Laplace transforms - System Analysis and Control

In this section, we will review the Laplace transform, a technique often covered in introductory differential equations courses. There are two reasons for why they are a relevant topic for controls:

LTI systems are precisely the systems whose differential equations can be solved using the method of Laplace transforms. Therefore, Laplace transforms will be our main tool for solving differential equations.
When dealing with control systems, it is often easier to work with the Laplace transforms of signals rather than their time-domain representations. This leads to simpler representations of LTI systems and easier manipulation of complex systems (interconnections of multiple LTI systems).

The Laplace transform¶

The Laplace transform is a transformation that converts a function of time $f(t)$ into a function of “ $s$ ”, $F(s)$ . As a matter of convention, we will use lower-case letters to denote functions of time and upper-case letters to denote the corresponding Laplace transform whenever possible. The definition is given by the integral:

F(s) = \int_0^\infty e^{-st}f(t)\,\dd t

(1)

Alternative notations

The Laplace transform is also denoted by the “script L” symbol $\Lap$ . Different ways of saying that $\Lap$ transforms $f(t)$ to $F(s)$ sometimes use functional notation:

f \overset{\Lap}{\longrightarrow} F, \qquad F = \Lap\{f\}, \qquad F = \Lap[f].

(2)

Sometimes $s$ or $t$ is explicitly included to remind us of the variables:

F(s) = \Lap\{f\}(s), \qquad F(s) = \Lap\{f(t)\}, \qquad F(s) = \Lap\{f(t)\}(s),

(3)

or the corresponding variants using $[\cdot]$ rather than $\{\cdot\}$ . The second notation is particularly strange, because the left-hand side depends on $s$ while the right-hand side appears to depend on $t$ . Nevertheless, it is common in engineering textbooks. All of the notations above mean the same thing; they are all equivalent to Eq. (1).

The thing to remember is that when you perform an integral $\int_0^\infty (\dots)\,\dd t$ , the result no longer depends on $t$ . This is why the definition (1) makes sense. The $t$ in the right-hand side is a dummy variable. The right-hand side is only a function of $s$ .

Since Eq. (1) involves an infinite integral, it is possible that the integral will not converge. In fact, there will always be values of $s$ for which the Laplace transform is undefined. This leads us to an important concept and convention.

Example: Heaviside step function¶

The unit step function, also known as the Heaviside step function, named after British mathematician Oliver Heaviside, is defined below.

Applying the definition, we have:

H(s) = \int_0^\infty e^{-st} H(t) \,\dd t = \int_0^\infty e^{-st}\,\dd t =\left[ -\frac{1}{s}e^{-st} \right]_{t=0}^\infty = \frac{1}{s}

(4)

In the last step, we picked an $s$ that would make the limits finite. In this case, it means picking $s>0$ so that $\lim_{t\to\infty} e^{-st} = 0$ .

Example: Simple exponential¶

Let’s consider $f(t) = e^{-at}$ . Applying the definition (1):

F(s) = \int_0^\infty e^{-st} e^{-at} \,\dd t = \int_0^\infty e^{-(a+s)t}\,\dd t =\left[ \frac{-1}{s+a}e^{-(s+a)t} \right]_{t=0}^\infty = \frac{1}{s+a}

(5)

Linearity of Laplace¶

The Laplace transform is a linear operator. The word linear means the same as in the context of systems: it satisfies additivity and homogeneity. The word operator means it is like a function, except instead of mapping numbers to numbers it maps signals to signals.

We can verify linearity by using the properties of the integral:

\begin{aligned} \Lap\{a f(t) + b g(t)\} &= \int_0^\infty e^{-st}\bigl( a f(t) + b g(t) \bigr) \,\dd t \\ &= a \int_0^\infty e^{-st} f(t) \,\dd t + b \int_0^\infty e^{-st} g(t)\, \dd t \\ &= a F(s) + b G(s) \end{aligned}

(6)

We can use linearity to evaluate Laplace transforms of more complicated functions by breaking it up into simpler parts for which we already know the Laplace transform.

