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签到天数: 235 天 连续签到: 1 天 [LV.7]常住居民III 累计签到：237 天
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发表于 2011-9-19 10:05:40
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本帖最后由 wangdc 于 2011-9-19 10:13 编辑
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回复 5# zhaoxiangen 7 |" S8 l7 \: H4 }) C
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0 r: V0 e3 s! g% e 那个你勾选的话，在计算结果LIS文件中就会列出最严重的值，不勾选就不列出；但是否勾选对PL14没有影响，电路图上只要有电压电流测量点，PL14都会有画图的,看一下我的结果
2 B# h+ Q$ J e8 I1 S( s13 ) -------------------------------------------------------------------------------------------------------------------------------! ]+ z1 L8 G8 o0 W
Statistical distribution of peak energy for branch "MOV3LA" to "X0017A". Base energy for per unit output = 2.50000000E+07 " s) V8 c' U; h! C! d) J X
Interval energy energy in Frequency Cumulative Per cent $ r+ e. k* X0 F' ?" N- j. c
number in per unit physical units (density) frequency .GE. current value
7 A! @ J% ]: h: f) Q6 v 21 1.0500000 2.62500000E+07 0 0 100.000000
) a- {: {) P7 [ 22 1.1000000 2.75000000E+07 74 74 26.000000
9 T# n0 s! h; d+ B2 R 23 1.1500000 2.87500000E+07 19 93 7.000000
6 m8 u* B( u: B2 A3 ` 24 1.2000000 3.00000000E+07 7 100 .000000
/ G. g$ e6 @3 ?- G; E% D' PSummary of preceding table follows: Grouped data Ungrouped data
% i* y1 @$ t- Z- v8 [ Mean = 1.09150000E+00 1.09587467E+00 9 r/ z8 p& d, n) J$ i' s
Variance = 9.11868687E-04 6.64562405E-04 ' c( x6 l& B( x! p2 b
Standard deviation = 3.01971636E-02 2.57791079E-02
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14 ) -------------------------------------------------------------------------------------------------------------------------------8 A, E- u O9 {
Statistical distribution of peak energy for branch "MOV3LB" to "X0017B". Base energy for per unit output = 2.50000000E+07 ( C" X$ \# w( ?1 ]0 e( r" h2 F
Interval energy energy in Frequency Cumulative Per cent
5 P2 e) n* s2 l/ ~+ S7 \ number in per unit physical units (density) frequency .GE. current value + m( @, h9 F8 ]& z' z' g) I: |
21 1.0500000 2.62500000E+07 0 0 100.000000 0 B" g. d @" o* E- M' A c
22 1.1000000 2.75000000E+07 83 83 17.000000 ! Z$ r4 p( F. \( N6 }
23 1.1500000 2.87500000E+07 11 94 6.000000 & |6 `0 _/ l6 t; P
24 1.2000000 3.00000000E+07 6 100 .000000 0 g% C! p, ~8 Y2 u! L5 C: }
Summary of preceding table follows: Grouped data Ungrouped data
5 u7 U1 _1 Y1 Q$ f, b4 F Mean = 1.08650000E+00 1.08919324E+00
( Q9 a+ w K* A, g* c! ^ Variance = 7.50252525E-04 6.91174224E-04
4 c1 Y: Z4 V a1 U/ T* ?! K Standard deviation = 2.73907379E-02 2.62901925E-02
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15 ) -------------------------------------------------------------------------------------------------------------------------------
% B r2 c% B# z& zStatistical distribution of peak energy for branch "MOV3LC" to "X0017C". Base energy for per unit output = 2.50000000E+07 * J9 P% b2 x2 e0 K7 @
Interval energy energy in Frequency Cumulative Per cent
* A* A6 e0 a/ Y& _, _* i0 E+ ? number in per unit physical units (density) frequency .GE. current value
; m q% x* }0 R 48 2.4000000 6.00000000E+07 0 0 100.000000 ( \* a- s8 i' I7 N: B& G) n
49 2.4500000 6.12500000E+07 21 21 79.000000 4 M6 }5 N, E! s% ?- ]2 u1 E7 y
50 2.5000000 6.25000000E+07 47 68 32.000000
. j8 z$ K4 b- v9 _" t8 F" x9 o! U* { 51 2.5500000 6.37500000E+07 32 100 .000000
9 @' E3 ]' c8 R% BSummary of preceding table follows: Grouped data Ungrouped data ! W- k v$ Z/ U3 O/ ~$ r9 F
Mean = 2.48050000E+00 2.47717072E+00 $ R8 R! |! x1 y% T$ y& k
Variance = 1.30782828E-03 9.07356635E-04
& b: R9 |2 I0 h- E1 R- F Standard deviation = 3.61639086E-02 3.01223610E-02
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( T0 F+ t. @- c9 x0 z- W SUMMARY SUMMARY SUMMARY SUMMARY SUMMARY SUMMARY SUMMARY SUMMARY SUMMARY SUMMARY SUMMARY SUMMARY SUMMARY
! C4 z$ x: w/ r16 ) -------------------------------------------------------------------------------------------------------------------------------
% ~7 Y: [: D4 D0 {+ {" sA distribution of peak values among all output branches of the last data card that have the same base value follows. This
1 ]$ y+ V# N9 tstatistical distribution is for the maximum of the peaks at all of these output branches having base energy = 2.50000000E+07 1 ]" h( n+ T/ H8 R: l# u# Z
Interval energy energy in Frequency Cumulative Per cent o. J' C( X2 l; s( B5 r4 P8 L* I; @
number in per unit physical units (density) frequency .GE. current value 0 d9 |4 N+ _) e+ \ v: e4 I
48 2.4000000 6.00000000E+07 0 0 100.000000
8 Z# [( s3 t' Y$ C# v( U' O 49 2.4500000 6.12500000E+07 21 21 79.000000
4 h0 f; M5 s) v 50 2.5000000 6.25000000E+07 47 68 32.000000 8 U9 N$ R) k: }! Q/ Z, t0 j2 m
51 2.5500000 6.37500000E+07 32 100 .000000
& @6 Q; L3 X u) uSummary of preceding table follows: Grouped data Ungrouped data
( |8 L3 w6 Q& E7 x Mean = 2.48050000E+00 2.47717072E+00
2 _( r B& s( a4 Q M Variance = 1.30782828E-03 9.07356635E-04 0 Z& q, r' C/ L( e
Standard deviation = 3.61639086E-02 3.01223610E-02
, R/ \8 S0 f: {( r9 UBlank card ends statistical tabulation requests. |BLANK STATISTICS
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+ M4 [7 |$ A9 b$ E) p1 y8 F/ i Sample curve symbols : "A", "B", and "C"
# G4 \& [- x2 G# C3 p- H9 ]& v Theoretical curve symbol : "*"( D2 g1 \0 w$ V) n3 X1 H
Common curve symbol : "X" |
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