1st Order Dynamic Systems

Response of 1st Order Systems

Definition

Systems modelled by a 1st order ODE

In General for a 1st Order Mass Damper System

Time function… $m{v}^{\prime }+cv=f$

LaPlace Transform… $msV\left(s\right)-mV\left(0\right)+cV\left(s\right)=F\left(s\right)$

For the zero input response… $m{v}^{\prime }+cv=0$

Applying a LaPlace transform $msV\left(s\right)-mV\left(0\right)+cV\left(s\right)=0$ → $v\left(t\right)=v\left(0\right)e^{\frac{-c}{m}t}$

Where we have the following cases…

• If $\frac{c}{m}>0$; $-\frac{c}{m}<0$ we have exponential decay
• If $\frac{c}{m}<0$; $-\frac{c}{m}>0$ we have exponential growth

Important terms here

The thing that we care about here is the $\tau$ term or the time constant to model decay or growth. You can use a similar approach to find the time constant of other first order systems.

Forced Response of 1st Order Systems

From $msV\left(s\right)-mV\left(0\right)+cV\left(s\right)=F\left(s\right)$

We can factor to the form of $V\left(s\right)\left(ms+c\right)=F\left(s\right)+mv\left(0\right)$
$V\left(s\right)=\frac{F\left(s\right)}{ms+c}+\frac{mv\left(0\right)}{ms+c}$
Where the first term is the zero state response and the second term is the zero input response.

Step Response of a 1st order system

Once again using the mass damper system as an example…

• $f\left(t\right)=Fu\left(t\right)$
• $V\left(s\right)=\frac{F}{ms}\frac{1}{s+\frac{c}{m}}$
• $v_{zsr}\left(t\right)=\frac{F}{m}{\int }_0^t{e}^{-\frac{c}{m}t}u\left(t\right)dt$
• Solving the integral…
• $v_{zsr}\left(t\right)=\frac{F}{c}\left(1-e^{-\frac{c}{m}t}\right)$
• $v\left(t\right)=v_{ZIR}+v_{ZSR}\left(t\right)$
• $v\left(t\right)=v\left(0\right)e^{-c/m}t+\frac{F}{C}\left(1-e^{-\frac{c}{m}t}\right)$
• $v\left(t\right)=\left(v\left(0\right)-\frac{F}{C}\right)e^{-\frac{c}{m}t}+\frac{F}{c}$

From MTE 360, the general equation of a first order dynamic system is…

• $Y\left(s\right)=G\left(s\right)U\left(s\right)=k\frac{a}{s\left(s+a\right)}=k\left(\frac{1}{s}-\frac{1}{s+a}\right)\Rightarrow y\left(t\right)=k\left(1-e^{-at}\right)$
  • $G\left(s\right)=k\frac{a}{s+a}=k\frac{1}{1+\frac{1}{a}s}$
• Where the unit step response is…
  • $Y\left(s\right)=k\frac{a}{s\left(s+a\right)}=k\left(\frac{1}{s}-\frac{1}{s+a}\right)\Rightarrow y\left(t\right)=k\left(1-e^{-at}\right)$
• And the DC gain is…
  • ${\lim }_{t\to \infty }y\left(t\right)={\lim }_{t\to \infty }k\left(1-e^{-at}\right)=k$ or by FVT $y\left(\infty \right)={\lim }_{s\to 0}sY\left(s\right)={\lim }_{s\to 0}\frac{a}{s+a}$
• And the rise time is…
  • $t=-\frac{1}{a}\ln \left(1-\frac{y}{k}\right)$ where y is a percentage of k

The Idea

We end up with a transient response and a steady state response which are the two terms in the above equation respectively.

Solving

Basically the idea here, is we want to transform a model into the LaPlace domain and then isolate the output, the transient response is the once with an e on it and the steady state term just has t’s

We probably need to use fractional decomposition here

Some LaPlace transformations to remember: $L\left[t^n\right]=\frac{n!}{s^{n+1}}$
$L\left[e^{-at}\right]=\frac{1}{s+a}$