Pixabay

Overview

This section will examine how to find solutions in the time domain, even though the relationship in the time domain may be too complicated to solve, unless Laplace Transforms are used to move in and out of the frequency domain.

Topic 1 – First Order Time Domain Solutions

Many engineering problems involve first order differential equations.  That is, equations which involve both a parameter and the first derivative (with respect to time) of that parameter.  These can be solved relatively easily using the Laplace Transform to move between time and frequency domains.   So long as the first order differential equation is of a common form, such as

$\text{y}={\text{y}}_0+\text{C}\text{dy}/\text{dt}$ 

where:

y = the parameter, which varies with time

y₀ = the value of the parameter at time t = 0

C = some relevant constant

dy/dt = the first derivative of the parameter with respect to time

If the equation is of this form, then it does not matter whether the problem is mechanical, thermal, electrical, or any other type.  The solution approach will be exactly the same.  The following chart indicates a wide variety of problems, whose response can be described by a similar first order differential equation of this form.

(Ogata)

To examine this further, we will work an example, using a thermal problem.

Application 1

A thermometer is at ambient temperature, ${\text{θ}}_{\text{A}}$, when it is placed in a water bath, of temperature ${\text{θ}}_{\text{A}}+{\text{θ}}_{\text{B}}$ at time $\text{t}=0$.   ${\text{θ}}_{\text{B}}$ is the difference in temperature between ambient and the temperature of the water bath.  All temperatures are in degrees C. 

The temperature reading of the thermometer at any point in time will be given by ${\text{θ}}_{\text{A}}+\text{θ}$ where $\text{θ}$ is the change in temperature with time. We want to find how $\text{θ}$ varies with time.

The thermal equation (as can be found on the previous chart), using the constants of Thermal Resistance (R) and Thermal Capacitance (C) would be

$\text{RCdθ}/\text{dt}+\text{θ}={\text{θ}}_{\text{B}}$      remember that $\text{θ}$ is a function of $\text{t}$, i.e. $\text{θ}\left(\text{t}\right)$ 

Transform using line 2 of the chart

$\text{RCsΘ(s)}+\text{Θ(s)}={\text{θ}}_{\text{B}}/\text{s}$ 

Solve for $\text{Θ(s)}$ 

$\text{Θ(s)}={\text{θ}}_{\text{B}}/\text{s}\text{(RCs + 1)}$ 

$={\text{θ}}_{\text{B}}[\text{(1/s)}-1/\text{(s + 1/RC)}]$ 

Inverse Transform using lines 2 and 6

$\text{θ(t)}={\text{θ}}_{\text{B}}\left[\right(1-{\text{e}}^{-\text{t/RC}}\left)\right]$ 

This says that at $\text{t = 0}$, $\text{θ(0)}={\text{θ}}_{\text{B}}\text{[1-1]}=\mathrm{0. }$

And it is possible to plot the variation in temperature with time as shown.  Note that T is called the Thermal Time Constant and is defined as  T = RC

(Ogata)

Note that the thermometer reading will equal 63.2% of the bath temperature in time equal to one T and will be within 2% of the bath temperature in time equal to 4T, after which it can reasonably be considered to be the true reading of the bath temperature.

Application 2

Let us consider the same thermal problem, but this time the water bath is being heated, such that ${\text{θ}}_{\text{B}}$ is not a constant, but rather varies with time as a ramp function,  ${\text{θ}}_{\text{B}}=\text{rt}$ 

So this time

$\text{RCdθ}/\text{dt}+\text{θ}={\text{θ}}_{\text{B}}=\text{rt}$

Transform using line 3 of the chart

$\text{RCsΘ(s)}+\text{Θ(s)}=\text{r}/{\text{s}}^2$ 

Solve for $\text{Θ(s)}$ 

$\text{Θ(s)}=\text{r}/{\text{s}}^2(\text{RCs}+1)$  

$=\text{r}\left[\right(1/{\text{s}}^2)-\text{RC/s}+\text{RC/(s + 1/RC)]}$  

Inverse Transform using lines 2, 3 and 6

$\text{θ(t)}=\text{r[t}-\text{RC}+\text{RC}{\text{e}}^{-\text{t/RC}}]$ 

This says that at $\text{t = 0}$ ,   $\text{θ(0)}={\text{θ}}_{\text{B}}[1-1]=0$  

And it is possible to plot the variation in temperature with time as shown

Remember that $\text{T = RC}$

(Ogata)

The temperature seen on the thermometer will basically lag behind the changing bath temperature by a constant amount determined by the time constant, T.

NOTE:   Any of the problems shown on the chart shown initially, whether they are mechanical, electrical, thermal or fluid related, take the same form and thus would be solved in exactly the same process.

Topic 2 - Higher Order Responses

Consider a very general system, that can be represented by the equation for response x (a function of time) under the influence of a forcing function p(t):

${\text{a}}_0{\text{x}}^{\text{n}}+{\text{a}}_1{\text{x}}^{\text{n}-1}+\text{ ...... }+{\text{a}}_{\text{n}-1}{\text{x}}^{\prime }+{\text{a}}_{\text{n}}\text{x}=\text{p(t)}$  

Where the n terms represent the order of derivative of the function x(t)

Recall from calculus that the equation

${\text{a}}_0{\text{x}}^{\text{n}}+{\text{a}}_1{\text{x}}^{\text{n}-1}+\text{ ...... }+{\text{a}}_{\text{n}-1}{\text{x}}^{\prime }+{\text{a}}_{\text{n}}\text{x}=\text{p(t)}$

has a solution comprised of two parts

$\text{x(t)}={\text{x}}_{\text{c}}\text{(t)}+{\text{x}}_{\text{p}}\text{(t)}$  

${\text{x}}_{\text{c}}\text{(t)}$ is called the “complementary solution” and can be found by setting p(t) = 0 and solving the differential equation.

