates the output value to a larger extent due to which an exact value cannot be obtained.
+
+disp('The decision threshold is '+string(lambda2)+' V');
+disp('The probability of error is approximately '+string(Pe1));
+
+
diff --git a/3161/CH11/EX11.4/Ex11_4.sce b/3161/CH11/EX11.4/Ex11_4.sce new file mode 100644 index 000000000..f0cd9a447 --- /dev/null +++ b/3161/CH11/EX11.4/Ex11_4.sce @@ -0,0 +1,31 @@ +clc;
+//page 559
+//problem 11.4
+
+//Part b
+
+//Voltage level V1 = 1
+V1 = 1;
+
+//Voltage level V2 = -1
+V2 = -1;
+
+//Prior probability of s1 P_s1 = 0.4
+P_s1 = 0.4;
+
+//Prior probability of s2 P_s2 = 1 - P_s1
+P_s2 = 1 - P_s1;
+
+//Cost of selecting s1 when s2 is transmitted C12 = 0.7
+C12 = 0.7;
+
+//Cost of selecting s2 when s1 is transmitted C21 = 1 - C12
+C21 = 1 - C12;
+
+//Noise Variance sigma = 10^-3
+sigma = 10^-3;
+
+//Descision Threshold lambda
+lambda = (V1+V2)/2 + (sigma)*log((C12*P_s2)/(C21*P_s1))/(V1-V2);
+
+disp('The decision threshold is '+string(lambda)+' V');
diff --git a/3161/CH11/EX11.5/Ex11_5.sce b/3161/CH11/EX11.5/Ex11_5.sce new file mode 100755 index 000000000..cd99b4750 --- /dev/null +++ b/3161/CH11/EX11.5/Ex11_5.sce @@ -0,0 +1,45 @@ +clc;
+//page 567
+//problem 11.5
+
+//The voltage level of reciever is V = 5 mV
+V = 5*10^-3;
+
+//The time required to transfer one bit is T = 1/9600 sec
+T = 9600^-1;
+
+//the signal energy of a bit be Es
+Es = (V^2)*T;
+
+//The power spectral density is n/2 = 10^-9 Watt/hertz
+n = 2*10^-9;
+
+//Probability error for optimal reciever is Pe
+Pe = 0.5*erfc((Es/n)^0.5);
+
+disp('The probability of error is '+string(Pe));
+
+//When the data rate is doubled, the new effective energy per bit is Es_new
+Es_new = (V^2)*(T/2);
+
+//The new probability of error is Pe_new
+Pe_new = 0.5*erfc((Es_new/n)^0.5);
+
+//Percentage increase in error rate is P
+P = 100*(Pe_new - Pe)/Pe;
+
+disp('Percentage increase in error rate is '+string(P));
+
+//Voltage required to restore probability levels, V_new
+V_new = V*2^0.5;
+
+disp('Voltage required to restore the probability levels is '+string(V_new)+' Volts');
+
+
+
+
+
+
+
+
+
diff --git a/3161/CH11/EX11.6/Ex11_6.sce b/3161/CH11/EX11.6/Ex11_6.sce new file mode 100755 index 000000000..f5f2c5533 --- /dev/null +++ b/3161/CH11/EX11.6/Ex11_6.sce @@ -0,0 +1,52 @@ +clc;
+//page 575
+//problem 11.6
+
+//Amplitude of signal is A = 10 mV
+A = 10*10^-3;
+
+//Power Spectral Density n = 2 * 10^(-9) W/Hz
+n = 2 * 10^(-9);
+
+//Frequency is f = 1 MHz
+f = 1*10^6;
+
+//Data rate is D = 10^4 bps;
+D = 10^4;
+
+//Time taken for a bit to traverse
+T = 1/D;
+
+//Energy per signal element is Es
+Es = A^2/(2*D);
+
+//Probability of error Pe
+Pe = 0.5*erfc((Es/n)^0.5);
+
+disp('Probability of error is '+string(Pe));
+
+//Phase shift phi = %pi/6
+phi = %pi/6;
+
+//Probability of error Pe_local_oscillator
+Pe_local_oscillator = 0.5*erfc(((Es/n)^0.5)*cos(phi));
+
+disp('Probability of error of local oscillator with phase shift is '+string(Pe_local_oscillator));
+
+//Timing error t
+t = 0.1*T;
+
+//Probability of error when there is a synchronization fault Pe_timing_error
+Pe_timing_error = 0.5*erfc(((Es/n)*(1 - 2*(t/T))^2)^0.5);
+
+disp('Probability of error with synchronization fault is '+string(Pe_timing_error));
+
+//Probability of error when both faults occur Pe_both
+Pe_both = 0.5*erfc(((Es/n)*(cos(phi)^2)*(1 - 2*(t/T))^2)^0.5);
+
+disp('Probability of error when both faults occur '+string(Pe_both));
+
+
+
+
+
diff --git a/3161/CH11/EX11.7/Ex11_7.sce b/3161/CH11/EX11.7/Ex11_7.sce new file mode 100755 index 000000000..846eadcba --- /dev/null +++ b/3161/CH11/EX11.7/Ex11_7.sce @@ -0,0 +1,31 @@ +clc;
+//page 575
+//problem 11.7
+
+//Amplitude of signal is A = 10 mV
+A = 10*10^-3;
+
+//Power Spectral Density n = 2 * 10^(-9) W/Hz
+n = 2 * 10^(-9);
+
+//Data rate is D = 10^4 bps;
+D = 10^4;
+
+//Time taken for a bit to traverse
+T = 1/D;
+
+//Energy per signal element is Es
+Es = A^2/(2*D);
+
+//Probability of error Pe_a
+Pe_a = 0.5*erfc((0.6*Es/n)^0.5);
+disp('Probability of error when offset is small is '+string(Pe_a));
+
+//Probability of error Pe_b
+Pe_b = 0.5*erfc((Es/(2*n))^0.5);
+disp('Probability of error when frequencies used are orthogonal is '+string(Pe_b));
+
+//Probability of error Pe_c
+Pe_c = 0.5*exp(-(Es/(2*n)));
+disp('Probability of error for non coherent detection is '+string(Pe_c));
+
diff --git a/3161/CH11/EX11.8/Ex11_8.sce b/3161/CH11/EX11.8/Ex11_8.sce new file mode 100755 index 000000000..f967df937 --- /dev/null +++ b/3161/CH11/EX11.8/Ex11_8.sce @@ -0,0 +1,30 @@ +clc;
+//page 588
+//problem 11.8
+
+//Energy aasosciated with each bit Eb = 5 * 10^(-8) J
+Eb = 5 * 10^(-8);
+
+//Power Spectral Density n = 2 * 10^(-9) W/Hz
+n = 2 * 10^(-9);
+
+//No of symbols M
+M = 16
+
+//No of bits N
+N = log2(M);
+
+//Error limit for 16-PSK is P_16_PSK
+P_16_PSK = erfc(((N*Eb*(%pi)^2)/(((M)^2)*n))^0.5);
+
+disp('Upper limit of error probability of 16 PSK system is '+string(P_16_PSK));
+
+//Error limit for 16-QASK is P_16_QASK
+P_16_QASK = 2*erfc(((0.4*Eb)/(n))^0.5);
+
+disp('Upper limit of error probability of 16 QASK system is '+string(P_16_QASK));
+
+//Error limit for 16-FSK is P_16_FSK
+P_16_FSK = ((2^4 - 1)/2)*erfc(((N*Eb)/(2*n))^0.5);
+
+disp('Upper limit of error probability of 16 FSK system is '+string(P_16_FSK)+', negligibly small');
diff --git a/3161/CH11/EX11.9/Ex11_9.sce b/3161/CH11/EX11.9/Ex11_9.sce new file mode 100755 index 000000000..a8edb9be3 --- /dev/null +++ b/3161/CH11/EX11.9/Ex11_9.sce @@ -0,0 +1,22 @@ +clc;
+//page 595
+//problem 11.9
+
+//Energy aasosciated with each bit Eb = 5 * 10^(-8) J
+Eb = 5 * 10^(-8);
+
+//Power Spectral Density n = 2 * 10^(-9) W/Hz
+n = 2 * 10^(-9);
+
+//Probability of error Pe
+Pe = 0.5*erfc(((Eb*(%pi)^2)/(16*n))^0.5);
+
+disp('Probability of error of QPR system is '+string(Pe));
+
+//Given Bandwidth of channel is BW
+BW = 10*10^3;
+
+D = 2*BW;
+
+disp('Data rate is '+string(D)+' bps');
+
diff --git a/3161/CH12/EX12.2/Ex12_2.sce b/3161/CH12/EX12.2/Ex12_2.sce new file mode 100644 index 000000000..dae239803 --- /dev/null +++ b/3161/CH12/EX12.2/Ex12_2.sce @@ -0,0 +1,71 @@ +clc;
+//page 608
+//problem 12.2
+
+//Baseband cutoff signal fM = 4 kHz
+fM = 4 * 10^3;
+
+//White noise power spectral density n
+n = 2*10^(-9);
+
+// Part (a)
+
+//Input Signal energy Si = 0.001
+Si_a = 0.001;
+
+//No of levels used for PCM Coding M = 8
+M_a = 8;
+
+N_a = log2(M_a);
+
+//Input SNR is SNR_ip
+SNR_ip = Si_a/(n*fM);
+
+//Output SNR is SNR_op
+SNR_op = (2^(2*N_a))/(1 + (2^(2*N_a + 1))*erfc((SNR_ip*(1/(2*N_a))))^0.5);
+
+disp('Input SNR for (a) is '+string(10*log10(SNR_ip))+' dB');
+disp('Output SNR (a) is '+string(10*log10(SNR_op))+' dB');
+
+// Part (b)
+
+//Input Signal energy Si = 0.001
+Si_b = 0.001;
+
+//No of levels used for PCM Coding M = 256
+M_b = 256;
+
+N_b = log2(M_b);
+
+//Input SNR is SNR_ip_b
+SNR_ip_b = Si_b/(n*fM);
+
+//Output SNR is SNR_op_b
+SNR_op_b = (2^(2*N_b))/(1 + (2^(2*N_b + 1))*erfc((SNR_ip_b*(1/(2*N_b))))^0.5);
+
+//Unfortunately in scilab the function erfc approximates the output value to a larger extent due to which an exact value cannot be obtained.
+//The difference in the textbook answer and obatined answer is significant because of converting the answer into dB.
