Wakar Taima: Me ke shi? (Tushen da Yadda ake Gano a MATLAB)

فیلڈ: Karkashin Kuliya da Dukkana

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Madda da Aikin Da Duk

Aikin da dukan wani na'ura mai yawa shine lokacin da muhimmanci ya shiga da zama da daraja ta bayanar da ake ba. Ana sanya da Ts. Aikin da duka tana da kusa da gaba-gaban da ake bukata da lokaci ga jirgin da ake bukata. Tana da lokacin da ake bukatar da aikin da duk zuwa darajan da ake bukata da tsarin da ake bukata.

Tsari da ake bukata shine yanayin da ake bukatar da aikin da duk za su. Amsa, tsarin da ake bukata suna 2% ko 5%.

Aikin da dukan na'urori mai yawa na'ura ta biyu ana nuna haka a cikin hoto na wannan.

Aikin da Duk

Tsarin Aikin da Duk

Aikin da duka tana da kusa da adadin sauti da amsa da ake bukata. Tsarin da ake bukata wa aikin da duka shine;

$T_S = \frac{ln(tolerance \, fraction)}{damping \, ratio \times Natural \, frequency}$

Amsa da ake bukata na'ura ta biyu an nuna haka;

$C(t) = 1 - \left( \frac{e^{-\zeta \omega_n t}}{\sqrt{1-\zeta^2}} \right) sin(\omega_d t + \theta)$

Zan iya kungiyar da biyu;

$exponential \, component = \left( \frac{e^{-\zeta \omega_n t}}{\sqrt{1-\zeta^2}} \right)$

$sinusoidal \, component = sin(\omega_d t + \theta)$

Don samun lokacin da yake, muna bukata da zan iya kungiya masu hankali da ba wani ba, domin ya kawo wani babban mutum da ke kan wani abu. Kuma yadda ake kawo cikin abubuwa ta shi ne ita ce.

$Tolerance \, fraction = \frac{e^{-\zeta \omega_n t}}{\sqrt{1-\zeta^2}}$

$t = T_S$

$Tolerance \, fraction \times \sqrt{1-\zeta^2} = e^{-\zeta \omega_n T_S}$

$ln \left( Tolerance \, fraction \times \sqrt{1-\zeta^2} \right) = -\zeta \omega_n T_S$

$T_S = - \frac{ ln \left( Tolerance \, fraction \times \sqrt{1-\zeta^2} \right)}{\zeta \omega_n}$

Yadda A Kula Waktar Da Zama Cikin Tsarin

Don kula waktar da zama, za mu iya duba sistem na farko da tashin yadda ake yi.

$\frac{C(s)}{R(s)} = \frac{\frac{1}{T}}{s+\frac{1}{T}}}$

Don tashin yadda ake yi,

$R(s) = \frac{1}{s}$

Saboda haka,

$C(s) = \frac{\frac{1}{T}}{s(s+\frac{1}{T})}}$

$C(s) = \frac{A_1}{s} + \frac{A_2}{s+\frac{1}{T}}$

A nan, kula a taka lalacewar A₁ da A₂.

$\frac{\frac{1}{T}}{s(s+\frac{1}{T})}} = \frac{A_1(s+\frac{1}{T}) + A_2s}{s(s+\frac{1}{T})}$

$\frac{1}{T} = A_1 (s+\frac{1}{T}) + A_2 s$

Tsanu s = 0;

$\frac{1}{T} = A_1( 0 + \frac{1}{T}) + A_2 (0)$

$\frac{1}{T} = A_1 \frac{1}{T}$

$A_1 = 1$

Tsanu s = -1/T;

$\frac{1}{T} = A_1 (0) + A_2 (\frac{-1}{T})$

$\frac{1}{T} = -A_2 \frac{1}{T}$

$A_2 = -1$

$C(s) = \frac{1}{s} - \frac{1}{s+\frac{1}{T}}$

$C(t) = L^{-1} C(s)$

$C(t) = 1 - e^{\frac{-t}{T}}$

$e^{\frac{-t}{T}} = 1 - C(t)$

Don samun ƙarfi 2%, 1-C(t) = 0.02;

$e^{\frac{-t_s}{T}} = 0.02$

$\frac{-t_s}{T} = ln(0.02)$

$\frac{-t_s}{T} = -3.9$

$t_s = 3.9T$

$t_s \approx 4T$

Tana da taka wani abu na tsawon kungiyar da ke nuna lokacin da za a yi aiki a cikin kungiyar da ke da rike mai yawa.

