lecture 1 sampling
TRANSCRIPT
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8/22/2019 Lecture 1 Sampling
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LECTURE 1: SAMPLING
Sampling Process = Transformation of analog signal into discrete time signal.
x(t)
=
0
).().(
n
Tsnttx
Fourrier spectrum of the sampled signal:
=
0
).().(
n
Tsnttx
=
=
0
).(.)()(
n
fsnffXfXs where fs=1/Ts.
)....2()2()()()()( fsfXfsfXfsfXfsfXfXfXs ++++++=
The spectrum of the sampled signal is the periodization of the spectrum of the analog
signal ( period= fs)
fmax0f
-fmax
)( fX
)( fXs
-fmax fmaxfs-fmax fs+ fmax-fs-fmax -fs+ fmax f
In order to avoid Overlap between different copies (replicas) of the spectrum, thesampling frequency should be chosen such that:
fs-fmax>fmax fs> 2.fmax where fmax: maximum frequency of the analog signal..
this condition is known as nyquist criterion( minimum sampling frequency).
If this condition is respected we can extract the spectrum of the analog signal from the
spectrum of the samples. So we say that the signal is correctly represented by its samples.
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Example1: what is the minimum sampling frequency required when :
t
ttx
.6280
).6280sin()( =
We need first the spectrum of x(t)
We use the duality property of Fourrier transform
0)( Trettx = 0..
)0..sin(.0)(Tf
TfTfX
= so we can say that
0..
)0..sin(0)(
Tt
TtTtx
= 0)( TrectfX = OR
0..
)0..sin(0)(
ft
ftftx
= 0)( frectfX =
Lets write our signal in this form:
tf
tf
f
f
tf
tf
t
ttx
.0.
).0.sin(
0
0
.0.
).0.sin(
.6280
).6280sin()(
=== 0
0
1)( frect
ffX =
The maximumfrequency fmax=f0/2 where 0.f =6280 so f0=2000 Hz.The Nyquist frequency=2.fmax=2.1000=2000 Hz
Example2: what is the minimum sampling frequency required when :
2]
.6280
).6280sin([)(
t
ttx =
We need X(f).
We use here the following fourrier transform property:
FT[ )().( txtx ] )()( fXfX =
== dfXXdfXX ))(().()().( where
00
1)( frect
ffX = = 0. frectA
The convolution value @ a given frequency f is the area between )(X and )( X shifted
by a value of f.
The amplitude of the convolution @ f=0 is:
=
2/0
2/0
)().(
f
f
dXX 0.22/0
2/0
2 fAdA
f
f
=
For f=f0 .-->
= dfXX ))0(().(
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A
f0/2-f0/2
A
f0/2-f0/2
)(X
))0(( fX+
f0
Graphically we calculate the area of the product X( ) and X(- ) shifted by f0.this integral
gives 0 .
For f>f0 the integral will be 0 no common area between the two part of the integral
product=0integral=0.
Same for f=-f0.the integral0
For f= f0/2 for example we have half the rectangle in common so the area= A
2
.f0/2.
So the convolution product twill look like the following figure
f00 f-f0
Convolution
0.2 fA
Now we know that the maximum frequency in the spectrum that is fmax=f0.
The required sampling frequency is at least =2.fmax=2.2000=4000 Hz.
We can use higher frequency rate but its unnecessarily high.