+
+disp('Input SNR for (b) is '+string(10*log10(SNR_ip_b))+' dB');
+disp('Output SNR for (b) is '+string(10*log10(SNR_op_b))+' dB');
+
+// Part (c)
+
+//Input Signal energy Si = 0.01
+Si_c = 0.01;
+
+//No of levels used for PCM Coding M = 256
+M_c = 256;
+
+N_c = log2(M_c);
+
+//Input SNR is SNR_ip_c
+SNR_ip_c = Si_c/(n*fM);
+
+//Output SNR is SNR_op_c
+SNR_op_c = (2^(2*N_c))/(1 + (2^(2*N_c + 1))*erfc((SNR_ip_c*(1/(2*N_c))))^0.5);
+
+disp('Input SNR for (c) is '+string(10*log10(SNR_ip_c))+' dB');
+disp('Output SNR for (c) is '+string(10*log10(SNR_op_c))+' dB');
+
+
diff --git a/3161/CH12/EX12.3/Ex12_3.sce b/3161/CH12/EX12.3/Ex12_3.sce new file mode 100644 index 000000000..eb896cee5 --- /dev/null +++ b/3161/CH12/EX12.3/Ex12_3.sce @@ -0,0 +1,71 @@ +clc;
+//page 609
+//problem 12.3
+
+//Baseband cutoff signal fM = 4 kHz
+fM = 4 * 10^3;
+
+//White noise power spectral density n
+n = 2*10^(-9);
+
+// Part (a)
+
+//Input Signal energy Si = 0.001
+Si = 0.001;
+
+//No of levels used for PCM Coding M = 8
+M = 8;
+
+N = log2(M);
+
+//Input SNR is SNR_ip
+SNR_ip = Si/(n*fM);
+
+//Output SNR is SNR_op
+SNR_op = (2^(2*N))/(1 + (2^(2*N + 1))*erfc((SNR_ip*(3/(10*N))))^0.5);
+
+disp('Input SNR for (a) is '+string(10*log10(SNR_ip))+' dB');
+disp('Output SNR (a) is '+string(10*log10(SNR_op))+' dB');
+
+// Part (b)
+
+//Input Signal energy Si = 0.001
+Si = 0.001;
+
+//No of levels used for PCM Coding M = 256
+M_b = 256;
+
+N_b = log2(M_b);
+
+//Input SNR is SNR_ip_b
+SNR_ip_b = Si/(n*fM);
+
+//Output SNR is SNR_op_b
+SNR_op_b = (2^(2*N_b))/(1 + (2^(2*N_b + 1))*erfc((SNR_ip_b*(3/(10*N_b))))^0.5);
+
+//Unfortunately in scilab the function erfc approximates the output value to a larger extent due to which an exact value cannot be obtained.
+//The difference in the textbook answer and obatined answer is significant because of converting the answer into dB.
+
+disp('Input SNR for (b) is '+string(10*log10(SNR_ip_b))+' dB');
+disp('Output SNR for (b) is '+string(10*log10(SNR_op_b))+' dB');
+
+// Part (c)
+
+//Input Signal energy Si = 0.01
+Si = 0.01;
+
+//No of levels used for PCM Coding M = 256
+M = 256;
+
+N = log2(M);
+
+//Input SNR is SNR_ip_c
+SNR_ip_c = Si/(n*fM);
+
+//Output SNR is SNR_op_c
+SNR_op_c = (2^(2*N))/(1 + (2^(2*N + 1))*erfc((SNR_ip_c*(3/(10*N))))^0.5);
+
+disp('Input SNR for (c) is '+string(10*log10(SNR_ip_c))+' dB');
+disp('Output SNR for (c) is '+string(10*log10(SNR_op_c))+' dB');
+
+
diff --git a/3161/CH12/EX12.4/Ex12_4.sce b/3161/CH12/EX12.4/Ex12_4.sce new file mode 100755 index 000000000..d24977f72 --- /dev/null +++ b/3161/CH12/EX12.4/Ex12_4.sce @@ -0,0 +1,31 @@ +clc;
+//page 618
+//problem 12.4
+
+//Upper cut off frequecny fb = 3200 Hz
+fM = 3200;
+
+//Lower cut off frequecny fl = 300 Hz
+fl = 300;
+
+//Data rate fb = 32000 bps
+fb = 32000;
+
+//White noise power spectral density n
+n = 2*10^(-9);
+
+//Input Signal energy Si = 0.001
+Si = 0.001;
+
+//Output SNR is SNR_op
+SNR_op = (0.6*(fb/fM)^3)/(1 + (0.3*(fb^2/(fl*fM)))*erfc(Si/(n*fb)));
+
+disp('Output SNR is '+string(10*log10(SNR_op))+'dB');
+
+//Data rate fb_n = 32000 bps
+fb_n = 2*32000;
+
+//Output SNR is SNR_op_n
+SNR_op_n = (0.6*(fb_n/fM)^3)/(1 + (0.3*(fb_n^2/(fl*fM)))*erfc(Si/(n*fb_n)));
+
+disp('Output SNR when data rate is doubled is '+string(10*log10(SNR_op_n))+'dB');
diff --git a/3161/CH13/EX13.1/Ex13_1.sce b/3161/CH13/EX13.1/Ex13_1.sce new file mode 100755 index 000000000..271ec7689 --- /dev/null +++ b/3161/CH13/EX13.1/Ex13_1.sce @@ -0,0 +1,19 @@ +clc;
+//page 631
+//problem 13.1
+
+//Given probabilities p1 = p4 = 1/8 & p2 = p3 = 3/8
+p1 = 1/8
+p4 = 1/8
+p2 = 3/8
+p3 = 3/8
+
+//The average information H is p1*log2 (1/p1)+p2*log2 (1/p2)+p3*log2 (1/p3)+p4*log2 (1/p4) bits/message
+H = p1*log2 (1/p1)+p2*log2 (1/p2)+p3*log2 (1/p3)+p4*log2 (1/p4)
+
+//information rate R is r*H bits/sec where r is 2*B
+//R1 = R/B
+R1 = 2*H
+
+disp('The average information is '+string(H)+' bits/message')
+disp('Information rate '+string(R1)+'*B bits/sec')
diff --git a/3161/CH13/EX13.11/Ex13_11.sce b/3161/CH13/EX13.11/Ex13_11.sce new file mode 100755 index 000000000..a30ead94a --- /dev/null +++ b/3161/CH13/EX13.11/Ex13_11.sce @@ -0,0 +1,38 @@ +clc;
+//page 696
+//problem 13.11
+
+//The output equations are as follows v1 = s1 xor s2 xor s3 & v2 = s1 xor s3
+//the no of bits in output mode(bits_out) is v*(L+K), v = no of outputs for commutatot = 2, L = length of input = 3 & K = no of memeory elements = 3
+v = 2
+L = 3
+K = 3
+bits_out = v*(L+K)
+
+//Taking in, s1, s2 , s3, v1 & v2 as row matrix where each column represents its corresponding input or output, in means input
+in = [0 1 0 1 0 0 0]
+
+s1 = zeros(1,7)
+s2 = zeros(1,7)
+s3 = zeros(1,7)
+v1 = zeros(1,7)
+v2 = zeros(1,7)
+
+
+for i = 2:7
+ s3(i) = s2(i-1)
+ s2(i) = s1(i-1)
+ s1(i) = in(i-1)
+ v1(i-1) = bitxor(s1(i),bitxor(s2(i),s3(i)))
+ v2(i-1) = bitxor(s1(i),s3(i))
+end
+
+//Output matrix is out
+out = zeros(1,12)
+for i = [1 3 5 7 9 11]
+ out(i) = v1((i+3)/2)
+ out(i+1) = v2((i+3)/2)
+end
+
+disp('output is')
+disp(out)
diff --git a/3161/CH13/EX13.12/Ex13_12.sce b/3161/CH13/EX13.12/Ex13_12.sce new file mode 100755 index 000000000..c92696c8c --- /dev/null +++ b/3161/CH13/EX13.12/Ex13_12.sce @@ -0,0 +1,58 @@ +clc;
+//page 697
+//problem 13.12
+
+//The qeneratr matrix requires impulse response of the coder.
+//This is the ourput generated when the initially reset coder is fed with a single 1.
+//The no of bits in the output code = 2(1+3) = 8
+
+//Taking in, s1, s2 , s3, v1 & v2 as row matrix where each column represents its corresponding input or output, in means input
+in = [0 1 0 0 0]
+
+s1 = zeros(1,5)
+s2 = zeros(1,5)
+s3 = zeros(1,5)
+v1 = zeros(1,5)
+v2 = zeros(1,5)
+
+
+for i = 2:5
+ s3(i) = s2(i-1)
+ s2(i) = s1(i-1)
+ s1(i) = in(i-1)
+ v1(i-1) = bitxor(s1(i),bitxor(s2(i),s3(i)))
+ v2(i-1) = bitxor(s1(i),s3(i))
+end
+
+//Output matrix is out
+out = zeros(1,8)
+for i = [1 3 5 7]
+ out(i) = v1((i+3)/2)
+ out(i+1) = v2((i+3)/2)
+end
+
+disp('impulse response is')
+disp(out)
+
+//Then generator matrix is G
+G = [1 1 1 0 1 1 0 0 0 0 0 0;0 0 1 1 1 0 1 1 0 0 0 0;0 0 0 0 1 1 1 0 1 1 0 0]
+
+//Note that, in G, impulse responses appear in staggered apper in a staggered manner in each row while the rest of the elements are 0.
+
+//Now, output is b_o = b_i*G where input b_i =[1 0 1]
+b_i = [1 0 1]
+
+b_o = b_i*G
+
+//Here multiplication means Exor operation so whereever two occurs it should be changed to 1
+
+for i = 1:12
+ if b_o(i) > 1 then
+ b_o(i) = 0;
+ end
+end
+
+disp('output is ')
+disp(b_o)
+disp('The output obtained is exactly same as example 13.1')
+
diff --git a/3161/CH13/EX13.14/Ex13_14.sce b/3161/CH13/EX13.14/Ex13_14.sce new file mode 100755 index 000000000..5094beefe --- /dev/null +++ b/3161/CH13/EX13.14/Ex13_14.sce @@ -0,0 +1,24 @@ +clc;
+//page 701
+//problem 13.14
+
+//Given, Tw = 10microsec, BCH(1023973) code is used implies k as 973 & n as 1023, P_A = 0.99, T1 = 40microsec & N = 4
+Tw = 10*10^-6
+k = 973
+n = 1023
+P_A = 0.99
+T1 = 40*10^-6
+N = 4
+
+//efficiency of Stop-and-Wait ARQ(n_SandW) = (k/n)*(P_A/(1+(T1/Tw)))
+n_SandW = (k/n)*(P_A/(1+(T1/Tw)))
+
+//efficiency of Go-Back-N ARQ(n_GBN) = (k/n)*(1/(1+(N*(1-P_A)/P_A)))
+n_GBN = (k/n)*(1/(1+(N*(1-P_A)/P_A)))
+
+//efficiency of Selective Repeat ARQ(n_SR) = (k/n)*P_A
+n_SR = (k/n)*P_A
+
+disp('efficiency of Stop-and-Wait ARQ is '+string(n_SandW))
+disp('efficiency of Go-Back-N ARQ is '+string(n_GBN))
+disp('efficiency of Selective Repeat ARQ is '+string(n_SR))
diff --git a/3161/CH13/EX13.15/Ex13_15.sce b/3161/CH13/EX13.15/Ex13_15.sce new file mode 100644 index 000000000..91bffdd71 --- /dev/null +++ b/3161/CH13/EX13.15/Ex13_15.sce @@ -0,0 +1,21 @@ +clc;
+//page 718
+//problem 13.15
+
+//Bit interval T = 1/10^6 = 10^-6 sec
+T = 10^-6
+
+//White Noise Power Spectral Density n/2 = 10^-9 W/Hz
+n = 2*10^-9
+
+//Power required Ps = Eb/T, where Eb = energy per bit
+
+//For information system feedback system Eb = n
+Ps = n/T
+
+disp('power required for information system feedback system is '+string(Ps)+' Watt')
+
+//For optimal system Ps = (0.69 * n)/T
+Ps = (0.69 * n)/T;
+
+disp('power required for optimal system is '+string(Ps)+' Watt')
diff --git a/3161/CH13/EX13.16/Ex13_16.sce b/3161/CH13/EX13.16/Ex13_16.sce new file mode 100755 index 000000000..53062aa2f --- /dev/null +++ b/3161/CH13/EX13.16/Ex13_16.sce @@ -0,0 +1,17 @@ +clc;
+//page 719
+//problem 13.16
+
+//Given, Eb = 10^-9Joule, n/2 = 10^-9 Watt/Hertz
+Eb = 10^-8
+n = 2*10^-9
+
+//Probability of error for trellis-decoded modulation(Pe) = (1/2)*erfc(sqrt(1.5*Eb/n))
+Pe = (1/2)*erfc(sqrt(1.5*Eb/n))
+
+disp('Probability of error for trellis-decoded modulation is '+string(Pe))
+
+//Probability of error for Qpsk modulation(Pe) = (1/2)*erfc(sqrt(Eb/n))
+Pe = (1/2)*erfc(sqrt(Eb/n))
+
+disp('Probability of error for Qpsk modulation is '+string(Pe))
diff --git a/3161/CH13/EX13.4/Ex13_4.sce b/3161/CH13/EX13.4/Ex13_4.sce new file mode 100755 index 000000000..e04563c2c --- /dev/null +++ b/3161/CH13/EX13.4/Ex13_4.sce @@ -0,0 +1,34 @@ +clc;
+//page 649
+//problem 13.4
+
+//Given bandwidth(B) = 4000Hertz & Noise PSD(n/2) = 10^-9 Watt/Hertz
+B = 4000
+n = 2*10^-9
+
+//Chanel capacity(C) = B*log2 (1+S/(n*B))
+
+//case 1
+//Signal energy(S) = 0.1Joule
+S = 0.1
+
+C = B*log2 (1+S/(n*B))
+
+disp('Channel capacity for bandwidth = 4000Hertz, Noise PSD = 10^-9 Watt/Hertz & Signal energy(S) = 0.1Joule is '+string(C)+' bits/sec')
+
+//case 2
+//Signal energy(S) = 0.001Joule
+S = 0.001
+
+C = B*log2 (1+S/(n*B))
+
+disp('Channel capacity for bandwidth = 4000Hertz, Noise PSD = 10^-9 Watt/Hertz & Signal energy(S) = 0.001Joule is '+string(C)+' bits/sec')
+
+//case 3
+//Signal energy(S) = 0.001Joule & incresed bandwidth(B) = 10000Hertz
+B = 10000
+S = 0.001
+
+C = B*log2 (1+S/(n*B))
+
+disp('Channel capacity for bandwidth = 10000Hertz, Noise PSD = 10^-9 Watt/Hertz & Signal energy(S) = 0.001Joule is '+string(C)+' bits/sec')
diff --git a/3161/CH13/EX13.8/Ex13_8.sce b/3161/CH13/EX13.8/Ex13_8.sce new file mode 100755 index 000000000..2b6f3b869 --- /dev/null +++ b/3161/CH13/EX13.8/Ex13_8.sce @@ -0,0 +1,20 @@ +clc;
+//page 675
+//problem 13.8
+
+//With single parity bit added, the code size = 4. An error evades parity check if any 2 or all symbols ofthe code arrives are erroneous.
