diff options
Diffstat (limited to 'Working_Examples/154/DEPENDENCIES')
74 files changed, 2027 insertions, 0 deletions
diff --git a/Working_Examples/154/DEPENDENCIES/ch10_4.sce b/Working_Examples/154/DEPENDENCIES/ch10_4.sce new file mode 100755 index 0000000..bd68900 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch10_4.sce @@ -0,0 +1,27 @@ +clc
+disp("Problem 10.4")
+printf("\n")
+
+//For V1
+Ro1=25
+Theta1=143.13
+//For V1
+Ro2=11.2
+Theta2=26.57
+//We need to find V1/V2
+//Let V=V1/V2
+Vmag=(Ro1/Ro2)
+Vph=Theta1-Theta2
+x=Vmag*cos((Vph*%pi)/180);
+y=Vmag*sin((Vph*%pi)/180);
+z=complex(x,y)
+//Let V1+V2=V12
+x1=Ro1*cos((Theta1*%pi)/180);
+y1=Ro1*sin((Theta1*%pi)/180);
+z1=complex(x1,y1)
+x2=Ro2*cos((Theta2*%pi)/180);
+y2=Ro2*sin((Theta2*%pi)/180);
+z2=complex(x2,y2)
+V12=z1+z2
+[R,Theta]=polar(V12)
+printf("V1/V2=%0.2f+i*%3.2f \nV1+V2=%3.2f(%3.2f deg)",x,y,R,(Theta*180)/%pi)
\ No newline at end of file diff --git a/Working_Examples/154/DEPENDENCIES/ch10_5.sce b/Working_Examples/154/DEPENDENCIES/ch10_5.sce new file mode 100755 index 0000000..87f47e8 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch10_5.sce @@ -0,0 +1,28 @@ +clc
+disp("Problem 10.5")
+printf("\n")
+
+printf("Given")
+disp("Voltage is 100(45 deg)")
+disp("Current is 5(15 deg)")
+//For V
+Ro1=100
+Theta1=45
+//For I
+Ro2=5
+Theta2=15
+//We need to find V/I=Z
+
+Zmag=(Ro1/Ro2)
+Zph=Theta1-Theta2
+x=Zmag*cos((Zph*%pi)/180);
+y=Zmag*sin((Zph*%pi)/180);
+z=complex(x,y)
+//Let Y=1/Z
+Ymag=(Ro2/Ro1)
+Yph=Theta2-Theta1
+x1=Ymag*cos((Yph*%pi)/180);
+y1=Ymag*sin((Yph*%pi)/180);
+z1=complex(x1,y1)
+
+printf("R=%3.2fohm XL=%3.2fH \nG=%0.3fS BL=%0.3fS",x,y,x1,abs(y1));
\ No newline at end of file diff --git a/Working_Examples/154/DEPENDENCIES/ch10_7.sce b/Working_Examples/154/DEPENDENCIES/ch10_7.sce new file mode 100755 index 0000000..13299d1 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch10_7.sce @@ -0,0 +1,45 @@ +clc
+disp("Problem 10.7")
+printf("\n")
+
+printf("Voltage v1=5*cos(w1*t)")
+printf("Voltage v2=10*cos(w2*t+60)")
+//The circuit is modeled as
+disp("Resistance is 10ohm and inductance is 5mH")
+R=10;L=5*10^-3;
+disp("a)")
+w1=2000;w2=2000;
+//Let Z be the impedance of the coil
+Z1=R+%i*L*w1
+Z2=R+%i*L*w2
+//Let V be phasor voltage between the terminals
+Vmag=10;
+Vph=60;
+x=Vmag*cos((Vph*%pi)/180);
+y=Vmag*sin((Vph*%pi)/180);
+z=complex(x,y)
+v=5-z;
+//Let I be the current
+I=v/Z1
+[R,Theta]=polar(I)
+printf("i=%0.2f*cos(%dt%d deg)",R,w1,(Theta*180)/%pi);
+
+disp("b)")
+R=10;L=5*10^-3;
+w1=2000;w2=4000;
+//Let Z be the impedance of the coil
+Z1=R+%i*L*w1
+Z2=R+%i*L*w2
+V1=5;
+//By applying superposition i=i1-i2
+I1=V1/Z1
+[R,Theta]=polar(I1)
+printf("i1=%0.2f*cos(%dt%d deg)\n",R,w1,(Theta*180)/%pi);
+V2mag=10;V2ph=60;
+I2=z/Z2
+[R1,Theta1]=polar(I2)
+printf("i2=%0.2f*cos(%dt%3.2f deg)\n",R1,w2,(Theta1*180)/%pi);
+//i=i1-i2
+printf("i=%0.2f*cos(%dt%d deg)-%0.2f*cos(%dt%3.2f deg)\n",R,w1,(Theta*180)/%pi,R1,w2,(Theta1*180)/%pi)
+
+
diff --git a/Working_Examples/154/DEPENDENCIES/ch11_1.sce b/Working_Examples/154/DEPENDENCIES/ch11_1.sce new file mode 100755 index 0000000..cb812c5 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch11_1.sce @@ -0,0 +1,20 @@ +clc
+disp("Problem 11.1")
+printf("\n")
+
+printf("Given")
+disp("Resistance =1000ohm")
+t=0:0.5:1;
+i=1;i1=-1;
+figure
+a=gca()
+plot(t,i,t+1,i1,t+2,i,t+3,i1)
+xtitle("i vs t",'t in ms','i in mA')
+i=1*10^-3;R=1000;
+//p=i^2*R
+p=i^2*R;
+figure
+a=gca()
+plot(t,p)
+xtitle("p vs t",'t in ms','p in mW')
+
diff --git a/Working_Examples/154/DEPENDENCIES/ch11_10.sce b/Working_Examples/154/DEPENDENCIES/ch11_10.sce new file mode 100755 index 0000000..0fd8a45 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch11_10.sce @@ -0,0 +1,56 @@ +clc
+disp("Problem 11.10")
+printf("\n")
+
+printf("Given")
+disp("Veff=10V v=10*sqrt(2)*cos(w*t)");
+Veff=10;vmag=10*1.414
+
+disp("a)")
+Z1=1+%i
+[R,Theta]=polar(Z1)
+printf("i1=%d*cos(w*t-%d)\n",(vmag/R),Theta)
+I1eff=(vmag/R)/1.414
+//p1(t)=100*sqrt(2)*cos(wt)*cos(wt-45)
+//On solving
+disp("p1(t)=50+50*sqrt(2)*cos(2*w*t-45) W")
+P1=Veff*I1eff*cos(Theta)
+Q1=Veff*I1eff*sin(Theta)
+S1=P1+%i*Q1
+S1mag=sqrt(P1^2+Q1^2)
+pf1=P1/S1mag
+printf("P1=%dW\nQ1=%dvar\npf1=%0.4f(lag)\n",P1,Q1,pf1)
+
+
+disp("b)")
+Z2=1-%i
+[R,Theta]=polar(Z2)
+printf("i2=%d*cos(w*t%d)\n",(vmag/R),Theta)
+I2eff=(vmag/R)/1.414
+//p2(t)=100*sqrt(2)*cos(wt)*cos(wt+45)
+//On solving
+disp("p2(t)=50+50*sqrt(2)*cos(2*w*t+45) W")
+P2=Veff*I2eff*cos(Theta)
+Q2=Veff*I2eff*sin(Theta)
+S2=P2+%i*Q2
+S2mag=sqrt(P2^2+Q2^2)
+pf2=P2/S2mag
+printf("P2=%dW\nQ2=%dvar\npf2=%0.4f(lag)\n",P2,Q2,pf2)
+
+disp("c)")
+Zmag=(Z1*Z2)/(Z1+Z2)
+printf("i=%d*cos(w*t)\n",(vmag/Zmag))
+Ieff=(vmag/Zmag)/1.414
+//p(t)=100*sqrt(2)*sqrt(2)*cos(wt)*cos(wt)
+//On solving
+disp("p2(t)=200*cos(w*t)^2 W")
+P=Veff*Ieff
+Q=0
+S=P+%i*Q
+Smag=sqrt(P^2+Q^2)
+pf=P/Smag
+printf("P=%dW\nQ=%dvar\npf=%0.4f\n",P,Q,pf)
+
+
+
+
diff --git a/Working_Examples/154/DEPENDENCIES/ch11_11.sce b/Working_Examples/154/DEPENDENCIES/ch11_11.sce new file mode 100755 index 0000000..143b3cf --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch11_11.sce @@ -0,0 +1,27 @@ +clc
+disp("Problem 11.11")
+printf("\n")
+
+printf("Given")
+disp("v=42.5*cos(1000*t+30 deg)V Z=3+i4 ohm")
+Vmag=42.5;
+Z=3+%i*4;
+R=sqrt(3^2+4^2)
+Theta=atan(4/3)*(180/%pi)
+Veffm=Vmag/sqrt(2)
+Veffph=30
+Ieffm=Veffm/R
+Ieffph=30-Theta
+
+Smag=Veffm*Ieffm
+Sph=Veffph-Ieffph
+x=Smag*cos((Sph*%pi)/180)
+y=Smag*sin((Sph*%pi)/180)
+z=complex(x,y)
+pf=cos((Theta*%pi)/180);
+
+printf("Real Power is %fW\n",x)
+printf("Reactive Power is %fvar(inductive)\n",y)
+printf("Complex Power is %fVA\n",Smag)
+printf("Power factor is %3.1f(lag)\n",pf)
+
diff --git a/Working_Examples/154/DEPENDENCIES/ch11_12.sce b/Working_Examples/154/DEPENDENCIES/ch11_12.sce new file mode 100755 index 0000000..346fdf4 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch11_12.sce @@ -0,0 +1,29 @@ +clc
+disp("Problem 11.12")
+printf("\n")
+
+printf("Given")
+disp("pf1=1 ; pf2=0.5 ; pf3=0.5")
+disp("P1=10kW;P2=20kW;P3=15kW")
+disp("Power supply is 6kV")
+P1=10000;P2=20000;P3=15000;
+Veff=6000;
+pf1=1 //implifies that theta1=0
+t1=0
+Q1=P1*t1
+
+pf2=0.5 //implifies that theta1=60
+t2=1.73;
+Q2=P2*t2
+
+pf3=1 //implifies that theta1=53.13
+t3=1.33;
+Q3=P3*t3
+
+PT=P1+P2+P3
+QT=Q1+Q2+Q3
+ST=sqrt(PT^2+QT^2)
+pfT=PT/ST
+Ieff=ST/Veff
+Ieffph=acos(pfT)*(180/%pi)
+printf("PT=%dW\nQT=%dvar\nST=%dVA\npf=%0.2f(lag)\nIeff=%3.1f(%3.2f deg)\n",PT,QT,ST,pfT,Ieff,Ieffph)
diff --git a/Working_Examples/154/DEPENDENCIES/ch11_13.sce b/Working_Examples/154/DEPENDENCIES/ch11_13.sce new file mode 100755 index 0000000..63dcd76 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch11_13.sce @@ -0,0 +1,20 @@ +clc
+disp("Problem 11.13")
+printf("\n")
+
+printf("Given")
+disp("Power factor is 0.95(lag)")
+vmag=240;Zmag=3.5;Zph=25;
+I1mag=vmag/Zmag;iph=0-Zph;
+//Smag=Veff*Ieff
+Smag=(vmag/sqrt(2))*(I1mag/sqrt(2))
+Sph=0+abs(iph)
+x=Smag*cos((Sph*%pi)/180)
+y=Smag*sin((Sph*%pi)/180)
+z=complex(x,y)
+pf=0.95
+theta=acos(0.95)*(180/%pi)
+//From fig 11.11
+//Solving for Qc
+Qc=y-(tan((theta*%pi)/180)*x)
+printf("\n Qc=%dvar(Capacitive )\n",Qc)
diff --git a/Working_Examples/154/DEPENDENCIES/ch11_14.sce b/Working_Examples/154/DEPENDENCIES/ch11_14.sce new file mode 100755 index 0000000..8b28778 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch11_14.sce @@ -0,0 +1,25 @@ +clc
