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- https://ukrayinska.libretexts.org/%D0%86%D0%BD%D0%B6%D0%B5%D0%BD%D0%B5%D1%80%D0%BD%D0%B0/%D0%9F%D1%80%D0%BE%D0%BC%D0%B8%D1%81%D0%BB%D0%BE%D0%B2%D0%B5_%D1%82%D0%B0_%D1%81%D0%B8%D1%81%D1%82%D0%B5%D0%BC%D0%BD%D0%BE%D0%B3%D0%BE_%D0%BC%D0%B0%D1%88%D0%B8%D0%BD%D0%BE%D0%B1%D1%83%D0%B4%D1%83%D0%B2%D0%B0%D0%BD%D0%BD%D1%8F/%D0%9A%D0%BD%D0%B8%D0%B3%D0%B0%3A_%D0%92%D1%81%D1%82%D1%83%D0%BF_%D0%B4%D0%BE_%D1%81%D0%B8%D1%81%D1%82%D0%B5%D0%BC_%D1%83%D0%BF%D1%80%D0%B0%D0%B2%D0%BB%D1%96%D0%BD%D0%BD%D1%8F_(Iqbal)/07%3A_%D0%9F%D1%80%D0%BE%D0%B5%D0%BA%D1%82%D1%83%D0%B2%D0%B0%D0%BD%D0%BD%D1%8F_%D1%81%D0%B8%D1%81%D1%82%D0%B5%D0%BC_%D0%B2%D0%B8%D0%B1%D1%96%D1%80%D0%BA%D0%BE%D0%B2%D0%B8%D1%85_%D0%B4%D0%B0%D0%BD%D0%B8%D1%85/7.05%3A_%D0%94%D0%B8%D0%B7%D0%B0%D0%B9%D0%BD_%D1%86%D0%B8%D1%84%D1%80%D0%BE%D0%B2%D0%BE%D0%B3%D0%BE_%D0%BA%D0%BE%D0%BD%D1%82%D1%80%D0%BE%D0%BB%D0%B5%D1%80%D0%B0_%D0%B7%D0%B0_%D0%B4%D0%BE%D0%BF%D0%BE%D0%BC%D0%BE%D0%B3%D0%BE%D1%8E_%D0%B5%D0%BC%D1%83%D0%BB%D1%8F%D1%86%D1%96%D1%97Емуляція контролера спрямована на отримання приблизного цифрового контролераK(z), реакція якого збігається з реакцією аналогового контролера,K(s) як вимірюється вибраною ...Емуляція контролера спрямована на отримання приблизного цифрового контролераK(z), реакція якого збігається з реакцією аналогового контролера,K(s) як вимірюється вибраною метрикою. Потім отримують відповідний цифровий контролер у вигляді:K(z)=T∑nk=1Ak1−pkz−1,pk=eskT, тобто полюси контролера відображаються на відповідні місця вz -площині.
- https://ukrayinska.libretexts.org/%D0%86%D0%BD%D0%B6%D0%B5%D0%BD%D0%B5%D1%80%D0%BD%D0%B0/%D0%9F%D1%80%D0%BE%D0%BC%D0%B8%D1%81%D0%BB%D0%BE%D0%B2%D0%B5_%D1%82%D0%B0_%D1%81%D0%B8%D1%81%D1%82%D0%B5%D0%BC%D0%BD%D0%BE%D0%B3%D0%BE_%D0%BC%D0%B0%D1%88%D0%B8%D0%BD%D0%BE%D0%B1%D1%83%D0%B4%D1%83%D0%B2%D0%B0%D0%BD%D0%BD%D1%8F/%D0%9A%D0%BD%D0%B8%D0%B3%D0%B0%3A_%D0%92%D1%81%D1%82%D1%83%D0%BF_%D0%B4%D0%BE_%D1%81%D0%B8%D1%81%D1%82%D0%B5%D0%BC_%D1%83%D0%BF%D1%80%D0%B0%D0%B2%D0%BB%D1%96%D0