Filtering x y.
-
Upload
bernard-lane -
Category
Documents
-
view
212 -
download
0
description
Transcript of Filtering x y.
![Page 1: Filtering x y.](https://reader036.fdocuments.in/reader036/viewer/2022083119/5a4d1af87f8b9ab0599825c4/html5/thumbnails/1.jpg)
x y
Filtering
Let ( ) ( ) be the system frequency response function.Then for any input ( ) with a Fourier transform ( ),
H i h tx t X i
F
( ) ( ) ( ) Y i H i X i Magnitudes are multiplied:
Amplify some frequencies, attenuate others. In decibels (dB), the effect becomes additive:
20 log 20 log + 20 log
( ) ( ) ( )
( ) ( ) ( )
i H i X i
i H i X i
Y
Y
( ) ( ) ( )Phases are added: Y i H i X i
![Page 2: Filtering x y.](https://reader036.fdocuments.in/reader036/viewer/2022083119/5a4d1af87f8b9ab0599825c4/html5/thumbnails/2.jpg)
Ideal lowpass filter
B B
( )H i
1
Is it physically realizable? Looking at the impulse response:
1 sin( )( ) ( ) sinc(B )Bt Bh t H i tt
F
( ) is not zero for 0. So theideal lowpassfilter is not causal!
h tt
![Page 3: Filtering x y.](https://reader036.fdocuments.in/reader036/viewer/2022083119/5a4d1af87f8b9ab0599825c4/html5/thumbnails/3.jpg)
for( ) sinc(B( )) ( )0 otherwise.
iB Bh t t H i e
This is still non-causal, but we approximate it by truncating
( ) to positive time: ( ) sinc(B( )) ( )Bh t h t t u t
An ideal lowpass filter cannot be built, but we can build approximations to it. If we only care about the magnitudeof ( ) (and not the phase), one way is to add a delay:H i
![Page 4: Filtering x y.](https://reader036.fdocuments.in/reader036/viewer/2022083119/5a4d1af87f8b9ab0599825c4/html5/thumbnails/4.jpg)
Causal approximation to ideal lowpass
( ) sinc( ( )) ( )Bh t B t u t
( )H i
B
This h(t) may be difficult to implement. Easier alternative?
![Page 5: Filtering x y.](https://reader036.fdocuments.in/reader036/viewer/2022083119/5a4d1af87f8b9ab0599825c4/html5/thumbnails/5.jpg)
“Butterworth” filter of order 10.
( )H i
This is a rational H(s), with denominator of order 10.
For more on filter design, see signal processing courses.Here we will focus on ideal filters.
![Page 6: Filtering x y.](https://reader036.fdocuments.in/reader036/viewer/2022083119/5a4d1af87f8b9ab0599825c4/html5/thumbnails/6.jpg)
Ideal highpass filter
( ) ( ) sinc(B )Bh t t t
1
1
Ideal bandpass filter
B B
2B
( )H i
( )H i
1B 1B 2B
![Page 7: Filtering x y.](https://reader036.fdocuments.in/reader036/viewer/2022083119/5a4d1af87f8b9ab0599825c4/html5/thumbnails/7.jpg)
Bandpass as cascade of lowpass and highpass
Lowpass(B2)
Highpass(B1)
1
2B
( )H i
1B 1B 2B
![Page 8: Filtering x y.](https://reader036.fdocuments.in/reader036/viewer/2022083119/5a4d1af87f8b9ab0599825c4/html5/thumbnails/8.jpg)
An equalizer
( )H i
Adjustable gains for each frequency band.
+x
y
Implementation by parallel bandpass filters.
![Page 9: Filtering x y.](https://reader036.fdocuments.in/reader036/viewer/2022083119/5a4d1af87f8b9ab0599825c4/html5/thumbnails/9.jpg)
Example: pure delay system: ( ) ( ).y t x t Impulse response function: ( ) ( ).Frequency response function: ( ) i
h t tH i e
( ) 1 for all : this is an "allpass" filt .erH i
2A rational allpass filter: ( ) .2iH ii
Both these systems do not affect the magnitude of thesignals (i.e. the input and output Fourier transformshave the same magnitude). They do, however, affect the phase.
