Biz foydalanishimiz mumkin bo'lgan birinchi eslatma #11 Buni qilish uchun stoldan tashqarida, bu bizning ishimizga qarshi yaxshi tekshiruv bo'ladi. Shuni ham yodda tutingki, bu erda konvolyutsiya integralidan foydalanish ushbu formulani jadvalimizdan olishning bir usuli hisoblanadi.

Endi biz bu yerda konvolyutsiya integralidan foydalanmoqchi bo‘lganimiz uchun uni shartlari teskari o‘zgarishlarni topish oson bo‘lgan mahsulot sifatida yozishimiz kerak bo‘ladi. Bunday holda buni qilish oson.

\[H\left( s \right) = \left( {\frac{1}{{{s^2} + {a^2}}}} \right)\left( {\frac{1}{{{s^2} + {a^2}}}} \right)\]

Shunday qilib, bu holatda bizda

\[F\left( s \right) = G\left( s \right) = \frac{1}{{{s^2} + {a^2}}}\hspace{0.25in} \Rightarrow \hspace{0.25in}\,\,\,\,\,\,f\left( t \right) = g\left( t \right) = \frac{1}{a}\sin \left( {at} \right)\]

Trig formulalaridan keng foydalanish bilan bir qatorda konvolyutsiya integralidan foydalanib, \(h(t)\),

\[\begin{align*}h\left( t \right) & = \left( {f * g} \right)\left( t \right)\\ & = \frac{1}{{{a^2}}}\int_{{\,0}}^{{\,t}}{{\sin \left( {at - a\tau } \right)\sin \left( {a\tau } \right)\,d\tau }}\\ & = \frac{1}{{{a^2}}}\int_{{\,0}}^{{\,t}}{{\left[ {\sin \left( {at} \right)\cos \left( {a\tau } \right) - \cos \left( {at} \right)\sin \left( {a\tau } \right)} \right]\sin \left( {a\tau } \right)\,d\tau }}\\ & = \frac{1}{{{a^2}}}\int_{{\,0}}^{{\,t}}{{\sin \left( {at} \right)\cos \left( {a\tau } \right)\sin \left( {a\tau } \right) - \cos \left( {at} \right){{\sin }^2}\left( {a\tau } \right)\,d\tau }}\\ & = \frac{1}{{2{a^2}}}\sin \left( {at} \right)\int_{{\,0}}^{{\,t}}{{\sin \left( {2a\tau } \right)d\tau }} - \frac{1}{{2{a^2}}}\cos \left( {at} \right)\int_{{\,0}}^{{\,t}}{{1 - \cos \left( {2a\tau } \right)\,d\tau }}\\ & = \frac{1}{{2{a^2}}}\sin \left( {at} \right)\left. {\left( { - \frac{1}{{2a}}\cos \left( {2a\tau } \right)} \right)} \right|_0^t - \frac{1}{{2{a^2}}}\cos \left( {at} \right)\left. {\left( {\tau - \frac{1}{{2a}}\sin \left( {2a\tau } \right)} \right)} \right|_0^t\\ & = \frac{1}{{4{a^3}}}\left( {\sin \left( {at} \right) - \sin \left( {at} \right)\cos \left( {2at} \right)} \right) - \frac{1}{{2{a^2}}}\cos \left( {at} \right)\left( {t - \frac{1}{{2a}}\sin \left( {2at} \right)} \right)\\ & = \frac{1}{{4{a^3}}}\sin \left( {at} \right) - \frac{1}{{2{a^2}}}t\cos \left( {at} \right) - \frac{1}{{4{a^3}}}\sin \left( {at} \right)\cos \left( {2at} \right) + \frac{1}{{4{a^3}}}\cos \left( {at} \right)\sin \left( {2at} \right)\\ & = \frac{1}{{4{a^3}}}\sin \left( {at} \right) - \frac{1}{{2{a^2}}}t\cos \left( {at} \right) - \frac{1}{{8{a^3}}}\left[ {\sin \left( {3at} \right) + \sin \left( { - at} \right)} \right] + \frac{1}{{8{a^3}}}\left[ {\sin \left( {3at} \right) - \sin \left( { - at} \right)} \right]\\ & = \frac{1}{{4{a^3}}}\sin \left( {at} \right) - \frac{1}{{2{a^2}}}t\cos \left( {at} \right) + \frac{1}{{8{a^3}}}\sin \left( {at} \right) + \frac{1}{{8{a^3}}}\sin \left( {at} \right)\\ & = \frac{1}{{2{a^3}}}\sin \left( {at} \right) - \frac{1}{{2{a^2}}}t\cos \left( {at} \right)\\ & = \frac{1}{{2{a^3}}}\left( {\sin \left( {at} \right) - at\cos \left( {at} \right)} \right)\end{align*}\]

Jadvaldagi №11 dan foydalangan holda biz aynan shu narsaga erishgan bo'lardik.

