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Unformatted text preview: ME575 Handouts Frequency Response
•
•
•
• Forced Response to Sinusoidal Inputs
Frequency Response of LTI Systems
Bode Plots
Quantify TF Modeling Errors and Bounds ME575 Session 7 – Frequency Response School of Mechanical Engineering
Purdue University Slide 1 Frequency Response ME575 Session 7 – Frequency Response School of Mechanical Engineering
Purdue University Slide 2 1 ME575 Handouts Forced Response to Sinusoidal Inputs
Ex: Let’s find the forced response of a stable first order system:
Let’
&
y + 3 y = 6u to a sinusoidal input: u (t ) = 5sin(4t )
Y (s) = G (s) ⋅U (s)
– Forced response:
where G ( s ) = and
∴ – PFE: Y (s) = Y ( s) = A1 U ( s ) = L [5sin(4t ) ] = + A2 ⋅ + A3 ⋅ – Compare coefficients to find A1, A2 and A3 : ME575 Session 7 – Frequency Response School of Mechanical Engineering
Purdue University Slide 3 Forced Response to Sinusoidal Inputs
Ex: (cont.)
– Use ILT to find y(t) :
⎡
y (t ) = L −1 [Y ( s ) ] = L −1 ⎢
⎣ ⋅ + ⋅ + ⋅ ⎤
⎥
⎦ = Useful Formula: A sin(ω ⋅ t ) + B cos(ω ⋅ t ) = A + B sin(ω ⋅ t + φ )
Where φ = atan2( B, A) = ∠( A + jB)
2 2 – Using this formula, the forced response can be represented by
−3
y (t ) = 1 24 t +
4 ⋅ e3 ME575 Session 7 – Frequency Response ⋅ sin(4t + φ )
1442443 School of Mechanical Engineering
Purdue University Slide 4 2 ME575 Handouts Frequency Response
Ex: Let’s revisit the same example where
Let’
&
y + 3y = 6u and the input is a general sinusoidal input: Asin(ω t).
6
Aω
6
Aω
⋅
=
⋅
s + 3 s2 +ω2 s + 3 (s − jω)(s + jω)
B
B
B
Y (s) = 1 + 2 + 3
s + 3 s − jω s + jω Y (s) = G(s) ⋅U(s) = – Instead of comparing coefficients, use the residue formula to find
residue
to
6
Aω
Bi’s: B = (s + 3)Y(s)
= (s + 3)
=
1
s =−3
(s + 3) s2 +ω2 s =−3
B2 = (s − jω)Y(s) s = jω = (s − jω)G(s) Aω
=
s +ω2 s = jω B3 = (s + jω)Y(s) s =− jω = (s + jω)G(s)
ME575 Session 7 – Frequency Response 2 Aω
=
s +ω2 s =− jω
2 School of Mechanical Engineering
Purdue University Slide 5 Frequency Response
Ex: (Cont.) 6 Aω
32 + ω 2
6
A
A
A
B2 =
⋅
=
⋅ G ( jω ) =
⋅
2 j jω + 3 2 j
2j
−A
6
−A
−A
B3 =
⋅
=
⋅ G (− jω ) =
⋅
2 j − jω + 3 2 j
2j
B1 = The steady state response YSS(s) is:
YSS ( s ) = B3
B2
+
s − jω s + jω ⇒ ⇒
ME575 Session 7 – ySS (t ) = L −1 [YSS ( s )] = B2 ⋅ e jω ⋅t + B3 ⋅ e − jω ⋅t ySS (t ) = A G ( jω ) ⋅ sin(ω ⋅ t + φ )
Frequency Response where φ = ∠G ( jω ) School of Mechanical Engineering
Purdue University Slide 6 3 ME575 Handouts Frequency Response
Frequency response is used to study the steady state output ySS(t) of a stable
is
system due to sinusoidal inputs at different frequencies.
In general, given a stable system:
&
&
y ( n ) + an −1 y ( n −1) + L + a1 y + a0 y = bm u ( m ) + bm −1u ( m −1) + L + b1u + b0 u bm s m + bm −1s m −1 + L + b1s+ b0 N ( s ) bm ( s − z1 )( s − z2 )L ( s − zm )
=
=
( s − p1 )( s − p2 )L ( s − pn )
D( s)
s n + an −1s n −1 + L + a1s + a0
If the input is a sinusoidal signal with frequency ω , i.e.
i.e.
