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4.2-bo'lim: Laplas o'zgarishlari

Biz oxirgi marta ko'rganimizdek bo'limi Laplas o'zgarishlarini to'g'ridan-to'g'ri hisoblash juda murakkab bo'lishi mumkin. Odatda biz faqat a dan foydalanamiz transformatsiyalar jadvali Laplas o'zgarishlarini aslida hisoblashda. Bu erda keltirilgan jadval hamma narsani o'z ichiga olgan jadval emas, lekin ko'p ishlatiladigan Laplas o'zgarishlarini va Laplas o'zgartirishlariga tegishli ko'p talab qilinadigan formulalarni o'z ichiga oladi.

Jadvaldan foydalanishni ko'rsatish uchun bir nechta misollar keltirishdan oldin, keling, tezda bir haqiqatni bilib olaylik.

Laplas transformatsiyasining chiziqliligi

\(f(t)\) va \(g(t)\) hisobga olingan holda,

\[\mathcal{L}\left\{ {af\left( t \right) + bg\left( t \right)} \right\} = a\,F\left( s \right) + b\,G\left( s \right)\]

har qanday \(a\) va \(b\) konstantalari uchun.

Boshqacha qilib aytadigan bo'lsak, biz Laplasni olishda doimiylar haqida qayg'urmaymiz va funktsiyalarning summalari yoki farqlari haqida qayg'urmaymiz. aylantiradi. Biz qilishimiz kerak bo'lgan yagona narsa - individual funktsiyalarning konvertatsiyasini olish, so'ngra har qanday konstantalarni qayta kiritish va natijalarni zaxiraga qo'shish yoki ayirish. Ushbu xususiyatga ega bo'lgan operatsiyalar (Laplas transformlari, limitlar, hosilalar, integrallar, va hokazo.) chiziqli deyiladi.

Shunday qilib, keling, bir nechta tezkor misollarni keltiraylik.

1-misol Berilgan funksiyalarning Laplas transformlarini toping.
  1. \(f\left( t \right) = 6{{\bf{e}}^{ - 5t}} + {{\bf{e}}^{3t}} + 5{t^3} - 9\)
  2. \(g\left( t \right) = 4\cos \left( {4t} \right) - 9\sin \left( {4t} \right) + 2\cos \left( {10t} \right)\)
  3. \(h\left( t \right) = 3\sinh \left( {2t} \right) + 3\sin \left( {2t} \right)\)
  4. \(g\left( t \right) = {{\bf{e}}^{3t}} + \cos \left( {6t} \right) - {{\bf{e}}^{3t}}\cos \left( {6t} \right)\)
Munozarani ko'rsatish

To'g'ri, bu erga borishdan boshqa ko'p ish yo'q stol, individual funktsiyalarni yuqoriga o'zgartiring, har qanday konstantalarni qayta kiriting va natijalarni qo'shing yoki ayirang.

Biz ushbu misollarni odatdagidan ko'ra biroz batafsilroq qilamiz, chunki biz jadvallardan birinchi marta foydalanmoqdamiz.


a \(f\left( t \right) = 6{{\bf{e}}^{ - 5t}} + {{\bf{e}}^{3t}} + 5{t^3} - 9\) Yechim
\[\begin{align*}F\left( s \right) & = 6\frac{1}{{s - \left( { - 5} \right)}} + \frac{1}{{s - 3}} + 5\frac{{3!}}{{{s^{3 + 1}}}} - 9\frac{1}{s}\\ & = \frac{6}{{s + 5}} + \frac{1}{{s - 3}} + \frac{{30}}{{{s^4}}} - \frac{9}{s}\end{align*}\]

b \(g\left( t \right) = 4\cos \left( {4t} \right) - 9\sin \left( {4t} \right) + 2\cos \left( {10t} \right)\) Yechim
\[\begin{align*}G\left( s \right) & = 4\frac{s}{{{s^2} + {{\left( 4 \right)}^2}}} - 9\frac{4}{{{s^2} + {{\left( 4 \right)}^2}}} + 2\frac{s}{{{s^2} + {{\left( {10} \right)}^2}}}\\ & = \frac{{4s}}{{{s^2} + 16}} - \frac{{36}}{{{s^2} + 16}} + \frac{{2s}}{{{s^2} + 100}}\end{align*}\]