%BD%D0%BD%D1%8F_(Iqbal)/07%3A_%D0%9F%D1%80%D0%BE%D0%B5%D0%BA%D1%82%D1%83%D0%B2%D0%B0%D0%BD%D0%BD%D1%8F_%D1%81%D0%B8%D1%81%D1%82%D0%B5%D0%BC_%D0%B2%D0%B8%D0%B1%D1%96%D1%80%D0%BA%D0%BE%D0%B2%D0%B8%D1%85_%D0%B4%D0%B0%D0%BD%D0%B8%D1%85/7.02%3A_%D0%92%D1%96%D0%B4%D0%BF%D0%BE%D0%B2%D1%96%D0%B4%D1%8C_%D1%81%D0%B8%D1%81%D1%82%D0%B5%D0%BC%D0%B8_%D0%B2%D0%B8%D0%B1%D1%96%D1%80%D0%BA%D0%BE%D0%B2%D0%B8%D1%85_%D0%B4%D0%B0%D0%BD%D0%B8%D1%85Вищевказане правило може бути запрограмовано на комп'ютері для вирішення відповіді системи вибіркових даних на вхідну послідовністьu(k), описану як: як\(u\left(k\right)={\left.u\left(t\...Вищевказане правило може бути запрограмовано на комп'ютері для вирішення відповіді системи вибіркових даних на вхідну послідовністьu(k), описану як: якu(k)=u(t)|t=kT. Ми використовуємо PFEy(z)/z для отримання:y(z)=(zz−1−zz−0.819). Використовуючи оберненеz -перетворення, послідовність крокової відповіді отримується як:y(kT)=1−(0.819)k,k=0,1,…
- https://ukrayinska.libretexts.org/%D0%86%D0%BD%D0%B6%D0%B5%D0%BD%D0%B5%D1%80%D0%BD%D0%B0/%D0%9F%D1%80%D0%BE%D0%BC%D0%B8%D1%81%D0%BB%D0%BE%D0%B2%D0%B5_%D1%82%D0%B0_%D1%81%D0%B8%D1%81%D1%82%D0%B5%D0%BC%D0%BD%D0%BE%D0%B3%D0%BE_%D0%BC%D0%B0%D1%88%D0%B8%D0%BD%D0%BE%D0%B1%D1%83%D0%B4%D1%83%D0%B2%D0%B0%D0%BD%D0%BD%D1%8F/%D0%9A%D0%BD%D0%B8%D0%B3%D0%B0%3A_%D0%92%D1%81%D1%82%D1%83%D0%BF_%D0%B4%D0%BE_%D1%81%D0%B8%D1%81%D1%82%D0%B5%D0%BC_%D1%83%D0%BF%D1%80%D0%B0%D0%B2%D0%BB%D1%96%D0%BD%D0%BD%D1%8F_(Iqbal)/02%3A_%D0%9C%D0%BE%D0%B4%D0%B5%D0%BB%D1%96_%D1%84%D1%83%D0%BD%D0%BA%D1%86%D1%96%D0%B9_%D0%BF%D0%B5%D1%80%D0%B5%D0%B4%D0%B0%D1%87%D1%96/2.04%3A_%D0%92%D1%96%D0%B4%D0%BF%D0%BE%D0%B2%D1%96%D0%B4%D1%8C_%D0%BD%D0%B0_%D0%BA%D1%80%D0%BE%D0%BAНехайG(s)=Kτs+1,u(s)=1s; потім,y(s)=Ks(τs+1)=Ks−Kττs+1. Нехай\(G(s)=\frac{K}{s\left(\tau s+1\right)}, \;\;u(s)=\f...НехайG(s)=Kτs+1,u(s)=1s; потім,y(s)=Ks(τs+1)=Ks−Kττs+1. НехайG(s)=Ks(τs+1),u(s)=1s; потім,y(s)=Ks2(τs+1). НехайG(s)=K(τ1s+1)(τ2s+1)τ1>τ2; потім, крок відповідь обчислюється як:y(s)=Ks(τ1s+1)(τ2s+1)=As+Bτ1s+1+Cτ2s+1.