G(s) ≡ u (t ) = A sin(ω ⋅ t )
then the steady state output ySS(t) is also a sinusoidal signal with the same
frequency as the input signal but with different magnitude and phase:
phase: ySS (t ) = G ( jω ) ⋅ A sin(ω ⋅ t + ∠G ( jω ))
where G(jω) is the complex number obtained by substitute jω for s in G(s) , i.e.
for G ( jω ) = G ( s ) s = jω ≡
ME575 Session 7 – Frequency Response bm ( jω ) m + bm −1 ( jω ) m −1 + L + b1 ( jω )+ b0
( jω ) n + an −1 ( jω ) n −1 + L + a1 ( jω ) + a0 School of Mechanical Engineering
Purdue University Slide 7 Frequency Response
Input u(t) Output y(t) U(s)
u LTI System
G(s) Y(s)
ySS 2π/ω 2π/ω t u (t ) = A sin(ω ⋅ t ) t ⇒ ySS (t ) = G ( jω ) ⋅ A sin(ω ⋅ t + ∠G ( jω )) − A different perspective on the role of the transfer function:
Amplitude of the steady state sinusoidal output
⎧
⎪ G ( jω ) =
Amplitude of the sinusoidal input
⎨
⎪ ∠G ( jω ) = Phase difference (shift) between y (t ) and the sinusoidal input
SS
⎩ ME575 Session 7 – Frequency Response School of Mechanical Engineering
Purdue University Slide 8 4 ME575 Handouts Frequency Response
G
Input u(t) Output y(t) G ME575 Session 7 – Frequency Response School of Mechanical Engineering
Purdue University Slide 9 Bode Diagrams (Plots)
• Magnitude Plot : plots the magnitude of G(jω) in decibels w.r.t. logarithmic
frequency, i.e. G ( jω ) dB = 20log10 G ( jω ) vs log10ω • Phase Plot : plots the linear phase angle of G(jω) w.r.t. logarithmic
frequency, i.e. ∠G ( jω ) vs log10ω To plot Bode diagrams, calculate the magnitude and phase of the
corresponding transfer function. ME575 Session 7 – Frequency Response School of Mechanical Engineering
Purdue University Slide 10 5 ME575 Handouts Bode Diagrams
Transfer Function: bm s m + bm −1 s m −1 + L + b1 s + b0 bm ( s − z1 )( s − z 2 ) L ( s − z m )
=
( s − p1 )( s − p2 ) L ( s − pn )
s n + an −1 s n −1 + L + a1 s + a0 G (s) = Frequency Response
G ( jω ) = bm ( jω − z1 )( jω − z 2 ) L ( jω − z m )
( jω − p1 )( jω − p2 ) L ( jω − pn ) = bm ⋅ 1
1
1
L⋅
⋅
⋅ ( jω − z1 ) ⋅ ( jω − z 2 ) L ⋅ ( jω − z m )
( jω − p1 ) ( jω − p2 )
( jω − pn ) Bode Magnitude: ⎛
1
20 log10 ( G ( jω ) ) = 20 log10 ( bm ) + 20 log10 ⎜
⎜ ( jω − p )
1
⎝ ⎛
⎞
⎞
1
⎟ + L + 20 log10 ⎜
⎟
⎜ ( jω − p ) ⎟
⎟
n
⎠
⎝
⎠ + 20 log10 ( ( jω − z1 ) ) + L + 20 log10 ( ( jω − z1 ) ) Bode Phase: ∠ G ( jω ) = ∠ bm ( jω − z1 )( jω − z 2 ) L ( jω − z m )
( jω − p1 )( jω − p 2 ) L ( jω − p n ) = ∠ bm + ∠ ( jω − z1 ) + L + ∠ ( jω − z m ) − ∠ ( jω − p1 ) − L − ∠ ( jω − p n )
ME575 Session 7 – Frequency Response School of Mechanical Engineering
Purdue University Slide 11 Bode Diagram Building Blocks
• 1st Order Real Poles 0
3 Transfer Function:
1 , τ s +1
Frequency Response:
1
,
G p1 ( jω ) =
τ jω + 1 τ >0
20 τ >0 1
⎧
G p 1 ( jω ) =
⎪
τ 2ω 2 + 1
⎨
⎪
⎩ ∠ G p 1 ( jω ) = − a tan2 (τω ,1)
