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Second Order Filters Overview
• What’s different about second order filters
• Resonance
• Standard forms
• Frequency response and Bode plots
• Sallen-Key filters
• General transfer function synthesis
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Prerequisites and New Knowledge
Prerequisite knowledge
• Ability to perform Laplace transform circuit analysis
• Ability to solve for the transfer function of a circuit
• Ability to generate and interpret Bode plots
New knowledge
• Ability to design second-order filters
• Knowledge of the advantages of transfer functions with complexpoles and zeros
• Understanding of resonance
• Familiarity with second-order filter properties
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Motivation
Y (s) =
(
∑Mk=0 bksk
∑N
k=0 aksk
)
X(s) = H(s)X(s)
• Second order filters are filters with a denominator polynomial thatis of second order
• Usually the roots (poles) are complex
p1,2 = −ζωn ± jωn
√
1 − ζ2
• In many applications, second order filters may be sufficient
• Are also the building blocks for higher-order filters
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Standard Form
H(s) =N(s)
s2
ω2n
+ sQωn
+ 1
• Second order filters can be designed as lowpass, highpass,bandpass, or notch filters
• All four types can be expressed in standard form shown above
• N(s) is a polynomial in s of degree m ≤ 2
– If N(s) = k, the system is a lowpass filter with a DC gain of k
– If N(s) = k s2
ω2n
, the system is a highpass filter with a high
frequency gain of k
– If N(s) = k sQωn
, the system is a bandpass filter with a
maximum gain of k
– If N(s) = k(
1 − s2
ω2n
)
, the system is a notch filter with a gain
of k
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Unity Gain Lowpass
A unity-gain lowpass second-order transfer function is of the form
H(s) =ω2
n
s2 + 2ζωns + ω2n
=1
1 + 2ζ sωn
+(
sωn
)2
• ωn is called the undamped natural frequency
• ζ (zeta) is called the damping ratio
• The poles are p1,2 = (−ζ ±√
ζ2 − 1) ωn
• If ζ ≥ 1, the poles are real
• If 0 < ζ < 1, the poles are complex
• If ζ = 0, the poles are imaginary: p1,2 = ±jωn
• If ζ < 0, the poles are in the right half plane (Rep > 0) and thesystem is unstable
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Unity Gain Lowpass Continued
The transfer function H(s) can also be expressed in the following form
H(s) =1
1 + 2ζ sωn
+(
sωn
)2=
1
1 + sQωn
+(
sωn
)2
where
Q ,1
2ζ
The meaning of Q, the Quality factor, will become clear in thefollowing slides.
H(jω) =1
1 −(
ωωn
)2
+ jωQωn
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Unity Gain Lowpass Magnitude Response
20 log10 |H(jω)| = 20 log10
∣
∣
∣
∣
∣
1 −(
ω
ωn
)2
+jω
Qωn
∣
∣
∣
∣
∣
−1
= −20 log10
√
(
1 − ω2
ω2n
)2
+
(
ω
Qωn
)2
For ω ≪ ωn,
20 log10 |H(jω)| ≈ −20 log10 |1| = 0 dB
For ω ≫ ωn,
20 log10 |H(jω)| ≈ −20 log10
ω2
ω2n
= −40 log10
ω
ωn
dB
At these extremes, the behavior is identical to two real poles.