c \(h\left( t \right) = 3\sinh \left( {2t} \right) + 3\sin \left( {2t} \right)\) Yechim
\[\begin{align*}H\left( s \right) & = 3\frac{2}{{{s^2} - {{\left( 2 \right)}^2}}} + 3\frac{2}{{{s^2} + {{\left( 2 \right)}^2}}}\\ & = \frac{6}{{{s^2} - 4}} + \frac{6}{{{s^2} + 4}}\end{align*}\]

d \(g\left( t \right) = {{\bf{e}}^{3t}} + \cos \left( {6t} \right) - {{\bf{e}}^{3t}}\cos \left( {6t} \right)\) Yechim
\[\begin{align*}G\left( s \right) & = \frac{1}{{s - 3}} + \frac{s}{{{s^2} + {{\left( 6 \right)}^2}}} - \frac{{s - 3}}{{{{\left( {s - 3} \right)}^2} + {{\left( 6 \right)}^2}}}\\ & = \frac{1}{{s - 3}} + \frac{s}{{{s^2} + 36}} - \frac{{s - 3}}{{{{\left( {s - 3} \right)}^2} + 36}}\end{align*}\]

"Oddiy" trig funktsiyasi va giperbolik funktsiyalar o'rtasidagi farqga e'tibor berganingizga ishonch hosil qiling. Ularning orasidagi yagona farq shundaki, "normal" trig funktsiyalari uchun "\( + {a^2}\)" giperbolik funksiyada "\( - {a^2}\)" ga aylanadi! Shoshilmaslik va e'tibor bermaslik va noto'g'ri formulani qo'lga kiritish juda oson. Agar siz giperbolik funktsiyalarning ta'rifini eslamasangiz, buning uchun eslatmalarga qarang stol.

Keling, yakuniy misollar to'plamini qilaylik.

2-misol Quyidagi funksiyalarning har birining konvertatsiyasini toping.
  1. \(f\left( t \right) = t\cosh \left( {3t} \right)\)
  2. \(h\left( t \right) = {t^2}\sin \left( {2t} \right)\)
  3. \(g\left( t \right) = {t^{\frac{3}{2}}}\)
  4. \(f\left( t \right) = {\left( {10t} \right)^{\frac{3}{2}}}\)
  5. \(f\left( t \right) = tg'\left( t \right)\)
a \(f\left( t \right) = t\cosh \left( {3t} \right)\) Yechim

Bu funksiya Laplas o'zgartirishlar jadvalida yo'q. Biroq, biz foydalanishimiz mumkin #30 uning o'zgarishini hisoblash uchun jadvalda. Agar oladigan bo'lsak, bu # 30 ga to'g'ri keladi n=1.

\[F\left( s \right) = \mathcal{L}\left\{ {tg\left( t \right)} \right\} = - G'\left( s \right),\hspace{0.25in}{\mbox{where }}g\left( t \right) = \cosh \left( {3t} \right)\]

Shunday qilib, bizda,

\[G\left( s \right) = \frac{s}{{{s^2} - 9}}\hspace{0.25in}G'\left( s \right) = - \frac{{{s^2} + 9}}{{{{\left( {{s^2} - 9} \right)}^2}}}\]

№30 dan foydalanib, bizda shunday bo'ladi:

\[F\left( s \right) = \frac{{{s^2} + 9}}{{{{\left( {{s^2} - 9} \right)}^2}}}\]

b \(h\left( t \right) = {t^2}\sin \left( {2t} \right)\) Yechim

Bu qism ham qo'llaniladi #30 jadvalda. Aslida, biz №30 dan ikkita usuldan birida foydalanishimiz mumkin. Biz uni \(n = 1\) bilan ishlatishimiz mumkin .

\[H\left( s \right) = \mathcal{L}\left\{ {tf\left( t \right)} \right\} = - F'\left( s \right),\hspace{0.25in}{\mbox{where }}f\left( t \right) = t\sin \left( {2t} \right)\]

Yoki biz uni \(n = 2\) bilan ishlatishimiz mumkin.