Example: More complicated exponential¶

Let’s consider $f(t) = \frac{1}{a}(1-e^{-at})$ . We can split this into two parts using linearity:

f(t) = \left(\frac{1}{a}\right) \cdot (1) + \left( -\frac{1}{a}\right) \cdot (e^{-at})

(7)

Applying linearity, we have:

\begin{aligned} F(s) &= \left(\frac{1}{a}\right)\Lap\{1\} + \left( -\frac{1}{a}\right)\Lap\{ e^{-at} \} \\ &= \left(\frac{1}{a}\right) \frac{1}{s} + \left(-\frac{1}{a}\right) \frac{1}{s+a} \\ &= \frac{1}{s(s+a)} \end{aligned}

(8)

Derivatives¶

A key feature of Laplace transforms is how they act on derivatives of signals. We can calculate this using integration by parts.

\begin{aligned} \Lap\{\dot f(t)\} &= \int_{0}^{\infty} \dot f(t)\,e^{-st}\,dt \\ &= \bigl[f(t)e^{-st}\bigr]_{0}^{\infty} - \int_{0}^{\infty} f(t)\,(-s)e^{-st}\,dt \qquad \textsf{(integration by parts)}\\ &= \bigl[f(t)e^{-st}\bigr]_{0}^{\infty} + s\int_{0}^{\infty} f(t)e^{-st}\,dt \\ &= -f(0) + s\,\Lap\{f(t)\} \\ &= sF(s) - f(0). \end{aligned}

(9)

We can also apply this formula recursively by noticing that

\Lap\{\ddot f(t)\}=\Lap\left\{\frac{d}{dt}\big(\dot f(t)\big)\right\} = s\,\Lap\{\dot f(t)\}-\dot f(0).

(10)

Substituting $\Lap\{\dot f(t)\}=sF(s)-f(0)$ gives

\Lap\{\ddot f(t)\} = s^2F(s)-s f(0)-\dot f(0).

(11)

Continuing in this way, for $n\ge 1$ ,

\Lap\{f^{(n)}(t)\} = s^n F(s) - s^{n-1}f(0) - s^{n-2}\dot f(0) - \cdots - f^{(n-1)}(0).

(12)

Here, $f^{(n)}(t)$ is short-hand notation for $n$ time derivatives. i.e., $\frac{\dd^n}{\dd t^n}f(t)$ .

Table of Laplace transforms¶

We can put all our results so far into a table:

Table 1:Table of Laplace transforms

Time-domain $f(t)$	Laplace transform $F(s)=\Lap\{f(t)\}$
$a f(t) + b g(t)$	$aF(s)+bG(s)$
$H(t)$ or 1	$\displaystyle \frac{1}{s}$
$e^{-at}$	$\displaystyle \frac{1}{s+a}$
$\dfrac{1}{a}(1-e^{-at})$	$\displaystyle \frac{1}{s(s+a)}$
$\dot f(t)$	$\displaystyle sF(s)-f(0)$
$\ddot f(t)$	$\displaystyle s^2F(s)-s f(0)-\dot f(0)$
$f^{(n)}(t)$	$\displaystyle s^nF(s)-\sum_{k=0}^{n-1} s^{n-1-k} f^{(k)}(0)$

Solving an ODE using Laplace¶

If a system is LTI, the method of Laplace transforms can be used to solve the corresponding differential equations. The basic steps are:

Take the Laplace transform of the ODE.
Solve for the desired output $Y(s)$ .
Take the inverse Laplace transform to obtain $y(t)$ .

Example: Cruise control¶

We will illustrate this approach to solve a cruise control example. Consider a car with mass $m$ moving along a flat road with initial velocity $v(0) = v_0$ . Starting at $t=0$ , we apply a constant force $f_a(t) = f_0$ . Assuming the drag force is proportional to velocity, so $f_d(t) = b v(t)$ ,^[1] find the resulting velocity $v(t)$ for $t\geq 0$ .

Free body diagram of a car with drag force f_d. — Figure 2:Free body diagram of a car with drag force $f_d$ .

Substituting $f_d(t)=b v(t)$ , the equation of motion is

m\dot v + b v = f_a,\qquad v(0)=v_0.

(14)

Solve using the Laplace transform¶

Take the Laplace transform of both sides, making use of Table 1:

m\bigl(sV(s)-v_0\bigr) + bV(s) = F(s).

(15)

Rearrange and group terms:

(ms+b) V(s) - m v_0 = F(s).

(16)

Solve for $V(s)$ :

V(s)=\left(\frac{1}{ms+b}\right)F(s) + \left(\frac{m}{ms+b}\right)v_0.

(17)

Step force input¶

Now substitute the constant force $f_0$ starting at $t=0$ , i.e.,

f(t)=f_0\, H(t) \qquad\implies\qquad F(s)=\frac{f_0}{s}.

(18)

Substitute into Eq. (17) and simplify:

V(s)=\frac{1}{ms+b}\cdot\frac{f_0}{s}+\frac{m v_0}{ms+b} =\frac{f_0/m}{s\left(s+\frac{b}{m}\right)}+\frac{v_0}{s+\frac{b}{m}}.

