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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%97
    Емуляція 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)=Tnk=1Ak1pkz1,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%85
    Вищевказане 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)=(zz1zz0.819). Використовуючи оберненеz -перетворення, послідовність крокової відповіді отримується як:y(kT)=1(0.819)k,k=0,1,
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    НехайG(s)=Kτs+1,u(s)=1s; потім,y(s)=Ks(τs+1)=KsKττs+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)=KsKττs+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%B8
    x<-1:10 Приклад вибірки в R x<-1:7 sample(x,2) Ділянка гістограми в R x<-rnorm(100) hist(x)
  • 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)/06%3A_%D0%9A%D0%BE%D0%BD%D1%81%D1%82%D1%80%D1%83%D0%BA%D1%86%D1%96%D1%8F_%D0%BA%D0%BE%D0%BC%D0%BF%D0%B5%D0%BD%D1%81%D0%B0%D1%82%D0%BE%D1%80%D0%B0_%D0%B7_%D0%BC%D0%B5%D1%82%D0%BE%D0%B4%D0%B0%D0%BC%D0%B8_%D1%87%D0%B0%D1%81%D1%82%D0%BE%D1%82%D0%BD%D0%BE%D1%97_%D1%85%D0%B0%D1%80%D0%B0%D0%BA%D1%82%D0%B5%D1%80%D0%B8%D1%81%D1%82%D0%B8%D0%BA%D0%B8/6.01%3A_%D0%93%D1%80%D0%B0%D1%84%D1%96%D0%BA%D0%B8_%D1%87%D0%B0%D1%81%D1%82%D0%BE%D1%82%D0%BD%D0%BE%D1%97_%D1%85%D0%B0%D1%80%D0%B0%D0%BA%D1%82%D0%B5%D1%80%D0%B8%D1%81%D1%82%D0%B8%D0%BA%D0%B8
    \[\phi (\omega )=\sum _{i=1}^{m} \angle \left(1+\frac{j\omega }{z_{i} } \right)-n_{0} (90^{\circ } )-\sum _{i=1}^{n_{1} } \angle \left(1+\frac{j\omega }{p_{i} } \right)-\sum _{i=1}^{n_{1} } \angle \le...ϕ(ω)=mi=1(1+jωzi)n0(90)n1i=1(1+jωpi)n1i=1(1ω2ω2n,i+j2ζiωωn,i).
  • 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)/05%3A_%D0%94%D0%B8%D0%B7%D0%B0%D0%B9%D0%BD_%D1%81%D0%B8%D1%81%D1%82%D0%B5%D0%BC%D0%B8_%D1%83%D0%BF%D1%80%D0%B0%D0%B2%D0%BB%D1%96%D0%BD%D0%BD%D1%8F_%D0%B7_%D0%BA%D0%BE%D1%80%D0%B5%D0%BD%D0%B5%D0%B2%D0%B8%D0%BC_%D0%BB%D0%BE%D0%BA%D1%83%D1%81%D0%BE%D0%BC/5.06%3A_%D0%A0%D0%B5%D0%B0%D0%BB%D1%96%D0%B7%D0%B0%D1%86%D1%96%D1%8F_%D0%BA%D0%BE%D0%BD%D1%82%D1%80%D0%BE%D0%BB%D0%B5%D1%80%D0%B0
    Отже, ми можемо вибратиRiCi>RfCf для фазового відведення таRiCi<RfCf дизайну фазового відставання. \[G_{\rm PID} (s)=-\frac{Z_{\rm...Отже, ми можемо вибратиRiCi>RfCf для фазового відведення таRiCi<RfCf дизайну фазового відставання. GPID(s)=Zfser(s)Zipar(s)=1RiCf(RiCis+1)(RfCfs+1)s. kp=RfRi+CiCf,ki=RfCi,kd=1RiCf
  • 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/C1/LR/L][vci]+[01/L]Vs
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    Природна реакція має вигляд:yn(t)=(C1cosωt +C2sinωt )eσt 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 )eσt може бути альтернативно виражений як:yn(t)=Ceσtsin(ωt+ϕ)  деC=C21+C22 іtanϕ =C1C2. Імпульсна характеристикаG(s)=ss2+2s+2=s(s+1)2+12=s+11(s+1)2+12 задається як:g(t)=(costsint)etu(t)=2sin(t+135)u(t)=2cos(t+45)u(t)
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    \[\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]=[1005510][iaω]+[1000]Va,ω=[01][iaω]
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    Для 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=ζ1ζ2. ζПостійні лінії в z-площині отримують за допомогою:z=eTs=eσTe±jωT, деσ=ζ1ζ2ω іωT коливається від0 доπ.
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    НехайG(s)=ω2ns2+2ζωns+ω2n; потімG(jω)=ω2nω2nω2+j2ζωωn, де\({\...НехайG(s)=ω2ns2+2ζωns+ω2n; потімG(jω)=ω2nω2nω2+j2ζωωn, деωn іζ представляють власну частоту і коефіцієнт загасання.