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-rwxr-xr-x929/CH1/EX1.16/Example1_16.sce378
-rwxr-xr-x929/CH3/EX3.1/Example3_1.sce48
-rwxr-xr-x929/CH6/EX6.7.a/Example6_7_a.sce123
3 files changed, 274 insertions, 275 deletions
diff --git a/929/CH1/EX1.16/Example1_16.sce b/929/CH1/EX1.16/Example1_16.sce
index 4a222071c..0326822bf 100755
--- a/929/CH1/EX1.16/Example1_16.sce
+++ b/929/CH1/EX1.16/Example1_16.sce
@@ -1,190 +1,190 @@
-//Example 1.16
-
-clear;
-
-clc;
-
-//Part(I)
-//This part of the program includes the plotting of the input wave (triangular wave). To plot the wave we have divided the time period(assuming T=2) into 3 time intervals t1, t2, t3 and then create voltage equation for each using the given conditions.
-
-VCC=13;
-
-VEE=-13;
-
-A=-2;//Gain
-
-t1=[0:10^(-4):0.5];
-
-t2=[0.5:10^(-4):1.5];
-
-t3=[1.5:10^(-4):2.5];
-
-vt1=20*t1;
-
-vt2=20*(1-t2);
-
-vt3=20*(t3-2);
-
-subplot(131);
-
-title(" Inverting Amplifier driven into saturation waveforms ","fontsize",6);
-
-subplot(334);
-
-plot(t1,vt1);
-
-plot(t2,vt2);
-
-plot(t3,vt3);
-
-xlabel("Time(t)","fontsize",3);
-
-ylabel("Input Voltage(Vin)","fontsize",3);
-
-title("Vin vs t","fontsize",4);
-
-//Part(II)
-//In this part we have plotted vo by using the conditions vo=-2vI for -6.5V<vI<6.5V, otherwise vo=-13.Again we have divided the time period into 5 parts to1, to2, to3, to1i, to2i depending upon the response in each interval.
-
-vIbor=VCC/2;
-
-to1min=0;
-
-to1max=6.5/20;
-
-to2min=1-(6.5/20);
-
-to2max=1+(6.5/20);
-
-to3min=2-(6.5/20);
-
-to3max=2+(6.5/20);
-
-to1=[to1min:10^(-4):to1max];
-
-to2=[to2min:10^(-4):to2max];
-
-to3=[to3min:10^(-4):to3max];
-
-to1imin=to1max;
-
-to1imax=to2min;
-
-to2imin=to2max;
-
-to2imax=to3min;
-
-to1i=[to1imin:10^(-4):to1imax];
-
-to2i=[to2imin:10^(-4):to2imax];
-
-vo1=-13*(to1-to1min)/(to1max-to1min);
-
-vo1i=-13;
-
-vo2=(((13+13)/(to2max-to2min))*(to2-to2min))-13;
-
-vo2i=13;
-
-vo3=(((13+13)/(to3min-to3max))*(to3-to3max))-13;
-
-subplot(335);
-
-plot(to1,vo1);
-
-plot(to1i,vo1i);
-
-plot(to2,vo2);
-
-plot(to2i,vo2i);
-
-plot(to3,vo3);
-
-ylabel("Output Voltage(Vout)","fontsize",3);
-
-xlabel("Time(t)","fontsize",3);
-
-title("Vout vs t","fontsize",4);
-
-//Part(III)
-//In this part we will plot vN for which we have divided the time period into 7 time intervals tNi1, tNi2, tNi3, tN11, tN12, tN21, tN22 depending upon the response in each cycle voltage equation is obtained and plotted.For -6.5<vI<6.5 vN=0 and when vI will peak at 10V vN will peak at 2.33 and for vI<6-.5 and vI>6.5, circuit behaviour is symmetric.