+//Probability of any symbol from n are erroneous = nCm*(p^m)*(1-p)^(n-m)
+
+//Thus, the probability of error undetected, P_undeterr = (4C2*(p^2)*(1-p)^2)+4C4*(p^4) = 6*(p^2)*(1-p)^2)+(p^4)
+
+//Probability of error in detection(P_undeterr1) for p = 0.1
+p = 0.1
+P_undeterr1 = 6*(p^2)*((1-p)^2)+(p^4)
+
+disp('Probability of error in detection for p = 0.1 is '+string(P_undeterr1))
+
+//Probability of error in detection(P_undeterr2) for p = 0.01
+p = 0.01
+P_undeterr2 = 6*(p^2)*((1-p)^2)+(p^4)
+
+disp('Probability of error in detection for p = 0.01 is '+string(P_undeterr2))
diff --git a/3161/CH14/EX14.1/Ex14_1.sce b/3161/CH14/EX14.1/Ex14_1.sce new file mode 100755 index 000000000..f7784bcdb --- /dev/null +++ b/3161/CH14/EX14.1/Ex14_1.sce @@ -0,0 +1,48 @@ +clc;
+//page 738
+//problem 14.1
+
+//Boltzman constant k = 1.3806488 × 10-23 m2 kg s-2 K-1
+k = 1.3806488 * 10^-23;
+
+//Let room temperature be 27 C
+T = 27 + 273;
+
+//Bandwidth BW = 10 MHz
+BW = 10 * 10 ^6;
+
+//For (a)
+//Let the equivalent resistance be Ra
+Ra = 10 + 10;
+
+//RMS Noise Voltage be Va
+Va = (4*k*T*Ra*BW)^0.5;
+
+disp('The rms voltage at output a is '+string(Va)+' Volt');
+
+//For (b)
+//Let the equivalent resistance be Rb
+Rb = (10 * 10)/(10+10);
+
+//RMS Noise Voltage be Vb
+Vb = (4*k*T*Rb*BW)^0.5;
+
+disp('The rms voltage at output b is '+string(Vb)+' Volt');
+
+//For (c)
+
+Rc = 10;
+C = 1*10^-9;
+
+//In the textbook, the author has forgotten to multiply the result with T, hence has obtained an erroneous result.
+//The given answer is 28.01uV but the correct answer is found out to be 1.2uV
+
+Vc_square = 2*k*integrate('Rc/(1 + (2*%pi*Rc*C*f)^2)','f',-10^7,10^7);
+Vc = Vc_square^0.5;
+
+disp('The rms voltage at output c is '+string(Vc)+' Volt');
+
+
+
+
+
diff --git a/3161/CH14/EX14.2/Ex14_2.sce b/3161/CH14/EX14.2/Ex14_2.sce new file mode 100755 index 000000000..55272601b --- /dev/null +++ b/3161/CH14/EX14.2/Ex14_2.sce @@ -0,0 +1,27 @@ +clc;
+//page 741
+//problem 14.2
+
+//The Antenna noise temperature is T_ant = 10 K
+T_ant = 10;
+
+//The reciever noise temperature is Te = 140 K
+Te = 140;
+
+//Midband available gain of reciever gao = 10^10
+gao = 10^10;
+
+//Noise bandwidth is BN = 1.5 * 10^5 Hz
+BN = 1.5 * 10^5;
+
+//Boltzman constant k = 1.3806488 × 10-23 m2 kg s-2 K-1
+k = 1.3806488 * 10^-23;
+
+//Available noise power at output is pao
+
+pao = gao*k*(T_ant + Te)*BN;
+
+disp('The available output noise power is '+string(pao)+' Watts');
+
+
+
diff --git a/3161/CH14/EX14.3/Ex14_3.sce b/3161/CH14/EX14.3/Ex14_3.sce new file mode 100755 index 000000000..ce5d545bc --- /dev/null +++ b/3161/CH14/EX14.3/Ex14_3.sce @@ -0,0 +1,39 @@ +clc;
+//page 748
+//problem 14.3
+
+//The distance d = 30 * 1.6 * 10^3 m;
+d = 30 * 1.6 * 10^3;
+
+//Frequency f = 4 * 10^9 Hz
+f = 4 * 10^9;
+
+//Wavelength w = c/f m
+w = 3*10^8 / f;
+
+//Transmitter gain KT = 40 dB
+KT = 10^4;
+
+//Reciever gain KT = 40 dB
+KR = 10^4;
+
+//Reciever power PR = 10^-6 Watt
+PR = 10^-6;
+
+//Transmitter power PT
+PT = PR*(4*%pi*d/w)^2/ (KT*KR);
+
+disp('The transmitter output is '+string(PT)+' Watt')
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3161/CH15/EX15.1/Ex15_1.sce b/3161/CH15/EX15.1/Ex15_1.sce new file mode 100644 index 000000000..be26fcd6e --- /dev/null +++ b/3161/CH15/EX15.1/Ex15_1.sce @@ -0,0 +1,40 @@ +clc;
+//page 764
+//problem 15.1
+
+//Signal Power data rate fb = 100 Kbps
+fb = 10^5;
+
+//Signal Strength Ps = 1 mW
+Ps = 1*10^(-3);
+
+//Chip frequency fs = 100 MHz
+fs = 10^8;
+
+//Noise Spectral Density n = 2*10^(-9) W/Hz
+n = 2*10^(-9);
+
+//Jamming Signal power is Pj = 1 W
+Pj = 1;
+
+//Processing Gain P
+P = fs/fb;
+disp('Processing Gain is '+string(P));
+
+//Bit Interval T
+T = 1/fb;
+disp('Bit Interval is '+string(T)+'s');
+
+//Energy per bit Eb
+Eb = Ps*T;
+disp('Energy per bit is '+string(Eb));
+
+//Error Probability without jamming E_without_jamming
+E_without_jamming = 0.5*erfc((Eb/(n))^0.5);
+disp('Error probability without jamming is '+string(E_without_jamming));
+
+//Error Probability with jamming E_jamming
+E_jamming = 0.5*erfc(((2*Ps*P)/(Pj))^0.5);
+disp('Error probability jamming is '+string(E_jamming));
+
+
diff --git a/3161/CH15/EX15.2/Ex15_2.sce b/3161/CH15/EX15.2/Ex15_2.sce new file mode 100644 index 000000000..fb437e618 --- /dev/null +++ b/3161/CH15/EX15.2/Ex15_2.sce @@ -0,0 +1,21 @@ +clc;
+//page 764
+//problem 15.2
+
+//Chip Rate fc = 110 MHz
+fc = 10*10^6;
+Tc = 1/fc;
+
+//Delay D = 0.1 ms
+D = 0.1*10^-3;
+
+//Speed of light c = 3*10^8 Kmps
+c = 3*10^8;
+
+//Estimated Distance d
+d = 0.5*c*D;
+
+//Tolerance Tol
+Tol = 0.5*c*Tc;
+
+disp('The target is between '+string(d-Tol)+' metres and '+string(d+Tol)+' metres of the source.');
diff --git a/3161/CH15/EX15.3/Ex15_3.sce b/3161/CH15/EX15.3/Ex15_3.sce new file mode 100755 index 000000000..b64f0375e --- /dev/null +++ b/3161/CH15/EX15.3/Ex15_3.sce @@ -0,0 +1,16 @@ +clc;
+//page 769
+//problem 15.3
+
+//Number of Flip Flops N
+N = 13;
+
+//Maximal length of sequence L
+L = 2^N - 1;
+
+//Upper Bound S
+S = (L - 1)/N;
+
+//No of basic sequences and mirror images
+disp('No of basic sequences and mirror images is '+string(S/2)');
+
diff --git a/3161/CH2/EX2.1/Ex2_1.sce b/3161/CH2/EX2.1/Ex2_1.sce new file mode 100755 index 000000000..6b9c2437c --- /dev/null +++ b/3161/CH2/EX2.1/Ex2_1.sce @@ -0,0 +1,24 @@ +clc;
+//page 85
+//problem 2.1
+
+//A & B are two events occured in sample space S, where P(A) & P(B) are their corresponding probability
+P_S=1
+
+//Given A&B are not mutually exclusive events,
+//Probability of A is 0.2 = P_A
+//Probability of B is 0.4 = P_B
+//Probability of either A or B is 0.5 = P_AUB
+P_A = 0.2
+P_B = 0.4
+P_AUB = 0.5
+
+//Probability of both of A&B jointly occur is P_AinterB = P_A+P_B-P_AUB where inter is intersection
+P_AinterB = P_A+P_B-P_AUB
+
+disp('Probability of both of A&B jointly occur is '+string(P_AinterB))
+
+//Probability of none of AorB are occur is P_NOAB = Total occurence(P_S) - Probability of either AorB(P_AUB)
+P_NOAB = P_S-P_AUB
+
+disp('Probability of none of AorB are occur is '+string(P_NOAB))
diff --git a/3161/CH2/EX2.10/Ex2_10.sce b/3161/CH2/EX2.10/Ex2_10.sce new file mode 100755 index 000000000..d09c35e86 --- /dev/null +++ b/3161/CH2/EX2.10/Ex2_10.sce @@ -0,0 +1,33 @@ +clc;
+//page 105
+//problem 2.10
+
+//We know that, Probabilty of error(P_error) for the signal correpted by Gaussian channel variance sigma^2 where signal having voltage levels as 0&V is (1/2)*erfc(V/(2*sqrt(2)*sigma))
+
+//P_error for V = 4 & sigma^2 =2
+V = 4
+sigma = sqrt(2)
+P_error = (1/2)*erfc(V/(2*sqrt(2)*sigma))
+
+disp('Probabilty of error for V = 4 & sigma^2 =2 is '+string(P_error))
+
+//P_error for V = 2 & sigma^2 =2
+V = 2
+sigma = sqrt(2)
+P_error = (1/2)*erfc(V/(2*sqrt(2)*sigma))
+
+disp('Probabilty of error for V = 2 & sigma^2 =2 is '+string(P_error))
+
+//P_error for V = 4 & sigma^2 =4
+V = 4
+sigma = sqrt(4)
+P_error = (1/2)*erfc(V/(2*sqrt(2)*sigma))
+
+disp('Probabilty of error for V = 4 & sigma^2 =4 is '+string(P_error))
+
+//P_error for V = 8 & sigma^2 =2
+V = 8
+sigma = sqrt(2)
+P_error = (1/2)*erfc(V/(2*sqrt(2)*sigma))
+
+disp('Probabilty of error for V = 8 & sigma^2 =2 is '+string(P_error))
