nd-order filters Second order filters: Have second order polynomials in the denominator of the transfer function, and can have zeroth-, first-, or second-order polyinomials in the numerator. Use two reactive components capacitor, inductors, or one of each. Can be used to make low-pass, high-pass, and band-pass frequency responses. There is also band-reject. Have sharper cut-offs than first-order for low-pass and high-pass types. (Clearly distinction between passband and cut-off band.) Provide more flexibility in shaping the frequency response. EE 30 second-order filters 1
nd-order filters Our approach is the same as the first-order circuits. Examine the transfer functions for low-pass, high-pass, and bandpass. Look at the details of the frequency response of each type of filter. Look at circuits that exhibit the low-pass, high-pass, or band-pass behavior. Try some numerical examples to get a feel for the numbers. Build and test some circuits in lab. EE 30 second-order 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) =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 cut-off frequency for the first-order 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 second-order filters 3
ω0 D (s) = s + s + ω0 Use the quadratic formula to determine the poles. P0,1 = ± Qp ωo = 1+ 1 4QP The key is the square-root. If the argument under the square-root 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 + 1800 rad/s) (s + 00 rad/s) = s + (000 rad/s) s + 3.6 X P X P1 σ 105 rad /s = 600 rad/s ; = 0.3 EE 30 second-order filters 4
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 + 1000 rad/s) = s +(000 rad/s) s + 10 6 rad /s ω o = 1000 rad/s ; Q P = 0.5 EE 30 second-order filters 5
> 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 + 1.5 j1000 rad/s)] 106 rad /s = 1118 rad/s ; = 1.18 EE 30 second-order filters 6
low-pass T (s) = N (s) s + ω o Q P s + ω o To achieve a low-pass 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 low-pass transfer function. (It may be worth memorizing.) EE 30 second-order filters 7
To examine the filtering properties of the low-pass 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 low-pass. Also, when ω =, Tlp = Go and θlp = 90. Exercise: Confirm all of the algebra leading to the above equations and results. EE 30 second-order filters 8
Linear-scale plots of -order low-pass magnitude, with Q P = 0.15, 0.5, and 1.5. The characteristic frequency, f o = 1 khz. Also shown for comparison is a first-order with f c = 1 khz. 1.50 = 0.15 = 0.5 = 1.5 first order cut-off 1.5 Q P = 1.5 1.00 T 0.75 0.50 Q P = 0.5 first-order 0.5 Q P = 0.15 0.00 10 100 1000 10000 100000 Frequency (Hz) What is going on with the magnitude curve for Q P = 1.5? EE 30 second-order filters 9
Bode plot version of second-order magnitudes from previous slide. 10.0 0.0 = 0.15 = 0.5 = 1.5 first order cut-off Q P = 1.5-10.0 Q P =0.5 T (db) -0.0 Q P = 0.15 first-order -30.0-40.0-40 db/dec -50.0 10 100 1000 10000 100000 Frequency (Hz) Note that for frequencies sufficiently high into the cut-off 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 second-order filters 10
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 ( ) -45.0-90.0 Q P =0.5 first-order -135.0-180.0 10 100 1000 10000 100000 Frequency (Hz) EE 30 second-order filters 11
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 + + 1 4Q 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 cut-off 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 second-order filters 1
The bump EE 30 second-order filters 13
high-pass T (s ) = N ( s) s + ω0 Q s + ω0 For a high-pass 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 second-order filters 14
Linear scale plots of second-order high-pass 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 first-order high-pass filter with f c = 1 khz. 1.50 = 0.15 = 0.5 = 1.5 first order cut-off Q P = 1.5 1.5 1.00 T 0.75 Q P = 0.5 0.50 0.5 first-order Q P = 0.15 0.00 10 100 1000 10000 100000 Frequency (Hz) EE 30 second-order filters 15
Bode plot version of second-order high-pass magnitudes from previous slide. 10.00 0.00-10.00 = 0.15 = 0.5 = 1.5 first order cut-off Q P = 1.5 Q P = 0.5 T -0.00-30.00 first-order Q P = 0.15-40.00-50.00-60.00 10 100 1000 10000 100000 Frequency (Hz) Note that for frequencies sufficiently low into the cut-off band, the magnitudes vary at rate of 40 db/dec twice the slope for first-order. This is due to T ω for ω << ω o. EE 30 second-order filters 16
Phase angle frequency responses corresponding to the high-pass plots on the previous slides. 180.0 = 0.15 = 0.5 = 1.5 first order Q P = 1.5 135.0 Q P = 0.15 Q P = 0.5 phase angle ( ) 90.0 first-order 45.0 0.0 10 100 1000 10000 100000 Frequency (Hz) EE 30 second-order filters 17
band-pass 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 second-order filters 18
Linear scale plots of second-order band-pass 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 = 0.15 0.80 T 0.60 Q P = 0.5 0.40 Q P = 1.5 0.0 0.00 10 100 1000 10000 100000 Frequency (Hz) EE 30 second-order filters 19
Bode plot version of second-order band-pass magnitudes from previous slide. 10.00 = 0.15 = 0.5 = 1.5 first order 0.00 Q P = 0.15-10.00 Q P =0.5 T Q P = 1.5-0.00-30.00-40.00 10 100 1000 10000 100000 Frequency (Hz) Note that above and below the pass-band, the plots all have slopes of 0 db/dec. EE 30 second-order filters 0
Phase angle frequency responses corresponding to the band-pass plots on the previous slides. 90.0 = 0.15 = 0.5 = 1.5 Q P = 1.5 60.0 Q P = 0.15 Q P = 0.5 phase angle ( ) 30.0 0.0-30.0-60.0-90.0 10 100 1000 10000 100000 Frequency (Hz) EE 30 second-order filters 1