2 Properties of causal functions

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

1. The sum of casusal functions is causal:

   $f\left(t\right)u\left(t\right)+g\left(t\right)u\left(t\right)=\left[f\left(t\right)+g\left(t\right)\right]u\left(t\right)$

2. The product of causal functions is causal:

   $\left\{f\left(t\right)u\left(t\right)\right\}\left\{g\left(t\right)u\left(t\right)\right\}=f\left(t\right)g\left(t\right).u\left(t\right)$

3. The derivative of a causal function is causal:

   $\frac{d}{dt}\left\{f\left(t\right)u\left(t\right)\right\}=\frac{df}{dt}.u\left(t\right)$

4. The definite integral of a causal function is a constant.

Calculating the definite integral of a causal function needs care.

Consider ${\int \; <!--nolimits\; -->}_a^{b}f\left(t\right)u\left(t\right)dt$ where $a<b$ . There are 3 cases to consider $\text{ (i)}b<0\text{ (ii)}a<0,b>0$ and $\text{ (iii)}a>0$ which are described in Figure 10:

Figure 10

• If $b<0$ then $t<0$ and so $u\left(t\right)=0$ $\therefore {\int \; <!--nolimits\; -->}_a^{b}f\left(t\right)u\left(t\right)dt=0$
• If $a<0,b>0$ then

  $F\left(t\right)={\int \; <!--nolimits\; -->}_a^{b}f\left(t\right)u\left(t\right)dt={\int \; <!--nolimits\; -->}_a^{0}f\left(t\right)u\left(t\right)dt+{\int \; <!--nolimits\; -->}_0^{b}f\left(t\right)u\left(t\right)dt=0+{\int \; <!--nolimits\; -->}_0^{b}f\left(t\right)u\left(t\right)dt={\int \; <!--nolimits\; -->}_0^{b}f\left(t\right)dt$

  since, in the first integral $t<0$ and so $u\left(t\right)=0$ whereas, in the second integral $t>0$ and so $u\left(t\right)=1$ .

• If $a>0$ then ${\int \; <!--nolimits\; -->}_a^{b}f\left(t\right)u\left(t\right)dt={\int \; <!--nolimits\; -->}_a^{b}f\left(t\right)dt$ since $t>0$ and so $u\left(t\right)=1.$

Task!

If $f\left(t\right)=\left({\text{e}}^{-t}+t\right)u\left(t\right)$ then find $\frac{df}{dt}$ and ${\int \; <!--nolimits\; -->}_{-3}^{4}f\left(t\right)dt$

Find the derivative first:

Answer

Answer

Now complete the integration:

Answer

Exercises

1. Find the derivative with respect to $t$ of $\left(t^3+sint\right)$ $u\left(t\right)$ .
2. Find the area under the curve $\left(t^3+sint\right)u\left(t\right)$ between $t=-3$ and $t=1$ .
3. Find the area under the curve $\frac{1}{\left(t+3\right)}\left[u\left(t-1\right)-u\left(t-3\right)\right]$ between $t=-2$ and $t=2.5$ .

Answer