diff --git a/3161/CH2/EX2.11/Ex2_11.sce b/3161/CH2/EX2.11/Ex2_11.sce new file mode 100755 index 000000000..08e05048f --- /dev/null +++ b/3161/CH2/EX2.11/Ex2_11.sce @@ -0,0 +1,40 @@ +clc;
+//page 106
+//problem 2.11
+
+//(a)
+//out of n attempts the probability of message reaching correctly for k times is given by binomial distribution pX(k) = nCk*(q^k)*(1-q)^(n-k) where q is probability of correctly reaching
+
+//Here n = 10, k = 1, q = 0.001
+n = 10
+k = 1
+q = 0.001
+
+//pX(k) is denoted as p_X_1
+//10C1 =10
+p_X_1 = 10*(q^k)*(1-q)^(n-k)
+
+disp('The probability that out of 10 transmissions 9 are corrent and 1 is incorrect is '+string(p_X_1))
+
+//probability that more than two erroneous out of 100 transmissions(p_100_2) = 1-probability of less than or equal to two error in transmission
+//p_100_2 = 1-pX(0)-pX(1)-pX(2)
+//p_100_2 =1-100C0*((0.001)^0)*((1-0.001)^100)-100C1*((0.001)^1)*((1-0.001)^99)-100C0*((0.001)^2)*((1-0.001)^98)
+
+//Since, calculation of above is cumbersome we may use Poisson ditribution to approximate above
+//Poisson distribution = pX(k) = (alfa^k)*(e^-alfa)/k!, where alfa = n*T
+
+//Here n = 100 & q = 0.001
+n = 100
+q = 0.001
+
+alfa = n*q
+
+p_100_2 = 1-(alfa^0)*(%e^-0.1)/factorial(0)-(alfa^1)*(%e^-0.1)/factorial(1)-(alfa^2)*(%e^-0.1)/factorial(2)
+
+disp('probability that more than two erroneous out of 100 transmissions is '+string(p_100_2))
+
+//(c)
+//from(b), required probability i.e probability of more than one are erroneous out of 100 transmission(p_100_1) is
+p_100_1 = 1-(alfa^0)*(%e^-0.1)/factorial(0)-(alfa^1)*(%e^-0.1)/factorial(1)
+
+disp('probability that more than one erroneous out of 100 transmissions is '+string(p_100_1))
diff --git a/3161/CH2/EX2.13/Ex2_13.sce b/3161/CH2/EX2.13/Ex2_13.sce new file mode 100755 index 000000000..1ccccd4a6 --- /dev/null +++ b/3161/CH2/EX2.13/Ex2_13.sce @@ -0,0 +1,22 @@ +clc;
+//page 115
+//problem 2.13
+
+//Given, Error probability is 10^-4 = P_e, no of ecperiments conducted = N = 4*10^5 & estimated probability of error p does not differ from P_e by more than 50%
+P_e = 10^-4
+N = 4*10^5
+
+//Tchebycheff's inequality is P(|p-Pe|>=E)<=P_e/(N*E^2)
+//From given values we can find that E = 50*10^-4
+E = 50*10^-4
+
+//Here R.H.S of Tchebycheff's inequality is denoted as Tc_RHS
+Tc_RHS = P_e/(N*E^2)
+
+//Tc_RHS in persentage is Tc_RHSper
+Tc_RHSper = Tc_RHS/100
+
+//disp(Tc_RHSper,Tc_RHS,'or P(|p-10^-4|>=0.5*10^-2)<=',Tc_RHS,'The probability of estimated probability of error p does not differ from P_e by more than 50% is less than equal to')
+
+//given solution has been computed wrong, obtaines solution is 10^-7
+disp('The probability of estimated probability of error p does not differ from P_e by more than 50% is less than equal to '+string(Tc_RHS)+'or P(|p-10^-4|>=0.5*10^-2)<='+string(Tc_RHS)+' = '+string(Tc_RHSper)+'%')
diff --git a/3161/CH2/EX2.2/Ex2_2.sce b/3161/CH2/EX2.2/Ex2_2.sce new file mode 100644 index 000000000..386dae4fb --- /dev/null +++ b/3161/CH2/EX2.2/Ex2_2.sce @@ -0,0 +1,38 @@ +clc;
+//page 86
+//problem 2.2
+
+//Probability that A will occur if B has already occurred(P_AB) = ratio of Probability of joint occurence of A&B (P_A_B) & Probability of B(P_B)
+//P_A_B(robability of joint occurence) = Probability that A&B both occur(P_AinterB)
+
+//From given values P_AinterB = 0.1 implies P_A_B = 0.1 & P_B = 0.4
+P_AinterB = 0.1
+P_A_B = P_AinterB
+P_B = 0.4
+
+P_AB = P_A_B/P_B
+
+//Similarly
+//Probability that B will occur if A has already occurred(P_BA) = ratio of Probability of joint occurence of A&B (P_A_B) & Probability of B(P_A)
+
+//From given values P_A = 0.2
+P_A = 0.2
+
+P_BA = P_A_B/P_A
+
+//Bayes theorem says that P_AB = (P_A/P_B)*P_BA
+//After Calculating LHS & RHS if both are equal then bayes theorem is satisfying
+
+//Calculating LHS
+LHS = P_AB
+
+//Calculating RHS
+RHS = (P_A/P_B)*P_BA
+
+disp('P(A/B) = '+string(P_AB) );
+
+if LHS == RHS then
+ disp('LHS = RHS, Hence Bayes theorem is verified' );
+end
+
+
diff --git a/3161/CH2/EX2.5/Ex2_5.sce b/3161/CH2/EX2.5/Ex2_5.sce new file mode 100755 index 000000000..7ef7b3888 --- /dev/null +++ b/3161/CH2/EX2.5/Ex2_5.sce @@ -0,0 +1,37 @@ +clc;
+//page 95
+//problem 2.5
+
+//Given, The probability that m0 is sent is 0.7 = P_m0
+P_m0 = 0.7
+
+//The probability that m0 is sent is 0.3 = P_m1
+P_m1 = 0.3
+
+//The probability that r0 is received given that m0 is sent is 0.9 = P_r0m0 where r is voltage & m is message
+P_r0m0 = 0.9
+
+//the probability that r1 is received given that m0 is sent is 0.1 = P_r1m0 where r is voltage & m is message
+P_r1m0 = 0.1
+
+//The probability that r1 is received given that m1 is sent is 0.6 = P_r1m1
+P_r1m1 = 0.6
+
+//the probability that r0 is received given that m1 is sent is 0.4 = P_r0m1 where r is voltage & m is message
+P_r0m1 = 0.4
+
+//With the given values check eqations P_r0m0*P_m0(P00) > P_r0m1*P_m1(P01)
+P00 = P_r0m0*P_m0
+P01 = P_r0m1*P_m1
+
+if P00>P01 then
+ disp('as P(r0|m0)*P(m0) > P(r0|m1)*P(m1) is valid, we whould select m0 whenever r0 is received')
+end
+
+//With the given values check eqations P_r1m1*P_m1(P11) > P_r1m0*P_m0(P10)
+P11 = P_r1m1*P_m1
+P10 = P_r1m0*P_m0
+
+if P11>P10 then
+ disp('as P(r1|m1)*P(m1) > P(r1|m0)*P(m0) is valid, we whould select m1 whenever r1 is received')
+end
diff --git a/3161/CH2/EX2.6/Ex2_6.sce b/3161/CH2/EX2.6/Ex2_6.sce new file mode 100755 index 000000000..3c23c0c15 --- /dev/null +++ b/3161/CH2/EX2.6/Ex2_6.sce @@ -0,0 +1,25 @@ +clc;
+//page 96
+//problem 2.6
+
+//Given, the probability that r0 is received given that m0 is sent is 0.9 = P_r0m0 where r is voltage & m is message
+P_r0m0 = 0.9
+
+//The probability that m0 is sent is 0.7 = P_m0
+P_m0 = 0.7
+
+//The probability that r1 is received given that m1 is sent is 0.6 = P_r1m1
+P_r1m1 = 0.6
+
+//The probability that m0 is sent is 0.3 = P_m1
+P_m1 = 0.3
+
+//The probability that the transmitted signal is correctly read at receiver is P(c)(P_c) = the probability that m0 was sent when r0 was read(P_r0m0*P_m0) + the probability that m1 was sent when r1 was read(P_r1m1*P_m1)
+
+P_c = P_r0m0*P_m0+P_r1m1*P_m1
+
+//P(e)(P_e) = 1-P(c)
+P_e = 1-P_c
+
+disp('P(e) = '+string(P_e))
+disp('P(c) = '+string(P_c))
diff --git a/3161/CH2/EX2.7/Ex2_7.sce b/3161/CH2/EX2.7/Ex2_7.sce new file mode 100644 index 000000000..07a72dee3 --- /dev/null +++ b/3161/CH2/EX2.7/Ex2_7.sce @@ -0,0 +1,48 @@ +clc;
+//page 96
+//problem 2.7
+
+//Here P(ra|mb) is denoted as P_ramb where a is 0,1,2 & b is 0,1
+//P(X) is denoted as P_X where X is m0, m1, C & E
+
+//From given values P_m0 = 0.6, P_m1 = 0.4, P_r0m0 =0.6, P_r1m1 = 0.7, P_r0m1 = 0, P_r1m0 = 0.2, P_r2m0 = 0.2 & P_r2m1 = 0.3
+P_m0 = 0.6
+P_m1 = 0.4
+P_r0m0 =0.6
+P_r1m1 = 0.7
+P_r0m1 = 0
+P_r1m0 = 0.2
+P_r2m0 = 0.2
+P_r2m1 = 0.3
+
+//(a)
+//Comaparing P(r0|m0)*P(m0) & P(r0|m1)*P(m1) gives result
+LHS = P_r0m0*P_m0
+RHS = P_r0m1*P_m1
+
+disp('As P(r0|m0)*P(m0)['+string(LHS)+'] > P(r0|m1)*P(m1)['+string(RHS)+']')
+disp('we select m0 whenever r0 is received')
+
+//Similarly compare P(r1|m1)*P(m1) & P(r1|m0)*P(m0)
+LHS = P_r1m1*P_m1
+RHS = P_r1m0*P_m0
+
+disp('As P(r1|m1)*P(m1)['+string(LHS)+'] > P(r1|m0)*P(m0)['+string(RHS)+']')
+disp('we select m1 whenever r1 is received')
+
