We have just seen that the DTFT can be viewed as a unitary operator between ℓ2(Z)ℓ2(Z) and L2[-π,π]L2[-π,π].
One can repeat this process for each Fourier transform pair. In fact due to
the symmetry between the DTFT and the CTFS, we have already established this
for CTFS, i.e.,
CTFS:
L
2
[
-
π
,
π
]
→
ℓ
2
(
Z
)
CTFS:
L
2
[
-
π
,
π
]
→
ℓ
2
(
Z
)
(1)is a unitary operator. Similarly, we have
CTFS:
L
2
(
R
)
→
L
2
(
R
)
CTFS:
L
2
(
R
)
→
L
2
(
R
)
(2)is a unitary operator as well. The proof of this fact closely mirrors the
proof for the DTFT. Finally, we also have
DFT:
C
N
→
C
N
.
DFT:
C
N
→
C
N
.
(3)This operator is also unitary, which can be easily verified by showing that
the DFT matrix is actually a unitary matrix: UHU=UUH=IUHU=UUH=I.
Note that this discussion only applies to finite-energy (ℓ2/L2ℓ2/L2)
signals. Whenever we talk about infinite-energy functions (things like the
unit step, delta functions, the all-constant signal) having a Fourier
transform, we need to be very careful about whether we are talking about a
truly convergent Fourier representation or whether we are merely using an
engineering “trick” or convention.
