• BME 444
  • First order response

    Assume that you are given a first-order system

    $$G(s)=\frac{a}{s+b}$$

    where \(a\) is a constant in the numerator and \(b\) is a different constant in the denominator. The response, \(C(s)\), of this system to a unit-step input, \(R(s)=\frac{1}{s}\), is

    $$C(s)=R(s)G(s)=\frac{1}{s}\left(\frac{a}{s+b}\right) \tag{1} \label{eq1}$$

    To solve this equation, we first need to perform partial fraction expansion

    $$C(s)=\frac{a}{s(s+b)}=\frac{k_1}{s}+\frac{k_2}{s+b}$$

    Solve for \(k_1\) using Heaviside cover-up

    $$\frac{a}{s+b}=k_1 + \frac{sk_2}{s+b}$$

    Let \(s\rightarrow 0 \) and we get

    $$k_1=\frac{a}{b} \tag{2} \label{eq2}$$

    Solve for \(k_2\) using Heaviside cover-up

    $$\frac{a}{s}=\frac{k_1(s+b)}{s}+k_2$$

    Let \(s\rightarrow -b \) and we get

    $$k_2=-\frac{a}{b} \tag{3} \label{eq3}$$

    Plug \(k_1\) and \(k_2\) back into Eq. \ref{eq1} and we get

    $$C(s)=\frac{a}{s(s+b)}=\frac{\frac{a}{b}}{s}-\frac{\frac{a}{b}}{s+b}$$

    Take the inverse LaPlace transform to get

    $$c(t)=\frac{a}{b}-\frac{a}{b}e^{-bt}$$

    Example with numbers

    If

    $$G(s)=\frac{5}{s+3}$$

    and we want to find the response to a unit-step input, we need to solve the equation

    $$C(s)=R(s)G(s)=\frac{1}{s}\left(\frac{5}{s+3}\right)$$

    Using Eq. \ref{eq2} and \ref{eq3} from above, we find \(k_1=\frac{5}{3}\) and \(k_2=-\frac{5}{3}\). The response of this first order system would therefore be

    $$c(t)=\frac{5}{3}-\frac{5}{3}e^{-3t}$$





    Last updated:
    March 21, 2019