+//compare P(r2|m0)*P(m0) & P(r2|m1)*P(m1)
+LHS = P_r2m0*P_m0
+RHS = P_r2m1*P_m1
+
+disp('As P(r2|m0)*P(m0)['+string(LHS)+'] = P(r2|m1)*P(m1)['+string(RHS)+']')
+disp('We can accordingly make either assignment and we arbitrarily associate r2 with m0')
+
+//(b)
+//The probability of being correct is P(C) = P(r0|m0)*P(m0)+P(r1|m1)*P(m1)+P(r2|m0)*P(m0)
+P_C = P_r0m0*P_m0+P_r1m1*P_m1+P_r2m0*P_m0
+
+//The probability of error is P(E) = 1-P(C)
+P_E = 1 - P_C;
+
+disp('Probability of being correct is P(C) = '+string(P_C)')
+disp('Probability of error is P(E) = '+string(P_E)')
diff --git a/3161/CH2/EX2.8/Ex2_8.sce b/3161/CH2/EX2.8/Ex2_8.sce new file mode 100755 index 000000000..c632c0220 --- /dev/null +++ b/3161/CH2/EX2.8/Ex2_8.sce @@ -0,0 +1,15 @@ +clc;
+//page 99
+//problem 2.8
+
+//Given, probability density function of X is fX_x where fX_x = a*e^(-0.2*x) for x greater than & equal to 0 & = 0 eleswhere
+
+//a = fX_x/(a*e^(-0.2*x))
+//from definition integration of fX_x with limits -infinity to +infinity is 1
+//As per given fX_x, integration of a*e^(-0.2*x) with limits 0 & +inffinity and obtained value be P
+//a = 1/p
+
+P = integrate('%e^(-0.2*x)','x',0,100)
+a = 1/P
+
+disp('a = '+string(a))
diff --git a/3161/CH3/EX3.4/Ex3_4.sce b/3161/CH3/EX3.4/Ex3_4.sce new file mode 100755 index 000000000..252f12be5 --- /dev/null +++ b/3161/CH3/EX3.4/Ex3_4.sce @@ -0,0 +1,16 @@ +clc;
+//page 163
+//problem 3.4
+
+//Transmission power effiency n = {(m^2)/[2+(m^2)]}*100% where m is modulated index
+
+//Given modulated indices are m1 = 0.25, m2 = 0.5 & m3 = 0.75
+
+//Transmission power effiencies are n1, n2 & n3 respectively for m1, m2 & m3
+n1 = {(0.25^2)/[2+(0.25^2)]}*100
+n2 = {(0.5^2)/[2+(0.5^2)]}*100
+n3 = {(0.75^2)/[2+(0.75^2)]}*100
+
+disp('Transmission power effiency for modulated index 0.25 is '+string(n1)+ '%')
+disp('Transmission power effiency for modulated index 0.5 is '+string(n2)+ '%')
+disp('Transmission power effiency for modulated index 0.75 is '+string(n3)+ '%')
diff --git a/3161/CH3/EX3.7/Ex3_7.sce b/3161/CH3/EX3.7/Ex3_7.sce new file mode 100644 index 000000000..3672376fa --- /dev/null +++ b/3161/CH3/EX3.7/Ex3_7.sce @@ -0,0 +1,36 @@ +clc;
+//page 185
+//problem 3.7
+
+//Given input inmedance of matching networkis R1 = 10 ohm & output impedance of matching networ is R2 = 50 ohm & carrier frequency is fc = 500 KHz
+R1 = 10
+R2 = 50
+fc = 500000
+
+//Wc = 2*pi*fc
+Wc = 2*%pi*fc
+
+//AS R1 = R2*(X2^2)/[(R2^2)+(X2^2)], X2 = 25ohm
+X2 = 25
+
+//AS X1 = (R2^2)*X2/[(R2^2)+(X2^2)] & R1>R2, X1 = -20ohm
+X1 = -20
+
+//|X1| = |jwL| = wL = 20 & |X2| = |1/jwC| = 1/wC = 25, so |X1*X2| = L/C = 500 denotes as LC_div
+LC_div = 500
+
+//Wc^2 = 1/(L*C). LC is denoted as LC_prod
+LC_prod = 1/(Wc^2)
+
+//In the textbook the calculated LC = 10^-3, in reality the value of LC = 1.013D-13
+
+L = sqrt(LC_div*LC_prod)
+
+//In the textbook the calculated L^2 = 50*10^-14, in reality the value of L^2 = 5.066D-11
+
+C = L/500
+
+//In the textbook the calculated C = 1.4*10^-9, in reality the value of C = 1.424D-08
+
+disp('Inductance '+string(L)+' H')
+disp('Capacitance '+string(C)+' F')
diff --git a/3161/CH3/EX3.8/Ex3_8.sce b/3161/CH3/EX3.8/Ex3_8.sce new file mode 100644 index 000000000..decf0bc65 --- /dev/null +++ b/3161/CH3/EX3.8/Ex3_8.sce @@ -0,0 +1,35 @@ +clc;
+//page 185
+//problem 3.8
+
+//Given ohmnic loss resistance is Ro = 12 Ohm,
+Ro = 12
+
+//radiation resistance is Rr = 48 Ohm,
+Rr = 48
+
+//directivity is D = 2
+D = 2
+
+//Input current = 0.1*cos[2*pi*(10^6)*t], Amplitude of input current is A = 0.1 Amp
+A = 0.1
+
+//Equivalent resistance = Re = Ro+Rr
+Re = Ro+Rr
+
+//Total power used in antenna = Pin = (A^2)*Re/2
+Pin = (A^2)*Re/2
+
+//Power used in radiation = Prad = (A^2)*Rr/2
+Prad = (A^2)*Rr/2
+
+//Efficiency of the antenna = n = Prad/Pin
+n = Prad/Pin
+
+//Gain of antenna = Ga = efficiency*directivity
+Ga = n*D
+
+disp('Total power used in antenna '+string(Pin)+' Watt')
+disp('Power used in radiation '+string(Prad)+' Watt')
+disp('Efficiency of the antenna '+string(n))
+disp('Gain of antenna '+string(Ga))
diff --git a/3161/CH4/EX4.2/Ex4_2.sce b/3161/CH4/EX4.2/Ex4_2.sce new file mode 100644 index 000000000..b622f9277 --- /dev/null +++ b/3161/CH4/EX4.2/Ex4_2.sce @@ -0,0 +1,40 @@ +clc;
+//page 199
+//problem 4.2
+
+//Given angle modulated signal is x = 3*cos[2*pi*(10^6)*t+2*sin(2*pi*10^3*t)]
+
+//So, phase of the angle modulates signal is Q = 2*pi*(10^6*t)+2*sin(2*pi*(10^3)*t)
+
+//Instantaneous frequency = dQ/dt = 2*pi*(10^6)+ 4*pi*(10^3)*sin(2*pi*(10^3)*t)
+
+//For Instantaneous frequency at 0.25ms, Substituting t = 0.25ms in Instantaneous frequency
+//Instantaneous frequency is expressed as f1_rad for frequency in radians per second
+f1_rad = 2*%pi*(10^6)+ 4*%pi*(10^3)*sin(2*%pi*(10^3)*0.00025)
+
+//Instantaneous frequency is expressed as f1_hz for frequency in hertz
+f1_hz = f1_rad/(2*%pi)
+
+disp('the Instantaneous frequency at time t=0.25ms is '+string(f1_rad)+' rad/sec = '+string(f1_hz)+' Hz')
+
+//For Instantaneous frequency at 0.25ms, Substituting t = 0.5ms in Instantaneous frequency
+//Instantaneous frequency is expressed as f2rad for frequency in radians per second
+f2_rad = 2*%pi*(10^6)+ 4*%pi*(10^3)*sin(2*%pi*(10^3)*0.0005)
+
+//Instantaneous frequency is expressed as f2hz for frequency in hertz
+f2_hz = f2_rad/(2*%pi)
+
+disp('the Instantaneous frequency at time t=0.5ms is '+string(f2_rad)+' rad/sec = '+string(f2_hz)+' Hz')
+
+//Maximum phase deviation = max[2*sin(2*pi*(10^3)*t)] = 2*1
+maxDp = 2;
+
+disp('Maximum phase deviation is '+string(maxDp)+' rad')
+
+//Maximum frequency deviation = max[4*pi*(10^3)*sin(2*pi*(10^3)*t)] = 4*pi*(10^3)*1
+maxDf = 4*%pi*(10^3)*1;
+
+disp('Maximum frequency deviation is '+string(maxDf)+' Hz')
+//disp('in rad',maxDf,'Maximum frequency deviation is')
+
+//In the textbook the calculated value of max frequency devaition is = 2000 Hz, in reality the value = 12566.371 Hz
diff --git a/3161/CH4/EX4.4/Ex4_4.sce b/3161/CH4/EX4.4/Ex4_4.sce new file mode 100755 index 000000000..c73555910 --- /dev/null +++ b/3161/CH4/EX4.4/Ex4_4.sce @@ -0,0 +1,62 @@ +clc;
+//page 208
+//problem 4.4
+
+//Given modulating signal m(t) = 2*sin(2*pi*(10^3)*t), B for phase modulation Bp = 10 & for fequency modulation Bf = 10
+Bp = 10
+Bf = 10
+
+//So Amplitude of modulating signal is Am=2 metres
+Am = 2
+
+//Frequency of modulating signal is fm = 1000 hertz
+fm=1000
+
+//Bandwidth = 2*(1+B)*fm
+
+//Bandwidth for phase modulation with modulating signal m(t) is bw_pm = 2*(1+Bp)*fm
+bw_pm = 2*(1+10)*1000
+
+//Bandwidth for frequency modulation with modulating signal m(t) is bw_fm = 2*(1+Bf)*fm
+bw_fm = 2*(1+10)*1000
+
+disp('Bandwidth for phase modulation '+string(bw_pm)+' Hz')
+disp('Bandwidth for frequency modulation '+string(bw_fm)+' Hz')
+
+//Bandwidth for phase & frequency modulation if frequency of modulating signal is doubled i.e fm = 2000 hertz
+
+//Bp & Bf after frequency of modulating signal is doubled
+
+//Bp = kp*Am, observing the equation as there is no change in amplitude Bp = 10
+Bp = 10
+
+//Bf = kf*Am/fm, observing the equation as there is change in frequency Bf = 10/2 = 5
+Bf = 5
+
+//Bandwidth for phase modulation if frequency of modulating signal is doubled is bw_double_pm = 2*(1+Bp)*fm
