A.1)
1.
The Fourier transform is just an exercise in math. It is: X
c
(
Ω
)=Sc(
Ω
)(1+a*e
j
τΩ
)
2.
The Fourier transform of the sequence x[n] is an application of the sampling theorem:
∑
∞
∞
=
+
+
+
=
n
T
n
w
j
c
jw
ae
T
n
w
j
S
T
e
X
]
1
)[
2
(
1
)
(
2
π
τ
π
3.
Result in 2.) is a repeating pattern. Since it nonoverlapping we can consider its first period (i.e. n=0).
That
gives us the desired result:
T
w
j
ae
w
H
τ
+
=
1
)
(
A.2)
We know that the ideal differentiator (from differentiation property) has the Fourier transform H(
Ω
)=j
Ω
. Its
equivalent, sampled version will be H(w)=jw/T (if we choose h[n]=Th
c
(nT)) given a sampling period T=1/10
sec. While the practical implementation of such a sampled differentiator is an entirely different story the
behavior should be identical to the ideal case IF the frequency of the input signal is within the bandwidth (via
Shannon’s theorem). So all we have to observe is if aliasing occurs:
1.
x(t)=cos(6*pi*t) => Sampled at T=1/10s this signal will be within [–
π,π
] and the result will be equal to
continuous case.
2.
x(t)=cos(14*pi*t) => Sampled at T=1/10s this signal will be aliased to frequency 0.4
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 Fall '06
 NA
 Digital Signal Processing, Signal Processing, sampling rate, 10 sec

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