6 files changed, 137 insertions, 0 deletions
diff --git a/196/CH2/EX2.1/example_2_1.sce b/196/CH2/EX2.1/example_2_1.sce
new file mode 100755
index 000000000..6058b655f
--- /dev/null
+++ b/196/CH2/EX2.1/example_2_1.sce
@@ -0,0 +1,91 @@
+//Chapter 2
+//Example 2-1
+//ProbOnOutputvoltage 
+//Page 19,figure 2-3
+clear;clc;
+//Given
+Vplus=15;Vminus=-15;Vsatp=13;Vsatm=-13;//All in Volts
+Aol=200000;//gain
+//Example 2-1(a)
+Vam=-10*(10^-6);//voltage at minus input
+Vap=-15*(10^-6);//voltage at plus input
+Ed1=Vap-Vam;//Differential Input Voltage
+Vout1=Ed1*Aol;//Output Voltage
+format(10);
+if(Vout1>15) then
+  disp("Value of o/p voltage1 = 13.0000V") //positive saturation voltage
+elseif(Vout1<-15) then
+  disp("Value of o/p voltage1 = -13.0000V")//negative saturation voltage
+else  
+  printf("\n\n Value of o/p voltage1 =  %.4f V \n\n",Vout1)
+end
+
+//Example 2-1(b)
+Vbm=-10*(10^-6);//voltage at minus input
+Vbp=+15*(10^-6);//voltage at plus input
+Ed2=Vbp-Vbm;//Differential Input Voltage
+Vout2=Ed2*Aol;//Output Voltage
+format(10);
+if(Vout2>15) then
+  disp("Value of o/p voltage2 = 13.0000V")//positive saturation voltage
+elseif(Vout2<-15) then
+  disp("Value of o/p voltage2 = -13.0000V")//negative saturation voltage
+else  
+  printf("\n\n Value of o/p voltage2 =  %.4f V \n\n",Vout2)
+end
+
+//Example 2-1(c)
+Vcm=-10*(10^-6);//voltage at minus input
+Vcp=-5*(10^-6);//voltage at plus input
+Ed3=Vcp-Vcm;//Differential Input Voltage
+Vout3=Ed3*Aol;//Output Voltage
+format(10);
+if(Vout3>15) then
+  disp("Value of o/p voltage3 = 13.0000V")//positive saturation voltage
+elseif(Vout3<-15) then
+  disp("Value of o/p voltage3 = -13.0000V")//negative saturation voltage
+else  
+  printf("\n\n Value of o/p voltage3 =  %.4f V \n\n",Vout3)
+end
+
+//Example 2-1(d)
+Vdm=+1.000001;//voltage at minus input
+Vdp=+1.000000;//voltage at plus input
+Ed4=Vdp-Vdm;//Differential Input Voltage
+Vout4=Ed4*Aol;//Output Voltage
+format(10);
+if(Vout4>15) then
+  disp("Value of o/p voltage4 = 13.0000V")//positive saturation voltage
+elseif(Vout4<-15) then
+  disp("Value of o/p voltage4 = -13.0000V")//negative saturation voltage
+else  
+  printf("\n\n Value of o/p voltage4 =  %.4f V \n\n",Vout4)
+end
+
+//Example 2-1(e)
+Vem=+5*(10^-3);//voltage at minus input
+Vep=0;//voltage at plus input
+Ed5=Vep-Vem;//Differential Input Voltage
+Vout5=Ed5*Aol;//Output Voltage
+format(10);
+if(Vout5>15) then
+  disp("Value of o/p voltage5 = 13.0000V")//positive saturation voltage
+elseif(Vout5<-15) then
+  disp("Value of o/p voltage5 = -13.0000V")//negative saturation voltage
+else  
+  printf("\n\n Value of o/p voltage5 =  %.4f V \n\n",Vout5)
+end
+
+//Example 2-1(f)
+Vfm=0;//voltage at minus input
+Vfp=+5*(10^-3);//voltage at plus input
+Ed6=Vfp-Vfm;//Differential Input Voltage
+Vout6=Ed6*Aol;//Output Voltage
+format(10);
+if(Vout6>15) then
+  disp("Value of o/p voltage6 = 13.0000V")//positive saturation voltage
+elseif(Vout6<-15) then
+  disp("Value of o/p voltage6 = -13.0000V")//negative saturation voltage
+else  
+  printf("\n\n Value of o/p voltage6 =  %.4f V \n\n",Vout6)
+end
diff --git a/196/CH2/EX2.1/result_2_1.txt b/196/CH2/EX2.1/result_2_1.txt
new file mode 100755
index 000000000..ecf76ec16
--- /dev/null
+++ b/196/CH2/EX2.1/result_2_1.txt
@@ -0,0 +1,11 @@
+Value of o/p voltage1 = -1.0000 V 
+
+Value of o/p voltage2 =  5.0000 V 
+
+Value of o/p voltage3 =  1.0000 V 
+
+Value of o/p voltage4 = -0.2000 V 
+ 
+Value of o/p voltage5 = -13.0000V   
+ 
+Value of o/p voltage6 =  13.0000V   
+\ No newline at end of file
diff --git a/196/CH2/EX2.2/example_2_2.sce b/196/CH2/EX2.2/example_2_2.sce
new file mode 100755
index 000000000..6608ff270
--- /dev/null
+++ b/196/CH2/EX2.2/example_2_2.sce
@@ -0,0 +1,18 @@
+//Chapter 2
+//Example 2-2
+//ProbOnPWM 
+//Page 34
+clear;clc;
+//Given
+f=50;//in Hz
+Vtemp=4; //input signal in volts
+Ecm=10; //maximum peak voltage of sawtooth carrier wave in volts 
+
+//Example 2-2(a)
+T=1/f;
+Th=(Vtemp*T)/Ecm;//High time in seconds
+printf("\n\n High Time = %.4f s \n\n",Th)
+
+//Example 2-2(b)
+d=(Th/T)*100;//duty cycle in percentage
+printf("\n\n Duty cycle = %.4f percent \n\n",d)
diff --git a/196/CH2/EX2.2/result_2_2.txt b/196/CH2/EX2.2/result_2_2.txt
new file mode 100755
index 000000000..57a3d0240
--- /dev/null
+++ b/196/CH2/EX2.2/result_2_2.txt
@@ -0,0 +1,4 @@
+
+High Time = 0.0080 s 
+
+Duty cycle = 40.0000 percent
+\ No newline at end of file
diff --git a/196/CH2/EX2.3/example_2_3.sce b/196/CH2/EX2.3/example_2_3.sce
new file mode 100755
index 000000000..dae12b6d2
--- /dev/null
+++ b/196/CH2/EX2.3/example_2_3.sce
@@ -0,0 +1,12 @@
+//Chapter 2
+//Example 2-3
+//ProbOnHighTime 
+//Page 34,35,figure 2-16(d)
+clear;clc;
+//Given
+Vtemp=4;//in volts
+Ecm=5;//maximum peak voltage of a sawtooth carrier wave
+T=0.01;//in seconds
+//calculate
+Th=T*(1-(Vtemp/Ecm));//High Time
+printf("\n\n High Time = %.4f s \n\n",Th)
diff --git a/196/CH2/EX2.3/result_2_3.txt b/196/CH2/EX2.3/result_2_3.txt
new file mode 100755
index 000000000..9e1188deb
--- /dev/null
+++ b/196/CH2/EX2.3/result_2_3.txt
@@ -0,0 +1 @@
+ High Time = 0.0020 s 
+\ No newline at end of file