Poles and Zeros of H(s), Analog Computers and...

Physics116A, Draft10/28/09

D. Pellett

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

Active Filters

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

LRC 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 + A2es2t

4

Proof 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)

i

Vin(t)=0: this is the response (output) for zero input driving voltage

Vin = 0

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 (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 transform

Page 2 of 5http://www.physics.ucdavis.edu/Classes/Physics116/Poles_rev.html

!0=5, Q=5 and

H(s)=1/(1+(s/5)2+s/25).

For these values, the poles (the points where the denominator of H(s) vanishes) are at

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 the pole s1 is shown below. The real part of s (") is plotted

along 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=5 10/30/2006 09:41 AMPoles of H(s) and Laplace transform


!0=5, Q=5 and

H(s)=1/(1+(s/5)2+s/25).


s1 = -1/2+j(3/2)(11)1/2, s2 = -1/2-j(3/2)(11)

1/2.








!0=5, Q=5 and

H(s)=1/(1+(s/5)2+s/25).


s1 = -1/2+j(3/2)(11)1/2, s2 = -1/2-j(3/2)(11)

1/2.








!0=5, Q=5 and

H(s)=1/(1+(s/5)2+s/25).


s1 = -1/2+j(3/2)(11)1/2, s2 = -1/2-j(3/2)(11)

1/2.




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.


ω

σ

From H(s) to V(t) via Laplace Transform



!0=5, Q=5 and

H(s)=1/(1+(s/5)2+s/25).


s1 = -1/2+j(3/2)(11)1/2, s2 = -1/2-j(3/2)(11)

1/2.




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.




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.

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 s2determines the time constant of the exponential falloff of the amplitude while the imaginary part sets theangular frequency of oscillation.

Critically Damped Case

Now we repeat this analysis for the critically damped case, Q=0.5, keeping !0=5.

H(s)=1/(1+(s/5)1/2+s/2.5).

The denominator has a double root at s=-5, leading to a double pole.

The magnitude of H(s) in the neighborhood of the double pole is shown below. Again, the real part of s (")is plotted along x and the imaginary part (!) along y.



V(t)=(50/(3*111/2))exp(-t/2)sin((3*111/2)t/2)u(t)






H(s)=1/(1+(s/5)1/2+s/2.5).



As seen earlier, pole location determines

angular frequency and rate of decay

Laplace transform allows use of H(s) with suitable nonsinusoidal inputs (see 116B)



Poles of H(s) and Laplace Transform

Contents:

1. Structure of |H(s)| in the complex frequency plane for an LRC bandpass filter

2. Relate H(s) to Vout(t) for delta function input through the Laplace transform

Modified 10/27/06

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!.

Underdamped Case

Choose some parameter values (underdamped case):



Again, observe that the value of Abs(H(y, x=0)) (i.e., along the imaginary axis) is the magnitude of theoutput voltage as a function of angular frequency for unit input AC amplitude (here, y is the angularfrequency).

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 polelocation 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.

Q=0.5, w0=5

ω

σ

Low pass filter behavior along the jω axis



V(t)=(50/(3*111/2))exp(-t/2)sin((3*111/2)t/2)u(t)






H(s)=1/(1+(s/5)1/2+s/2.5).





V(t)=(50/(3*111/2))exp(-t/2)sin((3*111/2)t/2)u(t)






H(s)=1/(1+(s/5)1/2+s/2.5).



Critically Damped Case (Q=0.5)

ω0 = 5:

The denominator has a double root at s = -5 leading to a double pole

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 theoutput voltage as a function of angular frequency for unit input AC amplitude (here, y is the angularfrequency).

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 polelocation 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.

Q=0.5, w0=5

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

General Case of Poles and Zeros(Following Bobrow, Sec. 5.4)

Circuit described by linear di!erential equation with constant,real coe"cients

andnydtn + an!1

dn!1ydtn!1 + . . . + a1

dydt + a0

= bmdmxdtm + bm!1

dm!1xdtm!1 + . . . + b1

dxdt + b0

with output (forced response) y(t) = Yest and input (forcingterm) 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 coef-ficients

H(s) "y

x=

bmsm + bm!1sm!1 + . . . + b1s + b0ansn + an!1sn!1 + . . . + a1s + a0

.

1

(see, for example, the series LRC circuit)

Pole and Zero Locations

It follows from the “fundamental theorem of algebra” that thepolynomials can be factored into products of polynomials withreal coe!cients of order 1 or 2. Thus, we can write

H(s) =K(s! z1)(s! z2) . . . (s! zm)

(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 (polesof H(s)). There can be repeated roots so the poles need notbe simple poles, as we saw before.

• Natural response in the case of distinct poles is of formynat(t) = A1ep1t + A2ep2t + . . . + Anepnt

2

Synthesis of Network Using Op-Amps

• Can implement network with transfer function H(s) usingintegrators, 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/(RCs)

– Use 4 integrators, two amplifiers to multiply by constantsand a summing amplifier to implement function (perhapsnot very e!cient since we started with a second orderdi"erential equation. . . )

– This is an analog computer, showing operational ampli-fiers in use for their original function: mathematical op-erations

3

(See another approach in a few minutes...)

Integrator and Network Block Diagram

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

More Efficient: State Variable Active Filter

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)

Reference: Millman and Grabel, Microelectronics, Sec. 16-7V ≡ VC in driven series LRC

State Variable Filter Outputs

Improved High Order Active Filters

Simple-Minded Approach: Buffered RC

Butterworth Filter Gives Sharper Cutoff

Amplitude Response; Implementation

Analysis: Part I

Analysis: Part II

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.

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

Oscillators• 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

• Historically important.

• Triode vacuum tube (voltage controlled current source) made electronic oscillators possible for radio transmitters, regenerative receiver (“blooper”), superheterodyne receiver of Edwin Armstrong.

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