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Lowpass Magnitude Continued
20 log10 |H(jω)| = −20 log10
√
(
1 − ω2
ω2n
)2
+
(
ω
Qωn
)2
For ω = ωn,
20 log10 |H(jωn)| = −20 log10
1
Q= 20 log10 Q = QdB
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Lowpass Magnitude
10−2
10−1
100
101
102
−60
−40
−20
0
20
40
Frequency (rad/sec)
Mag
(dB
)
Complex Poles
Q=0.100Q=0.500Q=0.707Q=1.000Q=2.000Q=10.000Q=100.000
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Matlab Code
function [] = BodeComplexPoles();
N = 1000;
wn = 1;
Q = [0.1 0.5 sqrt (1/2) 1 2 10 100];
w = logspace(-2,2,N);
nQ = length(Q);
mag = [];
phs = [];
legendLabels = cell(nQ ,1);
for c1 = 1:nQ
sys = tf([wn^2] ,[1 wn/Q(c1) wn ^2]);
[m, p] = bode(sys ,w);
mag = [mag reshape(m,N ,1)];
phs = [phs reshape(p,N ,1)];
legendLabels c1 = sprintf(’Q=%5.3f’,Q(c1));
end
figure(1);
FigureSet(1,’Slides’);
h = semilogx(w,20* log10(mag));
set(h,’LineWidth ’,1.5);
grid on;
xlabel(’Frequency (rad/sec)’);
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ylabel(’Mag (dB)’);
title(’Complex Poles’);
xlim ([w(1) w(end)]);
ylim ([-60 45]);
box off;
AxisSet (8);
legend(legendLabels );
print(’BodeMagComplexPoles’,’-depsc’);
figure(2);
FigureSet(2,’Slides’);
h = semilogx(w,phs);
set(h,’LineWidth ’,1.5);
grid on;
xlabel(’Frequency (rad/sec)’);
ylabel(’Phase (deg)’);
title(’Complex Poles’);
xlim ([w(1) w(end)]);
ylim ([ -190 10]);
box off;
AxisSet (8);
legend(legendLabels );
print(’BodePhaseComplexPoles’,’-depsc’);
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Lowpass Phase
∠H(jω) = ∠1
1 −(
ωωn
)2
+ jωQωn
For ω ≪ ωn,∠H(jω) ≈ ∠1 = 0
For ω ≫ ωn,
∠H(jω) ≈ ∠1
− ωQωn
= ∠ − Qωn
ω= ∠ − 1 = −180
For ω = ωn,
∠H(jω) ≈ ∠1
Qj= ∠
1
j= ∠ − j = −90
At the extremes, the behavior is identical to two real poles. At othervalues of ω near ωn, the behavior is more complicated.
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Unity Gain Lowpass Phase
10−2
10−1
100
101
102
−180
−160
−140
−120
−100
−80
−60
−40
−20
0
Frequency (rad/sec)
Pha
se (
deg)
Complex Poles
Q=0.100Q=0.500Q=0.707Q=1.000Q=2.000Q=10.000Q=100.000
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Lowpass Comments
H(jω) =1
1 +(
jωQωn
)
+(
jωωn
)2
• The second-order response attenuates twice as fast as a first-orderresponse (40 dB/decade)
• Generally better than the cascade of two first-order filters
• Offers additional degree of freedom (Q), which is the gain nearω = ωn
• Q may range from 0.5 to 100
• Is usually near Q = 1
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Lowpass Maximum Magnitude
What is the frequency at which |H(jω)| is maximized?
H(jω) =1
1 +(
jωQωn
)
+(
jωωn
)2|H(jω)| =
1√
(
1 − ω2
ω2n
)2
+(
ωQωn
)2
• For high values of Q, the maximum of |H(jω)| > 1
• This is called peaking
• The largest Q before the onset of peaking is Q = 1√2≈ 0.707
– Said to be maximally flat
– Also called a Butterworth response
– In this case, |H(jωn)| = −3 dB and ωn is the cutoff frequency
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Lowpass Maximum Continued
If Q > 0.707, the maximum magnitude and frequency are as follows:
ωr = ωn
√
1 − 1
2Q2|H(jωr)| =
Q
1 − 14Q2
• ωr is called the resonant frequency or the damped natural
frequency
• As Q → ∞, ωr → ωn
• For sufficiently large Q (say Q > 5)
– ωr ≈ ωn
– |H(jωr)| ≈ Q
• Peaked responses are useful in the synthesis of high-order filters
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Example 1: Passive Lowpass
v s ( t ) v o ( t )
-
+
50 mH R
200 nF
Generate the bode plot for the circuit shown above.