+disp("Problem 11.14")
+printf("\n")
+
+printf("Given")
+disp("Power =1000kW ; pf=0.5(lag)")
+disp("Voltage source is 5kV")
+disp("Improved power factor is 0.8")
+
+//Before improvement
+P=1000*10^3;
+pf=0.5;V=5*10^3;
+S=(P/pf)*10^-3
+I=S/V
+
+//After improvement
+P=1000*10^3;
+pf=0.8;V=5*10^3;
+S=(P/pf)*10^-3
+I1=S/V
+
+disp("Current is reduced by ")
+red=((I-I1)/I)*100
+printf("Percentage reduction in current is %3.1fpercent\n",red)
+
diff --git a/Working_Examples/154/DEPENDENCIES/ch11_16.sce b/Working_Examples/154/DEPENDENCIES/ch11_16.sce new file mode 100755 index 0000000..1389705 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch11_16.sce @@ -0,0 +1,46 @@ +clc
+disp("Problem 11.16")
+printf("\n")
+
+printf("Given")
+disp("Vg=100V(rms)")
+disp("Zg=1+i Z1=2")
+Vg=100;
+
+disp("a)")
+Zg=1+%i;
+Z1=2
+Z=Z1+Zg
+Zmag=sqrt(real(Z)^2+imag(Z)^2)
+I=Vg/Zmag
+PZ1=real(Z1)*(I^2)
+Pg=real(Zg)*(I^2)
+PT=PZ1+Pg
+printf("PZ=%dW\n Pg=%dW\n PT=%dW\n",PZ1,Pg,PT);
+
+disp("b)")
+//If Z2=a+i*b
+//Zg*=1-i
+//Given that
+//(Z1*Z2)/(Z1+Z2)=1-i
+//As Z1=2 and solving for Z2
+ disp(-%i,"Z2=")
+
+disp("c)")
+//If Z2 is taken the value as calculated in b) then Z=1-i
+Zg=1+%i;
+Z1=2;
+Z=1-%i;
+Zt=Z+Zg
+Zmag=sqrt(real(Zt)^2+imag(Zt)^2)
+I=Vg/Zmag
+PZ=real(Z)*(I^2)
+Pg=real(Zg)*(I^2)
+//To calculate PZ1 and PZ2 we need to first calculate IZ1 nad IZ2
+VZ=I*(1-%i)
+IZ1=VZ/Z1
+IZ1mag=sqrt(real(IZ1)^2+imag(IZ1)^2)
+PZ1=real(Z1)*(IZ1mag^2)
+PZ2=PZ-PZ1
+PT=PZ1+PZ2+Pg
+printf("PZ=%dW\n Pg=%dW\n PT=%dW\n",PZ,Pg,PT);
diff --git a/Working_Examples/154/DEPENDENCIES/ch11_17.sce b/Working_Examples/154/DEPENDENCIES/ch11_17.sce new file mode 100755 index 0000000..3fdef1c --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch11_17.sce @@ -0,0 +1,80 @@ +clc
+disp("Problem 11.17")
+printf("\n")
+
+printf("Given")
+disp("v1=5*cos(w1*t) v2=10*cos(w2*t+60)")
+//The circuit is modeled as
+disp("Resistance is 10ohm and inductance is 5mH")
+R=10;L=5*10^-3;
+//Let V be phasor voltage between the terminals
+Vmag=10;
+Vph=60;
+x=Vmag*cos((Vph*%pi)/180);
+y=Vmag*sin((Vph*%pi)/180);
+z=complex(x,y)
+
+disp("a)")
+w1=2000;w2=4000;
+//Let Z be the impedance of the coil
+Z1=R+%i*L*w1
+Z2=R+%i*L*w2
+V1=5;
+//By applying superposition i=i1-i2
+I1=V1/Z1
+[R1,Theta]=polar(I1)
+printf("i1=%0.2f*cos(%dt%d deg)\n",R1,w1,(Theta*180)/%pi);
+P1=(R*R1^2)/2
+
+V2mag=10;V2ph=60;
+I2=z/Z2
+[R2,Theta1]=polar(I2)
+printf("i2=%0.2f*cos(%dt%3.2f deg)\n",R2,w2,(Theta1*180)/%pi);
+P2=(R*R2^2)/2
+
+//i=i1-i2
+printf("i=%0.2f*cos(%dt%d deg)-%0.2f*cos(%dt%3.2f deg)\n",R1,w1,(Theta*180)/%pi,R2,w2,(Theta1*180)/%pi)
+
+printf("P1=%0.3fW\nP2=%3.1fW\nTotal power(P)=%3.3fW\n",P1,P2,(P1+P2))
+
+disp("b)")
+//From problem 10.7
+imagn=0.61
+P=(R*imagn^2)/2
+printf("Power dissipated in the coil=%3.3fW\n",P)
+
+disp("c)")
+w1=2000;w2=1414;
+//Let Z be the impedance of the coil
+Z1=R+%i*L*w1
+Z2=R+%i*L*w2
+V1=5;
+//By applying superposition i=i1-i2
+I1=V1/Z1
+[R1,Theta]=polar(I1)
+printf("i1=%0.2f*cos(%dt%d deg)\n",R1,w1,(Theta*180)/%pi);
+P1=(R*R1^2)/2
+
+V2mag=10;V2ph=60;
+x1=V2mag*cos((V2ph*%pi)/180);
+y1=V2mag*sin((V2ph*%pi)/180);
+z1=complex(x1,y1)
+I3=z1/Z2
+[R3,Theta3]=polar(I3)
+printf("i2=%0.2f*cos(%dt+%3.2f deg)\n",R3,w2,(Theta3*180)/%pi);
+P3=(R*R3^2)/2
+
+//i=i1-i2
+printf("i=%0.2f*cos(%dt%d deg)-%0.2f*cos(%dt+%3.2f deg)\n",R1,w1,(Theta*180)/%pi,R3,w2,(Theta3*180)/%pi)
+
+printf("P1=%0.3fW\nP2=%3.1fW\nTotal power(P)=%3.3fW\n",P1,P3,(P1+P3))
+
+
+
+
+
+
+
+
+
+
diff --git a/Working_Examples/154/DEPENDENCIES/ch11_2.sce b/Working_Examples/154/DEPENDENCIES/ch11_2.sce new file mode 100755 index 0000000..eeada8f --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch11_2.sce @@ -0,0 +1,37 @@ +clc
+disp("Problem 11.2")
+printf("\n")
+
+t=0:0.5:1;
+i=1;i1=-1;
+figure
+a=gca()
+plot(t,i,t+1,i1)
+xtitle("i vs t",'t in ms','i in mA')
+//Voltage across capacitor vC=(1/C)*integrate(i*dt)
+//On integration
+t=0:0.0005:0.001
+v=2000*t
+v1=2-v;
+figure
+a=gca()
+plot(t,v,t+0.001,v1,t+0.002,v,t+0.003,v1)
+xtitle("v vs t",'t in ms','v in V')
+
+//Power is p=v*i
+t=0:.0005:.001
+p=2000*t
+p1=p-2;
+figure
+a=gca()
+plot(t,p,t+0.001,p1,t+0.002,p,t+0.003,p1)
+xtitle("p vs t",'t in ms','p in W')
+
+//Work is (C*v^2)/2
+t=0:.0005:.001
+w=t^2
+w1=t^2+1*10^-6-(2*10^-3*t);
+figure
+a=gca()
+plot(t,w,t+0.001,w1,t+0.002,w,t+0.003,w1)
+xtitle("w vs t",'t in ms','w in J')
diff --git a/Working_Examples/154/DEPENDENCIES/ch11_4.sce b/Working_Examples/154/DEPENDENCIES/ch11_4.sce new file mode 100755 index 0000000..76ad594 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch11_4.sce @@ -0,0 +1,14 @@ +clc
+disp("Problem 11.4")
+printf("\n")
+
+printf("Given")
+disp("Veff=110V Z=10+i8 ohm")
+Veff=110;
+Z=10+%i*8
+Zmag=sqrt(10^2+8^2)
+Zph=(atan(8/10)*180)/%pi
+P=(Veff^2*R)/(Zmag^2)
+pf=cos((Zph*%pi)/180)
+
+disp(pf,"Power factor is")
\ No newline at end of file diff --git a/Working_Examples/154/DEPENDENCIES/ch11_5.sce b/Working_Examples/154/DEPENDENCIES/ch11_5.sce new file mode 100755 index 0000000..220644b --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch11_5.sce @@ -0,0 +1,14 @@ +clc
+disp("Problem 11.5")
+printf("\n")
+
+printf("Given")
+disp("Veff=110V Ieff=20(-50 deg)")
+Imagn=20;Iph =-50;
+Veff=110;
+
+P=Veff*Imagn*cos((abs(Iph)*%pi)/180)
+Q=Veff*Imagn*sin((abs(Iph)*%pi)/180)
+printf("Average power is %3.1fW\n",P)
+printf("Reactive power is %3.1fvar\n",Q)
+
diff --git a/Working_Examples/154/DEPENDENCIES/ch12_2.sce b/Working_Examples/154/DEPENDENCIES/ch12_2.sce new file mode 100755 index 0000000..a5eb3dc --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch12_2.sce @@ -0,0 +1,60 @@ +clc
+disp("Example 12.2")
+printf("\n")
+
+printf("Given")
+disp("The system ABC is DELTA connected")
+disp("Effective line voltage is 120V")
+disp("The three impedances are 5(45 deg)")
+Zmag=5;Zph=45;
+//Let maximum line voltage is Vmax
+Vmax=120*sqrt(2)
+//From fig 12.7(a)
+//VAB=Vmax(120 deg)
+//VBC=Vmax(0 deg)
+//VCA=Vmax(240 deg)
+
+//From figure 12.8
+IABmag=Vmax/Zmag
+IABph=120-Zph
+printf("IAB=%3.2f(%d deg)\n",IABmag,IABph);
+
+IBCmag=Vmax/Zmag
+IBCph=0-Zph
+printf("IBC=%3.2f(%d deg)\n",IBCmag,IBCph);
+
+ICAmag=Vmax/Zmag
+ICAph=240-Zph
+printf("ICA=%3.2f(%d deg)\n",ICAmag,ICAph);
+
+//Applying KCL equation
+//IA=IAB+IAC
+//IB=IBC+IBA
+//IC=ICA+ICB
+
+x=IABmag*cos((IABph*%pi)/180);
+y=IABmag*sin((IABph*%pi)/180);
+z=complex(x,y)
+
+x1=ICAmag*cos((ICAph*%pi)/180);
+y1=ICAmag*sin((ICAph*%pi)/180);
+z1=complex(x1,y1)
+
+x2=IBCmag*cos((IBCph*%pi)/180);
+y2=IBCmag*sin((IBCph*%pi)/180);
+z2=complex(x2,y2)
+
+IA=z-z1;
+[RA,ThetaA]=polar(IA)
+
+IB=z2-z;
+[RB,ThetaB]=polar(IB)
+
+IC=z1-z2
+[RC,ThetaC]=polar(IC)
+
+disp("Therefore")
+
+printf("\nIA=%3.2f(%d deg)A\n",RA,ThetaA*(180/%pi));
+printf("\nIB=%3.2f(%d deg)A\n",RB,ThetaB*(180/%pi));
+printf("\nIC=%3.2f(%d deg)A\n",RC,ThetaC*(180/%pi));
\ No newline at end of file diff --git a/Working_Examples/154/DEPENDENCIES/ch12_3.sce b/Working_Examples/154/DEPENDENCIES/ch12_3.sce new file mode 100755 index 0000000..1bb93b8 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch12_3.sce @@ -0,0 +1,31 @@ +clc
+disp("Example 12.3")
+printf("\n")
+
+printf("Given")
+disp("The system CBA is WYE connected")
+disp("Effective line voltage is 120V")