Don kungiyar na biyu, zan iya duba wannan tana da:

$C(t) = 1 - \frac{e^{- \zeta \omega_n t}}{\sqrt{1-\zeta^2}} sin(\omega_d t+\phi)$

A cikin wannan tana, tana da muhimmanci don samun balobi da ya shafi lokacin da za a yi aiki.

$C(t) = 1 - \frac{e^{- \zeta \omega_n t}}{\sqrt{1-\zeta^2}}$

$\frac{e^{- \zeta \omega_n t}}{\sqrt{1-\zeta^2}} = 1 - C(t)$

A nan, zaɓe ƙoƙari na ƙasa ɗaya. Saboda haka, 1 – C(t) = 0.02;

$\frac{e^{- \zeta \omega_n t}}{\sqrt{1-\zeta^2}} = 0.02$

Yadda ƙarfin ƙasa (ξ) ke shafi ne tana da yawa daga wata zuwa wata biyu. A nan, muna ƙara da ƙasar biyu mai ƙasa ta ƙarin. Kuma ƙarfin ƙasa (ξ) tana cikin 0 da 1.

Saboda haka, karamin ƙarfin ƙasa (ξ) tana cikin 0 da 1, karamin tsawon ƙarfin ƙasa (ξ) tana da yawan ƙarin ɗaya. Don in iya yi amfani, za mu iya fi ƙarfin ƙasa (ξ) ba ɗaya.

$e^{- \zeta \omega_n t_s} = 0.02$

$- \zeta \omega_n t_s = ln(0.02)$

$- \zeta \omega_n t_s = -3.9$

$t_s = \frac{3.9}{\zeta \omega_n}$

$t_s \approx \frac{4}{\zeta \omega_n}$

An harsuna zai iya amfani da shi kawai don bandin daji 2% da kuma na'urar da ta'addan tsawon biyu.

Duk da cewa, don bandin daji 5%; 1 – C(t) = 0.05;

$e^(- \zeta \omega_n t_s) = 0.05$

$- \zeta \omega_n t_s = ln(0.05)$

$- \zeta \omega_n t_s = -3$

$t_s \approx \frac{3}{\zeta \omega_n}$

A nan hanyar da system ta biyu, a baya da samun lokacin da yake cikin kadan, muna iya kula tsarin damping.

Wakar Zama na Tsawon Locus da Root

Zama na tsawo zai iya tafi kwa haka na harkokin locus da root. Zama na tsawo ta yi amfani da tsariyar damping da frequency mai kyau.

Wasu daban-daban za su iya samun da harkokin locus da root. Sannan muna iya samun zama na tsawo.

Yana da wani misali.

$G(s) = \frac{K}{(s+1)(s+2)(s+3)}$

Da Overshoot = 20%

$damping \, ratio \, \zeta = \frac{-ln(\%OS/100)}{\sqrt{\pi^2 + ln^2(\%OS/100)}}$

$\zeta = \frac{-ln(0.2)}{ \sqrt{\pi^2 + ln^2(0.2)}}$

$\zeta = \frac{1.609}{ \sqrt{\pi^2 + 2.59}}$

$\zeta = \frac{1.609}{3.529}$

$\zeta = 0.4559$

Daga diagram mai root locus, zaka iya samun kungiyoyi masu muhimmanci;

$P = -0.866 \pm j 1.691 = \sigma \pm j \omega_d$

$\omega_d = 1.691$

$\omega_d = \omega_n \sqrt{1-\zeta^2}$

$1.691 = \omega_n \sqrt{1-0.207}$

$\omega_n = \frac{1.691}{\sqrt{0.793}}$

$\omega_n = \frac{1.691}{0.890}$

$\omega_n = 1.9 \, rad/sec$

A nan, tana da ranar ξ da ωn,

$settling \, time \, t_s = \frac{4}{\zeta \omega_m}$

$t_s = \frac{4}{0.455 \times 1.9}$

$t_s = 4.62 sec$

Diagramma na root locus ita ce da MATLAB. Don haka amfani da 'yan “sisotool”. A kan, zaka iya hada kawo wata bayanin percentage overshoot ta 20%. Kuma samun dominant poles da kyau.

Tambayar da na biyu yana nuna diagramma na root locus daga MATLAB.

Misali na Root Locus

Za a iya samun lokacin da aka tabbatar da shi a kan MATLAB. A cikin haka, ake kawo bayanan tushen daidai na system ta haka.

