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 | \(\dfrac{1}{2\pi RC}\) | \( \dfrac{4}{\omega_0}= 4RC \) | \( \dfrac{1}{RC} \) |

In the time domain first order systems are the solution \(y(t)\) of the following differential equation \begin{equation}\label{foDef} \begin{cases} \dfrac{dy(t)}{dt}=-\omega_0 y(t)+Gu(t) \\ y(0)=y_0 \end{cases} \end{equation} 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 \begin{equation}\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} \end{equation} 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 \begin{equation}\label{foDefLap} Y(s)= H(s)U(s) \end{equation}

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 \( H(s)\).

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 \begin{equation} \label{mechSyst} M\frac{dv}{dt}=-Dv+u \implies \frac{dv}{dt}=-\frac{D}{M}v+\frac{1}{M}u \end{equation} 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 \begin{equation}\label{mecha} sMV=-DV+U \implies V=\dfrac{U}{sM+D} \implies V=\dfrac{1}{M}\frac{1}{s+\frac{D}{M}}U \end{equation}

In the electrical domain a first order system is represented by an RC circuit (figure 4) . Using Kirchhoff's voltage law we have \begin{equation}\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 \end{equation} From Laplace we obtain \begin{equation}\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 \end{equation} 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.

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) \begin{equation}\label{bodeCo} \lvert H(j\omega) \rvert =\frac{G}{\sqrt {\omega^2 +\omega^2_0}} \end{equation} \(\lvert H(j\omega) \rvert\) behaviour can be assessed considering that it is monotonically decreasing and it has two asymptotes, that is \begin{equation}\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} \end{equation} 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 \begin{equation} \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 \end{equation} 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}

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 \((t\geq) 0 \) \begin{equation}\label{tsol} y(t)=\frac{G}{\omega_0}\left(1 -e^{-\omega_0 t} \right) \end{equation} Now, to find the rise time we will need the settling value of \eqref{tsol} and to do this we take its limit as \begin{equation}\label{lim} \lim_{t\to\infty} \frac{G}{\omega_0}\left(1 -e^{-\omega_0 t} \right)=\frac{G}{\omega_0} \end{equation} 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 \begin{equation} t_s \simeq \frac{4}{\omega_0} \end{equation} 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}