Unformatted text preview: ME375 Handouts Case Study School of Mechanical Engineering
Purdue University ME375 Frequency Response  1 Case Study
SUPPORT POWER
WIRE DROPPERS Electric train derives power through a pantograph, which contacts the power
wire, which is suspended from a catenary. During highspeed runs between New
Haven, CT and New York City, the train experiences intermittent power loss at
42 km/hr and 100 km/hr.
School of Mechanical Engineering
Purdue University ME375 Frequency Response  2 1 ME375 Handouts Case Study – Pantograph Model
m1z1 + ( b1 + b2 ) z1 + ( k1 + k2 ) z1 − b2z2 − k2z2 = 0 Fc(t) m2z2 + b2z2 + k2z2 − b2 z1 − k2 z1 = −Fc (t) z2
m2
k2 b2
z1
m1 k1 b1 For m1 = 23.0 kg, b1 = 150 N/(m/s), k1 = 9600 N/m,
m2 = 11.5 kg, b2 = 75 N/(m/s), and k2 = 9580 N/m: Z 2 ( s)
−0.087( s 2 + 9.78s + 834)
= 4
Fc ( s ) s + 16.3s 3 + 1709 s 2 + 8155s + 347,705 School of Mechanical Engineering
Purdue University ME375 Frequency Response  3 Case Study – Frequency Response
60 Magnitude (dB) 70
80
90
100 Phase (deg) 110
180
135
90
45
0
0
10 1 10
Frequency (rad/sec) School of Mechanical Engineering
Purdue University 10 2 ME375 Frequency Response  4 2 ME375 Handouts Frequency Response
•
•
• Forced Response to Sinusoidal Inputs
Frequency Response of LTI Systems
Bode Plots School of Mechanical Engineering
Purdue University ME375 Frequency Response  5 Forced Response to Sinusoidal Inputs
Ex: Let’s find the forced response of a stable first order system:
Let’
y + 5 y = 10u to a sinusoidal input:
– Forced response: u (t ) = sin(2t )
Y (s) = G (s) ⋅U (s) where G ( s ) = and ∴ – PFE: Y (s) = Y ( s) = A1 U ( s ) = L [sin(2t ) ] = + A2 ⋅ + A3 ⋅ – Compare coefficients to find A1, A2 and A3 : School of Mechanical Engineering
Purdue University ME375 Frequency Response  6 3 ME375 Handouts 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
⋅ e −5 t + y (t ) = ⋅ sin(2t + φ ) School of Mechanical Engineering
Purdue University ME375 Frequency Response  7 Forced Response of 1st Order System
Input is sin(2t)
2
Output
1.5
Input 1
0.5 Response 0
0.5
1
1.5
2 0 2 4 6 8
Time (sec) School of Mechanical Engineering
Purdue University 10 12 14 ME375 Frequency Response  8 4 ME375 Handouts Forced Response to Sinusoidal Inputs
Ex: Given the same system as in the previous example, find the forced response
forced
to u(t) = sin(10 t).
Y ( s) = G ( s) ⋅U ( s)
where G ( s ) = and
∴ U ( s ) = L [sin(10t ) ] = Y ( s) = School of Mechanical Engineering
Purdue University ME375 Frequency Response  9 Forced Response of 1st Order Systems
Input is sin(10t)
Output
1
0.8
0.6 Response 0.4
0.2
0
0.2
0.4
0.6
0.8 Input 1
0 0.5 1 1.5 2 2.5
Time (sec) 3 School of Mechanical Engineering
Purdue University 3.5 4 4.5 5 ME375 Frequency Response  10 5 ME375 Handouts Frequency Response
Ex: Let’s revisit the same example where
Let’
y + 5y =10u and the input is a general sinusoidal input: sin(ω t).