+bw_double_pm = 2*(1+10)*2000
+
+//Bandwidth for frequency modulation if frequency of modulating signal is doubled is bw_double_fm = 2*(1+Bf)*fm
+bw_double_fm = 2*(1+5)*2000
+
+disp('Bandwidth for phase modulation for doubled frequency '+string(bw_double_pm)+' Hz')
+disp('bandwidth for frequency modulation for doubled frequency '+string(bw_double_fm)+' Hz')
+
+//Bandwidth for phase & frequency modulation if amplitude of modulating signal is halfed i.e Am = 1 metre
+
+//Bp & Bf after amplitude of modulating signal is halfed
+
+//Bp = kp*Am, observing the equation as there is change in amplitude Bp = 10/2 = 5
+Bp = 5
+
+//Bf = kf*Am/fm, observing the equation as there is change in amplitude Bf = 5/2 = 2.5
+Bf = 2.5
+
+//Bandwidth for phase modulation if frequency of modulating signal is doubled is bw_halfed_pm = 2*(1+Bp)*fm
+bw_halfed_pm = 2*(1+5)*2000
+
+//Bandwidth for frequency modulation if frequency of modulating signal is doubled is bw_halfed_fm = 2*(1+Bf)*fm
+bw_halfed_fm = 2*(1+2.5)*2000
+
+disp('Bandwidth for phase modulation for halfed amplitude '+string(bw_halfed_pm)+' Hz')
+disp('Bandwidth for frequency modulation for halfed amplitude '+string(bw_halfed_fm)+' Hz')
diff --git a/3161/CH5/EX5.11/Ex5_11.sce b/3161/CH5/EX5.11/Ex5_11.sce new file mode 100755 index 000000000..ce44fc830 --- /dev/null +++ b/3161/CH5/EX5.11/Ex5_11.sce @@ -0,0 +1,58 @@ +clc;
+//page 296
+//problem 5.11
+
+//case 1(a)
+//f = 400Hertz, fs = 8000Hertz
+f = 400
+fs = 8000
+
+//We know that maximum signal to noise ratio(SNR_max) = 3*(fs^2)/(8*(pi^2)*(f^2))
+SNR_max = 3*(fs^2)/(8*(%pi^2)*(f^2))
+//SNR_max in decibels is SNR_max_db
+SNR_max_db = 10*log10 (SNR_max)
+
+disp('Maximum signal to noise ratio for f = 400 & fs = 8000 is '+string(SNR_max)+' = '+string(SNR_max_db)+' db')
+
+//case 1(b)
+//f = 400Hertz, fs = 16000Hertz
+f = 400
+fs = 16000
+
+//We know that maximum signal to noise ratio(SNR_max) = 3*(fs^2)/(8*(pi^2)*(f^2))
+SNR_max = 3*(fs^2)/(8*(%pi^2)*(f^2))
+
+//SNR_max in decibels is SNR_max_db
+SNR_max_db = 10*log10 (SNR_max)
+
+//Given solution is 13.8385 dB obtained solution is 17.838515 dB
+
+disp('Maximum signal to noise ratio for f = 400 & fs = 16000 is '+string(SNR_max)+' = '+string(SNR_max_db)+' db')
+
+//case 2(a)
+//f = 400Hertz, fs = 8000Hertz & fc = 1000Hertz
+f = 400
+fs = 8000
+fc = 1000
+
+//If a 1kHz low pass post reconstruction filter is used then maximum signal to noise ratio(SNR_max) = 3*(fs^3)/(8*(pi^2)*(f^2)*fc)
+SNR_max = 3*(fs^3)/(8*(%pi^2)*(f^2)*fc)
+//SNR_max in decibels is SNR_max_db
+SNR_max_db = 10*log10 (SNR_max)
+
+disp('If a 1kHz low pass post reconstruction filter is used then')
+
+disp('Maximum signal to noise ratio for f = 400, fs = 8000 & fc = 1000 is '+string(SNR_max)+' = '+string(SNR_max_db)+' db')
+
+//case 2(b)
+//f = 400Hertz, fs = 16000Hertz & fc = 1000Hertz
+f = 400
+fs = 16000
+fc = 1000
+
+//If a 1kHz low pass post reconstruction filter is used then maximum signal to noise ratio(SNR_max) = 3*(fs^3)/(8*(pi^2)*(f^2)*fc)
+SNR_max = 3*(fs^3)/(8*(%pi^2)*(f^2)*fc)
+//SNR_max in decibels is SNR_max_db
+SNR_max_db = 10*log10 (SNR_max)
+
+disp('Maximum signal to noise ratio for f = 400, fs = 16000 & fc = 1000 is '+string(SNR_max)+' = '+string(SNR_max_db)+' db')
diff --git a/3161/CH5/EX5.2/Ex5_2.sce b/3161/CH5/EX5.2/Ex5_2.sce new file mode 100644 index 000000000..3b7d7152e --- /dev/null +++ b/3161/CH5/EX5.2/Ex5_2.sce @@ -0,0 +1,25 @@ +clc;
+//page 247
+//problem 5.2
+
+//Highest frequency(fH) = 10000/2 = 5000 Hz
+fH = 5000
+
+//Lowest frequency(fL) = 6000/2 = 3000 Hz
+fL = 3000
+
+//Minimum sampling frequency from low pass consideration(S_LOW) = 2*fH
+S_LOW = 2*fH
+
+disp('Minimum sampling frequency from low pass consideration is '+string(S_LOW)+' Hz')
+
+//B = fH-fL = 2000 Hz
+B = fH-fL
+
+//k = floor(fH/B) = 2, where floor(x) gives the largest integer that does not exceed x
+k = floor(fH/B)
+
+//The required sampling frequency from band pass consideration(S_BAND) = 2*fH/k
+S_BAND = 2*fH/k
+
+disp('Minimum sampling frequency from band pass consideration is '+string(S_BAND)+' Hz')
diff --git a/3161/CH5/EX5.3/Ex5_3.sce b/3161/CH5/EX5.3/Ex5_3.sce new file mode 100644 index 000000000..655bce9fe --- /dev/null +++ b/3161/CH5/EX5.3/Ex5_3.sce @@ -0,0 +1,25 @@ +clc;
+//page 259
+//problem 5.3
+
+//Given width of each pulse W = 150 us
+W = 150 * 10^-6
+
+//One cycle is a period,T = 1ms
+T = 1000 * 10^-6
+
+//There are 5 messages multiplexed each utilizeallocated time pulse width = s(T_5) = T/5
+T_5 = T/5
+
+//Gaurd time(GT_5) = allocated time-pulse width = T_5-W
+GT_5 = T_5-W
+
+disp('Gaurd time where 5 messages multiplexed is '+string(GT_5)+' seconds')
+
+//Here there are 10 messages multiplexed each utilizeallocated time pulse width = s(T_10) = T/10
+T_10 = T/10
+
+//Gaurd time(GT_10) = allocated time-pulse width = T_10-norrow pulses width = T_10 -50* 10^-6
+GT_10 = T_10 - 50 * 10^-6
+
+disp('Gaurd time where 10 messages multiplexed is '+string(GT_10)+' seconds')
diff --git a/3161/CH5/EX5.5/Ex5_5.sce b/3161/CH5/EX5.5/Ex5_5.sce new file mode 100755 index 000000000..d408fcde2 --- /dev/null +++ b/3161/CH5/EX5.5/Ex5_5.sce @@ -0,0 +1,14 @@ +clc;
+//page 272
+//problem 5.5
+
+//Let Abe the maximum value of the discrete samples.
+//Error tolerated is 0.1% i.e. 0.001A
+//If D is step size then possible maximum error is D/2
+//Thus D/2 = 0.001A or A/D = 500 = no of levels required(Levels)
+Levels = 500
+
+//minimum no of binary digits required(B) = rounded value to the next higher integer of log2 (Levels)
+B = round(log2 (Levels))
+
+disp('Minimum no of binary digits required '+string(B))
diff --git a/3161/CH5/EX5.6/Ex5_6.sce b/3161/CH5/EX5.6/Ex5_6.sce new file mode 100755 index 000000000..70eb06d50 --- /dev/null +++ b/3161/CH5/EX5.6/Ex5_6.sce @@ -0,0 +1,33 @@ +clc;
+//page 273
+//problem 5.6
+
+//The y axis is uniformly quantized with step size(step_size) = 1/[(2^8)/(2-1)] in both +ve & -ve direction between 1 & -1 when peak of input varies between 1 & -1.
+//The smallest step in x direction occurs nearest to x=0 i.e between y1 = 0 & y2 = step_size
+step_size = 1/[(2^8)/2-1]
+y1 = 0
+y2 = step_size
+
+//Then, y1 = [ln(1+255*x1)]/[ln(1+255)]
+
+x1 = (%e^(y1*log(256)) - 1)/255;
+
+//y2 = [ln(1+255*x2)]/[ln(1+255)]
+x2 = (%e^(y2*log(256)) - 1)/255;
+
+//The smallest step size is 10*(x2-x1)
+disp('The smallest step size is '+string(10*(x2-x1))+' Volts')
+
+//The largest step size occurs when x is at its extreme between y1 = 1-1/127 = 126/127 & y2 = 1
+y1 = 1-1/127
+y2 = 1
+
+//Then, y1 = [ln(1+255*x1)]/[ln(1+255)]
+
+x1 = (%e^(y1*log(256)) - 1)/255;
+
+//y2 = [ln(1+255*x2)]/[ln(1+255)]
+x2 = (%e^(y2*log(256)) - 1)/255;
+
+//The largest step size is 10*(x2-x1)
+disp('The largest step size is '+string(10*(x2-x1))+' Volts')
diff --git a/3161/CH5/EX5.9/Ex5_9.sce b/3161/CH5/EX5.9/Ex5_9.sce new file mode 100755 index 000000000..940d73e61 --- /dev/null +++ b/3161/CH5/EX5.9/Ex5_9.sce @@ -0,0 +1,36 @@ +clc;
+//page 296
+//problem 5.9
+
+//for error calculation e(n) = m(n) - [^hj(n)*m(n-1)+^hj(n)*m(n-2)+^hj(n)*m(n-3)+ ........+^hj(n)*m(n-N)]
+
+//for coefficient upgradation ^hj(n+1) = ^hj(n)+um(n-j)e(n) where u = learning parameter = 0.1.
