Signals and Systems - Electrical Engineering

(avery) #1

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)is

dI(t)
dt

=lim
h→ 0

I(t)−I(t−h)
h

=lim
h→ 0

1

h

∫t

t−h

x(τ)dτ

≈lim
h→ 0

x(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

∫t

t 0

x(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 written

D[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

∫^10

0

t 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

∑^9

n= 0

p[n]=

∑^9

n= 0

n= 0 + 1 + 2 +··· 9 =0.5

[ 9


n= 0

n+

∑^0

k= 9

k

]

=0.5

[ 9


n= 0

n+

∑^9

n= 0

( 9 −n)

]

=

9

2

∑^9

n= 0

1 =

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.

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