10
ω
10
ω
⋅
=
⋅
s + 5 s2 +ω2 s + 5 (s − jω)(s + jω)
A
A
A
Y (s) = 1 + 2 + 3
s + 5 s − jω s + jω Y (s) = G(s) ⋅U(s) = – Instead of comparing coefficients, use the residue formula to find Ai’s:
10
ω
A = (s + 5)Y(s) s=−5 = (s + 5)
=
1
(s + 5) s2 +ω2 s=−5
A2 = (s − jω)Y(s) s= jω = (s − jω)G(s) ω
=
s +ω2 s= jω A3 = (s + jω)Y(s) s=− jω = (s + jω)G(s) 2 ω
s2 +ω2 s=− jω School of Mechanical Engineering
Purdue University =
ME375 Frequency Response  11 Frequency Response
Ex: (Cont.) 10ω
52 + ω 2
1
10
1
1
A2 =
⋅
=
⋅ G ( jω ) =
⋅
2 j jω + 5 2 j
2j
10
−1
−1
−1
A3 =
⋅
=
⋅ G (− jω ) =
⋅
2 j − jω + 5 2 j
2j
A1 = The steady state response YSS(s) is:
YSS ( s ) = A3
A2
+
s − jω s + jω ⇒ ySS (t ) = L −1 [YSS ( s ) ] = A2 ⋅ e jω ⋅t + A3 ⋅ e− jω ⋅t ⇒ ySS (t ) = G ( jω ) ⋅ sin(ω ⋅ t + φ ) where φ = ∠G ( jω ) School of Mechanical Engineering
Purdue University ME375 Frequency Response  12 6 ME375 Handouts Frequency Response
• Frequency Response School of Mechanical Engineering
Purdue University ME375 Frequency Response  13 In Class Exercise
For the current example, y + 5 y = 10u Calculate the magnitude and phase shift of the steady state response when the
response
system is excited by (i) sin(2t) and (ii) sin(10t). Compare your result with the
sin(2t
sin(10t
steady state response calculated in the previous examples.
Note:
10
10
G(s) = s+5 G ( jω ) = ⇒ G ( jω ) = 10 ω 2 + 52 jω + 5 and φ = ∠G ( jω ) = −atan2(ω ,5) School of Mechanical Engineering
Purdue University ME375 Frequency Response  14 7 ME375 Handouts Frequency Response
•Frequency response is used to study the steady state output ySS(t) of a stable system
due to sinusoidal inputs at different frequencies.
In general, given a stable system:
an y ( n ) + an−1 y ( n −1) + + a1 y + a0 y = bm u ( m ) + bm−1u ( m−1) + + b1u + b0 u m −1 bm s + bm−1s + + b1s+ b0 N ( s ) bm ( s − z1 )( s − z2 ) ( s − zm )
=
=
an s n + an −1s n −1 + + a1s + a0 D( s ) an ( s − p1 )( s − p2 ) ( s − pn )
If the input is a sinusoidal signal with frequency ω , i.e.
