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UCD PHY 116A - Poles and Zeros of H(s), Analog Computers and Active Filters

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Physics116A, Draft10/28/09D. PellettPoles and Zeros of H(s), Analog Computers and Active FiltersLRC 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 filterLRC Filter Pole Locations and Natural Frequencies Poles at s1 and s2. If poles are distinct, the natural response is Natural response in the case of distinct poles is of formVout(t) = A1es1t+ A2es2t4Proof on next slide(Assumes there is no cancellation of a common pole and zero)LRC Natural Response•Natural response: solutions to homogeneous equation (Vin = 0) iVin(t)=0: this is the response (output) for zero input driving voltageVin = 0Comments 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 (half-power 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)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):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)


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UCD PHY 116A - Poles and Zeros of H(s), Analog Computers and Active Filters

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