Physics116A, Draft10/31/06D. PellettPoles and Zeros of H(s), Analog Computers and Active Filters1LRC 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 filter2LRC Filter Pole LocationsPoles at s1 and s2. If poles are distinct, the natural response is of formNatural response in the case of distinct poles is of formVout(t) = A1es1t+ A2es2t4310/30/2006 09:41 AMPoles of H(s) and Laplace transformPage 1 of 5http://www.physics.ucdavis.edu/Classes/Physics116/Poles_rev.htmlPoles of H(s) and Laplace TransformContents:1. Structure of |H(s)| in the complex frequency plane for an LRC bandpass filter2. Relate H(s) to Vout(t) for delta function input through the Laplace transformModified 10/27/06Introduction | Underdamped | Critical DampingIntroductionThe 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 networkanalysis with complex impedance, then substitute s for j!. Underdamped CaseChoose some parameter values (underdamped case):410/30/2006 09:41 AMPoles of H(s) and Laplace transformPage 2 of 5http://www.physics.ucdavis.edu/Classes/Physics116/Poles_rev.html!0=5, Q=5 andH(s)=1/(1+(s/5)2+s/25).For these values, the poles (the points where the denominator of H(s) vanishes) are ats1 = -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 the pole s1 is shown below. The real part of s (") is plottedalong x and the imaginary part (!) along y.The value of Abs(H(y, x=0)) (i.e., along the imaginary axis) is the magnitude of the output voltage as afunction of angular frequency for unit input AC (sinusoid) amplitude (here, y is the angular frequency, !).Notice the resonance peak.The linearity of the Laplace transform and its transformation of ordinary differential equations with constantcoefficients to algebraic equations allows a direct connection between H(s) and f(t). Essentially, f(t), theinverse 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 oruse Mathematica):s1 Pole and Resonance Peak for Q=510/30/2006 09:41 AMPoles of H(s) and Laplace transformPage 2 of 5http://www.physics.ucdavis.edu/Classes/Physics116/Poles_rev.html!0=5, Q=5 andH(s)=1/(1+(s/5)2+s/25).For these values, the poles (the points where the denominator of H(s) vanishes) are ats1 = -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 the pole s1 is shown below. The real part of s (") is plottedalong x and the imaginary part (!) along y.The value of Abs(H(y, x=0)) (i.e., along the imaginary axis) is the magnitude of the output voltage as afunction of angular frequency for unit input AC (sinusoid) amplitude (here, y is the angular frequency, !).Notice the resonance peak.The linearity of the Laplace transform and its transformation of ordinary differential equations with constantcoefficients to algebraic equations allows a direct connection between H(s) and f(t). Essentially, f(t), theinverse 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 oruse Mathematica):10/30/2006 09:41 AMPoles of H(s) and Laplace transformPage 2 of 5http://www.physics.ucdavis.edu/Classes/Physics116/Poles_rev.html!0=5, Q=5 andH(s)=1/(1+(s/5)2+s/25).For these values, the poles (the points where the denominator of H(s) vanishes) are ats1 = -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 the pole s1 is shown below. The real part of s (") is plottedalong x and the imaginary part (!) along y.The value of Abs(H(y, x=0)) (i.e., along the imaginary axis) is the magnitude of the output voltage as afunction of angular frequency for unit input AC (sinusoid) amplitude (here, y is the angular frequency, !).Notice the resonance peak.The linearity of the Laplace transform and its transformation of ordinary differential equations with constantcoefficients to algebraic equations allows a direct connection between H(s) and f(t). Essentially, f(t), theinverse 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 oruse Mathematica):10/30/2006 09:41 AMPoles of H(s) and Laplace transformPage 2 of 5http://www.physics.ucdavis.edu/Classes/Physics116/Poles_rev.html!0=5, Q=5 andH(s)=1/(1+(s/5)2+s/25).For these values, the poles (the points where the denominator of H(s) vanishes) are ats1 = -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 the pole s1 is shown below. The real part of s (") is plottedalong x and the imaginary part (!) along y.The value of Abs(H(y, x=0)) (i.e., along the imaginary axis) is the magnitude of the output voltage as afunction of angular frequency for unit input AC (sinusoid) amplitude (here, y is the angular frequency, !).Notice the resonance peak.The linearity of the Laplace transform and its transformation of ordinary differential equations with constantcoefficients to algebraic equations allows a direct connection between H(s) and f(t). Essentially, f(t), theinverse 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