- https://ukrayinska.libretexts.org/%D0%86%D0%BD%D0%B6%D0%B5%D0%BD%D0%B5%D1%80%D0%BD%D0%B0/%D0%9F%D1%80%D0%BE%D0%BC%D0%B8%D1%81%D0%BB%D0%BE%D0%B2%D0%B5_%D1%82%D0%B0_%D1%81%D0%B8%D1%81%D1%82%D0%B5%D0%BC%D0%BD%D0%BE%D0%B3%D0%BE_%D0%BC%D0%B0%D1%88%D0%B8%D0%BD%D0%BE%D0%B1%D1%83%D0%B4%D1%83%D0%B2%D0%B0%D0%BD%D0%BD%D1%8F/%D0%9A%D0%BD%D0%B8%D0%B3%D0%B0%3A_%D0%92%D1%81%D1%82%D1%83%D0%BF_%D0%B4%D0%BE_%D1%81%D0%B8%D1%81%D1%82%D0%B5%D0%BC_%D1%83%D0%BF%D1%80%D0%B0%D0%B2%D0%BB%D1%96%D0%BD%D0%BD%D1%8F_(Iqbal)/01%3A_%D0%9C%D0%B0%D1%82%D0%B5%D0%BC%D0%B0%D1%82%D0%B8%D1%87%D0%BD%D1%96_%D0%BC%D0%BE%D0%B4%D0%B5%D0%BB%D1%96_%D1%84%D1%96%D0%B7%D0%B8%D1%87%D0%BD%D0%B8%D1%85_%D1%81%D0%B8%D1%81%D1%82%D0%B5%D0%BC/1.09%3A_R_%D0%9F%D1%80%D0%B8%D0%BA%D0%BB%D0%B0%D0%B4%D0%B8x<-1:10 Приклад вибірки в R x<-1:7 sample(x,2) Ділянка гістограми в R x<-rnorm(100) hist(x)
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- https://ukrayinska.libretexts.org/%D0%86%D0%BD%D0%B6%D0%B5%D0%BD%D0%B5%D1%80%D0%BD%D0%B0/%D0%9F%D1%80%D0%BE%D0%BC%D0%B8%D1%81%D0%BB%D0%BE%D0%B2%D0%B5_%D1%82%D0%B0_%D1%81%D0%B8%D1%81%D1%82%D0%B5%D0%BC%D0%BD%D0%BE%D0%B3%D0%BE_%D0%BC%D0%B0%D1%88%D0%B8%D0%BD%D0%BE%D0%B1%D1%83%D0%B4%D1%83%D0%B2%D0%B0%D0%BD%D0%BD%D1%8F/%D0%9A%D0%BD%D0%B8%D0%B3%D0%B0%3A_%D0%92%D1%81%D1%82%D1%83%D0%BF_%D0%B4%D0%BE_%D1%81%D0%B8%D1%81%D1%82%D0%B5%D0%BC_%D1%83%D0%BF%D1%80%D0%B0%D0%B2%D0%BB%D1%96%D0%BD%D0%BD%D1%8F_(Iqbal)/01%3A_%D0%9C%D0%B0%D1%82%D0%B5%D0%BC%D0%B0%D1%82%D0%B8%D1%87%D0%BD%D1%96_%D0%BC%D0%BE%D0%B4%D0%B5%D0%BB%D1%96_%D1%84%D1%96%D0%B7%D0%B8%D1%87%D0%BD%D0%B8%D1%85_%D1%81%D0%B8%D1%81%D1%82%D0%B5%D0%BC/1.06%3A_%D0%9C%D0%BE%D0%B4%D0%B5%D0%BB%D1%96_%D0%B7%D0%BC%D1%96%D0%BD%D0%BD%D0%B8%D1%85_%D1%81%D1%82%D0%B0%D0%BD%D1%83\[\frac d{ dt} \left[\begin{array}{c} {v_ c } \\ {i} \end{array}\right]=\left[\begin{array}{cc} {0} & {1/C} \\ {-1/L} & {-R/L} \end{array}\right]\left[\begin{array}{c} {v_ c } \\ {i} \end{array}\right...ddt[vci]=[01/C−1/L−R/L][vci]+[01/L]Vs