+disp("The three impedances are 20(-30 deg)")
+Zmag=20;Zph=-30;
+//Let maximum line voltage is Vmax
+Vmax=120*sqrt(2)
+//Let the line to neutral voltage magnitude be Vn
+Vn=Vmax/sqrt(3)
+//From fig 12.7(b)
+//VAN=Vn(-90 deg)
+//VBN=Vn(30 deg)
+//VCN=Vn(150 deg)
+
+//From figure 12.10
+IAmag=Vn/Zmag
+IAph=-90-Zph
+printf("\nIA=%3.2f(%d deg)A\n",IAmag,IAph);
+
+IBmag=Vn/Zmag
+IBph=30-Zph
+printf("\nIB=%3.2f(%d deg)A\n",IBmag,IBph);
+
+ICmag=Vn/Zmag
+ICph=150-Zph
+printf("\nIC=%3.2f(%d deg)A\n",ICmag,ICph);
+
diff --git a/Working_Examples/154/DEPENDENCIES/ch12_4.sce b/Working_Examples/154/DEPENDENCIES/ch12_4.sce new file mode 100755 index 0000000..4bc71e2 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch12_4.sce @@ -0,0 +1,27 @@ +clc
+disp("Example 12.4")
+printf("\n")
+
+disp("Continued from Example 12.3")
+Vmax=120*sqrt(2)
+Zmag=20;Zph=-30;
+//Let the line to neutral voltage magnitude be VLN
+VLN=Vmax/sqrt(3)
+
+//From figure 12.14
+ILmag=VLN/Zmag
+ILph=0-Zph
+printf("\nIL=%3.2f(%d deg)\n",ILmag,ILph);
+
+//From fig 12.7(b)
+//VAN=Vn(-90 deg)
+//VBN=Vn(30 deg)
+//VCN=Vn(150 deg)
+IAph=ILph-90
+printf("\nIA=%3.2f(%d deg)A\n",ILmag,IAph);
+
+IBph=ILph+30
+printf("\nIB=%3.2f(%d deg)A\n",ILmag,IBph);
+
+ICph=ILph+150
+printf("\nIC=%3.2f(%d deg)A\n",ILmag,ICph);
diff --git a/Working_Examples/154/DEPENDENCIES/ch12_5.sce b/Working_Examples/154/DEPENDENCIES/ch12_5.sce new file mode 100755 index 0000000..d24df03 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch12_5.sce @@ -0,0 +1,63 @@ +clc
+disp("Example 12.5")
+printf("\n")
+
+printf("Given")
+disp("The system ABC is DELTA connected")
+disp("Maximum line voltage is 339.4V")
+disp("The three impedances are 10(0 deg),10(30 deg),15(-30 deg)")
+
+ZABmag=10;ZABph=0;
+ZBCmag=10;ZBCph=30;
+ZCAmag=15;ZCAph=-30;
+//Let maximum line voltage is Vmax
+Vmax=339.4
+//From fig 12.7(a)
+//VAB=Vmax(120 deg)
+//VBC=Vmax(0 deg)
+//VCA=Vmax(240 deg)
+
+//From figure 12.15
+IABmag=Vmax/ZABmag
+IABph=120-ZABph
+printf("IAB=%3.2f(%d deg)\n",IABmag,IABph);
+
+IBCmag=Vmax/ZBCmag
+IBCph=0-ZBCph
+printf("IBC=%3.2f(%d deg)\n",IBCmag,IBCph);
+
+ICAmag=Vmax/ZCAmag
+ICAph=240-ZCAph
+printf("ICA=%3.2f(%d deg)\n",ICAmag,ICAph);
+
+//Applying KCL equation
+//IA=IAB+IAC
+//IB=IBC+IBA
+//IC=ICA+ICB
+
+x=IABmag*cos((IABph*%pi)/180);
+y=IABmag*sin((IABph*%pi)/180);
+z=complex(x,y)
+
+x1=ICAmag*cos((ICAph*%pi)/180);
+y1=ICAmag*sin((ICAph*%pi)/180);
+z1=complex(x1,y1)
+
+x2=IBCmag*cos((IBCph*%pi)/180);
+y2=IBCmag*sin((IBCph*%pi)/180);
+z2=complex(x2,y2)
+
+IA=z-z1;
+[RA,ThetaA]=polar(IA)
+
+IB=z2-z;
+[RB,ThetaB]=polar(IB)
+
+IC=z1-z2
+[RC,ThetaC]=polar(IC)
+
+disp("Therefore")
+
+printf("\nIA=%3.2f(%d deg)A\n",RA,ThetaA*(180/%pi));
+printf("\nIB=%3.2f(%d deg)A\n",RB,ThetaB*(180/%pi));
+printf("\nIC=%3.2f(%d deg)A\n",RC,ThetaC*(180/%pi));
\ No newline at end of file diff --git a/Working_Examples/154/DEPENDENCIES/ch12_6.sce b/Working_Examples/154/DEPENDENCIES/ch12_6.sce new file mode 100755 index 0000000..039e4f5 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch12_6.sce @@ -0,0 +1,52 @@ +clc
+disp("Example 12.6")
+printf("\n")
+
+printf("Given")
+disp("The system CBA is WYE connected")
+disp("Maximum line voltage is 150V")
+disp("The three impedances are 6(0 deg),6(30 deg),5(45 deg)")
+ZAmag=6;ZAph=0;
+ZBmag=6;ZBph=30;
+ZCmag=5;ZCph=45;
+//Let maximum line voltage is Vmax
+Vmax=150
+//Let the line to neutral voltage magnitude be Vn
+Vn=Vmax/sqrt(3)
+//From fig 12.7(b)
+//VAN=Vn(-90 deg)
+//VBN=Vn(30 deg)
+//VCN=Vn(150 deg)
+
+//From figure 12.16
+IAmag=Vn/ZAmag
+IAph=-90-ZAph
+printf("\nIA=%3.2f(%d deg)A\n",IAmag,IAph);
+
+IBmag=Vn/ZBmag
+IBph=30-ZBph
+printf("\nIB=%3.2f(%d deg)A\n",IBmag,IBph);
+
+ICmag=Vn/ZCmag
+ICph=150-ZCph
+printf("\nIC=%3.2f(%d deg)A\n",ICmag,ICph);
+
+//Now to calculate IN
+//IN=-(IA+IB+IC)
+x=IAmag*cos((IAph*%pi)/180);
+y=IAmag*sin((IAph*%pi)/180);
+z=complex(x,y)
+
+x1=ICmag*cos((ICph*%pi)/180);
+y1=ICmag*sin((ICph*%pi)/180);
+z1=complex(x1,y1)
+
+x2=IBmag*cos((IBph*%pi)/180);
+y2=IBmag*sin((IBph*%pi)/180);
+z2=complex(x2,y2)
+
+IN=-(z+z1+z2)
+
+[R,Theta]=polar(IN)
+
+printf("\nIN=%3.2f(%d deg)A\n",R,Theta*(180/%pi));
diff --git a/Working_Examples/154/DEPENDENCIES/ch13_2.sce b/Working_Examples/154/DEPENDENCIES/ch13_2.sce new file mode 100755 index 0000000..b4b8409 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch13_2.sce @@ -0,0 +1,31 @@ +clc
+disp("Problem 13.2")
+printf("\n")
+
+printf("Given")
+disp("|Hv|=1/sqrt(2) (1)")
+disp("Resistance R1=5kohm")
+R1=5000;
+disp("Hv(w)=1/1+%i*(w/wx) (2)")
+//wx=1/(R1*C2)
+//On solving we get
+disp("wx=2*10^-4/C2 (3)")
+
+disp("a)")
+C2=10*10^-9;
+//Taking modulus of (2)
+disp("|Hv(w)|=1/sqrt(1+(w/wx)^2)")
+//Equating (1) and (2)
+wx=2*10^-4/C2;
+fx=(wx/(2*%pi))*10^-3
+printf("Frequency(a) is %3.2fkHz\n",fx)
+
+disp("b)")
+C2b=1*10^-9;
+//As frequency is inversely proportional to C2 (from (3))
+fx1=(C2/C2b)*fx
+printf("Frequency(b) is %3.2fkHz\n",fx1)
+
+
+
+
diff --git a/Working_Examples/154/DEPENDENCIES/ch13_7.sce b/Working_Examples/154/DEPENDENCIES/ch13_7.sce new file mode 100755 index 0000000..34dda83 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch13_7.sce @@ -0,0 +1,21 @@ +clc
+disp("Problem 13.7")
+printf("\n")
+
+s=%s;
+printf("Given")
+H=(10*s)/(s^2+300*s+10^6)
+disp(H,"H(s)=")
+//From the above transfer function
+//Comparing the denominator with s^2+a*s+b with w=sqrt(b)
+a=300;b=10^6;
+//Therefore center frequency is
+w0=sqrt(10^6)
+//The lower and upper frequencies are
+wl=sqrt(a^2/4+b)-a/2
+wh=sqrt(a^2/4+b)+a/2
+B=wh-wl //It can be inferred that B=a
+Q=sqrt(b)/a
+printf("\nCenter frequency= %drad/s\n",w0);
+printf("Low power frequency = %3.2frad/s\nHigh power frequency = %3.2frad/s\n",wl,wh);
+printf("Bandwidth= %drad/s\nQuality factor =%3.2f\n",B,Q)
diff --git a/Working_Examples/154/DEPENDENCIES/ch13_8.sce b/Working_Examples/154/DEPENDENCIES/ch13_8.sce new file mode 100755 index 0000000..a2b2331 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch13_8.sce @@ -0,0 +1,21 @@ +clc
+disp("Problem 13.8")
+printf("\n")
+
+s=%s;
+printf("Given")
+H=(10*s)/(s^2+30*s+10^6)
+disp(H,"H(s)=")
+//From the above transfer function
+//Comparing the denominator with s^2+a*s+b with w=sqrt(b)
+a=30;b=10^6;
+//Therefore center frequency is
+w0=sqrt(10^6)
+//The lower and upper frequencies are
+wl=sqrt(a^2/4+b)-a/2
+wh=sqrt(a^2/4+b)+a/2
+B=wh-wl
+Q=sqrt(b)/a
+printf("\nCenter frequency= %drad/s\n",w0);
+printf("Low power frequency = %3.2frad/s\nHigh power frequency = %3.2frad/s\n",wl,wh);
+printf("Bandwidth= %drad/s\nQuality factor =%3.2f\n",B,Q)
diff --git a/Working_Examples/154/DEPENDENCIES/ch14_1.sce b/Working_Examples/154/DEPENDENCIES/ch14_1.sce new file mode 100755 index 0000000..4737b79 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch14_1.sce @@ -0,0 +1,29 @@ +clc
+disp("Example 14.1")
+printf("\n")
+
+s=%s;
+//Applying KVL equation to the two loops we get
+//V1=2*I1+s*(I1+I2)
+//V2=3*I2+s*(I1+I2)
+
+//On solving we get
+disp("(s+2)*I1+s*I2=V1 (1)");
+disp("s*I1+(s+3)*I2=V2 (2)");
+
+//The equations which contain Z parameters are
+//V1=Z11*I1+Z12*I2
+//V2=Z21*I1+Z22*I2
+
+//On comparing (1) and (2) with above equations
+Z11=s+2;
+Z12=s;
+Z21=s;
+Z22=s+3;
+
+disp(Z11,"Z11=")
+disp(Z12,"Z12=")
+disp(Z21,"Z21=")
+disp(Z22,"Z22=")
+
+