H(s) =1
LC
s2 + RL
s + 1LC
=ω2
n
s2 + 2ζωns + ω2n
=ω2
n
s2 + ωn
Qs + ω2
n
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Example 1:Passive Lowpass Continued
ωn =
√
1
LC= 10 k rad/s
ζ =R
2L
√LC =
R
2
√CL = R × 0.001
Q =1√LC
L
R=
√
L
C
1
R=
500
R
R = 5 Ω ζ = 0.005 Q = 100 Very Light Damping
R = 50 Ω ζ = 0.05 Q = 10 Light Damping
R = 707 Ω ζ = 1.41 Q = 0.707 Strong Damping
R = 1 kΩ ζ = 1 Q = 0.5 Critical Damping
R = 5 kΩ ζ = 5 Q = 0.1 Over Damping
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Example 1: Lowpass Magnitude Response
102
103
104
105
106
−120
−100
−80
−60
−40
−20
0
20
40
Frequency (rad/sec)
Mag
(dB
)
Resonance Example
R = 5 ΩR = 50 ΩR = 707 ΩR = 1 kΩR = 5 kΩR = 50 kΩ
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Example 1: Lowpass Phase Response
102
103
104
105
106
−180
−160
−140
−120
−100
−80
−60
−40
−20
0
Frequency (rad/sec)
Pha
se (
deg)
Resonance Example
R = 5 ΩR = 50 ΩR = 707 ΩR = 1 kΩR = 5 kΩR = 50 kΩ
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Practical Second-Order Filter Tradeoffs
• There are many standard designs for second-order filters
• A library of first and second-order filters is all you need tosynthesize any transfer function
• The ideal transfer functions are identical
• Engineering design tradeoffs are practical
– Cost (number of op amps)
– Complexity
– Sensitivity to component value variation
– Ease of tuning
– Ability to vary cutoff frequencies and/or gains
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Practical Second-Order Filters
• Popular practical second-order filters include
– KRC/Sallen-key filters
– Multiple-feedback filters: more than one feedback path
– State-variable filters
– Biquad/Tow-Thomas filters filters
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Example 2: Lowpass Sallen-Key Filter
v o ( t )
-
+ v s ( t )
R 1 R 2
C 1
C 2
R b
R a
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Example 2: Questions
• Is the input impedance finite?
• What is the output impedance?
• How many capacitors are in the circuit?
• What is the DC gain?
• What is the “high frequency” gain?
• What is the probable order of the transfer function?
• Solve for the transfer function
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Example 2: Work Space
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Example 2: Work Space
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Example 2: Tuning
|H(0)| =Ra + Rb
Ra
= k
ωn =1√
R1C1R2C2
Q =1
√
R2C2
R1C1
+√
R1C2
R2C1
+ (1 − k)√
R1C1
R2C2
• 6 circuit parameters (four resistors, two capacitors)
• 3 constraints
– DC gain, k
– Natural frequency
– Q
• Prudent to pick K = 1, since gain is often added at in the firstand/or last stages
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Example 2: Tuning Tips
|H(0)| =RA + RB
RA
ωn =1√
R1C1R2C2
Q =1
√
R2C2
R1C1
+√
R1C2
R2C1
+ (1 − K)√
R1C1
R2C2
1. If you select R1 = R2 = R and C1 = C2 = C, the equationssimplify considerably
• Watch out for coarse tolerances on components
2. Helps guard against op amp limitations if you select k = 1
• Gain is often used at end or beginning of filter stages
3. Best to start of with two capacitances in a ratio C1
C2
≥ 4Q2
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Example 3: Design
Design a Sallen-Key lowpass filter with a Q = 1/√
2, corner
frequency of 1000 Hz, and DC gain of 1.