- https://ukrayinska.libretexts.org/%D0%86%D0%BD%D0%B6%D0%B5%D0%BD%D0%B5%D1%80%D0%BD%D0%B0/%D0%9F%D1%80%D0%BE%D0%BC%D0%B8%D1%81%D0%BB%D0%BE%D0%B2%D0%B5_%D1%82%D0%B0_%D1%81%D0%B8%D1%81%D1%82%D0%B5%D0%BC%D0%BD%D0%BE%D0%B3%D0%BE_%D0%BC%D0%B0%D1%88%D0%B8%D0%BD%D0%BE%D0%B1%D1%83%D0%B4%D1%83%D0%B2%D0%B0%D0%BD%D0%BD%D1%8F/%D0%9A%D0%BD%D0%B8%D0%B3%D0%B0%3A_%D0%92%D1%81%D1%82%D1%83%D0%BF_%D0%B4%D0%BE_%D1%81%D0%B8%D1%81%D1%82%D0%B5%D0%BC_%D1%83%D0%BF%D1%80%D0%B0%D0%B2%D0%BB%D1%96%D0%BD%D0%BD%D1%8F_(Iqbal)/02%3A_%D0%9C%D0%BE%D0%B4%D0%B5%D0%BB%D1%96_%D1%84%D1%83%D0%BD%D0%BA%D1%86%D1%96%D0%B9_%D0%BF%D0%B5%D1%80%D0%B5%D0%B4%D0%B0%D1%87%D1%96/2.02%3A_%D0%9F%D1%80%D0%B8%D1%80%D0%BE%D0%B4%D0%BD%D0%B0_%D1%80%D0%B5%D0%B0%D0%BA%D1%86%D1%96%D1%8F_%D1%81%D0%B8%D1%81%D1%82%D0%B5%D0%BC%D0%B8Природна реакція має вигляд:yn(t)=(C1cosωt +C2sinωt )e−σt може бути альтернативно виражений як:\[y_n\left(t\right)=Ce^{-\sigma t}{sin \left(\omega t+\p...Природна реакція має вигляд:yn(t)=(C1cosωt +C2sinωt )e−σt може бути альтернативно виражений як:yn(t)=Ce−σtsin(ωt+ϕ) деC=√C21+C22 іtanϕ =C1C2. Імпульсна характеристикаG(s)=ss2+2s+2=s(s+1)2+12=s+1−1(s+1)2+12 задається як:g(t)=(cost−sint)e−tu(t)=√2sin(t+135∘)u(t)=√2cos(t+45∘)u(t)
- https://ukrayinska.libretexts.org/%D0%86%D0%BD%D0%B6%D0%B5%D0%BD%D0%B5%D1%80%D0%BD%D0%B0/%D0%9F%D1%80%D0%BE%D0%BC%D0%B8%D1%81%D0%BB%D0%BE%D0%B2%D0%B5_%D1%82%D0%B0_%D1%81%D0%B8%D1%81%D1%82%D0%B5%D0%BC%D0%BD%D0%BE%D0%B3%D0%BE_%D0%BC%D0%B0%D1%88%D0%B8%D0%BD%D0%BE%D0%B1%D1%83%D0%B4%D1%83%D0%B2%D0%B0%D0%BD%D0%BD%D1%8F/%D0%9A%D0%BD%D0%B8%D0%B3%D0%B0%3A_%D0%92%D1%81%D1%82%D1%83%D0%BF_%D0%B4%D0%BE_%D1%81%D0%B8%D1%81%D1%82%D0%B5%D0%BC_%D1%83%D0%BF%D1%80%D0%B0%D0%B2%D0%BB%D1%96%D0%BD%D0%BD%D1%8F_(Iqbal)/08%3A_%D0%9C%D0%BE%D0%B4%D0%B5%D0%BB%D1%96_%D0%B7%D0%BC%D1%96%D0%BD%D0%BD%D0%B8%D1%85_%D1%81%D1%82%D0%B0%D0%BD%D1%83/8.04%3A_%D0%9B%D1%96%D0%BD%D1%96%D0%B9%D0%BD%D0%B5_%D0%BF%D0%B5%D1%80%D0%B5%D1%82%D0%B2%D0%BE%D1%80%D0%B5%D0%BD%D0%BD%D1%8F_%D0%B7%D0%BC%D1%96%D0%BD%D0%BD%D0%B8%D1%85_%D1%81%D1%82%D0%B0%D0%BD%D1%83\[\frac{\rm d}{\rm dt} \left[\begin{array}{c} {i_a } \\ {\omega } \end{array}\right]=\left[\begin{array}{cc} {-100} & {-5} \\ {5} & {-10} \end{array}\right] \left[\begin{array}{c} {i_a} \\ {\omega } \...ddt[iaω]=[−100−55−10][iaω]+[1000]Va,ω=[01][iaω]
