University:
California State University, Northridge
Course:
ECE 440/L | Electronics II and Lab
Academic year:

2013
Views:

162

Pages:

26
Author:

ueffenv0k

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. AMPLITUDE Low Pass Filter PASS BAND STOP BAND Cutoff Frequency Transition Band FREQUENCY High Pass Filter PASS BAND AMPLITUDE STOP BAND Cutoff Frequency Transition Band FREQUENCY AMPLITUDE Band Pass Filter Stop Band Stop Band Lower Cutoff Frequency Pass Band FREQUENCY Upper Cutoff Frequency Notch Filter AMPLITUDE Stop Band Pass Band Pass Band Upper Cutoff Frequency Lower Cutoff Frequency FREQUENCY Filter Transfer Function • General form of a transfer function am ( s  z1 )( s  z2 )    ( s  zm ) T ( s)  ( s  p1 )( s  p2 )    ( s  pn ) • 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 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. Only use these two roots for filter 2nd Order Example x x x x Four roots spaced every π/2 radians 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 j x π/4 1 π/4 Angle between roots =π/2 x Scaling • To scale roots to new frequency, simply multiply roots by new frequency. • Roots become: s1  0.707cutoff  0.707cutoff j s2  0.707cutoff  0.707cutoff j • Roots are complex conjugates. Higher Order Polynomial Roots N=3 Roots are: -1 -0.5+0.866j -0.5-0.866j j x π/3 1 x π/3 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: s  cutoff  cos( k )  j sin( k ) • N even: – Roots are: s  cutoff  cos( k )  j sin( k ) 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 + Vo R - Vi C 0 • Use op-amp to prevent circuit connected to output from affecting response. • Well known response of: 1 T ( s)  1  sRC First order circuit (cont’d) • Transfer function has a root of: 1 s RC • First order Butterworth has one root at –ωcutoff • Therefore to implement first order filter, set roots equal to each other: cutoff   1 1 or cutoff  RC RC A familiar equation. Second Order Filter • Cannot cascade first order circuits to make second order – poles are not in the right place – all real. • Use circuit below C + R Vo R - Vi C R1 0 R2 Second Order Filter • It can be shown through nodal analysis that the circuit has a response of: R2 k ( RC )2 where k  1  (3  k ) 1 R1 s2  s RC ( RC )2 • Remembering that the two poles for a Butterworth 2nd order transfer function are: s1  0.707cutoff  0.707cutoff j s2  0.707cutoff  0.707cutoff j Second Order Filter • This means the transfer function looks like: 1  s  s1  s  s2  • Inserting the vales for the roots and combining gives: 1 2 2 s  2 sc  c Second Order Filter • Normalizing gives:  2 s2  2 c  2 c s 1 • Equating this equation with the circuit’s transfer function:  2 ( RC ) 2  2 (3  k ) 1 2 s 2 s  s   s 1 2 2 RC ( RC ) c c Second Order Filter • This yields the following relationships: 1 RC 2  (3  k )  k  3  2  1.586 R2  .586 R1 c  • We can use these relationships to design a second order filter.

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