• 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