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найти переходную функцию звена заданного дифференц...

作成日: 2025年9月1日

使用モデル GPT-5 Thinking by Chat01

質問

найти переходную функцию звена заданного дифференциальным уравнением y'' - 5y' + 6y = u'

思考

Clarifying step response calculations

I’m working through a system where the right-hand side involves the derivative of u. The step input leads to a δ(t) response, and the system equation becomes y'' - 5y' + 6y = u. I’ll use Laplace transforms for the step input and find the transfer function.

Let’s continue with partial fractions for the inverse transform.

Analyzing system behavior

I’m solving for the inverse Laplace of the system's transfer function, which leads to an unstable solution with exponential growth from the poles at 2 and 3. This seems odd given the positive coefficients in the ODE. But the step response is still mathematically defined, and we find y(t) = e^{3t} - e^{2t} for t ≥ 0.

Understanding the terminology

The user asked for the "переходную функцию звена," which refers to the step response. This is typically denoted in textbooks as g(t). The transfer function W(s) relates the system output to input, and for a step response, we take the inverse Laplace of W(s)/s. I'll make sure to clarify and compute for u(t) = 1(t).

Deriving step response

We already computed the step response for u(t), but now I’m focusing on correctly deriving the transfer function. With the equation y'' - 5y' + 6y = u', taking the Laplace transform yields W(s) = s / (s^2 - 5s + 6). The step response is g(t) = e^{3t} - e^{2t}, which matches our previous result.

Thought for 38s

回答

Покажу коротко через Лапласа (нулевые начальные условия, переходная функция — отклик на единичный скачок u(t)=1(t)u(t)=1(t)).

  1. Лаплас уравнения y′′−5y′+6y=u′y''-5y'+6y=u':
(s2−5s+6)Y(s)=s U(s).(s^2-5s+6)Y(s)=s\,U(s).
  1. Передаточная функция:
W(s)=YU=ss2−5s+6=s(s−2)(s−3).W(s)=\frac{Y}{U}=\frac{s}{s^2-5s+6}=\frac{s}{(s-2)(s-3)}.
  1. Переходная функция (отклик на ступень):
g(t)=L−1 ⁣{W(s)s}=L−1 ⁣{1(s−2)(s−3)}=e3t−e2t,t≥0.g(t)=\mathcal{L}^{-1}\!\left\{\frac{W(s)}{s}\right\} =\mathcal{L}^{-1}\!\left\{\frac{1}{(s-2)(s-3)}\right\} = e^{3t}-e^{2t},\qquad t\ge 0.

(Система неустойчива, т.к. полюса при s=2,3>0s=2,3>0, поэтому g(t)g(t) возрастает без ограничений.)

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