G(s) ≡ m u (t ) = Au 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: ySS (t ) = G ( jω ) ⋅ Au sin(ω ⋅ t + ∠G ( jω ))
where G(jω) is the complex number obtained by substitute jω for s in G(s) , i.e. G ( jω ) = G ( s ) s = jω ≡ bm ( jω ) m + bm −1 ( jω ) m −1 + + b1 ( jω )+ b0
an ( jω ) n + an −1 ( jω ) n −1 + + a1 ( jω ) + a0 School of Mechanical Engineering
Purdue University ME375 Frequency Response  15 Frequency Response
Input u(t) Output y(t) U(s)
u LTI System
G(s) Y(s)
ySS 2π/ω t u (t ) = Au sin(ω ⋅ t ) 2π/ω t ⇒ ySS (t ) = G ( jω ) ⋅ Au sin(ω ⋅ t + ∠G ( jω )) − A different perspective of 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
⎩
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Purdue University ME375 Frequency Response  16 8 ME375 Handouts Frequency Response
G
Input u(t) Output y(t) G School of Mechanical Engineering
Purdue University ME375 Frequency Response  17 In Class Exercise
Ex: 1st Order System
(2) Calculate the steady state output of the
system when the input is
The motion of a piston in a cylinder can be
modeled by a 1st order system with force
Input f(t)
Steady State Output v(t)
as input and piston velocity as output:
sin(ω t) ⎢G(jω)⎥ sin(ω t + φ )
sin(0t +
sin(0t)
f(t) sin(10t) sin(10t + sin(20t) sin(20t + The EOM is:
Mv + Bv = f (t ) sin(30t) sin(30t + (1) Let M = 0.1 kg and B = 0.5 N/(m/s),
find the transfer function of the system: sin(40t) sin(40t + sin(50t) sin(50t + sin(60t) sin(60t + v School of Mechanical Engineering
Purdue University ME375 Frequency Response  18 9 ME375 Handouts In Class Exercise
(3) Plot the frequency response plot 20 1.4 30
Phase (deg) 0
10 1.6
Magnitude ((m/s)/N) 2
1.8 1.2
1
0.8 40
50
60 0.6 70 0.4 80 0.2 90 0 0 10 20
30
40
50
Frequency (rad/sec) 60 70 0 School of Mechanical Engineering
Purdue University 10 20
30
40
50
Frequency (rad/sec) 60 70 ME375 Frequency Response  19 Example  Vibration Absorber (I)
Without vibration absorber: EOM: M 1 z1 + B1 z1 + K1 z1 = f (t ) z1
M1
K1 f(t) B1 TF (from f(t) to z1): Let M1 = 10 kg, K1 = 1000 N/m, B1 = 4 N/(m/s).
Find the steady state response of the system for f(t) =
(a) sin(8.5t) (b) sin(10t) (c) sin(11.7t).
sin(8.5t
sin(10t
sin(11.7t Input f(t)
Steady State Output z1(t)
sin(ω t) ⎢G(jω)⎥ × sin(ω t +
φ)
sin(8.5t +
sin(8.5t)
sin(10t)
sin(11.7t) School of Mechanical Engineering
Purdue University sin(10t +
sin(11.7t + ME375 Frequency Response  20 10 ME375 Handouts Example  Vibration Absorber (I)
f(t) = sin(8.5 t) 0.01 z1 (m) 0.005
0
0.005
0.01 f(t) = sin(10 t) z1 (m) 0.04
0.02
0
0.02
0.04 f(t) = sin(11.7 t) z1 (m) 0.005 0 0.005
0 5 10 15 20 25 30 35 40 45 50 Time (sec)
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Purdue University ME375 Frequency Response  21 Example  Vibration Absorber (II)
With vibration absorber:
z2 M 2 z2 + B2 z2 + K 2 z2 − B2 z1 − K 2 z1 = 0 M2
K2 B2
z1
M1 K1 f(t) EOM:
M 1 z1 + ( B1 + B2 ) z1 + ( K1 + K 2 ) z1 − B2 z2 − K 2 z2 = f (t ) B1 TF (from f(t) to z1): Let M1 = 10 kg, K1 = 1000 N/m, B1 = 4 N/(m/s),
M2 = 1 kg, K2 = 100 N/m, and B2 = 0.1 N/(m/s). Find
the steady state response of the system for f(t) =
(a) sin(8.5t) (b) sin(10t) (c) sin(11.7t).