diff --git a/Working_Examples/154/DEPENDENCIES/ch14_10.sce b/Working_Examples/154/DEPENDENCIES/ch14_10.sce new file mode 100755 index 0000000..965dc13 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch14_10.sce @@ -0,0 +1,29 @@ +clc
+disp("Example 14.10")
+printf("\n")
+
+s=%s;
+//Applying KVL equation to the two loops we get
+//V1=3*I1+3*(I1+I2)
+//V2=7*I1+3*(I1+I2)+2*I2
+
+//On solving we get
+disp("6*I1+3*I2=V1 (1)");
+disp("10*I1+5*I2=V2 (2)");
+
+//The equations which contain Z parameters are
+//V1=Z11*I1+Z12*I2
+//V2=Z21*I1+Z22*I2
+
+//On comparing (1) and (2) with above equations
+Z11=6;
+Z12=3;
+Z21=10;
+Z22=5;
+
+disp(Z11,"Z11=")
+disp(Z12,"Z12=")
+disp(Z21,"Z21=")
+disp(Z22,"Z22=")
+
+disp("As DZZ results in zero(0) therefore Y parameters are not defined ")
diff --git a/Working_Examples/154/DEPENDENCIES/ch14_2.sce b/Working_Examples/154/DEPENDENCIES/ch14_2.sce new file mode 100755 index 0000000..bb23309 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch14_2.sce @@ -0,0 +1,27 @@ +clc
+disp("Example 14.2")
+printf("\n")
+
+s=%s;
+//Applying KVL equation to the two loops we get
+//V1=2*I1+s*(I1+I2)-I2
+//V2=3*I2+s*(I1+I2)
+
+//On solving we get
+disp("(s+2)*I1+(s-1)*I2=V1 (1)");
+disp("s*I1+(s+3)*I2=V2 (2)");
+
+//The equations which contain Z parameters are
+//V1=Z11*I1+Z12*I2
+//V2=Z21*I1+Z22*I2
+
+//On comparing (1) and (2) with above equations
+Z11=s+2;
+Z12=s-1;
+Z21=s;
+Z22=s+3;
+
+disp(Z11,"Z11=")
+disp(Z12,"Z12=")
+disp(Z21,"Z21=")
+disp(Z22,"Z22=")
diff --git a/Working_Examples/154/DEPENDENCIES/ch14_4.sce b/Working_Examples/154/DEPENDENCIES/ch14_4.sce new file mode 100755 index 0000000..5e72eb0 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch14_4.sce @@ -0,0 +1,33 @@ +clc
+disp("Example 14.4")
+printf("\n")
+
+s=%s;
+Ya=3/(5*s+6);
+Yb=2/(5*s+6);
+Yc=s/(5*s+6);
+
+//Writing KCL equations
+disp("I1=(Ya+Yc)*V1-Yc*V2 (1)")
+disp("I2=-Yc*V1+(Yb+Yc)*V2 (2)")
+
+//The equations which contain Y parameters are
+//I1=Y11*V1+Y12*V2
+//I2=Y21*V1+Y22*V2
+
+//On comparing (1) and (2) with above equations
+disp("Y11=Ya+Yc")
+disp("Y12=-Yc=Y21")
+disp("Y22=Yb+Yc")
+
+//Substituting Ya , Yb and Yc
+Y11=Ya+Yc
+Y12=-Yc
+Y21=-Yc
+Y22=Yb+Yc
+
+disp(Y11,"Y11=")
+disp(Y12,"Y12=")
+disp(Y21,"Y21=")
+disp(Y22,"Y22=")
+
diff --git a/Working_Examples/154/DEPENDENCIES/ch14_6.sce b/Working_Examples/154/DEPENDENCIES/ch14_6.sce new file mode 100755 index 0000000..6cef31e --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch14_6.sce @@ -0,0 +1,23 @@ +clc
+disp("Example 14.6")
+printf("\n")
+
+s=%s;
+//From example 14.4
+
+Y11=(3 + s)/(5*s+6)
+Y12=- s/(6 + 5*s)
+Y21=- s/(6 + 5*s)
+Y22=(2+s)/(6+5*s)
+
+DYY=Y11*Y22-Y12*Y21
+
+Z11=Y22/DYY;
+Z12=-Y12/DYY;
+Z21=-Y21/DYY;
+Z22=Y11/DYY;
+
+disp(Z11,"Z11=")
+disp(Z12,"Z12=")
+disp(Z21,"Z21=")
+disp(Z22,"Z22=")
diff --git a/Working_Examples/154/DEPENDENCIES/ch14_7.sce b/Working_Examples/154/DEPENDENCIES/ch14_7.sce new file mode 100755 index 0000000..ffb8c5c --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch14_7.sce @@ -0,0 +1,18 @@ +clc
+disp("Example 14.7")
+printf("\n")
+
+//From figure 14.9
+disp("V1=50*I1 (1)");
+disp("I2=300*I1 (2)");
+
+//The equations which contain h parameters are
+//V1=h11*I1+h12*V2
+//I2=h21*I1+h22*V2
+
+//On comparing (1) and (2) with above equations
+
+printf("\nh11=%d\n",50);
+printf("h12=%d\n",0);
+printf("h21=%d\n",300);
+printf("h22=%d\n",0);
diff --git a/Working_Examples/154/DEPENDENCIES/ch14_8.sce b/Working_Examples/154/DEPENDENCIES/ch14_8.sce new file mode 100755 index 0000000..059a397 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch14_8.sce @@ -0,0 +1,25 @@ +clc
+disp("Example 14.8")
+printf("\n")
+
+//From figure 14.10
+//By inspection
+//V1=10^9*I1
+//V2=10(I2-10^-3*V1)
+
+
+//On solving we get
+disp("I1=10^-9*V1 (1)");
+disp("V2=10*I2-10^-2*V1 (2)");
+
+
+//The equations which contain g parameters are
+//I1=g11*V1+g12*I2
+//V2=g21*V1+g22*I2
+
+//On comparing (1) and (2) with above equations
+
+printf("\ng11=%2.1e\n",10^-9);
+printf("g12=%d\n",0);
+printf("g21=%2.1e\n",-10^-2);
+printf("g22=%d\n",10);
diff --git a/Working_Examples/154/DEPENDENCIES/ch15_4.sce b/Working_Examples/154/DEPENDENCIES/ch15_4.sce new file mode 100755 index 0000000..a2703a4 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch15_4.sce @@ -0,0 +1,34 @@ +clc
+disp("Example 15.4")
+printf("\n")
+
+printf("Given")
+disp("L1=0.1H L2=0.2H")
+disp("i1=4A i2=10A")
+L1=0.1;L2=0.2
+i1=4;i2=10;
+//The energy stored in coupled coils is
+disp("W=(L1*i1^2)/2+(L2*i2^2)/2+M*i1*i2")
+
+disp("a)")
+M=0.1;
+W=(L1*i1^2)/2+(L2*i2^2)/2+M*i1*i2;
+printf("Total Energy in the coils=%3.2fJ\n",W);
+
+disp("b)")
+M=sqrt(2)/10;
+W=(L1*i1^2)/2+(L2*i2^2)/2+M*i1*i2;
+printf("Total Energy in the coils=%3.2fJ\n",W);
+
+disp("c)")
+M=-0.1;
+W=(L1*i1^2)/2+(L2*i2^2)/2+M*i1*i2;
+printf("Total Energy in the coils=%3.2fJ\n",W);
+
+disp("a)")
+M=-sqrt(2)/10;
+W=(L1*i1^2)/2+(L2*i2^2)/2+M*i1*i2;
+printf("Total Energy in the coils=%3.2fJ\n",W);
+
+
+
diff --git a/Working_Examples/154/DEPENDENCIES/ch15_7.sce b/Working_Examples/154/DEPENDENCIES/ch15_7.sce new file mode 100755 index 0000000..86d9339 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch15_7.sce @@ -0,0 +1,29 @@ +clc
+disp("Example 15.7")
+printf("\n")
+
+printf("Given")
+disp("N1=20 N2=N3=10")
+disp("I2=10(-53.13 deg) I3=10(-45 deg)")
+N1=20;N2=10;N3=10;
+I2mag=10;I2ph=-53.13;
+I3mag=10;I3ph=-45;
+//From figure 15.14
+disp("N1*I1-N2*I2-N3*I3=0")
+//Solving for I1
+Xmag=N2*I2mag
+Xph=I2ph
+x=Xmag*cos((Xph*%pi)/180);
+y=Xmag*sin((Xph*%pi)/180);
+z=complex(x,y)
+
+Ymag=N3*I3mag
+Yph=I3ph
+x1=Ymag*cos((Yph*%pi)/180);
+y1=Ymag*sin((Yph*%pi)/180);
+z1=complex(x1,y1)
+
+I1=(z+z1)/N1
+[R,Theta]=polar(I1);
+printf("I1=%3.2f(%3.2f deg) A\n",R,(Theta*180)/%pi);
+
diff --git a/Working_Examples/154/DEPENDENCIES/ch15_8.sce b/Working_Examples/154/DEPENDENCIES/ch15_8.sce new file mode 100755 index 0000000..73b8655 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch15_8.sce @@ -0,0 +1,22 @@ +clc
+disp("Example 15.8")
+printf("\n")
+
+printf("Given")
+disp("L1=0.2H L2=0.1H")
+disp("M=0.1H R=10ohm")
+disp("v1=142.3*sin(100*t)")
+L1=0.2;L2=0.1
+M=0.1;R=10;
+v1mag=142.3;
+w=100;
+//Let Input impedance be Z1 and can be calculated as
+//From the equations in 15.10
+disp("Z1=%i*w*L1+((M*w)^2)/(Z2+%i*w*L2)")
+Z1=%i*w*L1+((M*w)^2)/(R+%i*w*L2)
+[R,Theta]=polar(Z1)
+//If I1 is the input current
+I1mag=v1mag/R
+I1ph=-(Theta*180)/%pi
+//In time domain form
+printf("i1=%3.1f*sin(%d*t%3.1f deg) (A)",I1mag,w,I1ph);
diff --git a/Working_Examples/154/DEPENDENCIES/ch15_9.sce b/Working_Examples/154/DEPENDENCIES/ch15_9.sce new file mode 100755 index 0000000..0d9f2fa --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch15_9.sce @@ -0,0 +1,33 @@ +clc
+disp("Example 15.9")
+printf("\n")
+
+s=%s;
+printf("Given")
+disp("L1=0.2H L2=0.1H")
+disp("M=0.1H R=10ohm")
+disp("v1=u(t) a unit step function")
+L1=0.2;L2=0.1
+M=0.1;R=10;
+v1=1;
+w=100;
+//Let Input impedance be Z1 and can be calculated as
+//From the equations in 15.10
+disp("Z1(s)=L1*s-((M*s)^2)/(R+L2*s)")
+Z1=L1*s-(((M*s)^2)/(R+L2*s))
+//Proper rearranging of co-efficients
+Num=Z1('num')/0.01
+Den=Z1('den')*100
+
+disp(Num/Den,"Z1(s)")
+Y1=1/Z1
+disp(Den/Num,"Y1(s)")
+
+//As the input is unit step function the value is 1V for t>0
+//In exponential form the value is represented as exp(s*t) with s=0 as the pole of Y1(s)
