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+// chapter 13
+// example 13.3
+// page 278
+
+Vcc=15 // V
+Re=2 // kilo ohm
+Rc=1 // kilo ohm
+Rl=1 // kilo ohm
+Vbe=0.7 // V
+
+// dc load line
+
+    // when Ic=0, Vce=Vcc i.e. Vce=15 and when Vce=0, Ic=Vcc/(Rc+Re) i.e. Ic=15/3
+    // so equation of load line becomes Ic=-(1/3)*Vce+15
+
+       clf()
+    x=linspace(0,15,5)
+    y=-(1/3)*x+5
+    plot2d(x,y,style=3,rect=[0,0,16,6])
+    xtitle("dc load line-green         ac load line-blue","collector emitter voltage(volts)","collector current(mA)")
+
+    V2=5 // V
+    // since voltage across R2 is V2=5 V and V2=Vbe+Ie*Re we get
+    Ie=(V2-Vbe)/Re
+    Ic=Ie
+    Vce=Vcc-Ic*(Rc+Re)
+
+    printf("the operating point is %.3f V and %.3f mA \n",Vce,Ic)
+
+
+// ac load line
+    
+    R_AC=Rc*Rl/(Rc+Rl) // ac load
+    V_ce=Vce+Ic*R_AC // maximum collector emitter voltage
+    I_c=Ic+Vce/R_AC // maximum collector current
+    // the equation of ac load line in terms of V_ce and I_c becomes
+    y=-(I_c/V_ce)*x+I_c
+     plot2d(x,y,style=2,rect=[0,0,10,20])
+