Poles and Zeros of H(s), Analog Computers and Active Filters

Transcription

1 Poles and Zeros of H(s), Analog Computers and Active Filters Physics116A, Draft10/28/09 D. Pellett

2 LRC Filter Poles and Zeros Pole structure same for all three functions (two poles) HR has two poles and zero at s=0: bandpass filter HL has two poles and double zero at s=0: high-pass filter HC has two poles and no zeros: low-pass filter

3 LRC Filter Pole Locations and Natural Frequencies Poles at s1 and s2. If poles are distinct, the natural response is Natural response in the cas V out (t) = A 1 e s 1t + A 2 e s 2t Proof on next slide (Assumes there is no cancellation of a common pole and zero)

4 LRC Natural Response Natural response: solutions to homogeneous equation (Vin = 0) Vin = 0 i Vin(t)=0: this is the response (output) for zero input driving voltage

5 Comments on Natural Frequency vs. Resonant Frequency The resonant frequency is not the natural frequency Natural frequency si = σi+jωi : H denominator = 0 Resonant frequency ωr : reactance of H denominator = 0 However, for bandpass filter with Q>>1, ωr ωi Overdamped and critically damped cases (Q 0.5) don t have oscillatory natural response For high pass, low pass filters, the corner frequencies (halfpower points) are the key parameters; design to avoid strong resonance peaks in the frequency response A network of only passive components (no voltage or current sources) can have poles only in the left half of the complex frequency plane (including the imaginary axis in the ideal case of no resistance - actually not that easy: see oscillators)

6 s1 Pole and Resonance Peak for Q=5! 0 =5, Q=5 H(s)=1/(1+(s/5) 2 +s/25). s 1 = -1/2+j(3/2)(11) 1/2, s 2 = -1/2-j(3/2)(11) 1/2 ω σ

7 From H(s) to V(t) via Laplace Transform Laplace transform allows use of H(s) with suitable nonsinusoidal inputs (see 116B) The linearity of the Laplace transform and its transformation of ordinary differential equations with constant coefficients to algebraic equations allows a direct connection between H(s) and f(t). Essentially, f(t), the inverse Laplace transform of H(s), represents the output voltage of the circuit for a unit impulse function (delta function) input. This also assumes the voltages and currents are zero at t=0 (zero initial conditions). The Laplace transform method is studied in 116B. For this circuit, the inverse Laplace transform for H(s) is a damped sinusoid (see Table 5.1 in Bobrow or use Mathematica): V(t)=(50/(3*11 1/2 ))exp(-t/2)sin((3*11 1/2 )t/2)u(t) where u(t) is the unit step function. As seen earlier, pole location determines angular frequency and rate of decay Note that the expression is a linear combination of exp(s 1 t) and exp(s 2 t)

8 The transfer function H(s) = V out /V in for the above LRC low-pass filter circuit is:* H(s)=1/(1+(s/! 0 ) 2 +s/(q! 0 )) where! 0 =1/(LC) 1/2, Q=(1/R)*(L/C) 1/2. * You can work with s directly as in Sec. 5.3 of Bobrow, or you can find H(j!)=V out /V in using network analysis with complex impedance, then substitute s for j!.

9 Critically Damped Case (Q=0.5) H(s)=1/(1+(s/5) /2 ω0 = 5: +s/2.5) The denominator has a double root at s = -5 leading to a double pole ω σ Low pass filter behavior along the jω axis

10 From Frequency Domain to Time Domain Now, the inverse Laplace transform for H(s) is: V(t)=25 t exp(-5t) u(t) as expected for the pole of order 2 (i.e., double root of denominator) on the negative real axis. The pole location determines the time constant of the exponential. Again, this represents the output voltage of the circuit for a unit impulse function (delta function) input and zero initial conditions. Pole location determines angular frequency (0) and rate of decay Again, we don t need the Laplace transform to get the natural response - just use f(s) and initial conditions

11 General Case of Poles and Zeros (Following Bobrow, Sec. 5.4) Circuit described by linear differential equation with constant, real coefficients (see, for example, the series LRC circuit) d a n y d n dt n + a n 1 y n 1 dt n a 1 dy dt + a 0 = b d m x m dt m + b d m 1 x m 1 dt m b 1 dx dt + b 0 with output (forced response) y(t) = Ye st and input (forcing term) x(t) = Xe st. Substituting the exponentials for x and y gives (a n s n + a n 1 s n a 1 s + a 0 )Ye st = (b m s m + b m 1 s m b 1 s + b 0 )Xe st so the transfer function is a ratio of polynomials with real coefficients H(s) y x = b ms m + b m 1 s m b 1 s + b 0 a n s n + a n 1 s n a 1 s + a 0.