+
+//Therefore forced response
+k=1/L1;
+printf("Forced response i1,f=(%d*t) (A)\n",k);
+
+
diff --git a/Working_Examples/154/DEPENDENCIES/ch17_2.sce b/Working_Examples/154/DEPENDENCIES/ch17_2.sce new file mode 100755 index 0000000..e9ab99b --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch17_2.sce @@ -0,0 +1,3 @@ +syms t s ;
+ x=laplace ('3*%e^(2*t)' , t , s ) ;
+ disp (x , " X(s)=" )
diff --git a/Working_Examples/154/DEPENDENCIES/ch17_4.sce b/Working_Examples/154/DEPENDENCIES/ch17_4.sce new file mode 100755 index 0000000..6c4baab --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch17_4.sce @@ -0,0 +1,31 @@ +clc
+syms t
+s=%s;
+//Factorizing the denominator
+I=(s-10)/((s^2)*(s-%i)*(s+%i));
+disp(I,"I(s)=")
+//The principal part at s=0 is
+//B1/s+B2/s^2
+//Taking the limit s->0 to (s-10)/((s-%i)*(s+%i))
+
+B2=-10
+
+//Taking the limit s->0 to (s*(s-10))/(s^2)*(s^2+1)+(10/s)
+
+B1=1
+
+//The principal part at s=%i is
+//A/(s-%i)
+//Taking the limit s->%i to (s-10)/((s^2)*(s+%i))
+
+A=(-0.5-%i*5)
+
+//As the other co-efficient is conjugate of the above we can write the partial fraction expansion of I(s)
+I=(1/s)-(10/s^2)-(0.5+%i*5)/(s-%i)-(0.5-%i*5)/(s+%i);
+//Taking inverse of each term
+I1=ilaplace('1/s',s,t)
+I2=ilaplace('10/s^2',s,t)
+I3=ilaplace('(0.5+%i*5)/(s-%i)',s,t)
+I4=ilaplace('(0.5-%i*5)/(s+%i)',s,t)
+I=I1-I2-I3-I4
+disp(I,"i(t)=")
diff --git a/Working_Examples/154/DEPENDENCIES/ch1_1.sce b/Working_Examples/154/DEPENDENCIES/ch1_1.sce new file mode 100755 index 0000000..a0e4f4d --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch1_1.sce @@ -0,0 +1,20 @@ +clc
+disp("Example 1.1")
+printf("\n")
+printf("Given")
+disp("Acceleration is 2.0m/s^2")
+disp("Mass is 10kg")
+m=10;a=2;
+disp("a)")
+disp("F=m*a")
+F=m*a
+printf("Force is %dN\n",F)
+disp("b)")
+printf("time=4s\n")
+t=4;
+x=(a*t*t)/2
+KE=(F*x)
+P=KE/t
+printf("Position is %dm\n",x)
+printf("Kinetic energy =%3.1fJ\n",KE)
+printf("Power =%3.1fW\n",P)
diff --git a/Working_Examples/154/DEPENDENCIES/ch1_2.sce b/Working_Examples/154/DEPENDENCIES/ch1_2.sce new file mode 100755 index 0000000..eae2a42 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch1_2.sce @@ -0,0 +1,13 @@ +clc
+disp("Example 1.2")
+printf("\n")
+printf("Given")
+disp("Current flow is 5A")
+disp("Time is 1 minute")
+i=5;
+//As electroms/min is asked so we need to convert A(C/s) to C/min
+i1=5*60;
+//Let e be electronic charge
+e=1.602*10^-19
+n=(i1/e)
+printf("Number of electrons =%3.2f electrons/min\n",n)
\ No newline at end of file diff --git a/Working_Examples/154/DEPENDENCIES/ch1_3.sce b/Working_Examples/154/DEPENDENCIES/ch1_3.sce new file mode 100755 index 0000000..ffc73c5 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch1_3.sce @@ -0,0 +1,10 @@ +clc
+disp("Example 1.3")
+printf("\n")
+printf("Given")
+disp("Energy is 9.25uJ")
+disp("Charge to be transferred is 0.5uC")
+E=9.25*10^-6;q=0.5*10^-6;
+//1 volt is 1 joule per coulomb
+V=E/q;
+printf("Potential difference between two points a and b is %3.1fV\n",V)
diff --git a/Working_Examples/154/DEPENDENCIES/ch1_4.sce b/Working_Examples/154/DEPENDENCIES/ch1_4.sce new file mode 100755 index 0000000..4a00b66 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch1_4.sce @@ -0,0 +1,14 @@ +clc
+disp("Example 1.4")
+printf("\n")
+printf("Given")
+disp("Potential difference is 50V")
+disp("Charge per minute is 120C/min")
+V=50;x=120;
+//As Electrical energy is to be calculated charge per minute is to be converted in charge per second
+//Charge per second is nothing but the current
+i=x/60;
+P=i*V;
+//Since is 1W=1J/s
+printf("Rate of energy conversion is %dJ/s\n",P)
+
diff --git a/Working_Examples/154/DEPENDENCIES/ch2_1.sce b/Working_Examples/154/DEPENDENCIES/ch2_1.sce new file mode 100755 index 0000000..bbb83e9 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch2_1.sce @@ -0,0 +1,47 @@ +clc
+disp("Example 2.1")
+printf("\n")
+printf("Given")
+disp("Resistance used is 4 ohm")
+disp("Current flow is i=2.5*sin(w*t)")
+disp("Angular frequency(w)=500 rad/s")
+
+R=4;
+iamp=2.5;w=500;
+t=0:0.001:0.012566
+i=2.5*sin(w*t)
+
+
+Vamp=iamp*R;
+printf("v=%d*sin(%d*t)(V)\n",Vamp,w)
+
+pamp=iamp*iamp*R;
+printf("p=%d(sin(%d*t))^2(W)\n",pamp,w)
+p=pamp*sin(w*t)^2;
+
+//On integrating p with respect to t
+W=25*(t/2-sin(2*w*t)/(4*w))
+
+function p=f(t),p=pamp*sin(w*t)^2,endfunction
+w1=intg(0,2*%pi/w,f);
+
+figure
+a= gca ();
+plot(t,i)
+xtitle ('i vs wt','wt','i ');
+a. thickness = 2;
+
+figure
+a= gca ();
+plot(t,p)
+xtitle ('p vs wt','wt','p ');
+a. thickness = 2;
+
+figure
+a= gca ();
+plot(t,W)
+xtitle ('w vs wt','wt','w ');
+a. thickness = 2;
+
+
+
diff --git a/Working_Examples/154/DEPENDENCIES/ch2_2.sce b/Working_Examples/154/DEPENDENCIES/ch2_2.sce new file mode 100755 index 0000000..16684e3 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch2_2.sce @@ -0,0 +1,48 @@ +clc
+disp("Example 2.2")
+printf("\n")
+printf("Given")
+disp("Inductance used is 30mH")
+disp("Current flow is i=10*sin(50*t)")
+L=30*10^-3;iamp=10;
+t=0:0.01:0.06283;
+i=10*sin(50*t)
+//v=L*d/dt(i)
+//d/dt(sin 50t)=50*cos t
+vamp=L*iamp*50;
+v=vamp*cos(50*t)
+
+//sinA*cosB=(sin(A+B)+sin(A-B))/2
+
+pamp=vamp*iamp/2;
+p=pamp*sin(100*t)
+//On integrating 'p' w.r.t t
+
+wL=0.75*(1-cos(100*t));
+
+figure
+a= gca ();
+plot(t,i)
+xtitle ('i vs wt','wt','i');
+a. thickness = 2;
+
+figure
+a= gca ();
+plot(t,v)
+xtitle ('v vs wt','wt','v ');
+a. thickness = 2;
+
+figure
+a= gca ();
+plot(t,p)
+xtitle ('p vs wt','wt','p ');
+a. thickness = 2;
+
+figure
+a= gca ();
+plot(t,wL)
+xtitle ('wL vs wt','wt','wL ');
+a. thickness = 2;
+
+
+
diff --git a/Working_Examples/154/DEPENDENCIES/ch2_3.sce b/Working_Examples/154/DEPENDENCIES/ch2_3.sce new file mode 100755 index 0000000..cb92e89 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch2_3.sce @@ -0,0 +1,35 @@ +clc
+disp("Example 2.3")
+printf("\n")
+printf("Given")
+disp("Capacitance used is 20uF")
+disp("Voltage is v=50*sin(200*t)")
+C=20*10^-6;
+v=50*sin(200*t);
+vamp=50;
+t=0:0.001:0.015;
+//q=C*v
+qamp=vamp*C
+q=qamp*sin(200*t)
+//i=C*d/dt(v)
+//d/dt(sin 200t)=200*cos t
+iamp=C*vamp*200;
+i=iamp*cos(200*t)
+
+//sinA*cosB=(sin(A+B)+sin(A-B))/2
+
+pamp=vamp*iamp/2;
+p=pamp*sin(400*t)
+
+//On integrating 'p' w.r.t t
+
+wC=12.5*(1-cos(400*t));
+
+figure
+a= gca ();
+plot(t,wC)
+xtitle ('wC vs wt','wt','wC (mJ)');
+a. thickness = 2;
+
+
+
diff --git a/Working_Examples/154/DEPENDENCIES/ch2_4.sce b/Working_Examples/154/DEPENDENCIES/ch2_4.sce new file mode 100755 index 0000000..623b4b9 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch2_4.sce @@ -0,0 +1,17 @@ +clc
+disp("Example 2.4")
+printf("\n")
+printf("Given")
+disp("Current through diode is 30mA")
+//From the table the nearest value is at v=0.74V
+V=0.74;I=28.7*10^-3;
+R=V/I;
+delV=0.75-0.73
+delI=42.7*10^-3-19.2*10^-3
+r=delV/delI
+p=(V*I)*10^3
+printf("\n \n Static resistance is %3.2fohm\n",R)
+printf("Dynamic resistance is %3.2fohm\n",r)
+printf("Power consumption is %3.2fmW\n",p)
+
+
diff --git a/Working_Examples/154/DEPENDENCIES/ch2_5.sce b/Working_Examples/154/DEPENDENCIES/ch2_5.sce new file mode 100755 index 0000000..9d9c06c --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch2_5.sce @@ -0,0 +1,29 @@ +clc
+disp("Example 2.5")
+printf("\n")
+printf("Given")
+disp("a)")
