ECE 440 FILTERS

Review of Filters

• Filters are systems with amplitude and phase response that depends on frequency.

• Filters named by amplitude attenuation with relation to a transition or cutoff frequency.

• Designed to selectively attenuate range(s) of frequency.

• Some filters also designed to affect phase in a particular way

Review of Filter Types

• Low pass – passes frequencies below cutoff frequency

• High pass – passes frequencies above cutoff frequency

• Band pass – passes frequencies within a band between cutoff frequencies

• Notch – passes all frequencies outside a band between cutoff frequencies.

Terminology

• Pass band – band of frequencies with little or no attenuation. End marked by rise of attenuation above a specified limit. E.g. -3dB

• Transition band – band of frequencies where attenuation transitions from limit at edge of pass band to limit at the edge of the stop band.

• Stop band – band of frequencies where attenuation remains above a specified level.

Low Pass Filter

STOP BAND

PASS BAND

Transition Band

FREQUENCY

AM

PLI

TUD

E

Cutoff Frequency

High Pass Filter

STOP BAND PASS BAND

Transition Band

AM

PLI

TUD

E

FREQUENCY

Cutoff Frequency

Band Pass Filter

Pass Band

Stop Band Stop Band

AM

PLI

TUD

E

FREQUENCY

Lower Cutoff Frequency

Upper Cutoff Frequency

Notch Filter

Pass Band Pass Band

Stop Band

AM

PLI

TUD

E

FREQUENCY

Lower Cutoff Frequency

Upper Cutoff Frequency

Filter Transfer Function

• General form of a transfer function

• z’s are referred to as transmission zeroes

• p’s are referred to as transmission poles

• To be stable number of poles must be greater than or equal to number of zeroes

• Number of poles is referred to as the filter order

• In general, higher order = steeper/narrower transition band

)())((

)())(()(

21

21

n

mm

pspsps

zszszsasT

Filter Transfer Function (cont’d)

• To be stable, filter poles must all lie in left hand plane.

• If poles or zeroes are complex, they must be in conjugate pairs.

Transfer Function Types

• Placement of the poles and zeroes affects shape of filter.

• Mathematical relationships have been developed to calculate where poles and zeroes should fall.

– Butterworth

– Chebyshev

– Elliptical

Comparison of Filter Functions

• Low pass filter examples

Butterworth: Smooth, ever-increasing

or ever-decreasing attenuation with

frequency. Unlimited stop band attenuation.

Chebyshev: Ripple in pass band and in stop band, but steep transition band. Stop band has limited attenuation

Butterworth filters covered in this course.

Butterworth Filters • Poles are roots of the Butterworth polynomial.

• Poles equally spaced on circle in complex plane.

• Only roots in left hand plane are used in filters.

x x

x x 2nd Order Example

Four roots spaced every π/2 radians

Only use these two roots for filter

Plotting and finding roots

• Determine roots for a filter with ωcutoff=1:

– For ωcutoff=1 and the attenuation at cutoff = -3dB, the circle in complex plane has radius =1.

– Roots are symmetric about real axis and spaced π/N (N is the filter order) apart.

• Calculate root values based on geometry.

• Then scale roots based on actual cutoff frequency.

2nd Order Example

Roots are: -0.707+0.707j -0.707-0.707j x

x

π/4

π/4

1

j

Angle between roots =π/2

Scaling

• To scale roots to new frequency, simply multiply roots by new frequency.

• Roots become:

• Roots are complex conjugates.

1

2

0.707 0.707

0.707 0.707

cutoff cutoff

cutoff cutoff

s j

s j

Higher Order Polynomial Roots N=3

Roots are: -1 -0.5+0.866j -0.5-0.866j

x

x

π/3

π/3

1

j

Angle between roots =π/3

x

General Nth Order

• If N is odd, there will always be a root at -1

– Other roots spaced at angle of π/N from π/N to (N-2)π/N and from - π/N to -(N-2)π/N

– General formula for odd order:

angle θk =+/- k π/N with k= 0 to (N-1)/2

• If N is even, no real root. Roots symmetric about real axis at angle θk =

+/- (2k+1) π/2N with k= 0 to N/2-1

General Nth Order Roots (scaled)

• N odd:

– One root at -1*ωcutoff

– Other roots are:

• N even:

– Roots are:

)sin()cos( kkcutoff js

)sin()cos( kkcutoff js

Implementation

• Now that roots are calculated, need to implement them based on topology.

• Circuit topology yields a transfer function based on circuit components.

• Match transfer functions of desired filter shape and the circuit to determine circuit components.

First order circuit

• Use op-amp to prevent circuit connected to output from affecting response.

• Well known response of:

-

+

R

0

Vi

Vo

C

1( )

1T s

sRC

First order circuit (cont’d)

• Transfer function has a root of:

• First order Butterworth has one root at –ωcutoff

• Therefore to implement first order filter, set roots equal to each other:

A familiar equation.

1s

RC

1 1 cutoff cutoffor

RC RC

Second Order Filter • Cannot cascade first order circuits to make

second order – poles are not in the right place – all real.

• Use circuit below

-

+

R

0

Vi

Vo

C

R

C

R2R1

Second Order Filter

• It can be shown through nodal analysis that the circuit has a response of:

• Remembering that the two poles for a Butterworth 2nd order transfer function are:

2

2

2 12

( )where 1

(3 ) 1

( )

Rk RCk

k Rs s

RC RC

1

2

0.707 0.707

0.707 0.707

cutoff cutoff

cutoff cutoff

s j

s j

Second Order Filter

• This means the transfer function looks like:

• Inserting the vales for the roots and combining gives:

1 2

1

s s s s

2 2

1

2 c cs s

Second Order Filter

• Normalizing gives:

• Equating this equation with the circuit’s transfer function:

2

2

2

21

c c

ss

2 2

22

22

( )

(3 ) 121

( )c c

RC

ks s ssRC RC

Second Order Filter

• This yields the following relationships:

• We can use these relationships to design a second order filter.

2

1

1

2 (3 ) 3 2 1.586

.586

cRC

k k

R

R

Homework

• Design a second order low pass filter with a cutoff frequency of 1 kHz.

• Design a third order filter with a cutoff frequency of 1 kHz by cascading a first order with a second order filter. NOTE: The poles are at different locations for each section.

• Design a fourth order filter with a cutoff frequency of 1 kHz by cascading two second order filters. NOTE: The poles are in different locations for each section.

• Model your filters in PSPICE to verify their responses.

• Compare the transition bands of the filters by plotting them on the same graph.