- https://ukrayinska.libretexts.org/%D0%86%D0%BD%D0%B6%D0%B5%D0%BD%D0%B5%D1%80%D0%BD%D0%B0/%D0%9F%D1%80%D0%BE%D0%BC%D0%B8%D1%81%D0%BB%D0%BE%D0%B2%D0%B5_%D1%82%D0%B0_%D1%81%D0%B8%D1%81%D1%82%D0%B5%D0%BC%D0%BD%D0%BE%D0%B3%D0%BE_%D0%BC%D0%B0%D1%88%D0%B8%D0%BD%D0%BE%D0%B1%D1%83%D0%B4%D1%83%D0%B2%D0%B0%D0%BD%D0%BD%D1%8F/%D0%9A%D0%BD%D0%B8%D0%B3%D0%B0%3A_%D0%92%D1%81%D1%82%D1%83%D0%BF_%D0%B4%D0%BE_%D1%81%D0%B8%D1%81%D1%82%D0%B5%D0%BC_%D1%83%D0%BF%D1%80%D0%B0%D0%B2%D0%BB%D1%96%D0%BD%D0%BD%D1%8F_(Iqbal)/07%3A_%D0%9F%D1%80%D0%BE%D0%B5%D0%BA%D1%82%D1%83%D0%B2%D0%B0%D0%BD%D0%BD%D1%8F_%D1%81%D0%B8%D1%81%D1%82%D0%B5%D0%BC_%D0%B2%D0%B8%D0%B1%D1%96%D1%80%D0%BA%D0%BE%D0%B2%D0%B8%D1%85_%D0%B4%D0%B0%D0%BD%D0%B8%D1%85/7.06%3A_%D0%94%D0%B8%D0%B7%D0%B0%D0%B9%D0%BD_%D0%BA%D0%BE%D1%80%D0%B5%D0%BD%D0%B5%D0%B2%D0%BE%D0%B3%D0%BE_%D0%BB%D0%BE%D0%BA%D1%83%D1%81%D1%83_%D1%86%D0%B8%D1%84%D1%80%D0%BE%D0%B2%D0%B8%D1%85_%D0%BA%D0%BE%D0%BD%D1%82%D1%80%D0%BE%D0%BB%D0%B5%D1%80%D1%96%D0%B2Для прототипу передавальної функції другого порядку:T(s)=ω2ns2+2ζωns+ω2n константніζ рядки визначаються за допомогою:\(\frac{\sigma }{...Для прототипу передавальної функції другого порядку:T(s)=ω2ns2+2ζωns+ω2n константніζ рядки визначаються за допомогою:σω=∓ζ√1−ζ2. ζПостійні лінії в z-площині отримують за допомогою:z=eTs=eσTe±jωT, деσ=∓ζ√1−ζ2ω іωT коливається від0 доπ.
- https://ukrayinska.libretexts.org/%D0%86%D0%BD%D0%B6%D0%B5%D0%BD%D0%B5%D1%80%D0%BD%D0%B0/%D0%9F%D1%80%D0%BE%D0%BC%D0%B8%D1%81%D0%BB%D0%BE%D0%B2%D0%B5_%D1%82%D0%B0_%D1%81%D0%B8%D1%81%D1%82%D0%B5%D0%BC%D0%BD%D0%BE%D0%B3%D0%BE_%D0%BC%D0%B0%D1%88%D0%B8%D0%BD%D0%BE%D0%B1%D1%83%D0%B4%D1%83%D0%B2%D0%B0%D0%BD%D0%BD%D1%8F/%D0%9A%D0%BD%D0%B8%D0%B3%D0%B0%3A_%D0%92%D1%81%D1%82%D1%83%D0%BF_%D0%B4%D0%BE_%D1%81%D0%B8%D1%81%D1%82%D0%B5%D0%BC_%D1%83%D0%BF%D1%80%D0%B0%D0%B2%D0%BB%D1%96%D0%BD%D0%BD%D1%8F_(Iqbal)/02%3A_%D0%9C%D0%BE%D0%B4%D0%B5%D0%BB%D1%96_%D1%84%D1%83%D0%BD%D0%BA%D1%86%D1%96%D0%B9_%D0%BF%D0%B5%D1%80%D0%B5%D0%B4%D0%B0%D1%87%D1%96/2.05%3A_%D0%A1%D0%B8%D0%BD%D1%83%D1%81%D0%BE%D1%97%D0%B4%D0%B0%D0%BB%D1%8C%D0%BD%D0%B0_%D1%80%D0%B5%D0%B0%D0%BA%D1%86%D1%96%D1%8F_%D1%81%D0%B8%D1%81%D1%82%D0%B5%D0%BC%D0%B8НехайG(s)=ω2ns2+2ζωns+ω2n; потімG(jω)=ω2nω2n−ω2+j2ζωωn, де\({\...НехайG(s)=ω2ns2+2ζωns+ω2n; потімG(jω)=ω2nω2n−ω2+j2ζωωn, деωn іζ представляють власну частоту і коефіцієнт загасання.