# Poles and Zeros of H(s)

```Poles and Zeros of H(s),
Analog Computers and
Active Filters
Physics116A, Draft10/31/06
D. Pellett
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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 (&quot;) 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 (&quot;)
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)
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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
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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
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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
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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
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Integrator and Network Block Diagram
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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
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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)
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State Variable Filter Outputs
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Improved High Order Active Filters
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Simple-Minded Approach: Buffered RC
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Butterworth Filter Gives Sharper Cutoff
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Implementation
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Analysis: Part I
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Analysis: Part II
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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.
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