${\text{x}}_{\text{p}}\text{(t)}$ is called the “particular solution” and it depends on the input, p(t) when solving the differential equation.

If x(t) approaches any constant value as time gets very large, that is its steady state condition.

The response x(t) consists of a transient portion, which occurs soon after t >0

and a steady state portion that occurs when t becomes large enough that the system response ceases to change anymore.

Application 1

In the discussion just concluded, we examined a first order equation, with on a first derivative.  However, higher order differential equations can be solved using the Laplace Transform, it simply requires more time and mathematical manipulation.

Let us examine one more real world problem, that of a spring-mass-damper vibratory system, which can be represented by a second order equation involving mass (m), spring rate (k) and damping coefficient (c):

$\text{mx’’}+\text{cx’}+\text{kx}=\text{f(t)}$  

Let’s envision a simple model of a car on its suspension approaching a bump

Assume ${\text{x}}_0={\text{x}}_{0\text{’}}=0$      (i.e. vertical initial conditions are zero)

And $\text{m = 1}$, $\text{k = 16}$, $\text{c = 5}$, with the car about to encounter a unit impulse (i.e. bump)

Equation of motion:

$\text{x’’}+5\text{x’}+16\text{x}=\text{δ(t)}$ 

Laplace Transform

${\text{s}}^2\text{X(s)}+\text{5sX(s)}+\text{16X(s)}=1$ 

Solve for X(s)

$\text{X(s)}=\frac{1}{{\text{s}}^2+5\text{s}+16}$ 

$\text{X(s)}=\frac{1}{4^2}\frac{4^2}{{\text{s}}^2+5\text{s}+4^2}$ 

Where ${\text{ω}}_{\text{n}}=4$,   and $2\text{ ζ}{\text{ω}}_{\text{n}}=5$,   so damping ratio $=\text{ζ}=\text{5/8}$ 

Inverse Transform

Thus the bump will cause the car to oscillate, but damping will cause it to decay and die out.

Application 2

Return to the simple model, but this time about to drive up on a curb

Assume ${\text{x}}_0={\text{x}}_{0\text{’}}=0$      (i.e. vertical initial conditions are zero)

And $\text{m = 1}$, $\text{k = 16}$, $\text{c = 5}$, about to encounter a unit step (i.e. curb)

Equation of motion:

$\text{x’’}+5\text{x’}+16\text{x}=\text{1(t)}$ 

Laplace Transform

${\text{s}}^2\text{X(s)}+\text{5sX(s)}+\text{16X(s)}=1/\text{s}$ 

Solve for X(s)

$\text{X(s)}=\frac{1}{\text{s}({\text{s}}^2+5\text{s}+16)}$ 

rearrange:

$\text{X(s)}=\frac{1}{4^2}\frac{4^2}{\text{s}({\text{s}}^2+5\text{s}+4^2)}$ 

Where ${\text{ω}}_{\text{n}}=4$,   and $2\text{ ζ}{\text{ω}}_{\text{n}}=5$,   so damping ratio $=\text{ζ}=\text{5/8}$ 

Inverse Transform

Again, there is an oscillatory response which will die out with time, and leave a steady offset the height of the curb.

Application 3

Return again to the simple model, but this time on a bumpy Highland road that resembles a cosine wave

Assume ${\text{x}}_0={\text{x}}_{0\text{’}}=0$      (i.e. vertical initial conditions are zero)

$\text{m = 1}$, $\text{k = 16}$, $\text{c = 5}$, about to encounter a rough surface (i.e cosine wave)

Equation of motion:

$\text{x’’}+5\text{x’}+16\text{x}=\text{3 cos 6t}$ 

Laplace Transform

${\text{s}}^2\text{X(s)}+\text{5sX(s)}+\text{16X(s)}=3\text{s}/({\text{s}}^2+6^2)$ 

This gets complicated, and we can no longer use the table to do the Inverse Transform.

But if we did the integrations……………we would get:

Consider….as time gets large the ${\text{e}}^{\text{-5t/2}}$ terms go toward zero, so we get steady state oscillatory response $\text{x(t) = 3/[}{5}^2+{(4^2-6^2)}^2]$    $\left\{\right(4^2-6^2\text{ cos6t }+5\left(6\right)\text{sin6t}$  driven by the road’s forcing function.

Summary

Sometimes a differential equation (of any order) describes a time varying response (of any type) but in a relationship that is difficult or impossible to solve in the time domain due to the inclusion of higher order differential terms.  It is still possible to find a time domain solution using the Laplace Transform method to move to the frequency domain where the solution may be easier, and then return the solution to the time domain with an inverse transform so as to produce the response as a function of time.

References and Bibliography

Ogata, K. (2004).  System Dynamics, 4th edition. Upper Saddle River, USA: Pearson Prentice Hall.