12 Pole and Zero Locations It follows from the fundamental theorem of algebra that the polynomials can be factored into products of polynomials with real coefficients of order 1 or 2. Thus, we can write H(s) = K(s z 1)(s z 2 )... (s z m ) (s p 1 )(s p 2 )... (s p n ) where z i and p i are real or complex conjugate pairs. The z i are the points where H(s) = 0 (zeros of the function) The p i are the points where the denominator equals 0 (poles of H(s)). There can be repeated roots so the poles need not be simple poles, as we saw before. Natural response in the case of distinct poles is of form y nat (t) = A 1 e p 1t + A 2 e p 2t A n e p nt

13 Synthesis of Network Using Op-Amps Can implement network with transfer function H(s) using integrators, amplifiers and summing amplifiers Example: H(s) Y/X = (1 + s/25 + s 2 /25) 1 Rewrite as Y = (25/s 2 )X (25/s 2 )Y (1/s)Y Recall for op-amp integrator, H(s) = 1/(RCs) Use 4 integrators, two amplifiers to multiply by constants and a summing amplifier to implement function (perhaps not very efficient since we started with a second order differential equation... ) (See another approach in a few minutes...) This is an analog computer, showing operational amplifiers in use for their original function: mathematical operations

14 Integrator and Network Block Diagram

15 Why Use Integrators? Could in principle use differentiators Switch R and C in op-amp circuit, get H(s) = -RC s H(ω) increases with ω so performance is sensitive to high frequency noise and amplifier instabilities (for reasons to be discussed later) Amplifier more likely to be overloaded by rapidly changing waveforms Initial conditions easier to set for integrator by switching a constant voltage across each integrating capacitor at t=0 Thus, integrators are preferred

16 More Efficient: State Variable Active Filter V VC in driven series LRC Reference: Millman and Grabel, Microelectronics, Sec Analysis: start by defining the second derivative of V at point A. Then just follow through the circuit tracing the paths and the rest follows. Output at D is proportional to V and is a low pass filter. Output at B is a band pass filter and at A a high pass. Do you see why? Can get poles in right half of s plane by connecting R from B to -Sum2 (NG)

17 State Variable Filter Outputs

18 Improved High Order Active Filters

19 Simple-Minded Approach: Buffered RC

20 Butterworth Filter Gives Sharper Cutoff

21 Amplitude Response; Implementation

22 Analysis: Part I

23 Analysis: Part II

24 Other Filter Types and Comparisons The circuit can be modified to give high pass or bandpass filters (for example, the bandpass filter you made in lab) Can replicate any denominator function using proper choices of R, C and enough terms Another popular choice is Chebyshev filter - uses Chebyshev polynomials (specify corner freq., allowable passband variation) Gives even sharper cutoff but has more ripple in the passband response Bessel filter is not as sharp as Butterworth near the cutoff frequency but has smoother phase response and less overshoot in time response with step function input Reference: Horowitz and Hill, The Art of Electronics, Ch. 5. See this book for comparisons and construction information.

25 Filter Configurations, Response Comparisons See reference in Ch. 5 of Horowitz and Hill (copyrighted material) See notes posted for Wednesday of this week (Bode plots for various filter designs) Butterworth Chebychev Cascaded RC low pass filters

26 With active networks, one can arrange for the complex conjugate poles to be on the imaginary axis: sinusoidal output (σ = 0). Covered later (see last lab). Get oscillator with feedback amplifier when Loop gain AB = -1 Oscillators Historically important. Triode vacuum tube (voltage controlled current source) made electronic oscillators possible for radio transmitters, regenerative receiver ( blooper ), superheterodyne receiver of Edwin Armstrong.