\[H\left( s \right) = \mathcal{L}\left\{ {{t^2}f\left( t \right)} \right\} = F''\left( s \right),\hspace{0.25in}{\mbox{where }}f\left( t \right) = \sin \left( {2t} \right)\]

Bitta hosilani bajarish kamroq ish bo'lgani uchun, keling, buni birinchi usulda qilaylik. Shunday qilib, foydalanish #9 bizda bor,

\[F\left( s \right) = \frac{{4s}}{{{{\left( {{s^2} + 4} \right)}^2}}}\hspace{0.25in}F'\left( s \right) = - \frac{{12{s^2} - 16}}{{{{\left( {{s^2} + 4} \right)}^3}}}\]

Transformatsiya shundan iboratki,

\[H\left( s \right) = \frac{{12{s^2} - 16}}{{{{\left( {{s^2} + 4} \right)}^3}}}\]

c \(g\left( t \right) = {t^{\frac{3}{2}}}\) Yechim

Ushbu qism ikkalasi yordamida amalga oshirilishi mumkin #6 (\(n = 2\) bilan) yoki #32 (bilan birga #5). Buning misolini ko'rishimiz uchun biz №32 dan foydalanamiz. №32 dan foydalanish uchun biz buni sezishimiz kerak

\[\int_{{\,0}}^{{\,t}}{{\sqrt v \,dv}} = \frac{2}{3}{t^{\frac{3}{2}}}\hspace{0.25in} \Rightarrow \hspace{0.25in}{t^{\frac{3}{2}}} = \frac{3}{2}\int_{{\,0}}^{{\,t}}{{\sqrt v \,dv}}\]

Endi №5 dan foydalanib,

\[f\left( t \right) = \sqrt t \hspace{0.25in}F\left( s \right) = \frac{{\sqrt \pi }}{{2{s^{\frac{3}{2}}}}}\]

quyidagini olamiz.

\[G\left( s \right) = \frac{3}{2}\left( {\frac{{\sqrt \pi }}{{2{s^{\frac{3}{2}}}}}} \right)\left( {\frac{1}{s}} \right) = \frac{{3\sqrt \pi }}{{4{s^{\frac{5}{2}}}}}\]

Agar biz №6 dan foydalansak, shu narsaga erishgan bo'lardik.


d \(f\left( t \right) = {\left( {10t} \right)^{\frac{3}{2}}}\) Yechim

Ushbu qism uchun biz foydalanamiz #24 oldingi qismdagi javob bilan birga. Agar bu eslatmani ko'rish uchun

\[g\left( t \right) = {t^{\frac{3}{2}}}\]

keyin

\[f\left( t \right) = g\left( {10t} \right)\]

Shuning uchun transformatsiya.

\[\begin{align*}F\left( s \right) & = \frac{1}{{10}}G\left( {\frac{s}{{10}}} \right)\\ & = \frac{1}{{10}}\left( {\frac{{3\sqrt \pi }}{{4{{\left( {\frac{s}{{10}}} \right)}^{\frac{5}{2}}}}}} \right)\\ & = {10^{\frac{3}{2}}}\frac{{3\sqrt \pi }}{{4{s^{\frac{5}{2}}}}}\end{align*}\]

e \(f\left( t \right) = tg'\left( t \right)\) Yechim

Ushbu yakuniy qism yana ishlatiladi #30 stoldan, shuningdek #35.

\[\begin{align*}\mathcal{L}\left\{ {tg'\left( t \right)} \right\} & = - \frac{d}{{ds}}\mathcal{L}\left\{ {g'} \right\}\\ & = - \frac{d}{{ds}}\left\{ {sG\left( s \right) - g\left( 0 \right)} \right\}\\ & = - \left( {G\left( s \right) + sG'\left( s \right) - 0} \right)\\ & = - G\left( s \right) - sG'\left( s \right)\end{align*}\]

Esda tutingki, \(g(0)\) shunchaki doimiydir, shuning uchun biz uni farqlaganimizda nolga erishamiz!

Ushbu misollar to'plami bizga ko'rsatganidek, biz matndagi ba'zi umumiy formulalardan foydalanishni unuta olmaymiz stol Jadvalda aniq ko'rsatilmagan funktsiyalar uchun yangi Laplas o'zgarishlarini olish uchun!