+disp("Current through diode is 10mA")
+//From the table the value is at v=2.5V
+V=2.5;I=10*10^-3;
+R=V/I;
+delV=3-2
+delI=11*10^-3-9*10^-3
+r=delV/delI
+p=(V*I)*10^3
+printf("\n \n Static resistance is %3.2fohm\n",R)
+printf("Dynamic resistance is %3.2fohm\n",r)
+printf("Power consumption is %3.2fmW\n",p)
+
+disp("b)")
+disp("Current through diode is 15mA")
+//From the table the value is at v=5V
+V=5;I=15*10^-3;
+R=V/I;
+delV=5.5-4.5
+delI=16*10^-3-14*10^-3
+r=delV/delI
+p=(V*I)*10^3
+printf("\n \n Static resistance is %3.2fohm\n",R)
+printf("Dynamic resistance is %3.2fohm\n",r)
+printf("Power consumption is %3.2fmW\n",p)
\ No newline at end of file diff --git a/Working_Examples/154/DEPENDENCIES/ch3_3.sce b/Working_Examples/154/DEPENDENCIES/ch3_3.sce new file mode 100755 index 0000000..e72c536 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch3_3.sce @@ -0,0 +1,13 @@ +clc
+disp("Example 3.3")
+printf("\n")
+printf("Given")
+disp("Equivalent resistance of three resistors is 750 ohm")
+disp("values of two resistors are 40 ohm and 410 ohm")
+Req=750;R1=40;R2=410;
+
+//For series resistance
+disp("Req=R1+R2+R3")
+//On solving for R3
+R3=Req-R1-R2
+printf("Value of third ohmic resistor is %dohm\n",R3)
diff --git a/Working_Examples/154/DEPENDENCIES/ch3_4.sce b/Working_Examples/154/DEPENDENCIES/ch3_4.sce new file mode 100755 index 0000000..26e7932 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch3_4.sce @@ -0,0 +1,17 @@ +clc
+disp("Example 3.4")
+printf("\n")
+printf("Given")
+disp("values of two capacitors are 2uF and 10uF")
+C1=2*10^-6;C2=10*10^-6;
+//For two capacitors in series
+disp("Ceq=(C1*C2)/(C1+C2)")
+//On solving for Ceq
+Ceq=((C1*C2)/(C1+C2))*10^6
+printf("Value of equivalent capacitance is %3.2fuF\n",Ceq)
+
+disp("If C2=10pF")
+C2=10*10^-12;
+
+Ceq=((C1*C2)/(C1+C2))*10^12
+printf("Value of equivalent capacitance is %3.2fpF\n",Ceq)
\ No newline at end of file diff --git a/Working_Examples/154/DEPENDENCIES/ch3_5.sce b/Working_Examples/154/DEPENDENCIES/ch3_5.sce new file mode 100755 index 0000000..9ca5084 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch3_5.sce @@ -0,0 +1,18 @@ +clc
+disp("Example 3.5")
+printf("\n")
+printf("Given")
+disp("a)")
+disp("values of two resistors are 60 ohm and 60 ohm")
+R1=60;R2=60;
+disp("If resistors are parallel")
+Req=(R1*R2)/(R1+R2)
+printf("Value of equivalent resistance is %dohm\n",Req)
+
+disp("b)")
+disp("values of three equal resistors are 60 ohm")
+R1=60;R2=60;R3=60;
+disp("If resistors are parallel")
+x=1/R1+1/R2+1/R3
+Req=1/x;
+printf("Value of equivalent resistance is %dohm\n",Req)
diff --git a/Working_Examples/154/DEPENDENCIES/ch3_6.sce b/Working_Examples/154/DEPENDENCIES/ch3_6.sce new file mode 100755 index 0000000..5156eb9 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch3_6.sce @@ -0,0 +1,11 @@ +clc
+disp("Example 3.6")
+printf("\n")
+printf("Given")
+
+disp("values of two inductors are 3mH and 6 mH")
+L1=3*10^-3;L2=6*10^-3;
+disp("If inductors are parallel")
+Leq=((L1*L2)/(L1+L2))*10^3
+printf("Value of equivalent inductance is %3.1fmH\n",Leq)
+
diff --git a/Working_Examples/154/DEPENDENCIES/ch3_7.sce b/Working_Examples/154/DEPENDENCIES/ch3_7.sce new file mode 100755 index 0000000..72aa564 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch3_7.sce @@ -0,0 +1,19 @@ +clc
+disp("Example 3.7")
+printf("\n")
+printf("Given")
+disp("Total resistance of three resistors is 50 ohm")
+R=50;
+disp("Output voltage is 10 percent of the input voltage")
+//Let v be input voltage and v1 be output voltage
+//Let v1/v=V
+V=0.1;
+//As V=R1/(Total resistance)
+//Solving for R1
+R1=V*R;
+//As R=R1+R2
+//Solving for R2
+R2=R-R1;
+printf("R1=%dohm\n R2=%dohm\n",R1,R2)
+
+
diff --git a/Working_Examples/154/DEPENDENCIES/ch3_8.sce b/Working_Examples/154/DEPENDENCIES/ch3_8.sce new file mode 100755 index 0000000..f744f0f --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch3_8.sce @@ -0,0 +1,24 @@ +clc
+disp("Example 3.8")
+printf("\n")
+printf("Given")
+disp("Total current is 30mA")
+disp("Branch currents are 20mA and 10mA")
+disp("Equivalent resistance is equal to or greater than 10 ohm")
+
+//From Fig 3.6
+//Current flowing through R1 be i1 and let it be equal to 10mA
+//Current flowing through R2 be i2 and let it be equal to 20mA
+i1=10*10^-3;i2=20*10^-3;
+i=30*10^-3;
+
+//Let R1/(R1+R2)=X1 (1)
+//Let R2/(R1+R2)=X2 (2)
+X1=i1/i;
+X2=i2/i;
+//Let R1*R2(R1+R2)=Y (3)
+//Given that
+printf("\n Given")
+disp("R1*R2(R1+R2)>=10")
+//Solving (1),(2) and (3) we get
+printf("R1>=%dohm\nR2>=%dohm\n",15,30)
diff --git a/Working_Examples/154/DEPENDENCIES/ch4_1.sce b/Working_Examples/154/DEPENDENCIES/ch4_1.sce new file mode 100755 index 0000000..bbf2e06 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch4_1.sce @@ -0,0 +1,19 @@ +clc
+disp("Problem 4.1")
+printf("\n")
+
+//From figure 4.2
+//All the nodes and branches of the network are marked
+//The incidence matrix is written in a manner in which
+disp("1 is written if the direction of flow is away from the node")
+disp("-1 is written if the direction of flow is towards the node")
+disp("0 is written if the branch is not connected to the node")
+//Considering columns as branches with 1 to 6 from left to right
+//and rows as nodes with 0 to 3 from top to bottom
+A=[1 0 -1 -1 0 0
+ -1 1 0 0 1 0
+ 0 -1 1 0 0 1
+ 0 0 0 1 -1 -1]
+disp("The incidence matrix is ")
+disp(A,"Aa=")
+//It can be inferred from matrix Aa that the sum of elements in any row or a column is ZERO
\ No newline at end of file diff --git a/Working_Examples/154/DEPENDENCIES/ch4_2.sce b/Working_Examples/154/DEPENDENCIES/ch4_2.sce new file mode 100755 index 0000000..4692092 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch4_2.sce @@ -0,0 +1,22 @@ +clc
+disp("Problem 4.2")
+printf("\n")
+
+//From figure 4.2
+disp("The reduced incidence matrix is ")
+A=[-1 1 0 0 1 0
+ 0 -1 1 0 0 1
+ 0 0 0 1 -1 -1]
+ disp(A,"A=")
+//Here node D is chosen as reference
+//Let i=[ i1
+// i2
+// i3
+// i4
+// i5
+// i6 ]
+//Multiplying [A] with [i] we get the KCL equations
+disp("KCL equations are")
+disp("-i1+i2+i5=0")
+disp("-i2+i3+i6=0")
+disp("i4-i5-i6=0")
diff --git a/Working_Examples/154/DEPENDENCIES/ch4_3.sce b/Working_Examples/154/DEPENDENCIES/ch4_3.sce new file mode 100755 index 0000000..74e715c --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch4_3.sce @@ -0,0 +1,30 @@ +clc
+disp("Problem 4.3")
+printf("\n")
+
+//From figure 4.2
+disp("The reduced incidence matrix is ")
+A=[-1 1 0 0 1 0
+ 0 -1 1 0 0 1
+ 0 0 0 1 -1 -1]
+ disp(A,"A=")
+AT=[ -1 0 0
+ 1 -1 0
+ 0 1 0
+ 0 0 1
+ 1 0 -1
+ 0 1 -1 ]
+ disp(AT,"AT=")
+//Let e be the node to datum voltages
+//Let e=[ e1
+// e2
+// e3 ]
+//Multiplying [AT] with [e] we get the node voltages as
+disp("Node to datum voltages are")
+disp("v1=-e1")
+disp("v2=e1-e2")
+disp("v3=e2")
+disp("v4=e3")
+disp("v5=e1-e3")
+disp("v6=e2-e3")
+
diff --git a/Working_Examples/154/DEPENDENCIES/ch4_4.sce b/Working_Examples/154/DEPENDENCIES/ch4_4.sce new file mode 100755 index 0000000..9deb374 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch4_4.sce @@ -0,0 +1,59 @@ +clc
+disp("Problem 4.4")
+printf("\n")
+
+//From figure 4.7
+//Let us consider a tree with 5,6,7,8 as tree branches
+//Correspondingly links be 1,2,3,4
+//By adding links one after other Loops can be formed
+//The fundamental loops are..
