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+//chapter 5 example 7
+//=============================================================================
+clc;
+clear;
+//Given Data
+Va    =  1200;//Anode potential
+F     =  10*10^9;//Operating frequency in Hz 
+S     =  5*10^-2;//spacing b/w 2 cavities
+GS    = 1*10^-3;//gap spacing in either cavity
+e     =  1.6*10^-19;//charge of electron
+m     =  9.1*10^-31;//mass of electron in kg
+//Calculations
+//Condition of maximum output is (V1/Vo)max  =  (3.68)/((2*pi*n)-(pi/2);
+//(2*pi*n)-(pi/2) = Transit angle b/w two cavities
+//V1  = Peak amplitude of RF i/p
+//Vo  = accelarating potential
+
+Vo    = sqrt(2*e*Va/m);//velocity of the electrons 
+T     = S/Vo;//Transit time b/w the cavities
+TA    = 2*%pi*F*T;//transit angle in radians
+V1    = (3.68*Va)/TA;
+//Output
+mprintf('Required Peak Amplitude of i/p RF signal is %3.2f volts',V1);
+//=============================================================================