sin(8.5t
sin(10t
sin(11.7t Input f(t)
Steady State Output z1(t)
sin(ω t) ⎢G(jω)⎥ × sin(ω t +
φ)
sin(8.5t +
sin(8.5t)
sin(10t)
sin(11.7t) School of Mechanical Engineering
Purdue University sin(10t +
sin(11.7t + ME375 Frequency Response  22 11 ME375 Handouts Example  Vibration Absorber (II)
f(t) = sin(8.5 t) 0.04 z1 (m) 0.02
0
0.02
0.04 f(t) = sin(10 t) z1 (m) 0.004
0.002
0
0.002
0.004 f(t) = sin(11.7 t) 0.02
z1 (m) 0.01
0
0.01
0.02
0 5 10 15 20 25 30 35 40 45 50 Time (sec)
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Purdue University ME375 Frequency Response  23 Example  Vibration Absorber (II)
Take a closer look at the poles of the transfer function:
The characteristic equation
10 s 4 + 5.1s 3 + 2100.4 s 2 + 500 s + 100000 = 0
⇒ Poles:
p1,2 = −0.1 ± 8.5 j
p3,4 = −0.155 ± 11.7 j What part of the poles determines the rate of decay for the transient response?
transient
σt e jωt →
(Hint: when p = σ ± jω → the response is e School of Mechanical Engineering
Purdue University ) ME375 Frequency Response  24 12 ME375 Handouts Example  Vibration Absorbers
Frequency Response Plot
Absorber tuned at 10 rad/sec added Frequency Response Plot
No absorber added 0.025
Magnitude (m/N) Magnitude (m/N) 0.025
0.02
0.015
0.01
0.005
0 0 2 4 6 8 10 12 14 16 18 0.02
0.015
0.01
0.005
0 20 0 4 6 8
10 12 14
Frequency (rad/sec) 16 18 20 2 4 6 8
10 12 14
Frequency (rad/sec) 16 18 20 0
45 Phase (deg) Phase (deg) 0
45 2 0 Frequency (rad/sec) 90
135
180 90
135
180 0 2 4 6 8
10 12 14
Frequency (rad/sec) 16 18 20 School of Mechanical Engineering
Purdue University ME375 Frequency Response  25 Example  Vibration Absorbers
Bode Plot
No absorber added Bode Plot
Absorber tuned at 10 rad/sec added
30 40 40 50 50 60 Phase (deg); Magnitude (dB) Phase (deg); Magnitude (dB) 30 70
80
90
100
0
45
90 70
80
90
100
0
45
90 135 135 180
10 60 0 10 1
Frequency (rad/sec) 10 2 180
10 0 School of Mechanical Engineering
Purdue University 1 10
Frequency (rad/sec) 10 2 ME375 Frequency Response  26 13 ME375 Handouts Bode Diagrams (Plots)
• Bode Diagrams (Plots)
A unique way of plotting the frequency response function, G(jω), w.r.t. frequency ω of
systems.
Consists of two plots:
– Magnitude Plot : plots the magnitude of G(jω) in decibels w.r.t. logarithmic frequency,
i.e.
G ( j ω ) dB = 20 log 10 G ( j ω ) vs log 10ω
– Phase Plot : plots the linear phase angle of G(jω) w.r.t. logarithmic frequency, i.e.
∠G ( jω ) vs log 10ω To plot Bode diagrams, one needs to calculate the magnitude and phase of the
corresponding transfer function.
