2ndorder filters. EE 230 secondorder filters 1


 Earl Bridges
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1 ndorder filters Second order filters: Have second order polynomials in the denominator of the transfer function, and can have zeroth, first, or secondorder polyinomials in the numerator. Use two reactive components capacitor, inductors, or one of each. Can be used to make lowpass, highpass, and bandpass frequency responses. There is also bandreject. Have sharper cutoffs than firstorder for lowpass and highpass types. (Clearly distinction between passband and cutoff band.) Provide more flexibility in shaping the frequency response. EE 30 secondorder filters 1
2 ndorder filters Our approach is the same as the firstorder circuits. Examine the transfer functions for lowpass, highpass, and bandpass. Look at the details of the frequency response of each type of filter. Look at circuits that exhibit the lowpass, highpass, or bandpass behavior. Try some numerical examples to get a feel for the numbers. Build and test some circuits in lab. EE 30 secondorder filters
3 The general form for the transform function of a second order filter is that of a biquadratic (or biquad to the cool kids). T (s) =G o s + a 1 s + a 0 s = G o (s + Z 1)(s + Z 0 ) + b 1 s + b 0 (s + P 1 )(s + P 0 ) As before, G o is the gain of the transfer function it can be < 1. The poles of the transfer function determine the general characteristics and the location of the zeroes of the function determine the type of filter. We write the denominator using parameters that will better describe the general behavior. D (s) =s + ω o s + ω o Q P where ω o is the characteristic frequency, which determines where things are changing in the frequency response. Note that it is not (necessarily) the same as the cutoff frequency for the firstorder functions. (Details to follow.) Q P is the pole quality factor, and it determines the sharpness of the features in frequency response curve. Note that Q P has no dimensions. EE 30 secondorder filters 3
4 ω0 D (s) = s + s + ω0 Use the quadratic formula to determine the poles. P0,1 = ± Qp ωo = QP The key is the squareroot. If the argument under the squareroot is positive, there will be two real roots. If the argument is negative, the roots will be a complex conjugate pair. The dividing line is = 0.5. If < 0.5 two real, negative roots per the above formula D (s) = (s + P1 ) (s + P0 ) = s + s + ωo jω Example: D (s) = (s rad/s) (s + 00 rad/s) = s + (000 rad/s) s X P X P1 σ 105 rad /s = 600 rad/s ; = 0.3 EE 30 secondorder filters 4
5 Q P = 0.5 double real and negative roots. jω P 0 = P 1 = ω 0 Q D (s) =(s + P 0 ) = s + ω o Q P s + ω o X P o = P 1 σ Example: D (s) =(s rad/s) = s +(000 rad/s) s rad /s ω o = 1000 rad/s ; Q P = 0.5 EE 30 secondorder filters 5
6 > 0.5 complex conjugate roots ±j P0,1 = 4QP jω Po X 1 σ P1 = P0 P1 X D (s) = (s + P0 ) (s + P0 ) = s + s + ωo Example: D (s) = [s + (500 rad/s + j1000 rad/s)] [s + (500 rad/s = s + (1000 rad/s) s j1000 rad/s)] 106 rad /s = 1118 rad/s ; = 1.18 EE 30 secondorder filters 6
7 lowpass T (s) = N (s) s + ω o Q P s + ω o To achieve a lowpass response, the function should not go to zero as s 0. In other words, we don t want either the s or s terms in the numerator, so the numerator will be a constant = a 0. (Another way of saying this is that the zeros must be at infinity.) To make the magnitude of the transfer function go to G o as s goes to zero, choose a 0 = ω o. T lp (s) = ω o s + ω o Q P s + ω o This is the standard lowpass transfer function. (It may be worth memorizing.) EE 30 secondorder filters 7
8 To examine the filtering properties of the lowpass function, we need to look the AC behavior, so we shift to sinusoidal analysis by setting s = jω. Tlp (jω) = Go ωo ω + j ω + ωo = Go ωo [ωo ω ] + j ω Extracting the magnitude and phase from the complex function ωo Tlp = Go (ωo θlp = arctan ω ) + ω ω ωo ω We see that the magnitude goes to Go at low frequencies, as expected for a lowpass. Also, when ω =, Tlp = Go and θlp = 90. Exercise: Confirm all of the algebra leading to the above equations and results. EE 30 secondorder filters 8
9 Linearscale plots of order lowpass magnitude, with Q P = 0.15, 0.5, and 1.5. The characteristic frequency, f o = 1 khz. Also shown for comparison is a firstorder with f c = 1 khz = 0.15 = 0.5 = 1.5 first order cutoff 1.5 Q P = T Q P = 0.5 firstorder 0.5 Q P = Frequency (Hz) What is going on with the magnitude curve for Q P = 1.5? EE 30 secondorder filters 9