Shuni yodda tutingki, bu ko'pchilik eslay olmaydigan trig formulalaridan keng foydalanishni talab qiladi. Bundan tashqari, texnik jihatdan integral "oddiy" bo'lsa-da, aslida u juda uzoq va tartibsiz integral edi va nima uchun konvolyutsiya integrallari texnik jihatdan mumkin bo'lsa ham, har doim ham bajarilmasligini ko'rsatadi.

Bir fikrni aniq qilish uchun integral haqida oxirgi eslatma. To'rtinchi bosqichda biz \(\sin(a t)\) va \(\cos(a t)\) ni integrallardan ajratib oldik. Biz buni qilishimiz mumkin edi, bu holda, chunki integrallar \(\tau\) ga nisbatan va shuning uchun integrallarga kelsak, \(t\) ning har qanday funktsiyasi doimiy hisoblanadi. Biz, albatta, o'zgaruvchilarni integrallardan ajratib bo'lmaydi. Biz buni faqat o'zgaruvchilar hech qanday tarzda integratsiya o'zgaruvchisiga bog'liq bo'lmaganda qilishimiz mumkin.

Birinchidan, bu holatda majburlash funktsiyasi ko'rsatilmaganiga e'tibor bering. Ushbu bo'limdan oldin biz ushbu IVPga yechim topa olmagan bo'lardik. Konvolyutsiya integrallari yordamida biz ushbu turdagi IVPga yechim topishimiz mumkin. Yechim \(g(t)\) nuqtai nazaridan bo'ladi, ammo bu yechim bo'ladi.

Barcha shartlarning Laplas konvertatsiyasini oling va dastlabki shartlarni kiriting.

\[\begin{align*}4\left( {{s^2}Y\left( s \right) - sy\left( 0 \right) - y'\left( 0 \right)} \right) + Y\left( s \right) & = G\left( s \right)\\ \left( {4{s^2} + 1} \right)Y\left( s \right) - 12s + 28 & = G\left( s \right)\end{align*}\]

E'tibor bering, majburlash funktsiyasi uchun biz faqat \(G(s)\) ni o'zgartirish uchun yozishimiz mumkin edi. Endi \(Y(s)\) ni hal qiling.

\[\begin{align*}\left( {4{s^2} + 1} \right)Y\left( s \right) & = G\left( s \right) + 12s - 28\\ Y\left( s \right) & = \frac{{12s - 28}}{{4\left( {{s^2} + \frac{1}{4}} \right)}} + \frac{{G\left( s \right)}}{{4\left( {{s^2} + \frac{1}{4}} \right)}}\end{align*}\]

Biz teskari o'zgartirish jarayoniga tayyorgarlik ko'rishda maxrajdan 4 ni ajratib oldik. Teskari o'zgarishlarni olish uchun biz birinchi atamani ajratishimiz kerak va biz ikkinchi atamani ham biroz qayta yozamiz.

\[\begin{align*}Y\left( s \right) & = \frac{{12s - 28}}{{4\left( {{s^2} + \frac{1}{4}} \right)}} + \frac{{G\left( s \right)}}{{4\left( {{s^2} + \frac{1}{4}} \right)}}\\ & = \frac{{3s}}{{{s^2} + \frac{1}{4}}} - \frac{{7\frac{2}{2}}}{{{s^2} + \frac{1}{4}}} + \frac{1}{4}G\left( s \right)\frac{{\frac{2}{2}}}{{{s^2} + \frac{1}{4}}}\end{align*}\]

Endi birinchi ikkita atamani teskari aylantirish oson. Biz oxirgi hadda konvolyutsiya integralidan foydalanishimiz kerak. Biz foydalanadigan ikkita funksiya:

\[g\left( t \right)\hspace{0.25in}f\left( t \right) = 2\sin \left( {\frac{t}{2}} \right)\]

Konvolyutsiya integralidagi ikkita funktsiyadan birini siljitishimiz mumkin. Biz \(g(t)\) ni yechimimizda almashtiramiz. Teskari konvertatsiyani qabul qilish bizga beradi,

\[y\left( t \right) = 3\cos \left( {\frac{t}{2}} \right) - 14\sin \left( {\frac{t}{2}} \right) + \frac{1}{2}\int_{{\,0}}^{{\,t}}{{\sin \left( {\frac{\tau }{2}} \right)g\left( {t - \tau } \right)\,d\tau }}\]

Shunday qilib, biz \(g(t)\) haqida qaror qabul qilganimizdan so'ng, biz qilishimiz kerak bo'lgan yagona narsa integraldir va bizda yechim topiladi.