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Poles and Zeros of H(s), Analog Computers and Active Filters Physics116A, Draft10/31/06 D. Pellett 1 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 2 LRC Filter Pole Locations Poles at s1 and s2. If poles are distinct, the natural response is of form Natural response in the case of distinct poles is of form Vout(t) = A1es1t + A2es2t 3 Introduction | Underdamped | Critical Damping Introduction The transfer function H(s) = Vout/Vin 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!)=Vout/Vin using network analysis with complex impedance, then substitute s for j!. 4 ! =5, Q=5 and s 0= -1/2+j(3/2)(11)1/2 , 1 s2 = -1/2-j(3/2)(11)1/2 . s Pole!and Resonance Peak for Q=5 Poles H(s) andofLaplace transform Theof magnitude H(s) in the neighborhood of the pole s 1 is shown below. The real part of s (") is plotted 2 1 H(s)=1/(1+(s/5) 0 =5, along x and the imaginary part +s/25). (!) Q=5 along y. and ! andthe poles (the points where the denominator of H( 0 =5, For theseQ=5 values, 2 H(s)=1/(1+(s/5) +s/25). 1/2 , s = -1/2-j(3/2)(11)1/2 . 2 s = -1/2+j(3/2)(11) H(s)=1/(1+(s/5) +s/25). 2 1 For these values, the poles (the points where the The magnitude of H(s) in the (the neighborhood of thethe pole s 1 is shown For these values, the poles points1/2 where denominator s1imaginary = -1/2+j(3/2)(11) along x and the part (!) along y. , s2 = -1/2-j(3/2 s1 = -1/2+j(3/2)(11)1/2 , s2 = -1/2-j(3/2)(11)1/2 . The magnitude of H(s) in the neighborhood of th The magnitude of H(s) theimaginary neighborhood poley.s 1 is s along x andinthe part of (!)the along along x and the imaginary part (!) along y. ω σ The value of Abs(H(y, x=0)) (i.e., along the imaginary axis) is the magnitude of the output voltage as a function of angular frequency for unit input AC (sinusoid) amplitude (here, y is the angular frequency, !). Notice the resonance peak. 5 The value of Abs(H(y, x=0)) (i.e., along the imaginary axis) is the magnitude of the output voltage as a function of angular frequency for unit input AC (sinusoid) amplitude (here, y is the angular frequency, !). Notice the resonance peak. From H(s) to V(t) via Laplace Transform Poles of H(s) and Laplace transform 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 1/2 ))exp(-t/2)sin((3*11 1/2 )t/2)u(t) inverse Laplace transform of H(s), representsV(t)=(50/(3*11 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). where u(t) is the unit step function. The Laplace transform method is studied in 116B. Poles of H(s) and Laplace transform 10/30/2006 09:41 Note that the expression is a linear combination of exp(s t) and exp(s t 1 2 For this circuit, the inverse Laplace transform for H(s) is a damped sinusoid (see Table 5.1 in Bobrow or locations (complex conjugate poles in the left half of the complex plane). T use Mathematica): determines the time constant of the exponential falloff of the amplitude wh angular frequency of oscillation. V(t)=(50/(3*111/2 ))exp(-t/2)sin((3*111/2 )t/2)u(t) http://www.physics.ucdavis.edu/Classes/Physics116/Poles_rev.html Page 2 of 5 where u(t) is the unit step function. Note that the expression is a linear combination of exp(s1t) and exp(s2t) where s1 and s2 are the pole locations (complex conjugate poles in the left half of the complex plane). The real part of s1 and s2 Pole location determines determines the time constant of the exponential falloff of the amplitude while the imaginary part sets the Poles of H(s)frequency and Laplace transform 10/30/20 angular frequency angular of oscillation. and rate of decay V(t)=(50/(3*111/2 ))exp(-t/2)sin((3*111/2 )t/2)u(t) where u(t) is the unit step function. Critically Note that the expression is a linear combination of exp(s1t) and exp(s2t) where Damped s1 and s2Case are the p locations (complex conjugate poles in theNow leftwehalf of this the analysis complex Thedamped real part ofQ=0.5, s1 andkeepin s2 repeat for plane). the critically case, 6 determines the time constant of the exponential falloff of the amplitude while the imaginary part sets Critically Damped Case Case Now repeat thethecritically Nowwewe repeatthis thisanalysis analysisforfor criticalld Now we repeat this analysis Damped for the criticallyCase damped case, Q=0.5, keeping ! 0 =5. Critically (Q=0.5) Poles of H(s) and Laplace transform 10/30/2006 09 1/2 +s/2.5). 