Second-order filters Second order filters: Have second order polynomials in the denominator of the transfer function, and can have zeroth-, first-, or second-order polynomials in the numerator. Use two reactive components 2 capacitors, 2 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. (Clearer distinction between passband and cut-off band.) Provide more flexibility in shaping the frequency response. EE 230 second-order filters 1
Second-order filters Our approach is the same as the first-order circuits. Examine the transfer functions for low-pass, high-pass, and (now) band-pass. Look at the details of the frequency response of each type of filter cut-off frequency, pass-band gain, slope in the cut-off band. 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 230 second-order filters 2
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 s2 + a 1 s + a o s 2 + b 1 s + b o = G o ( s + Z 1 ) (s + Z 2 ) (s + P 1) (s + P 2) As before, Go is the gain of the transfer function. As seen with firstorder passive filters, Go 1. The poles of the transfer function determine the general characteristics, and the zeroes determine the filter type. We write the denominator using parameters that will better help us characterize the general behavior. D (s) = s 2 + ( ) s + ω2 o where is the characteristic frequency, which determines where things are changing in the frequency response. We will point out right now that is not (necessarily) the same as the cut-off frequency. (Details to follow.) is the pole quality factor, and it determines the sharpness of the features in frequency response curve. Note that has no dimensions. EE 230 second-order filters 3
D (s) = s 2 + ( ) s + ω2 o Use the quadratic formula to determine the poles. P 1,2 = 2 ± 2 ( 2 ) ω 2 o = 2 ( 1 ± 1 4Q 2 P ) 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 : QP < 0.5 two real, distinct, negative roots. two real, repeated negative roots. QP > 0.5 complex conjugate roots. EE 230 second-order filters 4
QP < 0.5 jω There will be two distinct real roots. The step function response would be an underdamped transient. X X σ P 1 = 2 + 2 ( 2 ) ω 2 o P 2 P 1 P 2 = 2 2 ( 2 ) ω 2 o Example: P1 = 200 s 1 and P2 = 1800 s 1. D (s) = (s + 200 s 1 ) (s + 1800 s 1 ) = s 2 + (2000 s 1 ) s + (360,000 s 2 ) D (s) = s 2 + ( ) s + ω2 o ωo = 600 s 1 ; QP = 0.3 EE 230 second-order filters 5
jω There will be two identical (repeated) real roots. The step function response would be a critically damped transient. X P 1 = P 2 σ P 1 = P 2 = 2 Example: P1 = P2 = 1000 s 1. D (s) = (s + 1000 s 1 ) 2 = s 2 + (2000 s 1 ) s + (10 6 s 2 ) D (s) = s 2 + ( ) s + ω2 o ωo = 1000 s 1 ; EE 230 second-order filters 6
QP > 0.5 The roots will be a complex-conjugate pair. The step response would be an underdamped transient. P 1 X jω σ P 1 = 2 + ω 2 o ( 2 2 ) X P 2 = P 1 * P 2 = 2 ω 2 o ( 2 2 ) Example: P1 = 1000 s 1 + j1000 s 1 and P2 = 1000 s 1 j1000 s 1 D (s) = (s + 1000 s 1 + j1000 s 1 ) (s + 1000 s 1 j1000 s 1 ) = s 2 + (2000 s 1 ) s + (2 10 6 s 2 ) D (s) = s 2 + ( ) s + ω2 o ωo = 1414 s 1 ; QP = 0.707 EE 230 second-order filters 7
low-pass T (s) = G o s 2 + a 1 s + a o s 2 + ( ) s + ω2 o For a low-pass response, the function should not go to zero as s 0, meaning that neither the s or s 2 terms can be in the numerator. The numerator reduces to just the constant term. (Another way of saying this is that the zeros must occur at infinity.) To make the magnitude of the transfer function go to Go as s goes to zero, choose ao = ωo 2. Then the standard form of the second-order low-pass filter is: T lp (s) = G o ω 2 o s 2 + ( ) s + ω2 o This may be worth memorizing. EE 230 second-order filters 8