• Redraw the circuit with a DC gain of 1
• Arbitrarily pick C2 = 1 nF, a common capacitor value that ischeaply and readily available
Hint: Recall that the solution to ax2 + bx + c = 0 is x = −b±√
b2−4ac2a
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Example 3: DC Gain Solution
|H(0)| =RA + RB
RA
= 1 +RB
RA
Pick RB = 0 and RA = ∞
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Example 3: Solution
Define R1 = mR2 and C1 = nC2
ωn =1√
R1C1R2C2
=1
R2C2
√nm
Q =1
√
R2C2
R1C1
+√
R1C2
R2C1
+ (1 − K)√
R1C1
R2C2
=1
√
1mn
+√
mn
√mn√mn
=
√mn
1 + m
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Example 3: Solution
Q2 =mn
1 + 2m + m2
m2 + 2m + 1 =mn
Q2
1
2m2 + m +
1
2=
mn
2Q2
1
2m2 +
(
1 − mn
2Q2
)
m +1
2= 0
β ,n
2Q2− 1
1
2m2 − βm +
1
2= 0
am2 + bm + c = 0
m =−b ±
√b2 − 4ac
2a= β +
√
β2 − 1
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Example 3: Solution
1. Arbitrarily pick C2 = 10 nF, a common capacitor value that ischeaply and readily available
2. Select n such that C1
C2
≥ 4Q2, to ensure solution to quadratic
equation for m = R1
R2
is real valued
• Pick n = 8
• Then C1 = 8Q2C2 = 80 nF
3. Solve for m = β +√
β2 − 1, given β = n2Q2 − 1
• m = 1
4. Solve for R2 = 1C2ωn
√nm
• R2 = 2.33 kΩ
5. Solve for R1 = mR2, given m and R2
• R1 = 13.6 kΩ
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Example 3: Magnitude Response
102
103
104
105
−60
−50
−40
−30
−20
−10
0
10
20
Frequency (rad/sec)
Mag
(dB
)
Complex Poles
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Example 3: Phase Response
102
103
104
105
−180
−160
−140
−120
−100
−80
−60
−40
−20
0
Frequency (rad/sec)
Pha
se (
deg)
Complex Poles
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Example 3: MATLAB Code
function [] = SallenKeyLowpass();
RA = inf;
RB = 0;
C2 = 10e-9;
Q = 1;
fn = 1000;
%==========================================================
% Preprocessing
%==========================================================
dCGain = 1 + RB/RA;
wn = 2*pi*fn;
n = 8*Q^2;
b = n/(2*Q^2) -1;
m = b + sqrt(b^2 -1);
C1 = n*C2;
R2 = 1/(wn*C2*sqrt(n*m));
R1 = m*R2;
fprintf(’C1: %5.3f nF\n’,C1*1e9);
fprintf(’C2: %5.3f nF\n’,C2*1e9);
fprintf(’R1: %5.3f kOhms\n’,R1*1e -3);
fprintf(’R2: %5.3f kOhms\n’,R2*1e -3);
N = 1000;
functionName = sprintf(’%s’,mfilename ); % Get the function name
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w = logspace(2,5,N);
mag = [];
phs = [];
fileIdentifier = fopen([ functionName ’.tex’],’w’);
sys = tf([ dCGain] ,[1/(wn^2) 1/(wn*Q) 1]);
[m, p] = bode(sys ,w);
bodeMagnitude = reshape(m,N,1);
bodePhase = reshape(p,N,1);
figure(1);
FigureSet(1,’Slides’);
h = semilogx(w,20* log10(bodeMagnitude ),’r’);
set(h,’LineWidth ’,1.5);
grid on;
xlabel(’Frequency (rad/sec)’);
ylabel(’Mag (dB)’);
title(’Complex Poles’);
xlim ([w(1) w(end)]);
ylim ([-60 20]);
box off;
AxisSet (8);
fileName = sprintf(’%s-Magnitude ’,functionName );
print(fileName ,’-depsc’);
fprintf(fileIdentifier ,’%%==============================================================================\n’
fprintf(fileIdentifier ,’\\ newslide\n’);
fprintf(fileIdentifier ,’\\ slideheading Example \\arabicexc: Magnitude Response\n’);
fprintf(fileIdentifier ,’%%==============================================================================\n’
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fprintf(fileIdentifier ,’\\ includegraphics[scale=1] Matlab/%s\n’,fileName);
figure(2);
FigureSet(2,’Slides’);
h = semilogx(w,bodePhase ,’r’);
set(h,’LineWidth ’,1.5);
grid on;
xlabel(’Frequency (rad/sec)’);
ylabel(’Phase (deg)’);
title(’Complex Poles’);
xlim ([w(1) w(end)]);
ylim ([ -190 10]);
box off;
AxisSet (8);
fileName = sprintf(’%s-Phase’,functionName );
print(fileName ,’-depsc’);
