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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);
//=============================================================================
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