Ex:
s +1
G(s) = s 2 + 10 s School of Mechanical Engineering
Purdue University ME375 Frequency Response  27 Bode Diagrams
Revisit the previous example: 50 ( jω ) + 1
s +1
⇒ G ( jω ) =
G(s) = 2
s + 10 s
jω ( jω + 10) 30 G ( jω ) = 10
0 ∠G ( jω ) = ω
0.1
0.2
0.5
1
2
5
10
20
50
100 G( jω ) 20log10 G( jω) ∠G( jω ) Phase (deg); Magnitude (dB) 10
30
50
100 50 0 50 100 1
10
School of Mechanical Engineering
Purdue University 10 0 10
Frequency (rad/sec) 1 10 2 ME375 Frequency Response  28 14 ME375 Handouts Bode Diagrams
Recall that if
G( s) = Then bm s m + bm −1s m −1 + + b1s + b0 bm ( s − z1 )( s − z2 ) ( s − zm )
=
an s n + an −1s n −1 + + a1s + a0
an ( s − p1 )( s − p2 ) ( s − pn ) G ( jω ) =
= bm ( jω − z1 )( jω − z2 ) ( jω − z m )
an ( jω − p1 )( jω − p2 ) ( jω − pn )
bm
1
1
⋅
⋅
an ( jω − p1 ) ( jω − p2 ) 20 log10 ( G ( jω ) ) = 20 log10 ⋅ 1
⋅ ( jω − z1 ) ⋅ ( jω − z2 )
( jω − pn ) F b I + 20 log F 1 I + +20 log F 1 I
GH ( jω − p ) JK
GH a JK
GH ( jω − p ) JK
c ( jω − z ) h+ +20 log c ( jω − z ) h
m 10 10 n +20 log10 and
∠ G ( jω ) = ∠ ⋅ ( jω − z m ) 1 1 10 n 1 bm ( jω − z1 )( jω − z 2 ) ( jω − z m )
an ( j ω − p1 )( jω − p2 ) ( jω − pn ) = ∠ ( j ω − z1 ) + ∠ ( jω − z 2 ) + +∠ ( jω − z m ) − ∠ ( jω − p1 ) − ∠ ( jω − p2 ) −∠ ( j ω − pn ) School of Mechanical Engineering
Purdue University ME375 Frequency Response  29 Example
Ex: Find the magnitude and the phase of the following transfer function:
function:
G( s ) = 3s 3 + 12 s 2 + 9s
=
2 s + 22 s 2 + 76 s + 80
3 ( School of Mechanical Engineering
Purdue University )( ) ME375 Frequency Response  30 15 ME375 Handouts 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 R G ( jω )

S
 ∠ G ( jω )
T
p1 p1 τ > 0 , = 20 , Phase (deg); Magnitude (dB) G p1 ( s ) = τ >0 1 τ 2ω 2 + 1
= − atan2 (τω ,1)
= − tan − 1 τω a f 40 0 45 Q: By just looking at the Bode diagram, can you
determine the time constant and the steady
state gain of the system ? 90
0.01/τ 0.1/τ 1/τ 10/τ 100/τ Frequency (rad/sec)
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Purdue University ME375 Frequency Response  31 Example
• 1st Order Real Poles
Transfer Function: 20 50
G (s) =
s+5
Plot the straight line approximation
of G(s)’s Bode diagram: 15
10 Phase (deg); Magnitude (dB) 5
0 0 45 90
1 10 0 10 1 10 2 10 Frequency (rad/sec)
School of Mechanical Engineering
Purdue University ME375 Frequency Response  32 16 ME375 Handouts Bode Diagram Building Blocks
40 • 1st Order Real Zeros
Transfer Function: τ >0 Frequency Response:
G z1 ( jω ) = τ jω + 1 , R G

S∠ G

T 20 τ >0 z1 ( jω ) = τ ω +1 z1 ( jω ) = atan2 ( τω ,1)
= tan − 1 τω 2 2 a f Phase (deg); Magnitude (dB) G z1 ( s ) = τ s + 1 , 3
0 90 45 0
0.01/τ 0.1/τ 1/τ 10/τ 100/τ Frequency (rad/sec)
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Purdue University ME375 Frequency Response  33 Example
• 1st Order Real Zeros 20
15
10
Phase (deg); Magnitude (dB) Transfer Function:
G ( s ) = 0 .7 s + 0 .7
Plot the straight line approximation
of G(s)’s Bode diagram: 5
0 90 45 0
1 10 School of Mechanical Engineering
Purdue University 0 10 1 10
Frequency (rad/sec) 2 10 ME375 Frequency Response  34 17 ME375 Handouts Example
• Lead Compensator
30
20
10
Phase (deg); Magnitude (dB) Transfer Function:
35s + 35
G (s) =
s+5
Plot the straight line approximation
of G(s)’s Bode diagram: 0
10 90 0 90
1 0 10 1 10 2 10 10 Frequency (rad/sec)
School of Mechanical Engineering
Purdue University ME375 Frequency Response  35 1st Order Bode Diagram Summary
• 1st Order Poles 1
,
τ s +1
– Break Frequency
1
ωb =
[ rad/s]
G p1 ( s ) = • 1st Order Zeros
Gz 1 ( s ) = τ s + 1 , τ > 0 τ >0 – Break Frequency
1
ωb =
[ rad/s] τ τ – 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 ω b . Between 1
ω and
10 b 10ωb , a straight line from 0 deg to −90 deg
(passing −45 deg at ωb). For frequency
higher than 10ωb , straight line on −90 deg. – 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 ω b . Between 1
ω
10 b and 10ωb , a straight line from 0 deg to 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
system,
break frequency, and DC (steadystate) gain. This process should also be reversible, i.e. given a
frequency,
(steadygain.
