0.3 Analog or Discrete? 13
which again coincides with the derivative. Finally, we consider a signal that changes faster thanx(t)
andx 1 (t)such asx 2 (t)=t^2. Samplingx 2 (t)withTs=1, we havex 2 [n]=n^2 and its forward finite
difference is given by
1 [x 2 [n]]= 1 [n^2 ]=(n+ 1 )^2 −n^2 = 2 n+ 1
which gives as an approximation to the derivative 1 [x 2 [n]]/Ts= 2 n+1. The derivative ofx 2 (t)
is 2t, which at 0 equals 0, and at 1 equals 2. On the other hand, 1 [n^2 ]/Tsequals 1 and 3 at
n=0 andn=1, respectively, which are different values from those of the derivative. Suppose
Ts=0.01, so thatx 2 [n]=x 2 (nTs)=(0.01n)^2 =0.0001n^2. If we compute the difference for this signal
we get
1 [x 2 (0.01n)]= 1 [(0.01n)^2 ]=(0.01n+0.01)^2 −0.0001n^2 = 10 −^4 ( 2 n+ 1 )
which gives as an approximation to the derivative 1 [x 2 (0.01n)]/Ts= 10 −^2 ( 2 n+ 1 ), or 0.01 when
n=0 and 0.03 whenn=1 which are a lot closer to the actual values of
dx 2 (t)
dt
|t=0.01n= 2 t|t=0.01n=0.02n
The error now is 0.01 for each case instead of 1 as in the case whenTs=1. Thus, whenever the
rate of change of the signal is faster, the difference gets closer to the derivative by makingTs
smaller.
It becomes clear that the faster the signal changes, the smaller the sampling periodTsshould be in order to
get a better approximation of the signal and its derivative. As we will learn in Chapters 4 and 5 the frequency
content of a signal depends on the signal variation over time. A constant signal has frequency zero, while a
signal that changes very fast over time would have high frequencies. The higher the frequencies in a signal,
the more samples would be needed to represent it with no loss of information, thus requiring thatTsbe
smaller.
0.3.3 Integrals and Summations.........................................................
Integration is the opposite of differentiation. To see this, supposeI(t)is the integration of a
continuous signalx(t)from some timet 0 tot(t 0 <t),
I(t)=
∫t
t 0
x(τ)dτ (0.6)