1/2 1/2 H(s)=1/(1+(s/5) 5, keeping ! 0 =5. H(s)=1/(1+(s/5) +s/2.5). H(s)=1/(1+(s/5) +s/2.5). The denominator has a double root at s=-5, leading to a double pole. The denominator has a double root at s=-5, The denominator has a double root at s=-5, The magnitude of H(s) in the neighborhood of the double pole is shown below. Again, t e pole. is plotted along x and the imaginary part (!) along y. The ofofH(s) of The magnitude H(s)ininthetheneighborhood neighborhood shown below. Again, the realmagnitude part of s (") is is plotted part (!) plottedalong alongx and x andthetheimaginary imaginary part (!a ω σ Again, observe that the value of Abs(H(y, x=0)) (i.e., along the imaginary axis) is the magnitude of the output voltage as a function of angular frequency for unit input AC amplitude (here, y is the angular frequency). We see low pass filter behavior along the jω axis Now, the inverse Laplace transform for H(s) is: http://www.physics.ucdavis.edu/Classes/Physics116/Poles_rev.html V(t)=25 t exp(-5t) u(t) 7 From Frequency Domain to Time Domain Again, observe that the value of Abs(H(y, x=0)) (i.e., along the imaginary axis) is the magnitude of the output voltage as a function of angular frequency for unit input AC amplitude (here, y is the angular frequency). Again, observe that the value of Abs(H(y, x=0)) (i.e., along the imagin output voltage as a function of angular frequency for unit input AC am Now, the inverse Laplace transform for H(s) is: frequency). V(t)=25 t exp(-5t) u(t) Now, the inverse Laplace transform for H(s) is: V(t)=25 t exp(-5t)on u(t) as expected for the pole of order 2 (i.e., double root of denominator) the negative real axis. The pole location determines the time constant of the exponential. as expected for the pole of order 2 (i.e., double root of denominator) on 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. Again, this represents the output voltage of the circuit for a unit impul and zero initial conditions. Q=0.5, w0=5 Q=0.5, w0=5 Pole location determines angular frequency (0) and rate of decay http://www.physics.ucdavis.edu/Classes/Physics116/Poles_rev.html http://www.physics.ucdavis.edu/Classes/Physics116/Poles_rev.html Page 4 of 5 8 General Case of Poles and Zeros (Following Bobrow, Sec. 5.4) Circuit described by linear differential equation with constant, real coefficients dny dn−1 y an dtn + an−1 dtn−1 + . . . + a1 dy dt + a0 mx m−1 x d d = bm dtm + bm−1 dtm−1 + . . . + b1 dx dt + b0 with output (forced response) y(t) = Yest and input (forcing term) x(t) = Xest. Substituting the exponentials for x and y gives (ansn + an−1sn−1 + . . . + a1s + a0)Yest = (bmsm + bm−1sm−1 + . . . + b1s + b0)Xest so the transfer function is a ratio of polynomials with real coefficients y bmsm + bm−1sm−1 + . . . + b1s + b0 . H(s) ≡ = n n−1 + . . . + a1 s + a0 x ans + an−1s 1 9 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 K(s − z1)(s − z2) . . . (s − zm) H(s) = (s − p1)(s − p2) . . . (s − pn) where zi and pi are real or complex conjugate pairs. • The zi are the points where H(s) = 0 (zeros of the function) • The pi 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 ynat(t) = A1ep1t + A2ep2t + . . . + Anepnt 2 10 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 + s2/25)−1 – Rewrite as Y = (25/s2)X − (25/s2)Y − (1/s)Y – Recall for op-amp integrator, H(s) = −1/(RC s) – 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. . . ) – This is an analog computer, showing operational amplifiers in use for their original function: mathematical operations 3 11 Integrator and Network Block Diagram 12 Why Use Integrators? • • Could in principle use differentiators– • • Switch R and C in op-amp circuit, get H(s) = -RC s • 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 |H(ω)| increases with ω so performance is sensitive to high frequency noise and amplifier instabilities (for reasons to be discussed later) Thus, integrators are preferred 13 More Efficient: State Variable Active Filter Reference: Millman and Grabel, Microelectronics, Sec. 16-7 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) 14 State Variable Filter Outputs 15 Improved High Order Active Filters 16 Simple-Minded Approach: Buffered RC 17 Butterworth Filter Gives Sharper Cutoff 18 Implementation 19 Analysis: Part I 20 Analysis: Part II 21 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. 22