Diffraction At High Frequencies By a Circular Disc
Diffraction At High Frequencies By a Circular Disc
Douglas S Jones
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„(, ) . -(i?) 1 / 2 I /(^ _o_ dt) which may be written as (118) Go ( T) . . (i?) i f ""/(i) 1 ^(tje-^fi - ^, at. The form of the solution needed is G (v) = A In v + /(v) where ?(v) is bounded and continuous in *'** l/2 . X r = TTo v tan V~ (1-|) 1/2 v = 0(v 2 ). Further 1 1/1 p 6 n "2 n ^ r £ / V n "2, , / vt m t dt = v / t — ic£vtl dt 4) t + v 4> t+i ni e / v -i -^ -i n "I = v n 2 J ft 1 "" 2 - t n "' 2 +, + (-) n - 1 t" 2 + ^^ ) ^n(vt)dt 1 ■-• 2(-J v / ^ dt + 0(vj H) t +1 = (-) n Jt v ' ^n v ...+ 0(v n+1 ^n v) + 0(v) •67- 00 since / —x — dt = 0. Hence J t + 1 o (119) 1 _ /, X . , 1 o e dt „ . 1/2, 2> -it Av' in v + 0(v ) ■- t o i7 W [7 Vv) l/2 2 ind the 0(v ' ) term is actually rt7o(0)v + 0(vin v).
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