Signals and Systems - Electrical Engineering

12 C H A P T E R 0: From the Ground Up!

signals is given. In Chapter 7, where we consider the problem of sampling, we will use this relation to

determine appropriate values for the sampling period.

0.3.2 Derivatives and Finite Differences

Differentiation is an operation that is approximated in finite calculus. The derivative operator

D[x(t)]=

dx(t)

dt

=lim

h→ 0

x(t+h)−x(t)

h

(0.2)

measures the rate of change of an analog signalx(t). In finite calculus theforward finite-difference

operator

1 [x(nTs)]=x((n+ 1 )Ts)−x(nTs) (0.3)

measures the change in the signal from one sample to the next. If we letx[n]=x(nTs), for a known

Ts, the forward finite-difference operator becomes a function ofn:

1 [x[n]]=x[n+1]−x[n] (0.4)

The forward finite-difference operator measures the difference between two consecutive samples: one

in the futurex((n+ 1 )Ts)and the other in the presentx(nTs). (See Problem 0.4 for a definition of

thebackward finite-difference operator.) The symbolsDand 1 are called operators as they operate on

functions to give other functions. The derivative and the finite-difference operators are clearly not the

same. In the limit, we have that

dx(t)

dt

|t=nTs=lim

Ts→ 0

1 [x(nTs)]

Ts

(0.5)

Depending on the signal and the chosen value ofTs, the finite-difference operation can be a crude or

an accurate approximation to the derivative multiplied byTs.

Intuitively, if a signal does not change very fast with respect to time, the finite-difference approximates

well the derivative for relatively large values ofTs, but if the signal changes very fast one needs very

small values ofTs. The concept of frequency of a signal can help us understand this. We will learn that

the frequency content of a signal depends on how fast the signal varies with time; thus a constant

signal has zero frequency while a noisy signal that changes rapidly has high frequencies. Consider a

constant signalx 0 (t)=2 having a derivative of zero (i.e., such a signal does not change at all with

respect to time or it is a zero-frequency signal). If we convert this signal into a discrete-time signal

using a sampling periodTs=1 (or any other positive value), thenx 0 [n]=2 and so

1 [x 0 [n]]= 2 − 2 = 0

coincides with the derivative. Consider then a signalx 1 (t)=twith derivative 1 (this signal changes

faster thanx(t)so it has frequencies larger than zero). If we sample it usingTs=1, thenx 1 [n]=n

and the finite difference is

1 [x 1 [n]]= 1 [n]=(n+ 1 )−n= 1