Description
Problem 1. (30 points) For the signal x[n] = n(u[n + 4] − u[n − 5]), compute and sketch the
following:
(a) (5 points) g1[n] = x[n − 3]
(b) (5 points) g2[n] = x[2n − 3]
(c) (5 points) g3[n] = x[−n]
(d) (5 points) g4[n] = x[2 − n]
(e) (5 points) g5[n] = (
x[n/2], if n/2 is integer
0, else
(f ) (5 points) g6[n] = x[n]δ[n]
Problem 2. (15 points) Consider the sequences
x[n] = (
1
2
n
, 0 ≤ n ≤ 3
0, otherwise
and
y[n] = (
1
4
n−1
, 0 ≤ n ≤ 5
0, otherwise
(a) (5 points) Plot the samples of x[n] and y[n].
(b) (5 points) Plot the samples of the sequence z[n] = x[n]y[n], which are obtained from the
point-wise product of the samples of x[n] and y[n].
(c) (5 points) What is the energy of the sequences x[n] and z[n]?
Problem 3. (10 points) For each of the discrete-time signals listed below, determine if the signal
is periodic or not. If the signal is periodic, determine the fundamental period.
(a) (5 points) x[n] = 3 cos( 5
17n +
π
4
)
(b) (5 points) x[n] = 2 cos(0.48πn) + 1.5 sin(0.8πn + π/3)
Problem 4. (15 points) Consider an arbitrary sequence x[n] with even and odd parts denoted by
xe[n] and xo[n], respectively. Show that the DC level of x[n] coincides with the DC level of its even
part.
The DC level of a sequence x[n] is defined as the average value of the sequence:
x¯ , lim
N→∞
1
2N + 1
·
X
N
n=−N
x[n]
!
Problem 5. (15 points) Let x[n] = x1[n]x2[n]. Show that:
(a) (5 points) If both x1[n] and x2[n] are even, then x[n] is even.
(b) (5 points) If both x1[n] and x2[n] are odd, then x[n] is even.
(c) (5 points) If x1[n] is even and x2[n] is odd, then x[n] is odd.
Problem 6. (15 points) Consider a discrete-time signal, x[n], defined over the interval n ≥ 0. At
each time n, let x¯[n] denote the average value of the samples x[n] from time k = 0 up to time k = n,
i.e.,
x¯[n] = 1
n + 1
Xn
k=0
x[k]
This is an example of one processing algorithm; it acts on the data and generates x¯[n]. We would
like to motivate an alternative processing algorithm that operates on the data in a recursive manner
to generate the same x¯[n]. Show that x¯[n] satisfies the recursion:
x¯[n] = n
n + 1
x¯[n − 1] + 1
n + 1
x[n]
with initial condition x¯[0] = x[0]. Note that the above algorithm is in terms of the previous value
x¯[n − 1] and the most recent term in the sequence, x[n]. In this way, the second procedure for
evaluating the mean of the sequence does not need to save all prior data; the history of the prior data
is incorporated into x¯(n − 1) and only the most recent sample, x[n], is needed along with x¯[n − 1] to
evaluate x¯[n].
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