Time Domain Analysis

Response of 1st Order Systems

Definition

Systems modelled by a 1st order ODE

For a first order mass damper system
$m{v}^{\prime }+cv=f$
$msV\left(s\right)-mV\left(0\right)+cV\left(s\right)=F\left(s\right)$
For the zero input response
$m{v}^{\prime }+cv=0$
Applying a LaPlace transform $msV\left(s\right)-mV\left(0\right)+cV\left(s\right)=0$
$v\left(t\right)=v\left(0\right)e^{\frac{-c}{m}t}$
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.

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…
$v_{zsr}\left(t\right)=\frac{F}{c}\left(1-e^{-\frac{c}{m}t}\right)$
$v\left(t\right)=v_{ZIR}+v_{ZSR}\left(t\right)$
$v\left(t\right)=v\left(0\right)e^{-c/m}t+\frac{F}{C}\left(1-e^{-\frac{c}{m}t}\right)$
$v\left(t\right)=\left(v\left(0\right)-\frac{F}{C}\right)e^{-\frac{c}{m}t}+\frac{F}{c}$

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

Response of Second Order Systems

For a similar mass damper system…
$m{x}^{{}^{\prime \prime }}+c{x}^{\prime }+kx=f\left(x\right)$
$m[s^{2}X\left(s\right)-sx\left(0\right)-x^{\prime }\left(0\right)]+c\left[sx\left(s\right)-x\left(0\right)\right]+kX\left(s\right)=F\left(s\right)$
Factoring out X(s) and F(s) we can derive the following to get to the transfer function
$X\left(s\right)=\frac{F\left(s\right)}{m{s}^2+cs+k}+\frac{msx\left(0\right)+m{x}^{\prime }\left(0\right)+cx\left(0\right)}{m{s}^2+cs+k}$
$\frac{X\left(s\right)}{F\left(s\right)}=\frac{1}{m{s}^2+cs+k}$
*You can use the auxiliary method from State Variable Models if you have a time differentiated input

For an undamped 2nd order response
c=0
$X\left(s\right)=\frac{F\left(s\right)}{m{s}^2+k}+\frac{m{x}^{\prime }\left(0\right)+msx\left(0\right)}{m{s}^2+k}$
$X\left(s\right)=\frac{\frac{F\left(s\right)}{m}}{s^2+\frac{k}{m}}+\frac{x^{\prime }\left(0\right)+sx\left(0\right)}{s^2+\frac{k}{m}}$
Where the terms above are out zero state response and zero input response

Zero Input Response
$X_{zir}\left(s\right)=\frac{x^{\prime }\left(0\right)}{s^2+\frac{k}{m}}+\frac{sx\left(0\right)}{s^2+\frac{k}{m}}\frac{}{}$
Transfer Function
$\frac{X\left(s\right)}{F\left(s\right)}=\frac{\frac{1}{m}}{s^2+\frac{k}{m}}$
Where the “characteristic equation” models our poles.

using the characteristic equation, we can derive the natural frequency of the system as…
${\omega }_n=\sqrt{\frac{k}{m}}$
$X\left(s\right)=\frac{x^{\prime }\left(0\right)}{s^2+{\omega }_n^2}+{\frac{sx\left(0\right)}{s^2+{\omega }_n^2}}_{}$
$X\left(s\right)=\frac{x^{\prime }\left(0\right)}{{\omega }_n}{\frac{{\omega }_n}{s^2+{\omega }_n^2}}_{}+x\left(0\right)\frac{s}{s^2+{\omega }_n^2}$
Using this inverse LaPlace transform…
$\cos \left(bt\right)u\left(t\right)=\frac{s}{s^2+b^2}$
$\sin \left(bt\right)u\left(t\right)=\frac{b}{s^2+b^2}$
Therefore, applying this concept…
$x\left(t\right)=\left[\frac{x^{\prime }\left(0\right)}{{\omega }_n}\right]\sin \left({\omega }_{n}t\right)+x\left(0\right)\cos \left({\omega }_{n}t\right)]u\left(t\right)$
Where this oscillates with the frequency of the poles

Also note the general solution for a 2nd order homogenous equation with $r=\pm \omega j$
$x\left(t\right)=B\sin \left(\omega t+\varphi \right)$
Using given boundary conditions, we can find $B$ and $\varphi$ by seeing which values satisfy the B.Cs