+disp("L1={1,5,6} L2={2,5,6,7,8} L3={3,6,7,8} L4={4,6,7}")
+disp("1 is written if the direction of flow is along the direction of loop")
+disp("-1 is written if the direction of flow is opposite to the direction of loop")
+disp("0 is written if the branch is not a part of loop")
+disp("The loop incidence matrix is")
+B=[1 0 0 0 -1 1 0 0
+ 0 1 0 0 1 -1 1 1
+ 0 0 1 0 0 -1 1 1
+ 0 0 0 1 0 -1 1 0]
+ disp(B,"B=")
+//The above matrix has branches as columns and the number of loops as rows
+//As we need to find branch currents(8 in number)in terms of loop currents(4 in number)
+//Let i=[ i1 also iL=[ iL1
+// i2 iL2
+// i3 iL3
+// i4 iL4 ]
+// i5
+// i6
+// i7
+// i8]
+
+
+//We know i=BT*iL
+//i=[ i1 [1 0 0 0 iL=[ iL1
+// i2 0 1 0 0 iL2
+// i3 0 0 1 0 * iL3
+// i4 = 0 0 0 1 iL4 ]
+// i5 -1 1 0 0
+// i6 -1 -1 -1 -1
+// i7 0 1 1 1
+// i8] 0 1 1 0 ]
+
+disp("The branch currents are")
+disp("i1=iL1")
+disp("i2=iL2")
+disp("i3=iL3")
+disp("i4=iL4")
+disp("i5=-iL1+iL2")
+disp("i6=iL1-iL2-iL3-iL4")
+disp("i7=iL2+iL3+iL4")
+disp("i8=iL2+iL3")
+
+
+
+
+
+
+
+
+
diff --git a/Working_Examples/154/DEPENDENCIES/ch4_5.sce b/Working_Examples/154/DEPENDENCIES/ch4_5.sce new file mode 100755 index 0000000..c9e2865 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch4_5.sce @@ -0,0 +1,35 @@ +clc
+disp("Problem 4.5")
+printf("\n")
+
+//From figure 4.8
+//The fundamental cut-sets are..
+disp("C1={1,2,5} L2={2,3,8} L3={1,2,3,4,6} L4={2,3,4,7}")
+disp("1 is written for the link if the direction of flow of tree branch is same as that of the link")
+disp("-1 is written for the link if the direction of flow of tree branch is opposite as that of the link")
+disp("0 is written if the branch or a link is not a part of cut set")
+disp("The cut set matrix is")
+C=[ 1 -1 0 0 1 0 0 0
+ 0 -1 -1 0 0 0 0 1
+ 1 -1 -1 -1 0 1 0 0
+ 0 -1 -1 -1 0 0 1 0 ]
+ disp(C,"C=")
+//The above matrix has branches as columns and the number of cut sets as rows
+
+//Let i=[ i1
+// i2
+// i3
+// i4
+// i5
+// i6
+// i7
+// i8 ]
+
+//As we need to write the KCL for the circuit
+//KCL=[C]*[i]
+//Multiplying [C] with [i] we get the KCL equations
+disp("KCL equations are")
+disp("i1-i2+i5=0")
+disp("-i2-i3+i8=0")
+disp("i1-i2-i3-i4+i6=0")
+disp("-i2-i3-i4+i7=0")
diff --git a/Working_Examples/154/DEPENDENCIES/ch5_8.sce b/Working_Examples/154/DEPENDENCIES/ch5_8.sce new file mode 100755 index 0000000..4c4e229 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch5_8.sce @@ -0,0 +1,19 @@ +clc
+//From figure 5.13(a)
+//Applying KVL equation to the loop
+I=(20+10)/(3+6)
+//As current will not flow in upper 3 ohm resistor so Thevenin voltage is equal to either of the two parallel branches
+V1=20-I*3
+printf("Thevenin voltage = %dV\n",V1)
+
+// Left 3 ohm and 6 ohm resistor are in parallel and their equivalent is in series with 3 ohm
+R1=3+(3*6)/(3+6)
+printf("Thevenin resistance =%dohm\n",R1)
+
+//Now to find Norton's equivalent
+I1=V1/R1
+printf("\n Norton current =%dA\n",I1)
+disp("The value of resistance in Norton equivalent will not change but will come in parallel with current source")
+
+
+
diff --git a/Working_Examples/154/DEPENDENCIES/ch6_8.sce b/Working_Examples/154/DEPENDENCIES/ch6_8.sce new file mode 100755 index 0000000..d971edf --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch6_8.sce @@ -0,0 +1,15 @@ +clc
+disp("Example 6.8")
+printf("Given")
+disp("R1= 10kohm R2=50kohm Ri=500kohm R0=0")
+disp("Open loop gain (A)=10^5")
+A=10^5;R1=10*10^3;R2=50*10^3;Ri=500*10^3;
+//From figure 6.11
+//Applying KCL equation at node B
+disp("(v1+vd)/10+ (v2+vd)/50+ vd/500=0 (1)")
+//Since R0=0
+disp("v2=A*vd")
+//Solving for vd
+disp("vd=10^-5*v2 (2)")
+//Substituting (2) in (1) we get
+printf("v2/v1=%d\n",-5)
\ No newline at end of file diff --git a/Working_Examples/154/DEPENDENCIES/ch7_1.sce b/Working_Examples/154/DEPENDENCIES/ch7_1.sce new file mode 100755 index 0000000..66f5b3d --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch7_1.sce @@ -0,0 +1,53 @@ +clc
+disp("Example 7.1")
+printf("\n")
+
+t1=-5:0.5:8
+v1=cos (t1)
+figure
+a= gca ();
+plot(t1,v1)
+xtitle ('v1 vs t1','t1','v1 ');
+a. thickness = 2;
+//From the graph
+printf("Time period1= %3.3fs\n Frequency 1=%0.3fHz\n",6.2832,0.159)
+
+t2=-4:0.5:10
+v2=sin (t2)
+figure
+a= gca ();
+plot(t2,v2)
+xtitle ('v2 vs t2','t2','v2 ');
+a. thickness = 2;
+//From the graph
+printf("Time period 2= %3.3fs\n Frequency 2=%0.3fHz\n",6.2832,0.159)
+
+t3=-1:0.05:1.5
+v3=2*cos (2*%pi*t3)
+figure
+a= gca ();
+plot(t3,v3)
+xtitle ('v3 vs t3','t3','v3 ');
+a. thickness = 2;
+//From the graph
+printf("Time period 3= %ds\n Frequency 3=%dHz\n",1,1)
+
+t4=-5:0.5:12
+v4=2*cos (%pi*t4/4-%pi/4)
+figure
+a= gca ();
+plot(t4,v4)
+xtitle ('v4 vs t4','t4','v4 ');
+a. thickness = 2;
+//From the graph
+printf("Time period 4= %ds\n Frequency 4=%0.3fHz\n",8,0.125)
+
+t5=-1:0.005:1
+v5=5*cos (10*t5+%pi/3)
+figure
+a= gca ();
+plot(t5,v5)
+xtitle ('v5 vs t5','t5','v5 ');
+a. thickness = 2;
+//From the graph
+printf("Time period 5= %0.3fs\n Frequency 5=%3.2fHz\n",.62832,1.59)
diff --git a/Working_Examples/154/DEPENDENCIES/ch7_13.sce b/Working_Examples/154/DEPENDENCIES/ch7_13.sce new file mode 100755 index 0000000..86523df --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch7_13.sce @@ -0,0 +1,24 @@ +clc
+disp("Example 7.13")
+printf("\n")
+
+printf("Given")
+disp("Capacitance is 1uF")
+C=1*10^-6;
+disp("a)")
+//Let k=1 which results in t=5ms
+t=5*10^-3;
+vac=(integrate('.004','t',0,0.003)-integrate('.002','t',0.003,0.005))/C;
+printf("vac=%dV\n",vac);
+
+//In general
+disp("At t=5k voltage follows as v=8k ms")
+
+disp("b)")
+//As vdc=1/C*integrate(Idc*dt)
+//On solving for Idc
+vdc=vac
+Idc=(1/((integrate('1/vac','t',0,0.005))/C))*10^3
+printf("Idc=%3.2fmA\n",Idc);
+disp("Idc is equal to <i(t)> in the period of 5ms")
+
diff --git a/Working_Examples/154/DEPENDENCIES/ch7_17.sce b/Working_Examples/154/DEPENDENCIES/ch7_17.sce new file mode 100755 index 0000000..de2f600 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch7_17.sce @@ -0,0 +1,39 @@ +clc
+disp("Example 7.17")
+printf("\n")
+
+printf("Given")
+disp("Capacitance is 100nF")
+disp("The voltage across capacitor increases linearly from 0 to 10V")
+C=100*10^-9;
+//From figure 7.10(a)
+disp("a)")
+//At t=T voltage across capacitor =10V
+vc=10;
+Q=C*vc;
+printf("Charge across capacitor is %fC\n",Q)
+disp("b)")
+//The waveform shown in fig 7.10(a) can be written as
+disp("0 t<0")
+disp("I0=10^-6/T 0<t<T")
+disp("0 t>T")
+
+
+//For T=1s;
+T=1;
+I0=10^-6/T;
+printf("I0(1s)=%fA\n",I0);
+
+//For T=1ms;
+T=1*10^-3;
+I0=10^-6/T;
+printf("I0(1ms)=%0.3fA\n",I0);
+
+//For T=1us;
+T=1*10^-6;
+I0=10^-6/T;
+printf("I0(1us)=%dA",I0);
+
+
+
+
diff --git a/Working_Examples/154/DEPENDENCIES/ch7_2.sce b/Working_Examples/154/DEPENDENCIES/ch7_2.sce new file mode 100755 index 0000000..07e77b0 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch7_2.sce @@ -0,0 +1,12 @@ +clc
+disp("Example 7.2")
+printf("\n")
+
+//Let wt=q
+q=-8:0.5:8
+v=5*cos (q)
+figure
+a= gca ();
+plot(q,v)
+xtitle ('v vs wt','wt','v ');
+a. thickness = 2;
diff --git a/Working_Examples/154/DEPENDENCIES/ch7_22.sce b/Working_Examples/154/DEPENDENCIES/ch7_22.sce new file mode 100755 index 0000000..b9cc022 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch7_22.sce @@ -0,0 +1,20 @@ +clc
+disp("Example 7.22")
+printf("\n")
+
+//The general equation of exponential decay function is given by
+disp("v(t)=A*e(-t/T)+B")
+//We need to solve A and B
+//At t=0 we get v(0)=A+B (1)
+//at t=inf we get B=1 (2)
+//Solving (1) and (2)
+A=4;B=1;
+T=3;
+t=0:0.05:10
+v=4*exp(-t/T)+1;
+figure
+a= gca ();
