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求fh3函数在[0.001 0.01]和[0.003 0.1]之间的最小值,估计在1至5之间。 , C' n" h. ]# g: C4 e
function f=fh3(n)7 t" p% a) ?3 |+ G; u
g2=2^0.5-1;3 S* V2 F4 F: j# I8 k5 k, N
mu0=4*pi*10^(-7);
8 T0 o8 G& [, Pgaf=7.8*10^3;
( k6 R5 r+ {+ l* f) j/ q& s3 Ygacu=8.9*10^3;
$ C8 Z( b# M0 v* o* q& g' L8 y1 _gat=4.5*10^3;5 D* k6 r' N; D" Q
gaa=2.7*10^3;
# ^- S: ~! M E7 ?" P( h1 ^p=2.7*10^4;
6 K# J& S+ g: Fq=3*10^(-5);
$ i" |* u0 l1 `, [4 D) d2 drou=0.7;! z. e# k& r; W) C. t4 ^! J+ X
kb=0.9;
1 | w" _8 _1 d% X1 p) e%j0=8*10^6;
- W5 K( U3 B9 m4 k; h Vj0=2*10^6;
" [# U! j, i1 Gkd=3;% \1 _- K( Z) U7 _, P; N
bm=2;
/ v7 @: k7 n4 q; t. Aht=2*10^(-3);
& b/ M% ^! }1 |$ Umu1=10^8/1.75;
6 M7 x) `" J) u$ p) _kt1=2*pi*gat*ht;
g+ Y3 e: d) Y( o1 S) w: `0 s5 x) a, vkt2=2*pi*gaa*ht;: j. _2 j. B* o9 y
kr=1.02;
# G2 R, ^6 p2 {2 l* ?/ ], eku=(mu0/2/bm)^0.5;
# b2 E7 J3 A, E, B4 mkx=mu0^0.5*kb*kd^2*j0;$ x5 o3 Y0 `, S! x: J' `- E* |# X
kp=(2*q/pi/p)^0.5;4 S9 m7 I$ ?0 g1 P6 I
kg=mu0^0.5*kb*kd*j0*g2;& C0 h2 W# z1 F1 m; Y
rx=ku*kx-kp*kg;
& v ?4 R) E6 T+ Y4 Ura=ku*kg;; k9 |( w4 f% S# H' N
rc=kp*kx;7 X0 y* ~* W6 X2 `7 f
kj=mu0*kb*kd*j0;/ g# U$ {( f+ W, E
kj=kj/kx;3 x* x* a" m# J5 S. ~ o' Q: ~0 z( K9 U
h0=3*ht;% E2 v; i+ H1 v2 z) J6 g$ z
h1=2*(1+1/rou)*q/pi;
0 L+ h. g0 D8 v+ E1 C+ ?% K Ah2=2*(kd-1);
: C1 k& ^0 L# Kh3=1/bm*(mu0*p*q/(2*pi))^0.5;
* c7 o' ~3 p. H, _+ uh4=1/kj*(mu0*p*q/(2*pi))^0.5;
$ s9 B! t6 h2 e e/ p Phg=2*pi*kr*ht;* e9 c: M7 I/ O. }" e3 Z* j Z
w1=gaf/bm*(pi*p*q*mu0/2)^0.5;" z; y, Z; b. Q! b& I. s F
w1p=8*g2*kd*w1;
; G% l! G, m6 P# Sw2=2*kd^2*gaf*(2*pi*mu0*p*q)^0.5/kj;
& Y; L# p/ `6 h) dw3p=2*g2*gacu/kj*(2*pi*p*q*mu0)^0.5;0 g4 q3 ]# ]/ s( V. ^' O7 }
w3pp=hg*h3;+ ~9 [# N' {4 Q9 d( k
w3=w3p+w3pp;
, X) a1 X1 Z' T; |, ~- b+ ww4=2*kd*gacu/kj*(2*pi*p*q*mu0);
: E0 ~2 S$ `1 Z+ {0 E2 s1 l3 Bw5=5*pi*ht*gat; A5 J( s) i B2 I# e4 d
w6=hg*h0*gat;
( Y7 y% U3 R7 ?& [5 |w7=hg*h1*gat;2 y7 }- F; K+ E; }9 }$ o+ N3 B
w8=hg*h0*gat;# |$ \9 X D. E% e& Y, n
w9=hg*h4*gat;
. {# o( o3 B1 t8 [0 r* D; q$ xf=w1p*r*x^0.5+w2*x^1.5/(g2*r-kd*x)+w3*r*x^(-0.5)+w4*x^0.5+w5*r^2+w6*r+w7/r+w8*r*x+w9*r*x^(-0.5)/(g2*r-kd*x);" b$ ]( L4 W0 a" l% h( d2 t. x
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