Aliasing
Challenge 03:
Sampling
Lesson 04
liasing
Aliasing
What is it?
How do you measure it?
How do you manage it?
Challenge 04
Lesson 04
epeat Challenge 02
Repeat Challenge 02
1
st
Order Impulse Invariant System
simple first rder
C
ircuit is show.
A simple first order
RC
circuit is show.
The relationship between the input forcing function
v
(
t
) and voltage
ld
t
h
i
tdt
d
i df
i db th 1
t
d
developed across the capacitor, denoted
v
o
(
t
), is defined by the 1
st
order
ordinary differential equation
v
hat is the equivalent discrete
me model of the circuit’s impulse
)
(
1
=
)
(
1
)
(
0
0
t
v
RC
t
v
RC
dt
t
dv
What is the equivalent discretetime model of the circuit s impulse
response based on a sample rate of f
s
=1 k Sa/s, and RC=10
2
?
Lesson 04
epeat Challenge 02
Repeat Challenge 02
It immediately follows that the system’s impulse response is given by:
1
=
/

RC
t
If the sampling period is
T
s
, the resulting discretetime impulse
esponse is:
)
(
)
(
t
u
e
RC
t
h
response is:
kT
h
T
h
or, for
k
0,
)
(
]
[
s
s
d
k
s
RC
kT
s
d
RC
T
e
RC
T
k
h
s
/

=
]
[
RC
T
s
e
/
;
α
Lesson 04
epeat Challenge 02
Repeat Challenge 02
or
C
10
and
1000
z the following results
For
RC
=10
2
, and
f
s
=1000 Hz, the following results.
0
0
0
1
.
0
10
/
10
/
;
904837
.
0
2
3
1
.
0
10
10
*
10
1
3
2
RC
T
e
e
e
s
s
s
d
C
kT
h
T
k
h
)
(
]
[
k
k
s
e
RC
T
)
.
(
.
)
(
/
.
904837
0
1
0
1
0
This represents the impulse response in the discrete time domain but
does not reduce the impulse response to a difference equation.
Lesson 04
epeat Challenge 02
Repeat Challenge 02
Where would you find a connection between a timedomain function
nd difference equation?
able of z
ansforms (signals and systems
and difference equation? – Table of ztransforms (signals and systems
stuff).
able 7 1 (Principles of Signals and System F Taylor McGraw
ill
Known, h
d
[k]
(T
s
/RC) e
kT
s
/RC
=(0.1) (0.904837)
k
for
RC
=10
2
, and
Table 7.1 (Principles of Signals and System, F. Taylor, McGrawHill
1
1
z
kT
u
e
k
h
s
kT
kT
akT
)
(
]
[
f
s
=1000 Hz, the following results.
1
1
904
0
1
1
z
z
e
e
z
s
s
akT
akT
s
d
)
.
(
Therefore:
1
1
904
0
1
1
0
1
1
0
1
0
1
0
z
z
e
e
z
z
kT
u
e
k
h
s
s
s
akT
akT
s
akT
d
)
.
(
.
.
.
)
(
.
]
[
Lesson 04
epeat Challenge 02
Repeat Challenge 02
rom
1
0
T
akT
.
From
1
904
0
1
1
0
z
kT
u
e
k
h
s
d
)
.
(
)
(
.
]
[
the following difference equation arises
h
d
[k]  0.904837h
d
[k1] = 0.1x[k]
or
h
d
k] = 0.904837h
d
[k1] + 0.1x[k]
x[k]=
] then y[k]:
[k]
y[k]
If x[k]=
[k], then y[k]:
y[0] = 0.1
1]=(01)(09)
x[k]
T
0.904837
y[1] = (0.1)(0.9)
y[2]= (0.1)(0.9)(0.9)=
=(0.1)(0.9)
2
0.1
Lesson 04
…
epeat Challenge 02
>> den=[1 0.9094837]; num=[0.1];
Repeat Challenge 02
>> x=[1 0 0 0 0 0 0 0 0 0 0 0 0 ] ;
% impulse
>> h=filter(num,den,x);
% impulse response
> plot(h)
>> plot(h)
h[k]
Lesson 04
hallenge 03
Challenge 03
A continuoustime signal x(t) = cos(
(t)), where
(t) =
0
+ sin(
0
t),
is to be sampled at a rate
f
s
.
What is the system’s Nyquist sample rate
if
0
=
(2
)
1000 rad/sec.?
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 Fall '11
 janisemcnair
 Digital Signal Processing, Computer Networks, Aliasing, Signal Processing, Baseband, baseband frequency

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