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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+//Ex:3.10
+clc;
+clear;
+close;
+xnp=35;// beam width in degree
+xnp1=(xnp/2)*(%pi/180);// half beam width in degree
+// T(m-1)(x)=0 or T(8-1)(x)=0, or T(7)(x)=0
+// cos((m-1)*acos(x))=0
+// (8-1)*acos(x)=cos(2k-1)*(%pi/2)
+// acos(x)=(2k-1)*pi/14
+// for first nulls , k=1
+// acos(x)=pi/14;
+x=cos(%pi/14);
+// but z=x/xo=cos(p/2)
+// p=Bd*sin(xnp1)
+// p/2=Bd*sin(xnp1)/2
+// x/xo=cos(Bd*sin(xnp1)/2)
+// and Bd*sin(a)=(2*%pi/y)*(y/2)*(1/2)*sin(xnp1)
+// and Bd*sin(xnp1)=90*sin(xnp1)
+xo=x/(cos((90*sin(xnp1)*(%pi/180))));
+// aoz+a1(4z^3-3z)+a2(16z^5-20z^3+5z)+a3(64z^7-112z^5+56z^3-7z)=64x^7-112x^5+56x^3-7x, where z=(x/xo)
+// Then on putting z=(x/xo), we get
+// ao(x/xo)+a1(4(x/xo)^3-3(x/xo))+a2(16(x/xo)^5-20(x/xo)^3+5(x/xo))+a3(64(x/xo)^7-112(x/xo)^5+56(x/xo)^3-7(x/xo))=64x^7-112x^5+56x^3-7x
+// on comparing the terms, we get ao=3.339,a1=2.919,a2=2.191,a3=1.886
+ao=3.339;
+a1=2.919;
+a2=2.191;
+a3=1.886;
+a33=a3/a3;// the ratio of the a3 to a3
+a23=a2/a3;// the ratio of the a2 to a3
+a13=a1/a3;// the ratio of the a1 to a3
+ao3=ao/a3;// the ratio of the ao to a3
+printf("The value of the parameter xo = %f", xo);
+printf("\n The value of the amplitude parameter ao= %f", ao);
+printf("\n The value of the amplitude parameter a1= %f", a1);
+printf("\n The value of the amplitude parameter a2= %f", a2);
+printf("\n The value of the amplitude parameter a3= %f", a3);
+printf("\n The value of the relative amplitude parameter a33= %f", a33);
+printf("\n The value of the relative amplitude parameter a23= %f", a23);
+printf("\n The value of the relative amplitude parameter a13= %f", a13);
+printf("\n The value of the relative amplitude parameter ao3= %f", ao3);
+printf("\n The five element array is shown in figure in the given textbook") \ No newline at end of file