(19)

Using Table 1 to find the inverse transform, we obtain

v(t)=\underbrace{\frac{1}{b}\left(1-e^{-\frac{b}{m}t}\right) f_0}_{\textsf{zero-state response}} \;\;+ \underbrace{e^{-\frac{b}{m}t}\,v_0\vphantom{\frac{1}{b}}}_{\textsf{zero-input response}}

(20)

The solution naturally splits into two parts:

The zero-state response, also called the forced response or particular solution, is how the system would respond if it started at rest $v_0=0$ and we just applied the input. This part depends on the input $f_a(t)$ .
The zero-input response, also called the natural response or homogeneous solution, is how the system would respond if there was no input $f_0=0$ and we just had an initial velocity. This part depends on the initial condition $v_0$ .

For our cruise control example, the zero-state response increases from zero and eventually reaches the steady state $v(\infty)=\frac{f_0}{b}$ . Meanwhile, the zero-input response decays exponentially to zero. Here is what the responses look like:

Left: zero-state response (with constant input f_a(t) = f_0). Right: zero-input response (with initial state v(0)=v_0). Both for the cruise control example of . — Figure 3:**Left:** zero-state response (with constant input $f_a(t) = f_0$ ). **Right:** zero-input response (with initial state $v(0)=v_0$ ). Both for the cruise control example of Figure 2.

When we sum the responses, the total $v(t)$ will increase or decrease to asymptotically match $\frac{f_0}{b}$ , depending on whether $v_0$ is smaller or larger than $\frac{f_0}{b}$ , respectively.

Possible total responses for the cruise control example of , with different relative sizes of v_0 and \tfrac{f_0}{b}. — Figure 4:Possible total responses for the cruise control example of Figure 2, with different relative sizes of $v_0$ and $\tfrac{f_0}{b}$ .

Test your knowledge¶

Solution to Exercise 1 #

The function has two nonzero pieces. Specifically,

f(t) = \begin{cases} 0, & t < 0 \\ 1, & 0 \leq t < 1 \\ 2-t, & 1 \leq t < 2 \\ 0, & t\geq 0 \end{cases}

(21)

Evaluating the Laplace transform, we obtain:

\begin{aligned} F(s) &= \Lap\{f\}(s) = \int_{0}^{\infty} f(t)e^{-st}\,\dd t \\ &= \int_{0}^{1} e^{-st}\,\dd t \;+\; \int_{1}^{2} (2-t)e^{-st}\,\dd t. \end{aligned}

(22)

The first integral is

\int_{0}^{1} e^{-st}\,\dd t = \left[-\frac{1}{s}e^{-st}\right]_{0}^{1} = \frac{1-e^{-s}}{s}.

(23)

For the second integral, use integration by parts with

u = 2-t,\quad \dd v = e^{-st}\,\dd t, \quad\implies\quad \dd u = -\dd t,\quad v = -\frac{1}{s}e^{-st}.

(24)

Then

\begin{aligned} \int_{1}^{2} (2-t)e^{-st}\,\dd t &= \Big[uv\Big]_{1}^{2} - \int_{1}^{2} v\,\dd u \\ &= \left[-\frac{2-t}{s}e^{-st}\right]_{1}^{2} - \int_{1}^{2}\left(-\frac{1}{s}e^{-st}\right)(-\dd t) \\ &= \left[-\frac{2-t}{s}e^{-st}\right]_{1}^{2} - \frac{1}{s}\int_{1}^{2} e^{-st}\,\dd t \\ &= \left[-\frac{2-t}{s}e^{-st}\right]_{1}^{2} - \frac{1}{s}\left[-\frac{1}{s}e^{-st}\right]_{1}^{2} \\ &= \frac{e^{-s}}{s} + \frac{e^{-2s}-e^{-s}}{s^{2}}. \end{aligned}

(25)

Putting the pieces together,

\begin{aligned} F(s) &= \frac{1-e^{-s}}{s} \;+\; \left(\frac{e^{-s}}{s} + \frac{e^{-2s}-e^{-s}}{s^{2}}\right) \\ &= \frac{1}{s} + \frac{e^{-2s}-e^{-s}}{s^{2}}. \end{aligned}

(26)

Therefore,

\boxed{F(s) = \frac{1}{s} + \frac{e^{-2s}-e^{-s}}{s^{2}}}

(27)

Footnotes¶

Drag is actually proportional to the square of velocity, as we saw in the cruise-control example from the linearization section. The linear model $f_d = b v$ used here is the linearization of that quadratic drag about a nominal speed $v_0$ , with $b = 2 c v_0$ . Working with the linear model keeps the system LTI so we can apply the Laplace transform.
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