ECE 3250
HOMEWORK II SOLUTIONS
Fall 2009
Note: I’ve expressed the solutions to the problems involving calculating convolutions in
particular ways that match up well with some stuﬀ we’ll be doing later in the course. Your
answers might not look exactly like mine, but perhaps if you check you’ll ﬁnd that your
answers are the same as mine but in a slightly diﬀerent form.
1.
I’ll recount the sequence of steps and then justify each one.
x
1
*
x
2
(
n
)
=
∞
X
k
=
∞
x
1
(
k
)
x
2
(
n

k
)
=
∞
X
k
=
∞
u
(
k
)3

(
n

k
)
u
(
n

k
)
=
∞
X
k
=0
3

(
n

k
)
u
(
n

k
)
=
P
n
k
=0
3

(
n

k
)
if
n
≥
0
0
if
n <
0
.
The ﬁrst line holds by deﬁnition of convolution. The second line results from plugging in
for
x
1
and
x
2
. The third line holds because
u
(
k
) = 0 for
k <
0 and
u
(
k
) = 1 for
k
≥
0.
The last line holds because
u
(
n

k
) = 1 when
k
≤
n
and
u
(
n

k
) = 0 when
k > n
. In
particular, when
n <
0,
u
(
n

k
) = 0 for all
k
in the range of summation 0
≤
k <
∞
.
You can simplify the sum on the last line using the geometricseries thing as follows:
n
X
k
=0
3

(
n

k
)
= 3

n
n
X
k
=0
3
k
=
3

n
(1

3
n
+1
)
1

3
=
3
2

1
2
3

n
.
Accordingly,
x
1
*
x
2
(
n
) =
3
2

1
2
3

n
if
n
≥
0
0
if
n <
0
,
which is the same as
x
1
*
x
2
(
n
) =
„
3
2

1
2
3

n
«
u
(
n
) for all
n
∈
Z
.
I’ve written the answer in a speciﬁc form for a reason that won’t become obvious until we
discuss
z
transforms.
2.
I’ll proceed as in the previous problem.
x
1
*
x
2
(
n
)
=
∞
X
k
=
∞
x
1
(
k
)
x
2
(
n

k
)
=
∞
X
k
=
∞
3
k
u
(
k
)7
n

k
u
(
n

k
)
=
∞
X
k
=0
3
k
7
n

k
u
(
n

k
)
=
P
n
k
=0
7
n
(3
/
7)
k
if
n
≥
0
0
if
n <
0
.
The ﬁrst line holds by deﬁnition of convolution. The second line results from plugging in
for
x
1
and
x
2
. The third line holds because
u
(
k
) = 0 for
k <
0 and
u
(
k
) = 1 for
k
≥
0.
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The last line holds because
u
(
n

k
) = 1 when
k
≤
n
and
u
(
n

k
) = 0 when
k > n
. In
particular, when
n <
0,
u
(
n

k
) = 0 for all
k
in the range of summation 0
≤
k <
∞
.
You can simplify the sum on the last line using the geometricseries thing as follows:
n
X
k
=0
7
n
(3
/
7)
k
= 7
n
n
X
k
=0
(3
/
7)
k
=
7
n
(1

(3
/
7)
n
+1
)
1

(3
/
7)
=
7
4
7
n

3
4
3
n
.
Accordingly,
x
1
*
x
2
(
n
) =
7
4
7
n

3
4
3
n
if
n
≥
0
0
if
n <
0
,
which is the same as
x
1
*
x
2
(
n
) =
„
7
4
7
n

3
4
3
n
«
u
(
n
) for all
n
∈
Z
.
I’ve written the answer in a speciﬁc form for a reason that won’t become obvious until we
discuss
z
transforms.
3.
I’ll proceed as in the previous problems.
x
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