-
-vIbor=VCC/2;
-
-tNi1min=0;
-
-tNi1max=6.5/20;
-
-tNi2min=1-(6.5/20);
-
-tNi2max=1+(6.5/20);
-
-tNi3min=2-(6.5/20);
-
-tNi3max=2+(6.5/20);
-
-tNi1=[tNi1min:10^(-4):tNi1max];
-
-tNi2=[tNi2min:10^(-4):tNi2max];
-
-tNi3=[tNi3min:10^(-4):tNi3max];
-
-tN11min=tNi1max;
-
-tN11max=(tNi2min+tNi1max)/2;
-
-tN12min=tN11max;
-
-tN12max=tNi2min;
-
-tN21min=tNi2max;
-
-tN21max=(tNi2max+tNi3min)/2;
-
-tN22min=tN21max;
-
-tN22max=tNi3min;
-
-tN11=[tN11min:10^(-4):tN11max];
-
-tN12=[tN12min:10^(-4):tN12max];
-
-tN21=[tN21min:10^(-4):tN21max];
-
-tN22=[tN22min:10^(-4):tN22max];
-
-vNi1=0;
-
-vN11=(2.33/(tN11max-tN11min))*(tN11-tN11min);
-
-vN12=-(2.33/(tN12max-tN12min))*(tN12-tN12max);
-
-vNi2=0;
-
-vN21=-(2.33/(tN21max-tN21min))*(tN21-tN21min);
-
-vN22=(2.33/(tN22max-tN22min))*(tN22-tN22max);
-
-vNi3=0;
-
-subplot(336);
-
-plot(tNi1,vNi1);
-
-plot(tN11,vN11);
-
-plot(tN12,vN12);
-
-plot(tNi2,vNi2);
-
-plot(tN21,vN21);
-
-plot(tN22,vN22);
-
-plot(tNi3,vNi3);
-
-xlabel("Time(t)","fontsize",3);
-
-ylabel("Vn","fontsize",3);
-
+//Example 1.16
+
+clear;
+
+clc;
+
+//Part(I)
+//This part of the program includes the plotting of the input wave (triangular wave). To plot the wave we have divided the time period(assuming T=2) into 3 time intervals t1, t2, t3 and then create voltage equation for each using the given conditions.
+
+VCC=13;
+
+VEE=-13;
+
+A=-2;//Gain
+
+t1=[0:10^(-4):0.5];
+
+t2=[0.5:10^(-4):1.5];
+
+t3=[1.5:10^(-4):2.5];
+
+vt1=20*t1;
+
+vt2=20*(1-t2);
+
+vt3=20*(t3-2);
+
+subplot(131);
+
+title(" Inverting Amplifier driven into saturation waveforms ","fontsize",6);
+
+subplot(334);
+
+plot(t1,vt1);
+
+plot(t2,vt2);
+
+plot(t3,vt3);
+
+xlabel("Time(t)","fontsize",3);
+
+ylabel("Input Voltage(Vin)","fontsize",3);
+
+title("Vin vs t","fontsize",4);
+
+//Part(II)
+//In this part we have plotted vo by using the conditions vo=-2vI for -6.5V<vI<6.5V, otherwise vo=-13.Again we have divided the time period into 5 parts to1, to2, to3, to1i, to2i depending upon the response in each interval.
