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

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

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

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?

“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.

Ideal highpass filter

( ) ( ) sinc(B )Bh t t t

1

1

Ideal bandpass filter

B B

2B

( )H i

( )H i

1B 1B 2B

Bandpass as cascade of lowpass and highpass

Lowpass(B2)

Highpass(B1)

1

2B

( )H i

1B 1B 2B

An equalizer

( )H i

Adjustable gains for each frequency band.

+x

y

Implementation by parallel bandpass filters.

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.

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

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

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

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

( )( ) (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