= − tan − 1 (τω )
Q: By just looking at the Bode diagram, can
you determine the time constant and the
steady state gain of the system ? Phase (deg); Magnitude (dB) G p1 ( s ) = 40 0 45 90
0.01/τ 0.1/τ 1/τ 10/τ 100/τ Frequency (rad/sec)
ME575 Session 7 – Frequency Response School of Mechanical Engineering
Purdue University Slide 12 6 ME575 Handouts 1st Order Bode Diagram Summary
• 1st Order Poles 1
,
τ s +1
– Break Frequency
1
ωb =
[ rad/s]
G p1 ( s ) = • 1st Order Zeros
Gz1 ( s ) = τ s + 1 , τ > 0 τ >0 – Break Frequency
1
ωb =
[ rad/s] τ τ – Mag. Plot Approximation – Mag. Plot Approximation 0 dB from DC to ωb and a straight line
with −20 dB/decade slope after ωb. – Phase Plot Approximation
1
0 deg from DC to 10 ωb . Between 1
10 0 dB from DC to ωb and a straight line
with 20 dB/decade slope after ωb. ωb and 10ωb , a straight line from 0 deg to
straight
−90 deg (passing −45 deg at ωb). For
frequency higher than 10ωb , straight
line on −90 deg. – Phase Plot Approximation
1
0 deg from DC to 10 ωb . Between 1
10 ωb and 10ωb , a straight line from 0 deg to
straight
90 deg (passing 45 deg at ωb). For
frequency higher than 10ωb , straight
line on 90 deg.
Note: By looking at a Bode diagram you should be able to determine the relative order of the system, its
break frequency, and DC (steadystate) gain. This process should also be reversible, i.e. given a
(steadytransfer function, be able to plot a straight line approximated Bode diagram.
ME575 Session 7 – Frequency Response School of Mechanical Engineering
Purdue University Slide 13 Bode Diagram Building Blocks
• 2nd Order Complex Poles 40 Transfer Function: 20 Frequency Response: 20 ωn2
G p2 (s) = 2
, 1≥ ζ ≥ 0
s + 2ζ ω n s + ω n 2 j G p 2 ( jω ) = 2ζ ω ωn Phase (deg); Magnitude (dB) G p 2 ( jω ) = 1
⎛
ω2 ⎞
+ ⎜1 − 2 ⎟
ωn ⎠
⎝ 1
4ζ ω
2 ωn 2 2 ⎛
ω2 ⎞
+ ⎜1 − 2 ⎟
ωn ⎠
⎝ ζ
∠ G p 2 ( jω ) = − tan −1 ⎡ 2ωnω
⎢
⎣ 0 2 40
60
0
45
90 (1 − )⎤⎥⎦
ω2
ωn 2 135
180
0.1ωn ωn 10ωn Frequency (rad/sec)
ME575 Session 7 – Frequency Response School of Mechanical Engineering
Purdue University Slide 14 7 ME575 Handouts 2nd Order Bode Diagram Summary
• 2nd Order Complex Poles ωn 2
G p 2 ( s) = 2
, 1≥ ζ > 0
s + 2ζ ω n s + ω n 2
– Break Frequency
ωb = ωn [ rad/s ]
– Mag. Plot Approximation • 2nd Order Complex Zeros
Gz 2 ( s ) = ωb = ωn [ rad/s] 0 dB from DC to ωn and a straight line
with 40 dB/decade slope after ωn. – Phase Plot Approximation 0 deg from DC to ( 1 5 ) ωn . Between ( 1 5 ) ωn
and 5ζ ωn , a straight line from 0 deg to −
straight
180 deg (passing −90 deg at ωn). For
frequency higher than 5ζ ωn , straight
line on −180 deg.
Frequency Response 1≥ ζ > 0 – Mag. Plot Approximation – Phase Plot Approximation ME575 Session 7 – , – Break Frequency 0 dB from DC to ωn and a straight line
with −40 dB/decade slope after ωn.