Damped Response

The Characteristic equation is defined as
$m{s}^2+cs+k=0$
$s^2+\frac{c}{m}s+\frac{k}{m}=0$
$\frac{c}{m}=2\zeta {\omega }_n$ or $\zeta =\frac{c}{2m{\omega }_n}\frac{c}{c_{cr}}$
$c_{cr}=2m{\omega }_n=2\sqrt{km}$ which means $\zeta =\frac{c}{2\sqrt{km}}$ and
$s=-\zeta {\omega }_n\pm {\omega }_n\sqrt{{\zeta }^2-1}$ are the poles.

The Free Damped Response

$X\left(s\right)=\frac{sx\left(0\right)+x^{\prime }\left(0\right)+\frac{c}{m}x\left(0_{}\right)}{s^2+2\zeta {\omega }_n+{\omega }_{n^2}}$
Where the poles are given above.

Case 1

If $\zeta >1$, then $c>2\sqrt{km}$ and $\sqrt{{\zeta }^2-1}>0$ which is the over-damped case.

and our discriminate is positive meaning we have two distinct real poles.
This yields…
$X\left(s\right)=\frac{sx\left(0\right)+x^{\prime }\left(0\right)+2\zeta {\omega }_{n}x\left(0\right)}{\left(s-s_1\right)\left(s-s_2\right)}$
$x\left(t\right)=C_1{e}^{s_{1}t}+c_2{e}^{s_{2}t}$
Where $\{\begin{array}{l}{\tau }_1=\frac{-1}{s_1}\\ {\tau }_2=\frac{-1}{s_2}\end{array}$
and ${\tau }_2>{\tau }_1$ meaning that $s_2$ is the dominant pole.

Solving

After coupling the system, solve for the s’s and the time constant above, then…
${\omega }_n=\sqrt{s_1{s}_2}$
$\zeta =\frac{s_1+s_2}{2{\omega }_n}$

Case 2

$\zeta =1$ then $c=c_{cr}$, this represents the critically damped case. This is the minimum damping to have no oscillation.

${\zeta }^2-1=0$ in this case and $s=-\zeta {\omega }_n\pm 0$ yields repeated real poles.

in this case, $s=-{\omega }_n$
Using the derivation from the lectures…
$x\left(t\right)=\left(c_1+c_2\right)e^{st}=\left(c_1+c_2\right)e_{}^{-{\omega }_{n}t}$

Note

No ${\omega }_n$ or ${\omega }_d$ since there is no oscillation

Case 3

$0<\zeta <1$ represents the underdamped case. This should yield imaginary roots.
$c<2\sqrt{\mathrm{km}}$

$s=-\zeta {\omega }_n\pm j{\omega }_n\sqrt{1-{\zeta }^2}$ where ${\omega }_d$ is the natural frequency and the square root term.

$e_{}^{-\zeta {\omega }_{n}t}\left[\cos {\omega }_{d}t+=\zeta \sin {\omega }_{d}t\right]$

From the line above $\tau =\frac{1}{\zeta {\omega }_n}=\frac{2m}{c}$

$x\left(t\right)=C_1{e}^{s_{1}t}+C_2{e}^{s_{2}t}$
$x\left(t\right)=e^{-\zeta {\omega }_{n}t}\left[A\sin \left({\omega }_{d}t\right)+x\left(0\right)\cos \left({\omega }_{d}t\right)\right]$ where $A=\frac{\zeta }{\sqrt{1-{\zeta }^2}}x\left(0\right)+\frac{1}{{\omega }_d}{x}^{\prime }\left(0\right)$

Where $s_1^{\ast }=s_2$ with maybe some error

Solving

Useful equations for this case…
${\omega }_n=\sqrt{Re{\left(s\right)}^2+Im{\left(s\right)}^2}$
${\omega }_d={\omega }_n\sqrt{1-{\zeta }^2}$ $\zeta =\frac{-Re\left(s\right)}{{\omega }_n}$ $\tau =-\frac{1}{\mathrm{Re}\left(s\right)}$
$\sin \theta -\sqrt{1-{\zeta }^2}=\sqrt{1-\cos {\theta }^2}$