+u = 0.1
+
+//Assign m values taking from m = -3 to 5
+//Denoting m(x) as matrix m where each element repesents from n = -3 to 5
+m = [0 0 0 1 2 3 4 5 6]
+
+//taking e(n) as matrix e, ^hj(n) as matrises h_j
+e = zeros(1,5)
+h_1 = zeros(1,6)
+h_2 = zeros(1,6)
+
+//given ^h1(0)= ^h2(0) = 0
+h_1(1) = 0
+h_2(1) = 0
+
+for i = 1:5
+ e(i) = m(i+3) - h_1(i)*m(i+2) - h_2(i)*m(i+1)
+ h_1(i+1) = h_1(i) + u*m(i+2)*e(i)
+ h_2(i+1) = h_2(i) + u*m(i+1)*e(i)
+end
+
+//here e(3) is given as 1.32 but it is displaying 0.92
+//here ^h2(3) is given as 0.26 but it is displaying 0.46
+
+for i = 1:5
+ disp('e('+string(i-1)+') = '+string(e(i)))
+ disp('^h1('+string(i)+') = '+string(h_1(i+1)))
+ disp('^h2('+string(i)+') = '+string(h_2(i+1)))
+end
diff --git a/3161/CH6/EX6.1/Ex6_1.sce b/3161/CH6/EX6.1/Ex6_1.sce new file mode 100755 index 000000000..683b13416 --- /dev/null +++ b/3161/CH6/EX6.1/Ex6_1.sce @@ -0,0 +1,120 @@ +clc;
+//page 341
+//problem 6.1
+
+//Given messages signal m = [1,0,1,1,0,1]
+m = [1,0,1,1,0,1];
+
+//Logical 0 corrsponds to pi i.e %pi and Logical 1 corresponds to 0
+
+////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
+
+//For BPSK, from the above deduction let the carrier phase be Carrier_Phase_BPSK
+for i = 1:5
+ if m(i)==1 then
+ Carrier_Phase_BPSK(i) = 0;
+ else
+ Carrier_Phase_BPSK(i) = %pi;
+ end
+end
+
+disp(Carrier_Phase_BPSK,'The Phase of the carrier signal for BPSK varies as ');
+
+////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
+
+//For DPSK
+//Let b represent the input to balance modulator
+
+//If the initial value of b be 0
+b = 0;
+
+for i = 2:5
+ b(i) = bitxor(m(i),b(i-1))
+end
+
+//Now the carrier phase, Carrier_Phase_DPSK
+for i = 1:5
+ if b(i)==1 then
+ Carrier_Phase_DPSK(i) = 0;
+ else
+ Carrier_Phase_DPSK(i) = %pi;
+ end
+end
+
+//Now the carrier amplitude, Carrier_Amplitude_DPSK
+for i = 1:5
+ Carrier_Amplitude_DPSK(i) = cos(Carrier_Phase_DPSK(i));
+end
+
+disp(Carrier_Phase_DPSK,'The Phase of the carrier signal for DPSK varies as follows, '+'when the initial value of b is 1');
+disp(Carrier_Amplitude_DPSK,'The Amplitude of the carrier signal for DPSK varies as follows, '+'when the initial value of b is 1');
+
+//If the initial value of b be 1
+b = 1;
+
+for i = 2:5
+ b(i) = bitxor(m(i),b(i-1))
+end
+
+//Now the carrier phase, Carrier_Phase_DPSK
+for i = 1:5
+ if b(i)==1 then
+ Carrier_Phase_DPSK(i) = 0;
+ else
+ Carrier_Phase_DPSK(i) = %pi;
+ end
+end
+
+//Now the carrier amplitude, Carrier_Amplitude_DPSK
+for i = 1:5
+ Carrier_Amplitude_DPSK(i) = cos(Carrier_Phase_DPSK(i));
+end
+
+disp(Carrier_Phase_DPSK,'The Phase of the carrier signal for DPSK varies as follows, '+'when the initial value of b is 0');
+disp(Carrier_Amplitude_DPSK,'The Amplitude of the carrier signal for DPSK varies as follows, '+'when the initial value of b is 0');
+
+////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
+
+//For DEPSK
+//The DEPSK transmitter output is same as that of DPSK
+
+//If the initial value of b be 0
+b = 0;
+
+for i = 2:5
+ b(i) = bitxor(m(i),b(i-1))
+end
+
+//Now the carrier phase, Carrier_Phase_DPSK
+for i = 1:5
+ if b(i)==1 then
+ Carrier_Phase_DEPSK(i) = 0;
+ else
+ Carrier_Phase_DEPSK(i) = %pi;
+ end
+end
+
+disp(Carrier_Phase_DEPSK,'The Phase of the carrier signal for DEPSK varies as follows, '+'when the initial value of b is 1');
+
+//If the initial value of b be 1
+b = 1;
+
+for i = 2:5
+ b(i) = bitxor(m(i),b(i-1))
+end
+
+//Now the carrier phase, Carrier_Phase_DPSK
+for i = 1:5
+ if b(i)==1 then
+ Carrier_Phase_DEPSK(i) = 0;
+ else
+ Carrier_Phase_DEPSK(i) = %pi;
+ end
+end
+
+disp(Carrier_Phase_DEPSK,'The Phase of the carrier signal for DEPSK varies as, '+'when the initial value of b is 0');
+
+
+
+
+
diff --git a/3161/CH6/EX6.2/Ex6_2.sce b/3161/CH6/EX6.2/Ex6_2.sce new file mode 100755 index 000000000..d637959eb --- /dev/null +++ b/3161/CH6/EX6.2/Ex6_2.sce @@ -0,0 +1,56 @@ +clc;
+//page 341
+//problem 6.2
+
+//From Ex6_1 the obtained carrier amplitude is c
+
+////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
+
+//For DPSK
+//Considering the initial value of the storage element to be 0 in polar and -1 in biploar
+c = [1,1,-1,1,1];
+y = -1;
+//Let the output be y
+for i = 2:5
+ y(i) = c(i)*c(i-1);
+end
+
+//Converting back to binary data
+for i = 1:5
+ if y(i)== -1 then
+ output_binary(i) = 0;
+ else
+ output_binary(i) = 1;
+ end
+end
+
+//Now inverting the output we get:
+for i = 1:5
+ output_binary(i) = ~output_binary(i);
+end
+
+disp(output_binary,'The DPSK output is');
+
+
+////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
+
+//For DEPSK
+
+//From example Ex6_1, we have b when initial storage value is assumed to be 1
+b = [1,1,0,1,1];
+
+//Output y
+y = 1;
+for i = 2:5
+ y(i) = bitxor(b(i),b(i-1));
+end
+
+disp(y,'The DEPSK output is');
+
+
+
+
+
+
+
+
diff --git a/3161/CH6/EX6.4/Ex6_4.sce b/3161/CH6/EX6.4/Ex6_4.sce new file mode 100644 index 000000000..b077cc0d6 --- /dev/null +++ b/3161/CH6/EX6.4/Ex6_4.sce @@ -0,0 +1,56 @@ +clc;
+//page 365
+//problem 6.4
+
+//Given energy per bit Eb = 0.01
+Eb = 0.01;
+
+//Given fundamental frequency is fb = 8 KHz
+fb = 8*10^3;
+
+//No of symbols M = 16
+M = 16;
+
+N = log2(M);
+
+BW_BPSK = 2*fb;
+disp('Bandwidth for BPSK is '+string(BW_BPSK)+'Hz');
+
+BW_QPSK = fb;
+disp('Bandwidth for QPSK is '+string(BW_QPSK)+'Hz');
+
+BW_16MPSK = fb/2;
+disp('Bandwidth for 16 MPSK is '+string(BW_16MPSK)+'Hz');
+
+BW_BFSK = 4*fb;
+disp('Bandwidth for BFSK is '+string(BW_BFSK)+'Hz');
+
+BW_MSK = 1.5*fb;
+disp('Bandwidth for MSK is '+string(BW_MSK)+'Hz');
+
+BW_16MFSK = 2*M*fb;
+disp('Bandwidth for 16 MFSK is '+string(BW_16MFSK)+'Hz');
+
+
+Min_dist_BPSK = 2*(Eb)^0.5;
+disp('Minimum distance in signal space in BPSK is '+string(Min_dist_BPSK));
+
+Min_dist_QPSK = 2*(Eb)^0.5;
+disp('Minimum distance in signal space in QPSK is '+string(Min_dist_QPSK));
+
+//The given answer in the textbook is 0.0152, which appears to be wrong. The correct answer is 0.078
+Min_dist_16MPSK = (4*N*Eb*(sin(%pi/16))^2)^0.5;
+disp('Minimum distance in signal space in 16 MPSK is '+string(Min_dist_16MPSK));
+
+Min_dist_BFSK = (2*Eb)^0.5;
+disp('Minimum distance in signal space in ortho BFSK is '+string(Min_dist_BFSK));
+
+Min_dist_MSK = 2*(Eb)^0.5;
+disp('Minimum distance in signal space in MSK is '+string(Min_dist_MSK));
+
+Min_dist_16MFSK = (2*N*Eb)^0.5;
+disp('Minimum distance in signal space in ortho 16 MFSK is '+string(Min_dist_16MFSK));
+
+disp('The best method that provides least noise susceptibility is 16 MFSK, then BPSK, then QPSK, then comes MSK, then orthogonal BFSK and finally 16 MPSK')
+
+
diff --git a/3161/CH6/EX6.5/Ex6_5.sce b/3161/CH6/EX6.5/Ex6_5.sce new file mode 100755 index 000000000..b91ae883a --- /dev/null +++ b/3161/CH6/EX6.5/Ex6_5.sce @@ -0,0 +1,89 @@ +clc;
+//page 381
+//problem 6.5
+
+//Given input signal is d
+d = [0,1,1,1,0,1,0,1,1];
+
+////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
+//The answers obtained here are different from the ones mentioned in the textbook.
+//The given answers have been checked rigorously and have been found out to be true.
+
+//When precoded
+
+//Signal b is initially assumed to be 0
+b = 0;
+
+for i = 2:9
+ b(i) = bitxor(b(i-1),d(i));
+end
+
+//Changing bit code to polar signal we get, 0 --> -1, 1 --> +1
+for i = 1:9
+ if b(i)==1 then
+ bp(i) = 1;
+ else
+ bp(i) = -1;
+ end
+end
+
+//Let initial value of Vd be 0
+//Vd = 0;
+
+for i = 2:9
+ Vd(i) = bp(i) + bp(i-1);
+end
+
+//Converting polar signal to bit code we get, -2 --> 0, 0 --> 1, 2 --> 0
+for i = 1:9
+ if Vd(i)== -2 then
+ da(i) = 0;
+ elseif Vd(i)== 2 then
+ da(i) = 0;
+ else
+ da(i) = 1;
+ end
+end
+
+disp(da,'Decoded output when precoded is ')
+
+////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
+
+//When not precoded exor gate is not there
+
+//Changing bit code to polar signal we get, 0 --> -1, 1 --> +1
+for i = 1:9
+ if d(i)==1 then
+ dp(i) = 1;
+ else
+ dp(i) = -1;
+ end
+end
+
+for i = 2:9
+ Vd(i) = dp(i) + dp(i-1);
+end
+
+//Converting polar signal to bit code we get, -2 --> 0, 0 --> 1, 2 --> 1
+for i = 2:9
+ if Vd(i)== -2 then
+ da(i) = 0;
+ elseif Vd(i)== 2 then
+ da(i) = 0;
+ else
+ da(i)= ~da(i-1);
+ end
+end
+
+disp(da,'Decoded output when not precoded is ')
+
+
+
+
+
+
+
+
+
+
+
diff --git a/3161/CH6/EX6.6/Ex6_6.sce b/3161/CH6/EX6.6/Ex6_6.sce new file mode 100644 index 000000000..21c695f42 --- /dev/null +++ b/3161/CH6/EX6.6/Ex6_6.sce @@ -0,0 +1,33 @@ +clc;
+//page 381
+//problem 6.6
+
+//Given Bandwidth BW = 4 kHz
+BW = 4*10^3;
+
+//Given data rate is fb = 6 kbps
+fb = 6*10^3;
+
+//The roll off factor alpha is
+alpha = ((2*BW)/fb) - 1;
+
+disp('The roll off factor is '+string(alpha));
+
+///////////////////////////////////////////////////////////////////////////////
+
+//The required data rate supported at alpha = 0.25 is D
+alpha = 0.25
+
+//The corresponding expression for D is
+D = (2*BW)/(1+alpha);
+
+disp('The supported data rate is '+string(D)+' kbps');
+
+//For full roll-off alpha = 1.0,
+alpha = 1;
+
