14 C H A P T E R 0: From the Ground Up!
or the sum of the area underx(t)fromt 0 tot. Notice that the upper bound of the integral istso the
integrand depends on a dummy variable.^2 The derivative ofI(t)isdI(t)
dt=lim
h→ 0I(t)−I(t−h)
h=lim
h→ 01
h∫tt−hx(τ)dτ≈lim
h→ 0x(t)+x(t−h)
2=x(t)where the integral is approximated as the area of a trapezoid with sidesx(t)andx(t−h)and height
h. Thus, for a continuous signalx(t),d
dt∫tt 0x(τ)dτ=x(t) (0.7)or if using the derivative operatorD[.], then its inverseD−^1 [.] should be the integration operator.
That is, the above equation can be writtenD[D−^1 [x(t)]]=x(t). (0.8)We will see in Chapter 3 a similar relation between the derivative and the integral. The Laplace trans-
form operatorssand 1/s(just likeDand 1/D) imply differentiation and integration in the time
domain.Computationally, integration is implemented by sums. Consider, for instance, the integral ofx(t)=t
from 0 to 10, which we know is equal to∫^100t dt=t^2
2∣
∣^10 t= 0 =50.That is, the area of a triangle with a base of 10 and a height of 10. ForTs=1, suppose we approximate
the signalx(t)by pulsesp[n] of widthTs=1 and heightnTs=n, or pulses of areanforn=0,..., 9.
This can be seen as a lower-bound approximation to the integral, as the total area of these pulses
gives a result smaller than the integral. In fact, the sum of the areas of the pulses is given by∑^9n= 0p[n]=∑^9
n= 0n= 0 + 1 + 2 +··· 9 =0.5[ 9
∑
n= 0n+∑^0
k= 9k]
=0.5
[ 9
∑
n= 0n+∑^9
n= 0( 9 −n)]
=
9
2
∑^9
n= 01 =
10 × 9
2
= 45
(^2) The integralI(t)is a function oftand as such the integrand needs to be expressed in terms of a so-calleddummy variableτthat takes
values fromt 0 totin the integration. It would be confusing to let the integration variable bet. The variableτis called adummy variable
because it is not crucial to the integration; any other variable could be used with no effect on the integration.