Peak value occurs at:
ω r = ω n 1 − 2ζ 2
1
⇒ G p 2 ( jωr )
=
MAX
2ζ 1 − ζ 2
ζ s 2 + 2ζωn s + ωn 2
ωn 2 ζ ζ
0 deg from DC to ( 1 5 ) ωn. Between ( 1 5 ) ωn
ζ ω , a straight line from 0 deg to
and 5 n
straight
180 deg (passing 90 deg at ωn). For
frequency higher than 5ζ ωn , straight
line on 180 deg. School of Mechanical Engineering
Purdue University ζ Slide 15 2nd Order System Frequency Response
A Few Observations:
• Three different characteristic frequencies:
different
– Natural Frequency (ωn)
– Damped Natural Frequency (ωd): ωd = ωn 1 − ζ 2
– Resonant (Peak) Frequency (ωr): ωr = ωn 1 − 2ζ 2 ωr ≤ ωd ≤ ωn
• When the damping ratio ζ > 0.707, there is no peak in the Bode
0.707
magnitude plot. DO NOT confuse this with the condition for overoverdamped and underdamped systems: when ζ < 1 the system is underunderthe
underdamped (has overshoot) and when ζ > 1 the system is overdamped (no
the
overovershoot).
• As ζ → 0 , ωr → ωn and G(jω)ΜΑΧ increases; also the phase transition
from 0 deg to −180 deg becomes sharper.
ME575 Session 7 – Frequency Response School of Mechanical Engineering
Purdue University Slide 16 8 ME575 Handouts Quantify TF Modeling Errors
• Nominal System Model Y ( s ) = Go ( s )U ( s )
• Actual (Calibration) System Model Y ( s) = G ( s)U ( s )
• Modeling Errors in terms of Transfer Functions Y ( s) = G ( s)U ( s ) = (Go ( s ) + Gε ( s ))U ( s )
=Go ( s )(1 + GΔ ( s ))U ( s )
• Bounds for Modeling Errors GΔ ( jω ) < ε (ω )
ME575 Session 7 – Frequency Response School of Mechanical Engineering
Purdue University Slide 17 Quantify TF Modeling Errors  Examples
• Missing HighFrequency Pole
HighBode Plots of GΔ ( s) for α = 0.2, 1, 20 Go (s) = F(s),
G(s) = 1
F(s)
α s +1 GΔ (s) = ME575 Session 7 – Frequency Response School of Mechanical Engineering
Purdue University −αs
αs +1 Slide 18 9 ME575 Handouts Quantify TF Modeling Errors  Examples
• Error in TimeDelay
TimeBode Plots of GΔ ( s) for τ = 0.2τ 0 , 5τ 0 Go (s) = e−τos F(s),
G(s) = e−τ s F(s)
−(τ −τo ) s GΔ (s) = e τ
τo (1− )τo s −1 = e −1 ⎛ ωτ ⎛ τ ⎞ ⎞
GΔ ( jω) = 2 sin ⎜ o ⎜1− ⎟ ⎟
⎜2
⎟
⎝ τo ⎠ ⎠
⎝
⎡⎛
⎛ τ ⎞⎞
∠GΔ ( jω) = atan2 ⎢sin ⎜ωτo ⎜1− ⎟ ⎟ ,
⎜
⎟
⎢⎝
⎝ τo ⎠ ⎠
⎣
⎛
⎛ τ ⎞ ⎞⎤
−1+ cos ⎜ωτo ⎜1− ⎟ ⎟⎥
⎜
⎟
⎝ τo ⎠ ⎠⎥
⎝
⎦
ME575 Session 7 – Frequency Response School of Mechanical Engineering
Purdue University Slide 19 Quantify TF Modeling Errors  Examples
• Missing Resonance Effect
Bode Plots of GΔ ( s) for ζ = 0.1, 0.2, 0.7 Go (s) = F(s),
G(s) = 2
ωn
F(s)
2
s2 + 2ζωn s + ωn GΔ (s) = ME575 Session 7 – Frequency Response School of Mechanical Engineering
Purdue University −s(s + 2ζωn )
2
s + 2ζωn s + ωn
2 Slide 20 10 ...
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 Fall '10
 Meckl
 Mechanical Engineering

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