Lokacin da aka tabbatar da shi a kan MATLAB

Yadda a Kaɗau Lokacin da aka Tabbatar Da Shi

Lokacin da aka tabbatar da shi shine lokaci da aka bukata don samun abubuwa. Don haka, wajen kontrolar da yawa, ya kamata a kaɗau lokacin da aka tabbatar da shi.

Kaɗau lokacin da aka tabbatar da shi ba shi ne aiki mai zurfi. Ana buƙata a yi controller don kaɗau lokacin da aka tabbatar da shi.

Sannan, akwai uku na controllers; proportional (P), Integral (I), derivative (D). Daga cikin hukumar da suka haɗa, za a iya samun abubuwan da muke so kuɗi game da system.

Gain na controllers (KP, KI, KD) ana zaba daidai a cikin abubuwan da muke so kuɗi game da system.

Zaɓi gain proportional KP, zai yi laifi a lokacin da aka tabbatar da shi. Zama gain integral KI, lokacin da aka tabbatar da shi zai ci. Sannan, zama gain derivative KD, lokacin da aka tabbatar da shi zai ci.

Saboda haka, zafayi na kammalun yana kanza don kawo rike wani lokaci. Idan an zaɓe abubuwa na kammalun ta PID, zai iya haɗa da muhimmanci da dama kamar lokacinin kawo rike, tsara, da kuma Tashin kasarwar.

Yadda a Koyar Lokacinin Rike a MATLAB

A MATLAB, lokacinin rike zai iya samun da aikace-aikacen step. Ba na fahimta tushen misali.

$G(s) = \frac{25}{s^2 + 6s + 25}$

Kadan, muna koyar lokacinin rike da aikace-aikacen. Don haka, kara wannan transfer function da transfer function na gaba-gabata na system na biyu.

$G(s) = \frac{\omega_n^2}{s^2 + 2 \zeta \omega_n s + \omega_n^2}$

Saboda haka,

$2 \zeta \omega_n = 6$

$\zeta \omega_n = 3$

$settling \, time \, (t_s) = \frac{4}{\zeta \omega_n}$

$t_s = \frac{4}{3}$

$t_s = 1.33 sec$

Wannan irin da ta ake kula shi ne kuma ake amfani da hukuma wajen kula shi. Amma a MATLAB, muna samun irin daidai. Saboda haka, wannan irin zai iya kasancewa wasu daga baya bayan duk biyu.

Sau, don kula irin daidai a MATLAB, muna amfani da hukumomin step.

clc; clear all; close all;
num = [0 0 25];
den = [1 6 25];
t = 0:0.005:5;
sys = tf(num,den);
F = step(sys,t);
H = stepinfo(F,t)

step(sys,t);

Output:

H =

RiseTime: 0.3708
SettlingTime: 1.1886
SettlingMin: 0.9071
SettlingMax: 1.0948
Overshoot: 9.4780
Undershoot: 0
Peak: 1.0948
PeakTime: 0.7850

Kuma za a samu grafikin takalmi kamar yadda aka nuna a cikin wannan siffofin.

Kula irin daidai a MATLAB

A MATLAB, bayanin cikakken kula irin daidai ya shafi 2%. Zan iya canza wannan a cikin grafiki don cikakken kula irin daidai masu wahala. Don haka, duba zuwa grafiki > properties > options > “show settling time within ___ %”.

Ediṭa na Maimakon MATLAB

Wani hikima da za a iya samun lokacin da yake shiga ita ce ta kula da tsabta. Idan an sani, don faduwar 2%, zan iya cewa taron da aka nuna yana kan 0.98 zuwa 1.02.

clc; clear all; close all;

num = [0 0 25];
den = [1 6 25];

t = 0:0.005:5;

[y,x,t] = step(num,den,t);

S = 1001;
while y(S)>0.98 & y(S)<1.02;
    S=S-1;
end
settling_time = (S-1)*0.005

Tsunafi:

settling_time = 1.1886

Bayanin: A yi amfani da asalin, babban lura mai gaskiya ya zama da ake baka, idandaza ba tare da kasa za a iya bayar don ake cire.

Ba da kyau kuma kara mai rubutu!

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Na'urar da Daukake	Dabbobi na Yawanci (ξ)	Wakar Kafa (TS)
Yadda yawancin dabbobi ya koma	0<ξ<1	$T_S = \frac{4}{\zeta \omega_n }$
Babu yawanci	ξ = 0	$T_S = \infty$
Yadda yawancin dabbobi ya haɗa	ξ = 1	$T_S = \frac{6}{\omega_n}$
Yadda yawancin dabbobi ya fiye	ξ > 1	Yana nufin kan poli mai yawa