Digital Signal Processing Reference
In-Depth Information
% Typical Test call:
% LVxCorrDelayMeasure(0.1)
22. For an input signal consisting of a unit step, using pencil and paper and the difference equation
y
[
n
]=
x
[
n
]+
ay
[
n
−
1
]
compute the first five output values of a single-pole IIR having the following pole values:
(a)
a
= 0.99
(b)
a
= 1.0
(c)
a
= 1.01
23. Repeat the previous exercise, but, using m-code, compute the first 100 values of output and
plot the results on a single figure having three subplots.
24. For an input signal consisting of a unit impulse, using pencil and paper, compute the first five
output values of a single-pole IIR having the following pole values:
(a)
a
= 0.98
(b)
a
= 1.0
(c)
a
= 1.02
25. Repeat the previous exercise, but, using m-code, compute the first 100 values of output and
plot the results on a single figure having three subplots.
26. Consider a cascade of two single-pole filters each of which has a pole at 0.95. Compute the
impulse response of the cascaded combination of IIRs the following two ways:
(a) Compute the first 100 samples of the impulse response of the first IIR, then, using the result
as the input to the second filter, compute the net output, which is the net impulse response.
(b) Compute the first 100 samples of the net output of the composite filter made by 1) convolving
the coefficient vectors for the two individual single pole filters to obtain a second-order coefficient vector,
then 2) use the function
filter
to process a unit impulse of length 100 to obtain the impulse response;
compare to the result obtained in (a).
27. Let two IIRs each be a properly-scaled leaky integrator with
β
= 0.1. Now construct a filter
as the cascade connection of the two leaky integrators. Filter a cosine of frequency 2500 Hz and unity
amplitude, sampled at 10,000 Hz with the cascade of two leaky integrators. Determine the steady state
amplitude of the output.
28. Determine the correlation (CZL) for the following two sequences, for the values of
N
indicated,
where
n
= 0:1:
N
−
1:
S
1
=
cos
(
2
πn
3
/N)
+
sin
(
2
πn
4
/N)
S
2
=
cos
(
2
πn
4
/N)
+
sin
(
2
πn
3
/N)
(a)
N
= 0.5;
(b)
N
=1;
(c)
N
=2;
(d)
N
=3;
(e)
N
=6;
(f )
N
=8;