+plot(t,v)
+xtitle ('v vs t','t','v');
+a. thickness = 2;
+
diff --git a/Working_Examples/154/DEPENDENCIES/ch7_23.sce b/Working_Examples/154/DEPENDENCIES/ch7_23.sce new file mode 100755 index 0000000..2109c67 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch7_23.sce @@ -0,0 +1,33 @@ +clc
+disp("Example 7.23")
+printf("\n")
+
+//Sketch voltage 'v'
+t=-.001:0.00005:0
+t1=0:0.00005:.001
+T=1*10^-3;
+V0=10;
+v=V0*exp(t/T)
+v1=V0*exp(-t1/T)
+figure
+a= gca ();
+plot(t,v)
+plot(t1,v1)
+xtitle ('v vs t','t (ms)','v ');
+a. thickness = 2;
+
+//Sketch current 'i'
+t=-.001:0.00005:0
+t1=0:0.00005:.001
+T=1*10^-3;
+I0=10*10^-3;
+i=I0*exp(t/T)
+i1=-I0*exp(-t1/T)
+figure
+a= gca ();
+plot(t,i)
+plot(t1,i1)
+xtitle ('i vs wt','t (ms)','i (mA)');
+a. thickness = 2;
+
+
diff --git a/Working_Examples/154/DEPENDENCIES/ch7_25.sce b/Working_Examples/154/DEPENDENCIES/ch7_25.sce new file mode 100755 index 0000000..4d0c5da --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch7_25.sce @@ -0,0 +1,9 @@ +clc
+disp("Example 7.25")
+printf("\n")
+
+Xavg=(2+4+11+5+7+6+9+10+3+6+8+4+1+3+5+12)/16;
+//Let X=X^2eff
+X=(2^2+4^2+11^2+5^2+7^2+6^2+9^2+10^2+3^2+6^2+8^2+4^2+1^2+3^2+5^2+12^2)/16
+Xeff=sqrt(X);
+printf("Xavg=%d\n Xeff=%3.2f\n",Xavg,Xeff)
\ No newline at end of file diff --git a/Working_Examples/154/DEPENDENCIES/ch7_26.sce b/Working_Examples/154/DEPENDENCIES/ch7_26.sce new file mode 100755 index 0000000..8b76377 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch7_26.sce @@ -0,0 +1,17 @@ +clc
+disp("Example 7.26")
+printf("\n")
+
+printf("Given")
+disp("Period =10s")
+disp("Interval is 1ms")
+disp("Voltage of binary signal is either 0.5 or -0.5")
+T=10;
+//During 10s period there are 10000 intervals of 1ms each
+//For calculating average equal number of intervals are considered at 0.5V and -0.5V
+vavg=(0.5*5000-0.5*5000)/10000
+//The effective value of v(t) is
+//Let V=V^2eff
+V=(0.5^2*5000+(-0.5)^2*5000)/10000
+Veff=sqrt(V)
+printf("vavg=%dV\nVeff=%3.2fV\n",vavg,Veff)
diff --git a/Working_Examples/154/DEPENDENCIES/ch7_3.sce b/Working_Examples/154/DEPENDENCIES/ch7_3.sce new file mode 100755 index 0000000..122832e --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch7_3.sce @@ -0,0 +1,12 @@ +clc
+disp("Example 7.3")
+printf("\n")
+
+t1=-10:0.05:10
+v=5*cos (%pi*t1/6+%pi/6)
+figure
+a= gca ();
+plot(t1,v)
+xtitle ('v vs %pi*t/6','%pi*t/6','v ');
+a. thickness = 2;
+
diff --git a/Working_Examples/154/DEPENDENCIES/ch7_5.sce b/Working_Examples/154/DEPENDENCIES/ch7_5.sce new file mode 100755 index 0000000..515f4b9 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch7_5.sce @@ -0,0 +1,14 @@ +clc
+disp("Example 7.5")
+printf("\n")
+
+printf("Given")
+disp("v(t)=cos5t+3sin(3t+45)")
+//Finding the periods of individual terms
+disp("Period of cos5t=2*%pi/5")
+disp("Period of 3*sin(3t+45)=2*%pi/3")
+//If T=2*%pi
+T=2*%pi;
+disp("Now T=5*T1=3*T2")
+//Now the relation for T is the smallest common integral multiple of T1 and T2
+printf("Period = %3.2fs\n",T)
\ No newline at end of file diff --git a/Working_Examples/154/DEPENDENCIES/ch8_1.sce b/Working_Examples/154/DEPENDENCIES/ch8_1.sce new file mode 100755 index 0000000..fdc51d2 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch8_1.sce @@ -0,0 +1,17 @@ +clc
+disp("Example 8.1")
+printf("\n")
+
+printf("Given")
+disp("Capacitance is 1uF")
+disp("Resistance is 1Mohm")
+disp("Voltage across capacitor is 10V")
+R=1*10^6;C=1*10^-6;V=10
+//Let T be time constant
+T=R*C
+//v(t)=V*exp(-t/T)
+disp("v(t)=10*exp(-t) (1)")
+//Substituting value of t=5 in (1)
+v5=10*exp(-5)
+printf("Time constant is %ds\n",T)
+printf("v(5)=%0.3fV\n",v5)
\ No newline at end of file diff --git a/Working_Examples/154/DEPENDENCIES/ch8_10.sce b/Working_Examples/154/DEPENDENCIES/ch8_10.sce new file mode 100755 index 0000000..3c081ca --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch8_10.sce @@ -0,0 +1,25 @@ +clc
+disp("Example 8.10")
+printf("\n")
+
+printf("Given")
+disp("vs= 5V t<0")
+disp("vs=5*sin(w*t) t>0")
+vs=5;
+R=5;L=10*10^-3;
+//At t<0
+//Inductor behaves as a short circuit
+//Let i(0-)=i
+i=vs/R;
+printf("i(0-)=%dA\n",i)
+//During the transition from t=0- to t=0+
+//Let i(0+)=i1
+i1=i
+printf("i(0+)=%dA\n",i1)
+//Applying KVL equation to the loop
+disp("vs=i*R+v")
+//Let v(0+)=v1 ; vs(0+)=vs1
+//From given vs(0+)=0
+vs1=0;
+v1=vs1-i*R
+printf("\nv(0+)=%dV\n",v1)
diff --git a/Working_Examples/154/DEPENDENCIES/ch9_11.sce b/Working_Examples/154/DEPENDENCIES/ch9_11.sce new file mode 100755 index 0000000..8563a90 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch9_11.sce @@ -0,0 +1,16 @@ +clc
+disp("Example 9.11")
+printf("\n")
+
+//From figure 9.16
+//H(s)=V(s)/I(s)=Z(s)
+//Let V(s)=1 the H(s)=Z(s)
+s=%s
+z1=(1/2.5)+(3/(5*s))+(s/20)
+Z=1/z1
+Dem=Z('den')
+//The roots are
+q=roots(Dem)
+disp(q,"Poles are")
+
+
diff --git a/Working_Examples/154/DEPENDENCIES/ch9_6.sce b/Working_Examples/154/DEPENDENCIES/ch9_6.sce new file mode 100755 index 0000000..a063207 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch9_6.sce @@ -0,0 +1,32 @@ +clc
+disp("Example 9.6")
+printf("\n")
+
+printf("Given")
+disp("Resistance is 10ohm and inductance is 2H")
+disp("Applied voltage is 10*exp(-2*t)*cos(10*t+30)")
+s=%s;
+//For a RL circuit
+//Applying KVL equation
+//v=i*R+L*d/dt(i) (1)
+//As v=10(30 deg) (2)
+//Equating (1) and (2)
+// Let i=I*exp(s*t) (3)
+// 10(30 deg)*exp(s*t)=10*I*exp(s*t)+2*s*I*exp(s*t)")
+//Solving for I
+disp("I=10(30 deg)/10+2*s")
+s=-2+%i*10
+a=10+2*s
+x=10*cos((30*%pi)/180);
+y=10*sin((30*%pi)/180);
+z=complex(x,y)
+I=z/a
+b=real(I);
+c=imag(I);
+magn=sqrt(b^2+c^2)
+ph=(atan(c/b)*180)/%pi
+//From (3)
+printf("\ni=%0.2f*exp(-2*t)*cos(10t%3.1f deg) (A)\n",magn,ph);
+
+
+
diff --git a/Working_Examples/154/DEPENDENCIES/ch9_7.sce b/Working_Examples/154/DEPENDENCIES/ch9_7.sce new file mode 100755 index 0000000..fe7c0cf --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch9_7.sce @@ -0,0 +1,31 @@ +clc
+disp("Example 9.7")
+printf("\n")
+
+printf("Given")
+disp("Resistance is 10ohm and Capacitance is 0.2F")
+disp("Applied voltage is 10*exp(-2*t)*cos(10*t+30)")
+s=%s;
+//For a RC circuit
+//Applying KVL equation
+//v=i*R+(1/C)*integrate(i*dt) (1)
+//As v=10(30 deg) (2)
+//Equating (1) and (2)
+// Let i=I*exp(s*t) (3)
+// 10(30 deg)*exp(s*t)=10*I*exp(s*t)+(5/s)*I*exp(s*t)")
+//Solving for I
+disp("I=10(30 deg)/10+(5/s)")
+s=-2+%i*10
+a=10+(5/s)
+x=10*cos((30*%pi)/180);
+y=10*sin((30*%pi)/180);
+z=complex(x,y)
+I=z/a
+b=real(I);
+c=imag(I);
+magn=sqrt(b^2+c^2)
+ph=(atan(c/b)*180)/%pi
+//From (3)
+printf("\ni=%0.2f*exp(-2*t)*cos(10t+%3.1f deg) (A)\n",magn,ph);
+
+
diff --git a/Working_Examples/154/DEPENDENCIES/ch9_8.sce b/Working_Examples/154/DEPENDENCIES/ch9_8.sce new file mode 100755 index 0000000..d9c67e9 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch9_8.sce @@ -0,0 +1,18 @@ +clc
+disp("Example 9.8")
+printf("\n")
+
+s=%s ;
+
+//From figure 9.13
+disp("Z(s)=(2.5+((5*s/3)*(20/s))/(5*s/3+20/s))")
+//On solving
+z1=poly([12 8 1],'s','coeff')
+z2=poly([12 0 1],'s','coeff')
+Z=2.5*(z1/z2)
+disp(Z,"Z(s)")
+//H(s)=I(s)/Z(s)
+//Let I(s)=1 the H(s)=1/Z(s)
+H=(1/2.5)*(z2/z1)
+disp(H,"H(s)")
+
diff --git a/Working_Examples/154/DEPENDENCIES/ch9_9.sce b/Working_Examples/154/DEPENDENCIES/ch9_9.sce new file mode 100755 index 0000000..668b268 --- /dev/null +++ b/Working_Examples/154/DEPENDENCIES/ch9_9.sce @@ -0,0 +1,11 @@ +clc
+disp("Example 9.9")
+printf("\n")
+
+s=%s ;
+H=syslin ( 'c',(0.4*(s^2+12))/((s+2)*(s+6) ) ) ;
+evans (H,1)
+//If s=1Np/s
+H1=0.4*(1+12)/((1+2)*(1+6))
+printf("H(1)=%0.3f\n",H1)
+
\ No newline at end of file |