fprintf(fileIdentifier ,’%%==============================================================================\n’
fprintf(fileIdentifier ,’\\ newslide\n’);
fprintf(fileIdentifier ,’\\ slideheading Example \\arabicexc: Phase Response\n’);
fprintf(fileIdentifier ,’%%==============================================================================\n’
fprintf(fileIdentifier ,’\\ includegraphics[scale=1] Matlab/%s\n’,fileName);
fprintf(fileIdentifier ,’%%==============================================================================\n’
fprintf(fileIdentifier ,’\\ newslide \n’);
fprintf(fileIdentifier ,’\\ slideheading Example \\arabicexc: MATLAB Code \n’);
fprintf(fileIdentifier ,’%%==============================================================================\n’
fprintf(fileIdentifier ,’\t \\ matlabcode Matlab/%s.m\n’,functionName );
fclose(fileIdentifier);
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Example 4: Highpass Sallen-Key Filter
v o ( t )
-
+ v s ( t ) R b
R a
R 1
C 1 C 2
R 2
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Example 4: Questions
• Is the input impedance finite?
• What is the output impedance?
• How many capacitors are in the circuit?
• What is the DC gain?
• What is the “high frequency” gain?
• What is the probable order of the transfer function?
• Solve for the transfer function
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Example 4: Work Space
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Example 4: Work Space
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Example 4: Solution
H(s) = K
(
sωn
)2
1 + sQωn
+(
sωn
)2
K = 1 +RB
RA
ωn =1√
R1C1R2C2
Q =1
√
R1C1
R2C2
+√
R1C2
R2C1
+ (1 − K)√
R2C2
R1C1
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Example 5: Bandpass Sallen-Key Filter
v o ( t )
-
+
v s ( t ) R b
R a
R 3
C 2
R 2
R 1
C 1
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Example 5: Questions
• Is the input impedance finite?
• What is the output impedance?
• How many capacitors are in the circuit?
• What is the DC gain?
• What is the “high frequency” gain?
• What is the probable order of the transfer function?
• Solve for the transfer function
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Example 5: Work Space
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Example 5: Work Space
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Transfer Function Synthesis
You can build almost any transfer function with integrators, adders, &subtractors
Y (s) = H(s)X(s) =N(s)
D(s)X(s) =
X(s)
D(s)N(s) = U(s)N(s)
where
U(s) ,1
D(s)X(s)
D(s) , bnsn + bn−1sn−1 + · · · + b1s + b0
X(s) = bnsnU(s) + bn−1sn−1U(s) + · · · + b1sU(s) + b0U(s)
snU(s) =1
bn
(
X(s) − bn−1sn−1U(s) − · · · − b1sU(s) − b0U(s)
)
Y (s) = U(s)N(s)
= ansnU(s) + an−1sn−1U(s) + · · · + a1sU(s) + a0U(s)
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Transfer Function Synthesis Diagram
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Transfer Function Synthesis Clean Diagram
-
+
y ( t )
x ( t )
ΣΣΣΣ
Σ
Σ
Σ
Σ1
bn
1
s
1
s
1
s
bn−1
b2
b1
b0
anan−1 a2 a1 a0
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Example 6: Transfer Function Synthesis
Draw the block diagram for the following transfer function:
H(s) =s3 + 5s2 + 2
s4 + 17s3 + 82s2 + 130s + 100
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Analog Filters Summary
• There are many types of filters
• Second-order filters can implement the four basic types
– Second-order
– Lowpass & Highpass
– Bandpass & Bandstop
– Notch
• For a given H(s), there are many implementations & tradeoffs
• Second order filters are fundamentally different than first-orderfilters because they can have complex poles
– Causes a resonance
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