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Kontrol Sistemleri · Dijital ders notu

Stability, Routh Criterion, Control Actions

Ders
Kontrol Sistemleri (ELEC322)
Dönem
6. Dönem · 2023–2024 Bahar
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Control-Systems-lecture-8.pdf

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Stability, Routh Criterion, Control Actions – dijital ders notunun ilk sayfası
Uyarı: Dijital notlar, el yazısı notlardan Claude (Anthropic) yardımıyla oluşturulmuştur; yardımcı kaynak olarak hazırlanmıştır ve hata içerebilir. Orijinal notları ve resmi ders kaynaklarını esas alın.
The digital notes were created from the handwritten notes with the help of Claude (Anthropic); they are meant as a supplementary resource and may contain errors. Always rely on the original notes and the official course materials.

Bu notta neler var?

Notun metni

Düz metin olarak göster (formüller ve çizimler yoktur)

1. Quality of response

Closed Loop TF characteristic roots (poles)

This is a characteristic of the system, and `not' governed by the input signal.

States of stability

Gives a way of determining stability state without the requirement of finding the poles

- No specific pole is given but the numbers of poles on the RHP is found.

zero pole cancellation is not happening

Requirement #1: all coefficients (a_n) must be positive and nonzero!!

Requirement #2: Construct the table of Routh

Conclusion: The number changes in signs = number of poles on the RHP (look to the first column)

Pass the necessary condition

2 changes of sign: two poles on the RHP.

it can be oscilatory or not, we don't know

The value of the cell in the first column at some specific row becomes zero

just replace zero with ε>0 and move on

→ Definitely unstable because there is negative coefficient

we have got two roots on the jω axis.Claude notu: muhtemelen ``RHP'' olmalı (iki işaret değişimi var).

getting a full zero row!

1 change in sign → 1 pole: it have to be real because complex pole have conjugate

construct an auxilary equation

Solution of the auxilary eq. gives us some of the poles of the main system

case 1 of exception → at least two poles on the imaginary axis

case 2 exception → some poles are symmetric with respect to the origin

values so that all poles are in the shaded area?

all elements have to be pozitive

Method #2 for the first exception case

This change of variable gives us the reciprocal of the root

Two poles are on the RHP

Solve this:Claude notu: üsler küçük yazıldığı için okunması zor; ardışık üsler (7'den 0'a) varsayıldı.

R(s) is the input and ``can not'' affect stability and accuracy of the system

k_d e → rate of change of error

Derivative operator is always used with `P' and `I' operations

→ responsible for stability

k_p is also responsible for the location of the pole

Steady state error Final value theorem

→ responsible for error

(A pole added to the open loop function)

(responsible for sensitivity and increases stability in that sense)

← 2nd/higher order systems, time-domain specsAccuracy of Systems: Steady-State Error (Ders 9) →

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