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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。 ; Q7 y& P @( P3 ` u/ r
function f=fh3(n)( X1 T; @ y) Z1 w% ?
g2=2^0.5-1;
0 i) G0 ]+ g- ~mu0=4*pi*10^(-7);' G* M6 S) K/ Q" x2 U4 H
gaf=7.8*10^3;
( o0 R* z5 }9 ~gacu=8.9*10^3;
4 h. S6 `( M0 x6 Z2 W4 V. Cgat=4.5*10^3;$ f7 \8 M6 z& \/ H8 X( e
gaa=2.7*10^3;( f" y9 r8 U4 x& C; ?' p
p=2.7*10^4;" {( Y6 T$ H% u7 H
q=3*10^(-5);8 l/ d: w7 |4 [6 {6 C: T. z
rou=0.7;
0 z! d4 a# C9 C" v8 ]6 bkb=0.9;
% c* y( ` g9 p%j0=8*10^6;/ [7 _4 q6 c5 \/ G, s8 b+ x {
j0=2*10^6;9 j: b0 q8 q! p) N7 T
kd=3;$ D. ]2 l# _2 t7 t7 y3 k
bm=2;
2 R( Q6 A6 D% J% b. p3 |' i: `9 R% W9 dht=2*10^(-3);
7 M; K& v" l' j: _ ?mu1=10^8/1.75;
" N" O# H7 s0 q6 O- ?* c+ Skt1=2*pi*gat*ht;
0 x! i1 W3 c& T8 {' Qkt2=2*pi*gaa*ht;! J5 Q* b/ e& B& Q6 K5 g1 i" a
kr=1.02;: L& x0 Y) `: C5 i0 T5 Y* G
ku=(mu0/2/bm)^0.5;8 |! j1 ]6 t+ s D0 O& Q2 R$ l+ Q
kx=mu0^0.5*kb*kd^2*j0;
5 G& v8 @+ ?: l7 s0 R9 c8 h2 Mkp=(2*q/pi/p)^0.5;" O4 Q: L- H" G1 h) l
kg=mu0^0.5*kb*kd*j0*g2;+ Y3 P+ d# n8 T# Z: z) U
rx=ku*kx-kp*kg;( Z0 F) M( `4 x* @! q0 v- [/ V
ra=ku*kg;/ c/ u! g' ~; j
rc=kp*kx;/ R4 c: U2 w8 Q
kj=mu0*kb*kd*j0;% w: O7 S) B: Z" X! Z4 X
kj=kj/kx;
" I6 D+ A4 D& M7 ~, j& N. a; Nh0=3*ht;
+ @9 I! n( y7 a5 f5 _' o8 }2 u2 w& rh1=2*(1+1/rou)*q/pi;
2 X5 s& [: u M! H3 |; Ah2=2*(kd-1);
# o% @! {, U, n4 j1 d% E+ ^h3=1/bm*(mu0*p*q/(2*pi))^0.5;, @" B! P& Z4 W' F8 i" `5 v
h4=1/kj*(mu0*p*q/(2*pi))^0.5;; ]# ]6 U# B2 D7 E: Z# x
hg=2*pi*kr*ht;
' W; F7 A& }1 G3 g5 ew1=gaf/bm*(pi*p*q*mu0/2)^0.5;
4 [4 o T1 A4 D& yw1p=8*g2*kd*w1;
. D% @; I% p4 L8 h1 W4 o# [w2=2*kd^2*gaf*(2*pi*mu0*p*q)^0.5/kj;: r9 M' n, |. }% n3 Z) x
w3p=2*g2*gacu/kj*(2*pi*p*q*mu0)^0.5;& d/ X; ^5 `( n, w, H& W7 Q
w3pp=hg*h3;
0 H3 ~" O/ Y8 E8 i" t# N- \7 rw3=w3p+w3pp;
2 S4 d5 A+ m8 `w4=2*kd*gacu/kj*(2*pi*p*q*mu0);. x, ?/ V' Y% }2 J
w5=5*pi*ht*gat;
4 L2 y- N" s" U$ D3 A ^# C9 Jw6=hg*h0*gat;4 n" @. {( f9 U
w7=hg*h1*gat;: H9 ?/ L$ t/ s& c6 s
w8=hg*h0*gat;
; G% b, Y, J2 o. P% @6 ]( [w9=hg*h4*gat;1 M6 y9 ?& v3 d5 P0 u0 ?
f=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);
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