Title: Default for 20.1 Casual Functions
Course: HELM Workbooks 2023-2024

Transforms and causal functions

Without perhaps realising it, we are used to employing transformations in mathematics. For example, we often transform problems in algebra to an equivalent problem in geometry in which our natural intuition and experience can be brought to bear. Thus, for example, if we ask:

"What are those values of \(x\) for which \(x(x-1)(x+2)>0\)" then perhaps the simplest way to solve this problem is to sketch the curve \(y=x(x-1)(x+2)\) and then, by inspection, find for what values of \(x\) it is positive. We obtain the following figure.

[Image: Sketch of the curve.]

We have transformed a problem in algebra into an equivalent geometrical problem.

Clearly, by inspection of the curve, this inequality is satisfied if

\(\qquad -2 < x < 0 \qquad \text{ or if } \qquad x > 1\)

and we have transformed back again to algebraic form.

The Laplace transform is a more complicated transformation than the simple geometric transformation considered above. What is done is to transform a function \(f(t)\) of a single variable \(t\) into another function \(F(s)\) of a single variable \(s\) through the relation:

\(\qquad F(s)=\int_0^\infty\ \text{e}^{-st}f(t)\,dt.\)

The procedure is to produce, for each \(f(t)\) of interest, the corresponding expression \(F(s).\) As a simple example, if \(f(t) = \text{e}^{-2t}\) then

\( \begin{eqnarray*} F(s)&=& \int_0^\infty \text{e}^{-st}\text{e}^{-2t}\,dt\\ &=& \int_0^\infty \text{e}^{-(s+2)t}\,dt\\ &=& \left[{\text{e}^{-(s+2)t}\over -(s+2)}\right]_0^\infty\\ &= &0-{\text{e}^0\over -(s+2)}= {1\over s+2} \end{eqnarray*} \)

(We remind the reader that \(\text{e}^{-kt}\rightarrow 0\) as \(t \rightarrow \infty\) if \(k>0\).)

Causal functions

As we have seen above, the Laplace transform involves an integral with limits \(t=0\) and \(t=\infty\). Because of this, the nature of the function being transformed, \(f(t)\), when \(t\) is negative is of no importance. In order to emphasize this we shall only consider so-called causal functions all of which take the value \(0\) when \(t<0.\)

The simplest causal function is the Heaviside or unit step function denoted by \(u(t)\) and defined by:

\(\qquad u(t)= \begin{cases} 1& \text{if}\quad t\ge 0\\ 0& \text{if}\quad t<0 \end{cases}\)

with graph as in the following figure.

[Image: Graph of Heaviside function]

Similarly we can consider other 'step-functions'. For example, from the above definition we deduce

\( u(t-3)= \begin{cases} 1& \text{if}\quad t-3\ge 0\\ 0& \text{if}\quad t-3<0 \end{cases}\)

or rearranging the inequalities:

\( u(t-3) = \begin{cases} 1& \text{if}\quad t\ge 3\\ 0& \text{if}\quad t<3 \end{cases}\)

with graph as in the following Figure:

[Image: Graph of u(t-3)]

The step function has a useful property: multiplying an ordinary function \(f(t)\) by the step function \(u(t)\) changes it into a causal function; e.g. if \(f(t)=\sin t\) then \( \sin t.u(t)\) is causal. This is illustrated in the change bellow:

[Image: Graph of the sine function]

[Image: Graph of the function h(t) = sin t.u(t)]

Key Point

Causal Functions

If \(u(t)\) is the unit step function and \(f(t)\) is any function then

\(f(t)u(t)\) is a causal function

The step function can be used to 'switch on' functions at other values of \(t\) (which we will normally interpret as time). For example \(u(t-1)\) has the value 1 if \(t\ge 1\) and \(0\) otherwise so that \(\sin t.u(t-1)\) is described by the (solid) curve in the following figure:

[Image: Graph of sin t.u(t−1)]

The step function can also be used to 'switch-off' signals. For example, the step function \(u(t-1)-u(t-3)\) in Figure 7 has the effect on \(f(t)\) such that \(f(t)\left[u(t-1)-u(t-3)\right]\) (described by the solid curve in Figure 8) switches on at \(t=1\) (because then \(u(t-1)-u(t-3)\) takes the value 1), remains 'on' for \(1\le t\le 3\), and then switches 'off' when \(t>3\) (because then \(u(t-1)-u(t-3)=1-1=0\)).