10 Bode plot version of secondorder magnitudes from previous slide = 0.15 = 0.5 = 1.5 first order cutoff Q P = Q P =0.5 T (db) 0.0 Q P = 0.15 firstorder db/dec Frequency (Hz) Note that for frequencies sufficiently high into the cutoff band, the magnitudes decrease at rate of 40 db/dec twice the slope for firstorder. This is due to T ω for ω >> ω o. EE 30 secondorder filters 10
11 Phase angle frequency responses corresponding to the magnitudes on the previous slides. 0.0 = 0.15 = 0.5 = 1.5 first order Q P = 1.5 Q P = 0.15 phase angle ( ) Q P =0.5 firstorder Frequency (Hz) EE 30 secondorder filters 11
12 From the magnitude plots, we see that the corner frequency is strongly dependent on the quality factor. We can calculate the corner frequency in the usual manner. G o = G o ω o (ω o ω c) + ω o Q P ω c Solving for ω c involves some tedious algebra with the quadratic equation. ω c = ω o 1 1 Q P Q 4 P 1 Q P Yikes! (Exercise: Derive this for yourself. It s not that hard just tedious.) The expression confirms what we noticed in the graphs for lower Q P, the cutoff frequency is lower than the characteristic frequency, ω c < ω o. At higher quality factors, ω c > ω o, due the bump. The dividing line is at Q P = 1/, where ω c = ω o. This is an important case that will prove useful. EE 30 secondorder filters 1
13 The bump EE 30 secondorder filters 13
14 highpass T (s ) = N ( s) s + ω0 Q s + ω0 For a highpass response we want T 0 as at s 0 and T constant as s. We accomplish that most easily by choosing N(s) = Go s. The polynomials give the correct limits, and the Go takes care of the constant value at high frequencies, as determined by any amplifiers or voltage dividers within the circuit. s Thp (s) = Go s + QωPo s + ωo Thp (jω) = Go ω (ωo Thp (jω) = Go θhp = 180 EE 30 arctan ω ) + j ω ω (ωo ω ) ω + ω ωo ω secondorder filters 14
15 Linear scale plots of secondorder highpass magnitude, with Q P = 0.15, 0.5, and 1.5. The characteristic frequency is f o = 1 khz. Also shown for comparison is a firstorder highpass filter with f c = 1 khz = 0.15 = 0.5 = 1.5 first order cutoff Q P = T 0.75 Q P = firstorder Q P = Frequency (Hz) EE 30 secondorder filters 15
16 Bode plot version of secondorder highpass magnitudes from previous slide = 0.15 = 0.5 = 1.5 first order cutoff Q P = 1.5 Q P = 0.5 T firstorder Q P = Frequency (Hz) Note that for frequencies sufficiently low into the cutoff band, the magnitudes vary at rate of 40 db/dec twice the slope for firstorder. This is due to T ω for ω << ω o. EE 30 secondorder filters 16
17 Phase angle frequency responses corresponding to the highpass plots on the previous slides = 0.15 = 0.5 = 1.5 first order Q P = Q P = 0.15 Q P = 0.5 phase angle ( ) 90.0 firstorder Frequency (Hz) EE 30 secondorder filters 17
18 bandpass T (s ) = N ( s) ω0 Q s + s + ω0 With bandpass, we want T to go to zero both at s = 0 and as s. To achieve this, we choose N(s) = Go(ω0/)s. Tbp (s) = Go s + Tbp (jω) = Go EE 30 s s + ωo j (ωo Thp (jω) = Go θbp = 90 arctan ω ω ) + j ω ω (ωo ω ) + ω ω ωo ω secondorder filters 18
19 Linear scale plots of secondorder bandpass magnitude, with Q P = 0.15, 0.5, and 1.5. The characteristic frequency is f o = 1 khz. 1.0 = 0.15 = 0.5 = 1.5 cut off 1.00 Q P = T 0.60 Q P = Q P = Frequency (Hz) EE 30 secondorder filters 19
20 Bode plot version of secondorder bandpass magnitudes from previous slide = 0.15 = 0.5 = 1.5 first order 0.00 Q P = Q P =0.5 T Q P = Frequency (Hz) Note that above and below the passband, the plots all have slopes of 0 db/dec. EE 30 secondorder filters 0
21 Phase angle frequency responses corresponding to the bandpass plots on the previous slides = 0.15 = 0.5 = 1.5 Q P = Q P = 0.15 Q P = 0.5 phase angle ( ) Frequency (Hz) EE 30 secondorder filters 1
Secondorder filters. EE 230 secondorder filters 1
Secondorder filters Second order filters: Have second order polynomials in the denominator of the transfer function, and can have zeroth, first, or secondorder polynomials in the numerator. Use two
More informationThe general form for the transform function of a second order filter is that of a biquadratic (or biquad to the cool kids).
ndorder filters The general form for the transform function of a second order filter is that of a biquadratic (or biquad to the cool kids). T (s) A p s a s a 0 s b s b 0 As before, the poles of the transfer
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