2 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:

    f ( t ) u ( t ) + g ( t ) u ( t ) = f ( t ) + g ( t ) u ( t )

  2. The product of causal functions is causal:

    { f ( t ) u ( t ) } { g ( t ) u ( t ) } = f ( t ) g ( t ) . u ( t )

  3. The derivative of a causal function is causal:

    d d t { f ( t ) u ( t ) } = d f d t . u ( t )

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

Calculating the definite integral of a causal function needs care.

Consider a b f ( t ) u ( t ) d t where a < b . There are 3 cases to consider  (i) b < 0  (ii) a < 0 , b > 0 and  (iii) a > 0 which are described in Figure 10:

Figure 10

No alt text was set. Please request alt text from the person who provided you with this resource.

Task!

If f ( t ) = ( e t + t ) u ( t ) then find d f d t and 3 4 f ( t ) d t

Find the derivative first:

d f d t = ( e t + 1 ) u ( t ) Now obtain another integral representing 3 4 f ( t ) d t :

You should obtain 0 4 ( e t + t ) d t since

3 4 f ( t ) d t = 3 4 ( e t + t ) u ( t ) d t = 0 4 ( e t + t ) d t

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 .

Now complete the integration:

You should obtain 8.9817 (to 4 d.p.) since

0 4 ( e t + t ) d t = e t + t 2 2 0 4 = ( e 4 + 8 ) ( 1 ) = e 4 + 9 8.9817

Exercises
  1. Find the derivative with respect to t of ( t 3 + sin t ) u ( t ) .
  2. Find the area under the curve ( t 3 + sin t ) u ( t ) between t = 3 and t = 1 .
  3. Find the area under the curve 1 ( t + 3 ) [ u ( t 1 ) u ( t 3 ) ] between t = 2 and t = 2.5 .
  1. ( 3 t 2 + cos t ) u ( t )
  2. 0.7097
  3. 0.3185