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diff --git a/2579/CH3/EX3.1/Ex3_1.sce b/2579/CH3/EX3.1/Ex3_1.sce
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+//Ex:3.1

+clc;

+clear;

+close;

+// the path difference, x=dcos(a)

+// therefore, phase difference , w=(2*%pi/y)*dcos(a)=Bdcos(a)

+// from the geometry of the figure in the far field r>>d

+// r1=r+x=r+dcos(a)

+// r2=r-x=r-dcos(a)

+// Hence, Et=I1exp^(-jB(r+dcos(a)))+I2exp^(-jB(r-dcos(a)))

+// Et=exp^(-jBr)(I1exp^(-jBdcos(a))+I2exp^(-jBdcos(a)))

+// case (a): in case I, we have I1=I2=I

+// Hence, Et=Iexp^(-jBr)*(exp^(-jBdcos(a))+exp^(-jBdcos(a)))=2exp^(-jBr)*cos(Bdcos(a))

+// Et will be max when cos(Bdcos(a)) will be max. therefore

+// cos(Bd*cos(a))=1

+// Bd*cos(a)=0

+// a_max=n*%pi/2, where n=1,2,3,........

+// hence , for the half power point a_HPPD

+// cos(Bd*cos(a))=1/(sqrt(2))

+// Bd*cos(a)=%pi/4

+// cos(a_HPPD)=%pi/4Bd= %pi/(4*2%pi*0.75y/y)=1/6

+// a_HPPD=acos(1/6)

+a_HPPD=(acos(1/6)*180/%pi);// the half power point in degree

+a_m=2*a_HPPD;// the half power beam width in degree

+// In case I1=I and I2=Iexp^(j540)=Iexp^(j180)=-I

+// therefore, Et2=Iexp^(-jBr)*(exp^(-jBdcos(a))+exp^(-jBdcos(a)))

+// =2j*I*exp^(-jBd)*sin(Bdcos(a))

+// The max value of sin(Bdcos(a)) is at a=%pi. When

+// sin(Bd*cos(a))=sin(Bd*cos(%pi))=sin(-Bd)=sin(-2*%pi*3y/(y*4))=sin(-3%Pi/2)=1

+// Hence at the half power point a_HPPD2

+// sin(Bd*cos(a))=1/(sqrt(2))

+// Bd*cos(a_HPPD2)=%pi/4

+// cos(a_HPPD2)=%pi/(4*2%pi*0.75y/y)=1/6

+a_HPPD2=(acos(1/6)*180/%pi);// the half power point in degree

+a_m2=2*a_HPPD2;// the half power beam width in degree

+printf("The half power beam width for broad side array = %f degree", a_m);

+printf("\n The half power beam width for end fire array = %f degree", a_m2);
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