[Image: Graph of u(t − 1) − u(t − 3)]

[Image: Graph of f(t)[u(t − 1) − u(t − 3)]]

If we have an expression \(f(t − a)u(t − a)\) then this is the function \(f(t)\) translated along the \(t\)-axis through a time a. For example \(\sin(t − 2).u(t − 2)\) is simply the causal sine curve \(\sin t.u(t)\) shifted to the right by two units as described in the following Figure.

[Image: Graph of sin(t − 2).u(t − 2)]

Example

Sketch the curve \(f(t)=e^t (u(t-1)-u(t-2)).\)

Solution

You should obtain

[Image: The graph of the above function, which is a curve between x=1 and x=2]

This is botained since, if \(t < 1\) then \(t − 1 < 0\) and \(t − 2 < 0\) and so

\( \quad u(t − 1) = 0, \quad u(t − 2) = 0 \quad \) leading to \(f(t) = 0.\)

Also if \(1 < t < 2\) then \(t − 1 > 0\) and \(t − 2 < 0\) so

\( \quad u(t − 1) = 1 \quad \) and \(\quad u(t − 2) = 0 \quad\) implying \(f(t) = e^t \quad\)for this range of \(t-\)values.

Finally if \(t > 2\) then \(t − 1 > 0\) and \(t − 2 > 0\) and so

\(\quad u(t − 1) = 1, \quad u(t − 2) = 1 \quad \) giving \( \quad f(t) = e^t (1 − 1) = 0.\)

Properties of causal functions

Even though a function \(f(t)\) may be causal we shall still often use the step function \(u(t)\) to emphasize its causality and write \(f(t)\ u(t).\) The following properties are easily verified.

1. The sum of casusal functions is causal:
\(\qquad f(t)u(t)+g(t)u(t)=\left[f(t)+g(t)\right] u(t)\)
2. The product of causal functions is causal:
\(\qquad \{f(t)u(t)\}\ \{g(t)u(t)\} = f(t)g(t).u(t)\)
3. The derivative of a causal function is causal:
\(\qquad {d\over dt}\{f(t)u(t)\}={df\over dt}.u(t)\)
4. The definite integral of a causal function is a constant.

Calculating the definite integral of a causal function needs care.

Consider \(\,\,\int_a^bf(t)u(t)\,dt\) where \(a < b.\)

There are 3 cases to consider \(\text{ (i)}\ \ b < 0\quad \text{ (ii)}\ \ a < 0,\ \ b>0\) and \( \text{ (iii)}\ \ a>0\) which are described in the figure bellow:

[Image: Figure of the 3 cases]

1. If \(b < 0\) then \(t < 0\) and so \(u(t)=0\) \(\quad \therefore \qquad \int_a^b f(t)u(t)\,dt = 0\)
2. If \(a < 0,\ b>0\) then
\(F(t) = \int_a^b f(t)u(t)\,dt = \int_a^0 f(t)u(t)\,dt + \int_0^b f(t)u(t)\,dt =0+\int_0^b f(t)u(t)\,dt = \int_0^b f(t)\,dt\)
since, in the first integral \(t < 0\) and so \(u(t)=0\) whereas, in the second integral \(t>0\) and so \(u(t)=1.\)
3. If \(a>0\) then \(\quad \int_a^b f(t)u(t)\,dt = \int_a^b f(t)\,dt \quad\) since \(t>0\) and so \(u(t)=1.\)

The integral \(\int_0^\infty \text{e}^{-st}f(t)\,dt\) is called the Laplace transform of \(f(t)\) and is denoted by \({\cal L}\{f(t)\}.\)

Key Point

The Laplace Transform

\[{\cal L}\{f(t)\} \equiv \int_0^\infty \text{e}^{-st}f(t)\,dt = F(s)\]

Find \(F(s)\) if \(f(t)=t\) using \(F(s)=\displaystyle \int_0^\infty \text{e}^{-st}t\,dt\)

\(F(s)=\)
Feedback: You should obtain \(F(s)=1/s^2\). You do this by integrating by parts: \[\begin{eqnarray*} F(s)=\int_0^\infty \text{e}^{-st}t\,dt &=&\left[t{\text{e}^{-st}\over(-s)}\right]_0^\infty-\int_0^\infty {\text{e}^{-st}\over (-s)}\,dt \\ &=&0+\int_0^\infty {\text{e}^{-st}\over s}\,dt\\ &=&\left[-{\text{e}^{-st}\over s^2}\right]^\infty_0={1\over s^2}. \end{eqnarray*}\]

Exercise

If \(f(t)=(\text{e}^{-t}+t)u(t)\) then find \(\displaystyle {df\over dt}\) and \(\displaystyle \int_{-3}^4 f(t)\,dt\).