+
+vIbor=VCC/2;
+
+to1min=0;
+
+to1max=6.5/20;
+
+to2min=1-(6.5/20);
+
+to2max=1+(6.5/20);
+
+to3min=2-(6.5/20);
+
+to3max=2+(6.5/20);
+
+to1=[to1min:10^(-4):to1max];
+
+to2=[to2min:10^(-4):to2max];
+
+to3=[to3min:10^(-4):to3max];
+
+to1imin=to1max;
+
+to1imax=to2min;
+
+to2imin=to2max;
+
+to2imax=to3min;
+
+to1i=[to1imin:10^(-4):to1imax];
+
+to2i=[to2imin:10^(-4):to2imax];
+
+vo1=-13*(to1-to1min)/(to1max-to1min);
+
+vo1i=-13;
+
+vo2=(((13+13)/(to2max-to2min))*(to2-to2min))-13;
+
+vo2i=13;
+
+vo3=(((13+13)/(to3min-to3max))*(to3-to3max))-13;
+
+subplot(335);
+
+plot(to1,vo1);
+
+plot(to1i,vo1i*ones(1,length(to1i)));
+
+plot(to2,vo2);
+
+plot(to2i,vo2i*ones(1,length(to2i)));
+
+plot(to3,vo3);
+
+ylabel("Output Voltage(Vout)","fontsize",3);
+
+xlabel("Time(t)","fontsize",3);
+
+title("Vout vs t","fontsize",4);
+
+//Part(III)
+//In this part we will plot vN for which we have divided the time period into 7 time intervals tNi1, tNi2, tNi3, tN11, tN12, tN21, tN22 depending upon the response in each cycle voltage equation is obtained and plotted.For -6.5<vI<6.5 vN=0 and when vI will peak at 10V vN will peak at 2.33 and for vI<6-.5 and vI>6.5, circuit behaviour is symmetric.
+
+vIbor=VCC/2;
+
+tNi1min=0;
+
+tNi1max=6.5/20;
+
+tNi2min=1-(6.5/20);
+
+tNi2max=1+(6.5/20);
+
+tNi3min=2-(6.5/20);
+
+tNi3max=2+(6.5/20);
+
+tNi1=[tNi1min:10^(-4):tNi1max];
+
+tNi2=[tNi2min:10^(-4):tNi2max];
+
+tNi3=[tNi3min:10^(-4):tNi3max];
+
+tN11min=tNi1max;
+
+tN11max=(tNi2min+tNi1max)/2;
+
+tN12min=tN11max;
+
+tN12max=tNi2min;
+
+tN21min=tNi2max;
+
+tN21max=(tNi2max+tNi3min)/2;
+
+tN22min=tN21max;
+
+tN22max=tNi3min;
+
+tN11=[tN11min:10^(-4):tN11max];
+
+tN12=[tN12min:10^(-4):tN12max];
+
+tN21=[tN21min:10^(-4):tN21max];
+
+tN22=[tN22min:10^(-4):tN22max];
+
+vNi1=0;
+
+vN11=(2.33/(tN11max-tN11min))*(tN11-tN11min);
+
+vN12=-(2.33/(tN12max-tN12min))*(tN12-tN12max);
+
+vNi2=0;
+
+vN21=-(2.33/(tN21max-tN21min))*(tN21-tN21min);
+
+vN22=(2.33/(tN22max-tN22min))*(tN22-tN22max);
+
+vNi3=0;
+
+subplot(336);
+
+plot(tNi1,vNi1*ones(1,length(tNi1)));
+
+plot(tN11,vN11);
+
+plot(tN12,vN12);
+
+plot(tNi2,vNi2*ones(1,length(tNi2)));
+
+plot(tN21,vN21);
+
+plot(tN22,vN22);
+
+plot(tNi3,vNi3*ones(1,length(tNi3)));
+
+xlabel("Time(t)","fontsize",3);
+
+ylabel("Vn","fontsize",3);
+
title("Vn vs t","fontsize",4); \ No newline at end of file
