Poles, Zeros, and Frequency Response

Transcription

1 Complex Poles Poles, Zeros, and Frequency esponse With only resistors and capacitors, you're stuck with real poles. If you want complex poles, you need either an op-amp or an inductor as well. Complex poles allow you to build better low pass filters, high pass filters, band pass filters, and band reject filters. Graphically, this works as follows. Consider the transfer function s+a The frequency response is obtained by letting s jω: jω+a Graphically, the gain is equal to the vector '' divided by the vector ' jω + a'. The latter term is equal to the vector from the pole at -a to the origin (a) plus the vector jω. jw+a imag jw -a a real Since you're dividing by jω+a, the gain is A maximum when you're closest to the pole (i.e. at w = 0). Zero when you're far away from the pole (at infinity), and Down by 2 when the frequency is ja. If the pole is complex, this still works. The gain is A maximum when you're closest to the pole (i.e. at w = imag(a)). Zero when you're far away from the pole (at infinity), and Down by when the frequency is real(a) away from the resonance (imag(a). JSG - - rev February 26, 2009

2 For example, the frequency response of s+ j0 looks like the following. Note that the maximum gain is at w=0 (the complex part of the pole). The gain is down by 2 when you're rad/sec away from the resonance. Of course, you'll never (?) have a complex pole without it's conjugate. There should be a similar pole with a resonance at -0 rad/sec. The net result will be the above plot, slightly shifted left due to the conjugate pole. JSG rev February 26, 2009

3 Passive LC Filters: One type of LC filter uses a voltage source and LC in series: C L X Vr Vc Vl LC Filter. Make the voltage across, C, or L the output to get different types of filters. Y=Vc: Low Pass Filter Y = V c = /Cs +Ls+/Cs = LCs 2 +Cs+ Note that at DC (s=0), the gain is one (it's a low pass filter). The poles can be complex with,l,c determining the coefficients of the denominator. One element is arbitrary, the other two determine the poles. Y=Vr: Band Pass Filter Y = V r = +Ls+/Cs = Cs LCs 2 +Cs+ Note that the gain at DC (s=0) is zero due to the numerator 's'. The gain at infinity is also zero due to the denominator having an s 2 term. At resonance, LCs 2 + = 0 LCω 2 + = 0 ω 2 = LC At this frequency, the gain is one. Y = V L = High Pass Filter Y = V L = Ls +Ls+/Cs = LCs 2 LCs 2 +Cs+ Note that the gain at DC is zero while the gain as s goes to infinity is one. This makes a high pass filter. When you have complex poles, you can make the angle of the poles whatever you want. The angle affects how much resonance you get, however. Typical angles are: 0 degrees: 'C filter'. Both poles are real. JSG rev February 26, 2009

4 45 degrees: 'Butteworth Filter': The gain is maximally flat with no resonance degrees: 'Chebychev Filter' with a resonance of 3dB. For example, for the low-pass filter, the gain vs. frequency for these three angles are as follows: Note that there is a tradeoff. As you increase the angle of the poles (they become more complex), the bandwidth increases (good). The resonance also increases as well (bad). This is a common design decision in making speakers for stereos. The deeper the bass, the better the speaker sells. If you push the design too far, the speaker will have a resonance similar to the red curve above. When the bass guitar hits this note, the speaker 'booms' and plays it too loud - hence the name 'boom box.' In case you're interested, the relationship between the angle of the poles and the resonance is: where M m = 2ζ ζ 2 ζ = cos (θ) θ is the angle of the complex poles, and the transfer function can be written as? s ζω o s+ω 0 You'll see more about this in ECE 46. Example : Design a LC low pass filter with a corner at 250Hz. Solution: Take the output across the capacitor. Let ω o = 2π 250 = 500π for a corner at 250Hz. Let ζ = cos (45 0 ) = for a Butteworth filter. Then JSG rev February 26, 2009

