initial commit / add all books

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diff --git a/2459/CH21/EX21.8/Ex21_8.PNG b/2459/CH21/EX21.8/Ex21_8.PNG
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diff --git a/2459/CH21/EX21.8/Ex21_8.sce b/2459/CH21/EX21.8/Ex21_8.sce
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+//chapter21

+//example21.8

+//page473

+

+Rl=1d3 // ohm

+R=200 // ohm

+

+// for positive half cycle, diode is forward biased and since load is in parallel with diode we get

+V_out_p=0.7 // V

+

+// for negative half cycle, diode is reverse biased so it is open. Hence

+V_in=-10 // V

+V_out_n=V_in*Rl/(Rl+R)

+

+printf("output voltage for positive cycle = %.3f V \nand for negative cycle = %.3f V",V_out_p,V_out_n)

+

+// plotting input and output waveforms in same graph using following code instead of using xcos

+clf()

+t=linspace(0,%pi,100)

+Vin=V_in*sin(t)

+Vout=-V_out_n*sin(t)

+subplot(2,2,1)

+plot2d(t,-Vin,style=3,rect=[0,0,10,11])

+xtitle("Vin +ve","t","volts")

+subplot(2,2,2)

+plot2d(t,Vout,style=2,rect=[0,-5,10,0.7])

+xtitle("Vout","t","volts")

+t=linspace(%pi,2*%pi,100)

+Vin=V_in*sin(t)

+subplot(2,2,3)

+plot2d(t,-Vin,style=3,rect=[0,-11,10,0])

+xtitle("Vin -ve","t","volts")

+subplot(2,2,4)

+plot2d(t,-Vout,style=2,rect=[0,-11,10,0])

+xtitle("Vout","t","volts")

diff --git a/2459/CH21/EX21.8/Figure21_8.JPG b/2459/CH21/EX21.8/Figure21_8.JPG
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+++ b/2459/CH21/EX21.8/Figure21_8.JPG