The Math $$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$$

First order systems

System Cutoff Frequency
( $$f_c$$, Hz)
Settling Time
($$t_s$$, s)
Gain
($$G$$)
Mechanical $$\dfrac{D}{2\pi M}$$ $$\dfrac{4}{\omega_0} = 4\dfrac{M}{D}$$ $$\dfrac{1}{M}$$
Electrical $$\dfrac{1}{2\pi RC}$$ $$\dfrac{4}{\omega_0}= 4RC$$ $$\dfrac{1}{RC}$$

Definition

In the time domain first order systems are the solution $$y(t)$$ of the following differential equation $$\label{foDef} \begin{cases} \dfrac{dy(t)}{dt}=-\omega_0 y(t)+Gu(t) \\ y(0)=y_0 \end{cases}$$ where all the quantities are scalars, $$y_0$$ is the initial condition of $$y(t)$$, $$u(t)$$ is the input of the system and $$G$$ and $$\omega_0$$ are constants.

In the Laplace domain \eqref{foDef} becomes $$\label{foDefLapIni} Y(s)= H(s)U(s)+\frac{H(s)}{G}y_0, \quad \text{with} \quad H(s)= \frac{G}{s+\omega_0}$$ where $$H(s)$$ is the transfer function of $$y(t)$$.

From now on we will consider $$y_0=0$$ and under this condition \eqref{foDefLapIni} becomes $$\label{foDefLap} Y(s)= H(s)U(s)$$

If $$\omega_0 > 0$$, which is usually the case for physical systems of this type, \eqref{foDef} is asymptotically/BIBO stable (asymptotic and BIBO stability are equivalent for first order systems) and the time response to the step function $$u(t) = \begin{cases} 1 & \text{if } t \geq 0 \\ 0 & \text{if } t < 0 \end{cases}$$ is shown in figure 1 while figure 2 represents the Bode plot of $$H(s)$$.

Examples

Mechanical Domain

In the mechanical domain a first order system can be built using a mass force system with viscous damping, ($$M$$, $$u$$ and $$D$$ respectively, see figure 3). From Newton we have $$Ma=u-Dv$$ which can be represented through velocity as $$\label{mechSyst} M\frac{dv}{dt}=-Dv+u \implies \frac{dv}{dt}=-\frac{D}{M}v+\frac{1}{M}u$$ Comparing \eqref{mechSyst} with \eqref{foDef} we have that for a mechanical system $$y=v$$, $$G=\frac{1}{M}$$ and $$\omega_0=\frac{D}{M}$$. A small note is made on $$\omega_0$$ which is positive implying that the system is stable.

In the Laplace domain \eqref{mechSyst} becomes $$\label{mecha} sMV=-DV+U \implies V=\dfrac{U}{sM+D} \implies V=\dfrac{1}{M}\frac{1}{s+\frac{D}{M}}U$$

Electrical Domain

In the electrical domain a first order system is represented by an RC circuit (figure 4) . Using Kirchhoff's voltage law we have $$\label{eleSys} u=iR+v_c \xRightarrow{i=C\frac{dv_c}{dt}} RC\frac{dv_c}{dt}=u-v_c \implies \frac{dv_c}{dt}=-\frac{1}{RC}v_c+\frac{1}{RC} u$$ From Laplace we obtain $$\label{electro} sV_c=-\frac{1}{RC}V_c+\frac{1}{RC}U \implies V_c=\frac{1}{RC}\frac{1}{s+\frac{1}{RC}}U$$ and comparing \eqref{eleSys} with \eqref{foDef} (or equivalently \eqref{electro} with \eqref{foDefLap}) it can be seen that $$y=v_c$$, $$G=\frac{1}{RC}$$ and $$\omega_0=\frac{1}{RC}$$. A small note is made on $$\omega_0$$ which is positive implying that the system is stable.

Cut off frequency

Also known as corner or break frequency, the cutoff frequency decreases the energy flow through a system and obstructs it from working as intended.

