ECE 2200 Signals and Information – Section 5 (Week 5 Mar 14)
Page 1
1. The FIR ﬁlter with impulse response
h
[
n
] = 2
δ
[
n

2]
has output
y
[
n
] =
u
[
n

3]

u
[
n

6]
.
Determine the input x[n].
—————————————
±
—————————————
Solution
(see “Section 56.2.1 Convolution as an Operator”):
The
deconvolution
of
y
[
n
] is as follows:
±
±
±
y
[
n
] =
x
[
n
]
*
h
[
n
]
±
±
±
= 2
x
[
n

2]
±
±
±
⇒
x
[
n
] =
1
2
u
[
n

1]

1
2
u
[
n

4]
2. The FIR ﬁlter has a step starting at sample zero as its input, i.e.
x
[
n
] =
u
[
n
], and the output
is the Kronecker delta function, i.e.
y
[
n
] =
δ
[
n
]. Determine the impulse response of the ﬁlter.
—————————————
±
—————————————
Solution
: The output
y
[
n
] =
δ
[
n
] is achieved by convolving the unitstep input
x
[
n
] =
u
[
n
]
with the ﬁrst diﬀerence
h
[
n
] =
δ
[
n
]

δ
[
n

1]
.
3. Consider the linear, timeinvariant system described by the diﬀerence equation
y
[
n
] = 21
x
[
n
]

37
x
[
n

1] + 21
x
[
n

2]
(i) Find the frequency response of
H
(
e
j
ˆ
ω
) as a mathematical formula in polar form.
(ii) Determine the period of
H
(
e
j
ˆ
ω
) which is a periodic function of ˆ
ω
.
(iii) Plot the magnitude and phase of
H
(
e
j
ˆ
ω
) as a function of ˆ
ω
for

π <
ˆ
ω <
5
π
. Check
your answer using the MATLAB function
freqz
.
(iv) Find ALL frequencies ˆ
ω
, if any, for which the output response to the input
e
j
ˆ
ωn
is zero.
(v) For the input
x
[
n
] = sin(2
πn/
17)
determine the output in the form
y
[
n
] =
M
cos(
γn
+
θ
)
by specifying
M
,
γ
, and
θ
.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Page 2
—————————————
±
—————————————
Solution
(read “Section 61 Sinusoidal Response of FIR Systems”):
(i) The ﬁlter coeﬃcients are
b
0
= 21,
b
1
=

37 and
b
2
= 21. By Equation (6.4), the
frequencyresponse function
(or frequencyresponse) for the system after factoring out
e

j
ˆ
ω
and reducing complex exponentials using Euler’s formula is
±
±
±
H
(
e
j
ˆ
ω
) =
e

j
ˆ
ω
(42 cos ˆ
ω

37)
(ii) 2
π
(iii) As indicted in the MATLAB documentation,
freqz
returns a frequency response
h
and
frequency vector
w
with values ranging from 0 to
π
. By interpreting this result, we
obtain the magnitude and phase plot shown in Figure 1.
0
20
40
60
80
−
3
−
2
−
1
0
1
2
3

π
0
π
2
π
3
π
4
π
5
π

π
0
π
2
π
3
π
4
π
5
π

H
(
e
j
ˆ
ω
)

∠
H
(
e
j
ˆ
ω
)
Figure 1: (top) Magnitude and (bottom) phase for
H
(
e
j
ˆ
ω
) over three periods.
(iv)
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '05
 JOHNSON
 LTI system theory, Impulse response

Click to edit the document details