To examine the frequency response, set s = jω. T lp (jω) = G o ω 2 o ω 2 + j ( ) ω + ω2 o = G o ω 2 o (ω 2 o ω 2 ) + j ( ) ω Extracting the magnitude and phase. T lp = G o ω 2 o (ω 2 o ω 2 ) 2 + ( ω ) 2 θ lp = arctan ( ) ( ω ω 2 o ω 2 ) At low frequencies, Tlp Go, as expected. At high frequencies, the magnitude varies inversely with the square of the frequency. The phase ranges from 0 at low frequencies to 180 at high frequencies. When ω = ωo, Tlp = QP Go and θlp = 90. Here is an interesting observation: If Qp > 1, then at ω = ωo, Tlp > Go!! This requires some further exploration. EE 230 second-order filters 9
The corner frequency is defined in exactly the same manner as with first-order filters, T lp (ω c ) = G o / 2 G o ω 2 o (ωo 2 ωc) 2 2 ω 2 + o ( ω Q c P ) = G o 2 Unfortunately, the math isn t as simple as in the first-order case. After some tedious algebra that includes choosing the correct root when applying the quadratic equation: ω c = 1 1 2Q 2 P + 1 + 1 1 2 ( 2QP 2 ) Yikes! (Exercise: Derive this for yourself.) EE 230 second-order filters 10
A plot of ωc / ωo vs. QP. (Function on the previous slide.) 2.0 1.5 ω c 1.0 0.5 0.0 0 1 2 3 4 5 As QP become large, ωc = 1.55ωo. Note that ωc = ωo. when = 1/ 2 EE 230 second-order filters 11
Linear-scale plots of the second-order low-pass magnitude, with = 0.125, 0.5, and 1.25. For all plots, the characteristic frequency is f o = 1 khz and the gain is G o = 1. Also shown for comparison is a first-order response with f c = 1 khz. What is going on with the magnitude curve for = 1.25? 1.50 1.25 1.00 first order cut-off T 0.75 cut-off level 0.50 0.25 first-order 0.00 10 100 1000 10000 100000 Frequency (Hz) EE 230 second-order filters 12
Bode versions of magnitude plots from previous slide. Note that for frequencies sufficiently high into the cut-off band, the magnitudes decreases at rate of 40 db/dec twice the slope for first-order. This is due to T ω 2 for ω >>. 10 0-10 -3 db T (db) -20-30 -40 first-order 3 db first-order -50-60 10 100 1000 10000 100000 Frequency (Hz) EE 230 second-order filters 13
Phase angle frequency responses for the various cases from the previous slides. 0-45 first-order phase (degrees) -90 first-order -135-180 10 100 1000 10000 100000 Frequency (Hz) EE 230 second-order filters 14
The bump EE 230 second-order filters 15
high-pass T (s) = G o s 2 + a 1 s + a o s 2 + ( ) s + ω2 o For a high-pass response we want T 0 as s 0 and T Go as s. We accomplish that most easily by setting a1 = ao = 0 in the numerator of the function. T hp (s) = G o s 2 s 2 + ( ) s + ω2 o The polynomial ratio varies from 0 to 1 as s increases from 0 to. The gain, Go, takes care of the constant value at high frequencies, as determined by any amplifiers or voltage dividers within the circuit. This also may be worth memorizing. EE 230 second-order filters 16
To examine the frequency response, set s = jω. T hp (jω) = G o ω 2 ω 2 + j ( ) ω + ω2 o = G o ω 2 (ω 2 o ω 2 ) + j ( ) ω Extracting the magnitude and phase. T hp = G o ω 2 (ω 2 o ω 2 ) 2 + ( ω ) 2 θ hp = 180 arctan ( ) ( ω ω 2 o ω 2 ) The 180 comes from the negative sign in the numerator of the T.F. These expressions are very similar to the corresponding low-pass functions. The resulting plots are mirror images of the low-pass plots. The expressions for the cut-off frequency and the properties of the bump in the high-qp plots are similarly symmetric to those from the low-pass case. EE 230 second-order filters 17