![Page 10: Filtering x y.](https://reader036.fdocuments.in/reader036/viewer/2022083119/5a4d1af87f8b9ab0599825c4/html5/thumbnails/10.jpg)
Fourier and LaplaceLaplace:
0
converges for Re[ ] when
Applies to large classes of functions, including those that "blow up" exponentially as time :
( ) ( ) ( )
Considers only positive time.
st tsF s f t dt f t Ce e
Fourier: More restrictive convergence.
One sufficient condition is integrability, i.e.
Generalizes to sinusoids, but not to increasing exponentials.Includes negative time.
( ) < .f t dt
![Page 11: Filtering x y.](https://reader036.fdocuments.in/reader036/viewer/2022083119/5a4d1af87f8b9ab0599825c4/html5/thumbnails/11.jpg)
Examples:
Re[ ] 1.
11) ( ) ( ) . ( ) , DOC1
( ) is not defined.
( )t sf t u t f ts
f t
F se
FL
2) ( ) Laplace transform would apply only to the 0 portion.
( ) well defined, and considers all time.
.tf tt
f t
e
F
( ) 0 for 0.Rule: if
DOC of ( ) ( ) contains Re[ ] 0.
Then the Fourier transform exists, and ( ) ( ) .s i
f t tF s f t s
f t F s
FL
DOC Re[ ] 1.
3) ( ) ( )1 1( ) , ( )
1 1
.t
s
f t u t
f t f ts i
e
FL
![Page 12: Filtering x y.](https://reader036.fdocuments.in/reader036/viewer/2022083119/5a4d1af87f8b9ab0599825c4/html5/thumbnails/12.jpg)
What happens if the DOC of ( ) ( ) is Re[ ] 0?
The Fourier transform exists, but need be equal to ( )not s i
F s f t s
F s
L
0
Re[ ] 0.
( )
1 , DOC
( ) = does not converge absolutely,
but it can be defin
Example:
ed in a generalized sense. What do we get
( ) ( )
?
( )
i t
s
U i
f t us
t
t
f
F s
dte
F
?
11
One approach: use the derivative property
( ) = ( ) ( ) .
This would correspond to setting in the Laplace transform.i
dut i U i U idt
s i
F F
Problem: the same approach would apply to ( )
for any constant , since ( ) ( )
C u tdC C u t tdt
F
![Page 13: Filtering x y.](https://reader036.fdocuments.in/reader036/viewer/2022083119/5a4d1af87f8b9ab0599825c4/html5/thumbnails/13.jpg)
1For example, let ( ) ( ), ( )2
drr t u t tdt
12
12
?11 = ( ) ( ) .i
dr i R i R idt
F
Now 1 1( ) ( ) = 1 + ( ) = ( ) ( )2 2
1So both ( ) and ( ) can't be equal to . Which one is right?
R i u t u t U i
R i U ii
F F F
1 1Answer: ( ) ( ) = . Why? a purely imaginary 2
transform must correspond to an function of time, such a ( .s )odd
R i u ti
r t
F
, Correspondingly is the transform of the st1( ) ( ) ep.U ii
0
1Note that ( ) ( ) 1 ( ) 1
Problem with the previous derivation: dividing by misses th .e
i U i i ii
i
( )r t
t
![Page 14: Filtering x y.](https://reader036.fdocuments.in/reader036/viewer/2022083119/5a4d1af87f8b9ab0599825c4/html5/thumbnails/14.jpg)
( )( ) (0) ( )
Application to provingthe integration property:
t F if d Fi
F
Proof: Consider a system with impulse response ( ) ( ),
and step response ( ) ( ) . The main theorem says that
1 ( )( ) ( ) ( ) ( ) ( ) (0) ( )
th t f t
g t f d
F iG i H i U i F i Fi i
0 2 2
0 0 02 20
0
( follows from ( ) by modulation property):
Setting on the Laplace transform ( )cos( )
gives the first term only
Another delicate transform
( )cos( ) ( )2
,
( )2
i
U i
ss i u
t
t ts
u t
L
F
again misses the ' .s