We've seen that if
ft
f
t
is bandlimited to
−π
π
,
we can reconstruct
it from its samples at the integers exactly. The key to this
was the fact that the DTFT of
f
s
n
f
s
n
is just a periodic version of the CTFT of
ft
f
t
Let's examine the situation in general
What is the relationship of the CTFT of
ft
f
t
to the DTFT of
f
s
n=ft
f
s
n
f
t
sampled at
nT
n
T
?
We will use the following notation. Note that we will use
different variables, ΩΩ
and ωω, so that we do not
get confused.
ft
CTFT
FiΩ
f
t
CTFT
F
Ω
fnT=
f
s
n
CTFT
F
s
eiω
f
n
T
f
s
n
CTFT
F
s
ω
ft=12π∫−∞∞FiΩeiωnd
n
f
t
1
2
n
F
Ω
ω
n
F
s
eiω=∑n=−∞∞
f
s
ne−(iωn)=∑n=−∞∞fnTe−(iωn)=∑n=−∞∞12π∫−∞∞FiΩeiΩnTdΩe−(iωn)=12π∫−∞∞FiΩ∑n=−∞∞ein(ΩT−ω)dΩ=12π∫−∞∞FiΩ2π∑r=−∞∞δω−(ΩT−2πr)dΩ=∫−∞∞FiΩ∑r=−∞∞δω−(ΩT−2πr)dΩ=∫−∞∞FiΩ∑r=−∞∞δ(−T)(Ω−ω−2πrT)dΩ
F
s
ω
n
f
s
n
ω
n
n
f
n
T
ω
n
n
1
2
Ω
F
Ω
Ω
n
T
ω
n
1
2
Ω
F
Ω
n
n
Ω
T
ω
1
2
Ω
F
Ω
2
r
δ
ω
Ω
T
2
r
Ω
F
Ω
r
δ
ω
Ω
T
2
r
Ω
F
Ω
r
δ
T
Ω
ω
2
r
T
(1)
Recall, property of
δt
δ
t
δaΩ=1|a|δΩ
δ
a
Ω
1
a
δ
Ω
(2)
F
s
eiω=∫−∞∞FiΩ∑r=−∞∞1TδΩ−ω−2πrTdΩ=1T∑r=−∞∞∫−∞∞FiΩδΩ−ω−2πrTdΩ=1T∑r=−∞∞Fiω−2πrT
F
s
ω
Ω
F
Ω
r
1
T
δ
Ω
ω
2
r
T
1
T
r
Ω
F
Ω
δ
Ω
ω
2
r
T
1
T
r
F
ω
2
r
T
(3)
F
s
eiω
F
s
ω
is a sum of shifted (and scaled)
Fiω
F
ω
's. In order to get to our final results in
Equation 3, we used the
sifting
property.
Fs
eiω=1T∑r=,∞∞Fiω−2πrT
Fs
ω
1
T
r
F
ω
2
r
T
Let us take a minute to examine this relation in 2 steps:
FiΩ→FiωT
→
F
Ω
F
ω
T
Ω=πT→ω=π
→
Ω
T
ω
, this covers the
r=0
r
0
case in the sum above.
Fs
eiω=1T∑r=−∞∞Fiω−2πrT
Fs
ω
1
T
r
F
ω
2
r
T
What this equation says is that we "plop down" a copy of
1TFiωT
1
T
F
ω
T
at every multiple of
2π
2
(and sum up)..
Fs
eiω
Fs
ω
is
2π
2
-periodic, just like we know it should be.
Recall,
T=sample period
T
sample period
(
i.e. each sample in the
discrete-time has a distance of
TT between them). Before, we
saw that if
T=1
T
1
(sample on the integers) and
ft
f
t
is bandlimited to
−π
π
we can reconstruct
ft
f
t
exactly from its samples
fs
n
fs
n
.
Say
ft
f
t
is bandlimited to
−ΩB
ΩB
ΩB
ΩB
. Then if we sample with period
T≤π
ΩB
T
ΩB
, we can reconstruct
ft
f
t
exactly from its samples
fs
n
fs
n
.
Show this following the exact same steps as in the
−π
π
,
T=1
T
1
case.
-
If
ft
f
t
is bandlimited to
−ΩB
ΩB
ΩB
ΩB
, then
FiωT
F
ω
T
,
T≤π
ΩB
T
ΩB
is zero outside of
−π
π
-
then
1T∑r=,∞∞Fiω−2πrT
1
T
r
F
ω
2
r
T
= DTFT of
fs
n=fnT
fs
n
f
n
T
is just a periodic version of
1TFiωT
1
T
F
ω
T
Fs
eiω=1T∑r=−∞∞Fiω−2πrT
Fs
ω
1
T
r
F
ω
2
r
T
-
turning
fs
n
fs
n
into an impulse sequence
fimp
t=∑n=−∞∞
fs
nδt−nT
fimp
t
n
fs
n
δ
t
n
T
-
From here we make some calculations and we get
fimp
t↔∫−∞∞
fimp
te−(iωt)dt=∫−∞∞∑n=−∞∞
fs
nδt−nTe−(iωt)dt
↔
fimp
t
t
fimp
t
ω
t
t
n
fs
n
δ
t
n
T
ω
t
so,
fimp
t=∑n=−∞∞
fs
n∫−∞∞δt−nTe−(iωt)dt=∑n=−∞∞
fs
ne−(iωTn)=
Fs
ei(ωT)
fimp
t
n
fs
n
t
δ
t
n
T
ω
t
n
fs
n
ω
T
n
Fs
ω
T
-
to recover
f
~
t=ft
f
~
t
f
t
from
fimp
t
fimp
t
, we lowpass filter with cutoff frequency
πT
T
(also the filter has a gain of
1T
1
T
).
If
Giω
G
ω
is "ideal", then
F
~
iω
F
~
ω
and the signal is reconstructed
perfectly.
Starting with the following signal,
ft
ft,
and its CTFT,
Fiω
F
ω
,
use the same steps above to sample the signal, convert it
into an impulse train, and then filter it to recreate the
original signal.