Linear-scale plots of the second-order high-pass magnitude, with = 0.125, 0.5, and 1.25. For all plots, the characteristic frequency is f o = 1 khz and the gain is G o = 1. Also shown for comparison is a first-order response with f c = 1 khz. 1.50 1.25 first order cut-off 1.00 T 0.75 cut-off level 0.50 first-order 0.25 0.00 10 100 1000 10000 100000 Frequency (Hz) EE 230 second-order filters 18
Bode versions of magnitude plots from previous slide. As seen with the highpass Bode plot, there is a slope of 40 db/dec in the cut-off region, indicative of an inverse-squared relationship. 10 0-10 3 db T (db) -20-30 -40 first-order first-order 3 db -50-60 10 100 1000 10000 100000 Frequency (Hz) EE 230 second-order filters 19
Phase angle frequency responses for the various cases of the secondorder high-pass functions. 180 135 first-order phase (degrees) 90 first-order 45 0 10 100 1000 10000 100000 Frequency (Hz) EE 230 second-order filters 20
band-pass T (s) = G o s 2 + a 1 s + a o s 2 + ( ) s + ω2 o For a band-pass response, the function should go to zero both for s = 0 and s. This can be accomplished by keeping only the linear term in the numerator. (Check it.) To make the magnitude of the transfer function go to Go in the pass-band region, choose a1 = ωo /QP. Then the standard form of the second-order band-pass filter is: T bp (s) = G o ( ) s s 2 + ( ) s + ω2 o This may be worth memorizing. (But if you note the symmetry of the various expressions, remembering them all is easy.) EE 230 second-order filters 21
To examine the frequency response, set s = jω. T bp (jω) = G o j ( ) ω ω 2 + j ( ) ω + ω2 o = G o j ( ) ω (ω 2 o ω 2 ) + j ( ) ω Extracting the magnitude and phase. T bp = G o ( ) ω (ω 2 o ω 2 ) 2 + ( ω ) 2 θ bp = 90 arctan ( ) ( ω ω 2 o ω 2 ) The 90 comes from the imaginary value in the numerator of the T.F. Again, very similar expressions to the low-pass and high-pass cases. EE 230 second-order filters 22
Linear-scale plots of the second-order band-pass magnitude, with = 0.125, 0.5, and 1.25. For all plots, the characteristic frequency is f o = 1 khz and the gain is G o = 1. (There is no first-order curve for comparison first-order BP filters do not exist.) 1.20 1.00 cut-off 0.80 T 0.60 cut-off level 0.40 0.20 0.00 10 100 1000 10000 100000 Frequency (Hz) EE 230 second-order filters 23
Bode versions of magnitude plots from previous slide. In the high- and lowfrequency cut-off bands, the slope is ±20 db/dec. 10 0-10 3 db T (db) -20-30 -40-50 3 db -60 10 100 1000 10000 100000 Frequency (Hz) EE 230 second-order filters 24
Phase angle frequency responses for the various cases of the secondorder high-pass functions. 90 68 45 phase (degrees) 23 0-23 -45-68 -90 10 100 1000 10000 100000 Frequency (Hz) EE 230 second-order filters 25
The peak of the bandpass magnitude occurs the characteristic frequency and falls off at both higher and lower frequencies. Thus there are two corner frequencies, defined in the usual manner. The difference between the two frequencies is the bandwidth of the filter. To calculate the two corners and the bandwidth, we start in the usual manner: G o ( ) ω c (ωo 2 ωc) 2 2 ω 2 + o ( ω Q c P ) = G o 2 After a fair amount of tedious and sometimes tricky algebra, we find the high and low corner frequencies, which we denote as ω c+ and ω c. ω c+ = 1 + 1 4Q 2 P + 1 2 ω c = 1 + 1 4Q 2 P 1 2 The bandwidth (BW) is the difference between these corners. It turns out to have a surprisingly simple relationship to and : bandwidth (BW): Δω = ω c+ ω c = EE 230 second-order filters 26 The quality factor directly controls the bandwidth.