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0
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Problem 2.3
x
1
[n] =
1 and
x
2
[n] = cos(
2
πn) =
1, for all
n
. This shows that two signals can be different but
their sampled versions be the same.
Problem 2.4
Let
x
1
[n]
and
x
2
[n]
be two periodic signals with periods
N
1
and
N
2
, respectively, and let
N =
LCM(N
1
, N
2
)
, and define
x[n] = x
1
[n] +x
2
[n]
. Then obviously
x
1
[n +N] = x
1
[n]
and
x
2
[n +N] =
x
2
[n], and hence x[n] = x[n + N], i.e., x[n] is periodic with period N.
For continuous-time signals
x
1
(t)
and
x
2
(t)
with periods
T
1
and
T
2
respectively, in general we
cannot find a
T
such that
T = k
1
T
1
= k
2
T
2
for integers
k
1
and
k
2
. This is obvious for instance if
T
1
=
1 and
T
2
= π
. The necessary and sufficient condition for the sum to be periodic is that
T
1
T
2
be a
rational number.
Problem 2.5
Using the result of problem 2.4 we have:
1.
The frequencies are 2000 and 5500, their ratio (and therefore the ratio of the periods) is
rational, hence the sum is periodic.
2. The frequencies are 2000 and
5500
π
. Their ratio is not rational, hence the sum is not periodic.
3. The sum of two periodic discrete-time signal is periodic.
4.
The fist signal is periodic but
cos[
11000
n]
is not periodic, since there is no
N
such that
cos[11000(n +N)] = cos(11000n) for all n. Therefore the sum cannot be periodic.
5
Proakis/Salehi/Fundamentals of Communications Systems 2E
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