diff --git a/929/CH3/EX3.1/Example3_1.sce b/929/CH3/EX3.1/Example3_1.sce
index 920cacc19..0abfe1c48 100755
--- a/929/CH3/EX3.1/Example3_1.sce
+++ b/929/CH3/EX3.1/Example3_1.sce
@@ -1,25 +1,25 @@
-//Example 3.1
-
-clear;
-
-clc;
-
-R=10;
-
-C=40*10^(-6);
-
-L=5*10^(-3);
-
-Hsnum=(R/L)*%s;
-
-Hsden=((%s^(2))+(R/L)*%s+(1/(L*C)));
-
-Hs=Hsnum/Hsden;//Transfer Function
-
-h=syslin('c',Hs);
-
-plzr(h);
-
-zeroes=roots(Hsnum);
-
+//Example 3.1
+
+clear;
+
+clc;
+
+R=10;
+
+C=40*10^(-6);
+
+L=5*10^(-3);
+
+Hsnum=(R/L)*%s;
+
+Hsden=((%s^(2))+(R/L)*%s+(1/(L*C)));
+
+Hs=Hsnum/Hsden;//Transfer Function
+
+h=syslin('c',Hs);
+
+plzr(h);
+
+zeroes=roots(Hsnum);
+
poles=roots(Hsden); \ No newline at end of file
diff --git a/929/CH6/EX6.7.a/Example6_7_a.sce b/929/CH6/EX6.7.a/Example6_7_a.sce
index 9aef3664e..ac48bfa30 100755
--- a/929/CH6/EX6.7.a/Example6_7_a.sce
+++ b/929/CH6/EX6.7.a/Example6_7_a.sce
@@ -1,63 +1,62 @@
-//Example 6.7(a)
-
-clear;
-
-clc;
-
-IA=19.6*10^(-6);
-
-Cc=30*10^(-12);
-
-SR=0.633*10^6;
-
-R1=3*10^3;
-
-R2=12*10^3;
-
-A0=-(R2/R1);
-
-b=R1/(R1+R2);
-
-a0=2*10^5;
-
-ft=1*10^6;
-
-ro=100;
-
-Vim=-0.5;
-
-tau=1/(2*%pi*b*ft);
-
-Vomcrit=SR*tau;
-
-Voinf=A0*Vim;
-
-V1=Voinf-Vomcrit;
-
-t=[0:2*10^(-8):7*10^(-6)];
-
-t1=V1/SR;
-
-t12=[0:2*10^(-8):tau]
-
-vo3=0;
-
-plot(t12,vo3);
-
-t11=[tau:2*10^(-8):t1+tau];
-
-vo1=SR*(t11-tau);
-
-t22=[t1+tau:2*10^(-8):7*10^(-6)];
-
-vo2=Voinf+((V1-Voinf)*exp(-(t22-t1-tau)/tau));
-
-plot(t11,vo1);
-
-plot(t22,vo2);
-
-xlabel("time(t)","fontsize", 6);
-
-ylabel("vo(t)","fontsize", 6);
-
+//Example 6.7(a)
+
+clear;
+
+clc;
+
+IA=19.6*10^(-6);
+
+Cc=30*10^(-12);
+
+SR=0.633*10^6;
+
+R1=3*10^3;
+
+R2=12*10^3;
+
+A0=-(R2/R1);
+
+b=R1/(R1+R2);
+
+a0=2*10^5;
+
+ft=1*10^6;
+
+ro=100;
+
+Vim=-0.5;
+
+tau=1/(2*%pi*b*ft);
+
+Vomcrit=SR*tau;
+
+Voinf=A0*Vim;
+
+V1=Voinf-Vomcrit;
+
+t=[0:2*10^(-8):7*10^(-6)];
+
+t1=V1/SR;
+
+t12=[0:2*10^(-8):tau]
+
+vo3=0*ones(1,length(t12));
+plot(t12,vo3);
+
+t11=[tau:2*10^(-8):t1+tau];
+
+vo1=SR*(t11-tau);
+
+t22=[t1+tau:2*10^(-8):7*10^(-6)];
+
+vo2=Voinf+((V1-Voinf)*exp(-(t22-t1-tau)/tau));
+
+plot(t11,vo1);
+
+plot(t22,vo2);
+
+xlabel("time(t)","fontsize", 6);
+
+ylabel("vo(t)","fontsize", 6);
+
title("Step Response of the Circuit","fontsize", 8); \ No newline at end of file
generated by cgit (git 2.34.1) at 2025-03-13 10:07:30 +0000