5 LCs 2 +Cs+ = (57) 2 s 2 +0s+(57) 2 ewriting it so it's in the same form: So Let Then: /LC s 2 +/Ls+/LC LC = 572 L = 222 L = 0. H = 222 Ohms C = 4.05 uf = (57) 2 s s+(57) 2 (57) 2 Y = s s+(57) 2 To plot the gain vs. frequency the following MATLAB code works: >f = [0:000]'; >w = 2*pi*f; >s = j*w; >G = (57^2)./ (s.^ *s + 57^2); >plot(f,abs(g)) >xlabel('hz') >ylabel('gain') JSG rev February 26, 2009

6 Example 2: Design a LC high pass filter with a corner at 250Hz. Solution: Use the same design as before, just take the voltage across the inductors as the output: s 2 Y = s 2 +0s+(57) 2 >G2 = (s.^2)./ (s.^ *s + 57^2); >plot(f,abs(g2)) >xlabel('hz') >ylabel('gain') Example 3: Design a LC band pass filter which passes frequencies between 240Hz and 260Hz. Solution: Use the LC filter with the output being the resistor. For a maximum gain of 250Hz, let the complex part of the poles be imag(pole)=250hz = 500π For a bandwidth of +/-0Hz from the resonance, pick the real part of the pole to be real(pole)=0hz = 20π Thus, the transfer function should be So,? (s+20π+j500π)(s+20π j500π)? s 2 +(40π)s+ (40π)2 +(500π) 2 Matching terms: 25s s 2 +25s+2,483,92 25s s 2 +25s+2,483,92 = X Cs LCs 2 +Cs+ JSG rev February 26, 2009

7 25s s 2 +25s+2,483,92 /L = 25 /LC = 2,483,92 Let L = 0.H = 2.5 Ohms C = 4.02 uf In MATLAB: = /Ls s 2 +/Ls+/LC >G3 = (25*s)./ (s.^2 + 25*s ); >plot(f,abs(g3)) >xlabel('hz') >ylabel('gain') Band eject Filters A slight variant uses an inductor and a capacitor in series and parallel, The parallel combination is the band pass filter above (where the output is across the resistor). Note that the impedance of L and C are: or Z = jωc + jωl Z = j ωl ωc At DC, the impedance is infinity due to the capacitor. At infinity, the impedance is infinity due to the inductor. At some frequency, however, the impedance of L and C cancel. In this case, their combined resistance is zero at that particular frequency. You can do the same thing with L and C in parallel. In this case, the admittance is JSG rev February 26, 2009

8 Z = (jωl) z = jωl + jωc jωc = j Z ωc + ωl Now, the admittance is infinity at ω=0 (DC) and ω = (meaning the resistance is zero at ω=0 and ω= ). At resonance, the C and L terms cancel, resulting in the admittance being zero (or the resistance being infinity). So, LC in series has a resistance of zero at resonance LC in parallel has a resistance of infinity at resonance. This lets you build networks to pass or block different frequencies. Example 4: Design a filter to reject frequencies between 58Hz and 62Hz. Solution: Use a voltage divider like we've been doing. In the top part, use a resistor. In the part to ground, add a LC in series with a resonance of 60Hz to short out the output at 60Hz. C + X + - L Y - By voltage division, the gain is Ls+ Cs Ls+ Cs + LCs 2 + LCs 2 +Cs+ s 2 + LC s 2 + L s+ LC For a resonance of 60Hz (20 π rad/sec), let LC = (20π)2 JSG rev February 26, 2009

9 For a bandwidth of 4Hz (8 π rad/sec), let Let L = 8π L = 0. H = 2.53 Ohms C = 70.3uF The gain vs. frequency is then s 2 +(20π) 2 Y = s 2 +8πs+(20π) 2 In MATLAB: >f = [50:0.:70]'; >w = 2*pi*f; >s = j*w; >G5 = (s.^ ^2)./ (s.^ *s + 377^2); >plot(f,abs(g5)) >xlabel('hz'); >ylabel('gain') Note: You can do better if you replace the resistor with an LC parallel network. This is what the utilities do at their substations. High-frequency signals are placed on the power lines so you can talk to and from the substations. To prevent the 60Hz power from interfering with this communication, LC series / parallel networks are used to pass / block the 60Hz signal. This is more efficient that a resistor, which would dissipate energy as heat, costing the utilities money. JSG rev February 26, 2009