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Lecture 27: Tue Nov 17, 2020 Reminder: • HW11 due Thursday Lecture: 2nd-order systems • frequency-domain characteristics three frequencies: natural, resonant, damped
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Lecture 27: Tue Nov 17, 2020
Reminder:
• HW11 due Thursday
Lecture: 2nd-order systems
• frequency-domain characteristics
three frequencies: natural, resonant, damped
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Quiz 2 Results
2
3084 Reading
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A 2nd-Order System
[ under- ][ critically ][ over- ] damped?
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A 2nd-Order System
[ under- ][ critically ][ over- ] damped?
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A 2nd-Order System
[ under- ][ critically ][ over- ] damped?
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A 2nd-Order System
radius r =
n
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A 2nd-Order System
radius n
let = cos
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A 2nd-Order System
n 1 2–
–n let = cos
radius n
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A 2nd-Order System
n
p1,2 = –n ± jn 1 2–
1 2–
–n let = cos
radius n
⇒ denominator = (s – p1)(s – p2)
= s2 + 2ns + n2
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2nd-Order System: Standard FormGet in the habit — whenever possible, put denominator in “standard form”:
where
• n = “natural” frequency > 0
• = damping coefficient ∈[0, ∞)
• H( 0 ) = H0 = dc gain
H0n2
s2 2n s n2+ +
-----------------------------------------------H( s ) =
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What Kind of Filters are These?
H0n2
s2 2ns n2+ +
-------------------------------------------- Hn 2n s
s2 2n s n2+ +
-------------------------------------------- H∞ s2
s2 2n s n2+ +
--------------------------------------------
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What Kind of Filters are These?
H0n2
s2 2ns n2+ +
-------------------------------------------- Hn 2n s
s2 2n s n2+ +
-------------------------------------------- H∞ s2
s2 2n s n2+ +
--------------------------------------------
HLPF( s ) HBPF( s ) HHPF( s )
0
1
2
3
n 2n
0
1
0
1
2
3
n 2n n 2n
= 0.2
0.3
0.7
0.7
0.2
= 0.2
0.7
0.3
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Recall Example
H( s ) = = 48
s2 2s 16+ +-------------------------------- H0n
2
s2 2n s n2+ +
-----------------------------------------------H0 = 3n = 4 = 0.25
½where
|H(j)|
0 1 2 3 4 5 6 7 8 9 100
1
2
H0 = 3
4
5
6
7
Whe
re is
the
peak
?
How high?
w = 0:0.01:10;
s = j*w;
H = 48./(s.^2 + 2*s + 16);
r
Hpk
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Magnitude Response
• For what values of is there a peak?
• Where are the peaks?
• How high are the peaks?
|H(j)| = 0.1
0.2
0.3
0.7
n2n0
0
1
3
5
0.60.5
0.4
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Where is Peak in Mag Response?Peak in |H(j)|2 happens when magnitude-squared of denom is minimized:
(n2 – 2)2 + (2n)2
=
⇒ 2 – n2 + 22n
2 = 0
⇒ = “resonant frequency”
Resonant peak exists only when < ≈0.707!
Value of peak? Evaluating at r yields .
Depends only on
d
d-------
2(n2 – 2)(–2) + 2(2n)2n = 0
r = n 1 22–
1/2
Hpk
H0
--------- 1
2 1 2–-------------------------=
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Resonant Peak Pole Locations
RESONANT PEAK ONLY
= 0
= 0.707
= 1
IN SHADED REGION
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From Mag Response to n and
|H(j)|
00
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From Mag Response to n and
|H(j)|
0
H0
Hpk 1) Solve Hpk/H0 = for 1
2 1 2–-------------------------
H0
Hpk-------- = sin( sin–1( ))1
2---
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From Mag Response to n and
|H(j)|
r0
H0
Hpk 1) Solve Hpk/H0 = for 1
2 1 2–-------------------------
2) Solve r = n for n1 22–
H0
Hpk-------- = sin( sin–1( ))1
2---
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Peak vs Damping Coeff
0 0.1 0.2 0.3 0.4 0.5 0.6 0.71
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Hpk/H
0=
1
2
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–----
--------
--------
-----
H0
Hpk-------- = sin( sin–1( ))1
2---
2
3
4
5
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Relationship to Pole Locationspeak height Hpk/H0 depends only on = cos
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Relationship to Pole Locationspeak height Hpk/H0 depends only on = cos
SAME Hpk/H0
Hpk/H0
Hpk/H0
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Pop Quiz
0 100 200 300 4000
2
4
6
8
10
12
14
16
18
|H(j)|
⇒ H( s ) = where
H0 =
n =
=
H0n2
s2 2n s n2+ +
-----------------------------------------------
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Time Domain Properties of
2nd Order Systems
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Underdamped LPF Step ResponseCan’t measure n and directly, but can derive them from measurements of:
steady-state value (determines H0)
percentage overshoot (determines )
peak time
settling time (given , determines n)
0 t
y( t )
tpk
STEP RESPONSE
tsettle
1.02H0
0.98H0
H0
(1 + )H0
