3U]\NáDGURGHNFL *NRFLNU]\ZHM

=QDOH(üURGHNFL *NRFLNU]\ZHMSU]HGVWDZLRQHMQDSRQL*V]\PU\VXQNX

Rysunek 1

URGHN FL *NRFL NU]\ZHM PR*QD REOLF]Dü QD SRGVWDZLH Z]RUyZ L ]DVDG SRGREQ\FK GR

RERZL]XMF\FK SU]\ REOLF]HQLDFK URGND FL *NRFL ILJXU SáDVNLFK -HG\Q ]PLDQ MHVW

]DVWSLHQLHRGSRZLHGQLRQDZV]\VWNLFKPLHMVFDFKZHZ]RUDFKSU]HVWDZLRQ\FKZ

]DGDQLXSRODSRZLHU]FKQLILJXU$±GáXJRFLNU]\Z\FK/

3U]\NáDGRZRZ]RU\QDVWDW\F]QHPRPHQW\EH]ZáDGQRFLSR]PLDQLHZ\JOGDMQDVW SXMFR

Q

Q

/

/

6

=

/ \

6

∑

=

/ [

∑

(1)

[

L

&

\

L

&

L=

L=

Rysunek 2

:\]QDF]DMF URGHN FL *NRFL NU]\ZHM SU]HGVWDZLRQHM QD U\VXQNX SRG]LHORQR M QD JyUQ

SRáRZ L GROQ üZLDUWN RNU JX RUD] SU]\M WR Z FHOX XSURV]F]HQLD REOLF]H XNáDG RVL Z

URGNXSRáRZ\RNU JXSDWU]U\VXQHN

URGHNFL *NRFLSRáRZ\RNU JX&1OH*\QDMHMRVLV\PHWULLMHMZVSyáU] GQDSLRQRZD\C1 jest

WDNDVDPDMDNGODüZLDUWNL

: FHOX SR]QDQLD WHM ZVSyáU] GQHM SROLF]\P\ SRáR*HQLH URGND FL *NRFL üZLDUWNL RNU JX

XV\WXRZDQHMMDNQDU\VXQNX2ELHNU]\ZHE GPLDá\W VDPZVSyáU] GQSLRQRZURGND

FL *NRFL

Rysunek 3

5yZQDQLHQDVWDW\F]Q\PRPHQWEH]ZáDGQRFLGODNU]\ZHMZ\QRVL

/

6

= [ G/

∫

\

/

: FHOX SROLF]HQLD Z\VW SXMFHM Z SRZ\*V]\P Z]RU]H FDáNL ZSURZDG]LP\ ZVSyáU] GQ

ELHJXQRZ . SU]HGVWDZLRQ QD U\VXQNX :\UD*HQLD VWRMFH SRG FDáN PDM QDVW SXMFH

wzory:

[ = 5 FRVα

G/ = 5 Gα

/

3RSRGVWDZLHQLXLFKGRFDáNLVWDW\F]Q\PRPHQWEH]ZáDGQRFL 6 wynosi:

\

π

π

/

6

= [ G/ =

5 FRV α 5Gα = 5 VLQ α

= 5

∫

∫

\

/

'áXJRüüZLDUWNLRNU JXSROLF]\P\Z\NRU]\VWXMFZ]yUQDGáXJRüFDáHJRRNU JX

π5

/ =

π5 =

URGHNFL *NRFLüZLDUWNLRNU JXPD]DWHPZVSyáU] GQH

/

6

\

5

5

[

=

=

=

≅ 5

&

π

/

5

π

2

5

\

= [ =

≅ 5

&

&

π

2GPLHU]DMFZXNáDG]LHRVLSRZ\*V]HZDUWRFL]QDMGXMHP\URGHNFL *NRFLNU]\ZHM

Rysunek 4

=QDP\ MX* ]DWHP SRáR*HQLH URGND FL *NRFL üZLDUWNL RNU JX RUD] V]XNDQ ZVSyáU] GQ

URGNDFL *NRFLSyáRNU JX

5

\

=

&

π

'OD NU]\ZHM ] U\VXQNX PRPHQW\ VWDW\F]QH EH]ZáDGQRFL ]JRGQLH ] Z]RUDPL WHJR

]DGDQLDZ\QRV]

5

 5

/

6

= π5

+ π5 −

5

[

π





π  =

 

5 

/

6

= 5

π + 5

π − 5 −

(π )5

\





π 



 = − −

'áXJRüFDáHMNU]\ZHM/Z\QRVL

π5

/ =

π5 =

3RáR*HQLHURGNDFL *NRFLNU]\ZHMRNUHODP\SRGREQLHMDNGODILJXUSáDVNLFK

/

6

− π −

− π

\

(

)5

(

)

[

=

=

=

5 ≅ −

5

&

π

/

5

π

/

6

5

[

\

=

=

=

5 ≅

5

&

π

/

5

π

:\]QDF]DMFSRáR*HQLHURGNDFL *NRFLQDU\VXQNXRWU]\PXMHP\

3

Rysunek 5

URGHN FL *NRFL NU]\ZHM ]áR*RQHM ] GZyFK VNáDGRZ\FK NU]\Z\FK OH*\ ]DZV]H QD RGFLQNX

áF]F\P URGNL FL *NRFL W\FK NU]\Z\FK SDWU] U\VXQHN 3RGREQD ]DVDGD RERZL]XMH

UyZQLH*GODILJXULEU\á

4