The cutoff frequency of a first order system can be found by analysing the magnitude of $$H(j\omega)$$ (Figure 2) $$\label{bodeCo} \lvert H(j\omega) \rvert =\frac{G}{\sqrt {\omega^2 +\omega^2_0}}$$ $$\lvert H(j\omega) \rvert$$ behaviour can be assessed considering that it is monotonically decreasing and it has two asymptotes, that is $$\label{asy} \lvert H(j\omega) \rvert\simeq \begin{cases} H_0=\dfrac{G}{\omega_0} & \text{if } \omega \ll \omega_0 \\ H_R=\dfrac{G}{\omega} & \text{if } \omega \gg \omega_0 \end{cases}$$ These properties guaranty that there is a cutoff frequency and it can be calculated by finding where $$H_0$$ and $$H_R$$ intersects (from \eqref{asy}, $$H_0$$ is constant while $$H_R$$ is a function of $$\omega$$), that is $$H_0=H_R\left( \omega_c \right) \implies \frac{G}{\omega_0}=\frac{G}{\omega_c}$$ and the cutoff frequency is $$\begin{cases} \omega_c =\omega_0 & \left [ \text{rad/s} \right] \\ f_c =\dfrac{\omega_0}{2\pi} & \left [ \text{Hz} \right] \end{cases}$$ From here we can also find that $$\lvert H(j\omega_c) \rvert =\frac{1}{\sqrt {2}}\frac{G}{\omega_0} \implies \lvert H(j\omega_c) \rvert =\frac{1}{\sqrt {2}} H_0$$ which shows that in $$\omega=\omega_c$$, $$\lvert H(j\omega) \rvert$$ loses 3dB $$\left( =20\text{log}\left( \sqrt {2} \right) \right)$$.

Figure 2 shows an example where $$\omega_0=1$$.

For the two systems described in the above examples the cutoff frequency (Hz) is \begin{align} \text{Mechanical:}\quad & f_c=\frac{D}{2\pi M} \\ \text{Electrical: } \quad & f_c=\frac{1}{2\pi RC} \end{align}

Settling time

The settling time of \eqref{foDef} can be calculated in the Laplace domain applying the step function $$u(t) = \begin{cases} 1 & \text{if } t \geq 0 \\ 0 & \text{if } t < 0 \end{cases} \xRightarrow[]{\mathscr{L}\text{ }} \frac{1}{s}$$ to \eqref{foDefLap}. The response becomes $$Y= \frac{G}{s(s+\omega_0)}$$ which can be written as $$Y= \frac{G}{\omega_0}\left( \dfrac{1}{s}-\frac{1}{s+\omega_0} \right)$$ and returning back to the time domain we obtain $$(t\geq) 0$$ $$\label{tsol} y(t)=\frac{G}{\omega_0}\left(1 -e^{-\omega_0 t} \right)$$ Now, to find the rise time we will need the settling value of \eqref{tsol} and to do this we take its limit as $$\label{lim} \lim_{t\to\infty} \frac{G}{\omega_0}\left(1 -e^{-\omega_0 t} \right)=\frac{G}{\omega_0}$$ Defining $$(t_s)$$ as the time required to reach 98% of \eqref{tsol}'s final value then from \eqref{tsol} and \eqref{lim} we obtain the following equation $$0.98\frac{G}{\omega_0}=\frac{G}{\omega_0}\left(1 -e^{-\omega_0 t_s} \right) \implies 0.98=1 -e^{-\omega_0 t_s} \implies t_s = -\frac{\ln (0.02)}{\omega_0}$$ and from here we finally calculate the settling time as $$t_s \simeq \frac{4}{\omega_0}$$ For the two systems described in the above examples the settling time is \begin{align} \text{Mechanical:}\quad & t_s = 4\frac{M}{D} \\ \text{Electrical: } \quad & t_s = 4RC \end{align}