Geology Reference
In-Depth Information
AO(I)=A(I)
40 CONTINUE
C Find remaining elements of new auxiliary vector.
LM2=L-2
DO 50 J=1,LM2
JP1=J+1
K=L-J
A(JP1)=A(JP1)+AK*DCONJG(AO(K))
50 CONTINUE
C Update alpha.
20 ALPHA=ALPHA+AK*DCONJG(BETA)
C Find q for L-1.
Q=(G(L)-GAMMA)/DCONJG(ALPHA)
C Find last element of solution vector.
F(L)=Q
C Find remaining elements of the solution vector.
DO 60 J=1,LM1
K=L-J+1
F(J)=F(J)+Q*DCONJG(A(K))
60 CONTINUE
C Return if solution is complete.
IF(L.EQ.LR) RETURN
C Find new values of beta and gamma.
BETA=(0.D0,0.D0)
GAMMA=(0.D0,0.D0)
DO 10 I=1,L
K=L-I+2
BETA=BETA+A(I)*R(K)
GAMMA=GAMMA+F(I)*R(K)
10
CONTINUE
RETURN
END
2.3 Frequency domain analysis
Time domain sequences can equally well be represented in the frequency domain.
The representations in the two domains have a one-to-one correspondence, and any
operation on the representation in one domain has an equivalent counterpart in the
other. Some operations, such as filtering, are more easily done in the time domain,
while the e
ects of discrete sampling and finite record length, for example, are
more clearly understood in the frequency domain. We begin with a description of
the discrete Fourier transform (DFT) , the main tool for going from the time domain
to the frequency domain.
ff
2.3.1 The discrete Fourier transform
The time sequence is taken to be of length N
=
2 n
+
1, and we write it as
g n ,g n + 1 ,...,g 1 ,g 0 ,g 1 ,...,g n 1 ,g n ,
consisting of equally spaced samples at intervals
Δ
t of a record of length T ,as
illustrated in Figure 2.2.
Search WWH ::




Custom Search