1. Find the derivative first:
\(\displaystyle \frac{\mathrm{d}f}{\mathrm{d}t}=\)
2. Now obtain another integral representing \(\displaystyle \int_{-3}^4 f(t)\,dt\):

\(I = \displaystyle \int_{-3}^4 f(t)\,dt = \displaystyle \int\) |  | \(\mathrm{d}t\)

3. Now complete the integration: (Give your answer correct to 4 d.p.)
\(I =\)
Feedback: 1. \[\displaystyle {\mathrm{d}f\over \mathrm{d}t}=(-\text{e}^{-t}+1)u(t)\] 2. You should obtain \(\displaystyle \int_0^4 (\text{e}^{-t}+t)\,dt\) since \[\displaystyle \int_{-3}^4f(t)\,dt=\int_{-3}^4 (\text{e}^{-t}+t)u(t)\,dt=\int_0^4(\text{e}^{-t}+t)\,dt.\] This follows because in the range \(t=-3\) to \(t=0\) the step function \(u(t)=0\) and so that part of the integral is zero. In the other part of the integral \(u(t)=1\). 3. You should obtain 8.9817 (to 4 d.p.) since \[\begin{align}\displaystyle \int_0^4(\text{e}^{-t}+t)\,dt &= \left[-\text{e}^{-t}+\displaystyle {t^2\over 2}\right]_0^4 \\ &= (-\text{e}^{-4}+8)-(-1) \\ &=-\text{e}^{-4}+9 \\ &\approx 8.9817.\end{align}\]

Exercise

Find the derivative with respect to \(t\) of \((t^[value]+[value]\sin t)\) \(u(t)\).
Feedback: Since the derivative of a causal function is casual, the derivative is: \[\frac{\mathrm{d}}{\mathrm{d}t} (t^[value]+[value]\sin t) u(t) = ([value]t^[value] + [value]\cos(t)) u(t).\]

Exercise

Find the area under the curve \( (t^[value] + \sin(t)) u(t)\) between \(t=[value]\) and \(t=[value]\). (Give your answer correct to 4 d.p. )
Feedback: The area under the curve is given by: \[\displaystyle \int_{[value]}^[value] (t^[value] + \sin(t)) u(t) \,dt =\int_0^[value] (t^[value] + \sin(t))\,dt.\] This follows because in the range \(t=[value]\) to \(t=0\) the step function \(u(t)=0\) and so that part of the integral is zero. In the other part of the integral \(u(t)=1\). You should obtain [value] (to 4 d.p.) since: \[ \begin{align} \int_0^[value] (t^[value] + \sin(t))\,dt &= \left[\frac{t^[value]}{[value]} - \cos(t) \right]_0^1 \\ &= \frac{1}{[value]} - \cos(1) + 1 \\ &\approx [value]. \end{align}\]

Exercise

Find the area under the curve \(\displaystyle {1\over (t+[value])} [u(t-1)-u(t-3)]\) between \(t=[value]\) and \(t=[value]\). (Give your answer correct to 4 d.p. )
Feedback: The area under the curve is given by: \[\displaystyle \int_{[value]}^[value] {1\over (t+[value])} [u(t-1)-u(t-3)] \,dt =\int_1^[value] {1\over (t+[value])}\,dt.\] This follows because in the range \( 1 \leq t < 3 \) the step function \(u(t-1)-u(t-3) = 1\) and it is zero elsewhere. You should obtain [value] (to 4 d.p.) since: \[ \begin{align} \int_1^[value] {1\over (t+[value])} \,dt &= \left[ \ln(t + [value]) \right]_1^[value] \\ &= \ln([value]) - \ln([value]) \\ &\approx [value]. \end{align}\]