+fb = 2*BW/(1+alpha);
+
+disp('The data rate is '+string(fb)+' kbps');
+
+
diff --git a/3161/CH7/EX7.3/Ex7_3.sce b/3161/CH7/EX7.3/Ex7_3.sce new file mode 100644 index 000000000..db5106fab --- /dev/null +++ b/3161/CH7/EX7.3/Ex7_3.sce @@ -0,0 +1,48 @@ +clc;
+//page 413
+//problem 7.3
+
+//The resistance R = 1000 Ohm
+R = 10^3;
+
+//The capacitance C = 0.5*10^-6 F
+C = 0.1*10^-6;
+
+//Cutoff frequency for RC filter is f
+f = 1/(2*%pi*R*C)
+
+//White noise power spectral density n
+n = 10^(-9);
+
+//Noise power at filter output P
+P = (%pi/2)*n*f;
+
+disp('Noise power at output filter is '+string(P)+' Watt');
+
+//Noise power at filter output P_new when cutoff frequency is doubled
+P_new = (%pi/2)*n*2*f;
+
+disp('Noise power at output filter when cutoff frequency is doubled is '+string(P_new)+' Watt');
+
+//Ideal Low Pass filter Bandwidth B = 1000 Hz
+B = 1000;
+
+disp('Output Noise Power is '+string(n*B)+' Watt');
+
+disp('Output Noise Power when cut-off frequency is doubled is '+string(2*n*B)+' Watt');
+
+//Proportionality constant T = 0.01
+T = 0.01;
+
+//Output noise power O
+O = n*(B^3)*(T^2)*(4/3)*(%pi)^2;
+
+disp('Output Noise Power when signal is passed through a differentiator passed through ideal low pass filter '+string(O)+' Watt');
+
+O_new = 8*n*(B^3)*(T^2)*(4/3)*(%pi)^2;
+
+disp('Output Noise Power when signal is passed through a differentiator passed through ideal low pass filter and when cut-off frequency is doubled is '+string(O_new)+' Watt');
+
+
+
+
diff --git a/3161/CH7/EX7.4/Ex7_4.sce b/3161/CH7/EX7.4/Ex7_4.sce new file mode 100644 index 000000000..a9ce92a1c --- /dev/null +++ b/3161/CH7/EX7.4/Ex7_4.sce @@ -0,0 +1,21 @@ +clc;
+//page 413
+//problem 7.4
+
+//Given signal strength S = 0.001 W
+S = 0.001;
+
+//Gaussian Noise Magnitude n
+n = 10^(-8);
+
+//Frequency of signal f = 4000 Hz
+F = 4000;
+
+//Noise at equalizer output N
+N = integrate('n*(1+(f^2)/F^2)','f',-F,F);
+
+//Signal to Noise Ratio value is SNR
+SNR = S/N;
+
+disp('SNR value is '+string(10*log10(SNR))+' dB');
+
diff --git a/3161/CH8/EX8.1/Ex8_1.sce b/3161/CH8/EX8.1/Ex8_1.sce new file mode 100644 index 000000000..05b7ce0fc --- /dev/null +++ b/3161/CH8/EX8.1/Ex8_1.sce @@ -0,0 +1,36 @@ +clc;
+//page 436
+//problem 8.1
+
+//Given frequency range fc= 1MHz to fc = 1.0005Mhz
+//Single side message bandwidth is fM
+fM= (1.0005 - 1)*10^6;
+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 are erroneous
+
+//Given input signal strength Si= 1mW
+//Let output signal strength be So
+//So=Si/4
+Si= 10^(-3);
+So= Si/4;
+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)/4;
+disp('Output Noise Strength is '+string(No)+' dB');
+
+//Let SNR at filter output be SNR
+SNR= So / No;
+disp('Output SNR 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');
+
+
diff --git a/3161/CH8/EX8.2/Ex8_2.sce b/3161/CH8/EX8.2/Ex8_2.sce new file mode 100644 index 000000000..b9feabfd0 --- /dev/null +++ b/3161/CH8/EX8.2/Ex8_2.sce @@ -0,0 +1,37 @@ +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');
+
+
diff --git a/3161/CH8/EX8.3/Ex8_3.sce b/3161/CH8/EX8.3/Ex8_3.sce new file mode 100644 index 000000000..042dfa9ce --- /dev/null +++ b/3161/CH8/EX8.3/Ex8_3.sce @@ -0,0 +1,24 @@ +clc;
+//page 446
+//problem 8.3
+
+//Given bandwidth of signal is fM = 4kHZ
+fM = 4*10^3;
+//Given power spectral density of white noise n = 2*10^-9 W/Hz
+n = 2*10^-9;
+//Also given that minimum output SNR is 40dB
+//Signal undergoes a loss of 30dB
+
+//For SSB:
+// Required minimum output SNR = Si_min_SSB / (n*fM) = 40 dB = 10^4
+Si_min_SSB = (10^4)*n*fM;
+// Required minimum signal strength at transmitter output Si_tran = Si_min * 30 dB
+Si_tran_SSB = Si_min_SSB * 10^3;
+disp('Required minimum SSB signal strength at transmitter output is'+string(Si_tran_SSB)+' W');
+
+//For DSB-SC:
+// Required minimum output SNR = (Si_min_DSB/3) / (n*fM) = 40 dB = 10^4
+Si_min_DSB = 3*(10^4)*n*fM;
+// Required minimum signal strength at transmitter output Si_tran = Si_min * 30 dB
+Si_tran_DSB = Si_min_DSB * 10^3;
+disp('Required minimum DSB signal strength at transmitter output is'+string(Si_tran_DSB)+' W');
diff --git a/3161/CH8/EX8.4/Ex8_4.sce b/3161/CH8/EX8.4/Ex8_4.sce new file mode 100644 index 000000000..7bf00c03b --- /dev/null +++ b/3161/CH8/EX8.4/Ex8_4.sce @@ -0,0 +1,33 @@ +clc;
+//page 447
+//problem 8.4
+
+//Given bandwidth of signal is fM = 60 kHZ
+fM = 60*10^3;
+
+//Given power spectral density of white noise n = 2*10^-6 W/Hz
+n = 2*10^-6;
+
+//Given time average of square of mssg signal P = 0.1W
+P = 0.1;
+
+//Noise power at input baseband range NM
+NM = n * fM;
+
+//Threshold occurs at carrier power Pc = 2.9 * NM
+Pc_Threshold = 2.9 * NM;
+
+//For carrier power Pc = 10W, output SNR
+Pc = 10;
+SNRo = Pc * P / NM ;
+disp('Output SNR is '+string(SNRo)+' dB');
+
+//Carrier power is reduced by 100 times making the new power Pc_new
+Pc_new = Pc / 100;
+
+//In the given solutions the NM value is 1.2W instead of 0.12W
+//The corect answer is 0.0925926 instead of 0.000926
+SNR_new = (4/3) * P * (Pc_new/NM)^2;
+disp('Output SNR when carrier power is reduced is '+string(SNR_new)+' dB');
+
+
diff --git a/3161/CH9/EX9.1/Ex9_1.sce b/3161/CH9/EX9.1/Ex9_1.sce new file mode 100644 index 000000000..f2e27b7a3 --- /dev/null +++ b/3161/CH9/EX9.1/Ex9_1.sce @@ -0,0 +1,33 @@ +clc;
+//page 463
+//problem 9.1
+
+//Input signal strength Si = 0.5 W
+Si = 0.5;
+
+//Gaussian Power Spectral Density n = 10^(-10) W/Hz
+n = 10^(-10);
+
+//Baseband cutoff signal fM = 15 kHz
+fM = 15 * 10^3;
+
+//Maximum frequency deviation Df = 60 kHz
+Df = 60 * 10^3;
+
+//Average power of the modulating signal mt = 0.1 W
+mt = 0.1;
+
+SNR = (3/(4*%pi^2))*((Df/fM)^2)*mt^2*(Si/(n*fM));
+
+disp('SNR is '+string(10*log10(SNR))+' dB');
+
+//Part b
+
+//Required SNR at output>40 dB = 10000
+
+//From (a), required Si/0.5 > 10000/4052.8
+//Or, required Si > 1.2337 W
+//Since, channel loss is 20 dB (=100),
+//Required transmitter power > 1.2337*100 = 123.37
+
+disp('Required transmitter power > 1.2337 x 100 = 123.37 ');
diff --git a/3161/CH9/EX9.2/Ex9_2.sce b/3161/CH9/EX9.2/Ex9_2.sce new file mode 100755 index 000000000..7c1da149b --- /dev/null +++ b/3161/CH9/EX9.2/Ex9_2.sce @@ -0,0 +1,30 @@ +clc;
+//page 464
+//problem 9.2
+
+//Baseband cutoff signal fM = 15 kHz
+fM = 15 * 10^3;
+
+//Maximum frequency deviation Df = 60 kHz
+Df = 60 * 10^3;
+
+//Figure of Merit for FM is G_FM
+G_FM = (3/2)*(Df/fM)^2;
+
+disp('Figure of Merit for FM system is '+string(G_FM));
+
+//Ratio of Figure of Merits of FM and AM systems is R
+R = G_FM/(1/3);
+
+disp('Ratio of Figure of Merits for FM and AM systems is '+string(R));
+
+Df_new = 2*Df;
+
+//Figure of Merit for FM when bandwidth is doubled is G_FM_new
+G_FM_new = (3/2)*(Df_new/fM)^2;
+
+//Ratio of Figure of Merits of FM and AM systems when bandwidth is doubled is R_new
+R_new = G_FM_new/(1/3);
+
+disp('Ratio of Figure of Merits for FM and AM systems when bandwidth is doubled is '+string(R_new));
+
diff --git a/3161/CH9/EX9.3/Ex9_3.sce b/3161/CH9/EX9.3/Ex9_3.sce new file mode 100644 index 000000000..f3773ff67 --- /dev/null +++ b/3161/CH9/EX9.3/Ex9_3.sce @@ -0,0 +1,27 @@ +clc;
+//page 475
+//problem 9.3
+
+//Resistance R = 1000 Ohm
+R = 10^3;
+
+//Capacitance C = 0.1 * 10^-6 F
+C = 0.1*10^-6;
+
+//Break point for RC filter is f1
+f1 = 1/(2*%pi*R*C)
+
+//Baseband bandwidth of signal fM = 15 kHz
+fM = 15 * 10^3;
+
+Gain = atan(fM/f1)/(3*(f1/fM)*[1 - (f1/fM)*atan(fM/f1)]);
+
+disp('Initial Gain is '+string(10*log10(Gain))+' dB');
+
+//New Baseband bandwidth of signal fM_new = 15 kHz
+fM_new = 2*15 * 10^3;
+
+Gain_new = atan(fM_new/f1)/(3*(f1/fM_new)*[1 - (f1/fM_new)*atan(fM_new/f1)]);
+
+disp('Final Gain is '+string(10*log10(Gain_new))+' dB');
+
diff --git a/3161/CH9/EX9.6/Ex9_6.sce b/3161/CH9/EX9.6/Ex9_6.sce new file mode 100755 index 000000000..6fcea73f4 --- /dev/null +++ b/3161/CH9/EX9.6/Ex9_6.sce @@ -0,0 +1,43 @@ +clc;
+//page 495
+//problem 9.6
+
+//Baseband cutoff signal fM = 15 kHz
+fM = 15 * 10^3;
+
+//Carrier filter bandwidth is B = 60 kHz
+B = 60 * 10^3;
+
+//RMS frequency division Df_RMS = 30 kHz
+Df_RMS = 30 * 10^3;
+
+//Let a = Df_RMS/fM for substitution
+a = Df_RMS/fM;
+
+//Let b = fM/B for substitution
+b = fM/B;
+
+//Let input SNR 1 be I_SNR1 = 10 dB = 10
+I_SNR1 = 10;
+
+//Output SNR is O_SNR1
+O_SNR1 = (3*(a^2)*I_SNR1)/(1+6*((2/%pi)^0.5)*I_SNR1*exp(-(b)*I_SNR1));
+
+disp('Output SNR is '+string(10*log10(O_SNR1))+' dB');
+
+//Let input SNR 2 be I_SNR2 = 20 dB = 100
+I_SNR2 = 100;
+
+//Output SNR is O_SNR2
+O_SNR2 = (3*(a^2)*I_SNR2)/(1+6*((2/%pi)^0.5)*I_SNR2*exp(-(b)*I_SNR2));
+
+//Solution given in the book is 13.5431 which is fallacious, the correct answer is 24.32444
+disp('Output SNR is '+string(10*log10(O_SNR2))+' dB');
+
+//Let input SNR 3 be I_SNR3 = 30 dB = 1000
+I_SNR3 = 1000;
+
+//Output SNR is O_SNR3
+O_SNR3 = (3*(a^2)*I_SNR3)/(1+6*((2/%pi)^0.5)*I_SNR3*exp(-(b)*I_SNR3));
+
+disp('Output SNR is '+string(10*log10(O_SNR3))+' dB');
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