transfer function, be able to plot a straight line approximated Bode diagram.
School of Mechanical Engineering
Purdue University ME375 Frequency Response  36 18 ME375 Handouts Bode Diagram Building Blocks
• Integrator (Pole at origin) 20 Transfer Function:
1
G p0 (s) =
s
Frequency Response: 0 Phase (deg); Magnitude (dB) 20 1
G p 0 ( jω ) =
jω R G ( jω )

S
 ∠ G ( jω )
T = p0 p0 1 ω = − 90 ° 40 60
0 45 90 135
0.1 1 10 100 1000 Frequency (rad/sec) School of Mechanical Engineering
Purdue University ME375 Frequency Response  37 Bode Diagram Building Blocks
• Differentiator (Zero at origin) 60 Transfer Function: 40 Gz0 (s) = s
G z 0 ( jω ) = jω R G

S∠ G

T z0 ( jω ) =ω z0 ( jω ) = 90 ° Phase (deg); Magnitude (dB) 20 Frequency Response: 0 20
135 90 45 0
0.1 1 10 100 1000 Frequency (rad/sec) School of Mechanical Engineering
Purdue University ME375 Frequency Response  38 19 ME375 Handouts Example
• Combination of Systems
Transfer Function: 30 G (s) = 20 35s + 35
s(s + 5) 10
0
Phase (deg); Magnitude (dB) Plot the straight line approximation of
G(s)’s Bode diagram: 10 90 0 90
1 0 10 1 10 2 10 10 Frequency (rad/sec)
School of Mechanical Engineering
Purdue University ME375 Frequency Response  39 Example
• Combination of Systems 40 Transfer Function:
2500
s(s2 + 55s + 250 ) 0 Plot the straight line approximation of
G(s)’s Bode diagram: 40
Phase (deg); Magnitude (dB) G (s) = 80 120 0 90 180 270
1 10 10 0 1 10 2 10 3 10 Frequency (rad/sec)
School of Mechanical Engineering
Purdue University ME375 Frequency Response  40 20 ME375 Handouts Bode Diagram Building Blocks
• 2nd Order Complex Poles 20 Transfer Function: 0 ω n2
Gp2 (s) = 2
s + 2ζω n s + ω n 2 , 1≥ζ ≥ 0 40 j G p 2 ( jω ) = 2ζω ωn Phase (deg); Magnitude (dB) Frequency Response:
G p 2 ( jω ) = 1
⎛
ω2 ⎞
⎟
+ ⎜1 −
⎜
ω n2 ⎟
⎠
⎝ 1
4 ζ 2ω 2 ωn 2 20 ⎛
ω2 ⎞
+ ⎜1 −
⎜ ω 2⎟
⎟
⎝
n ⎠ 2 60
80 0 90 ⎡ ζ
ω2 ⎟⎤
⎜
∠ G p 2 ( jω ) = − tan −1 ⎢ 2ω ω ⎛ 1 − ω 2 ⎞ ⎥
n ⎠
⎣ n ⎝
⎦ 180
0.01 ωn 0.1 ωn ωn 10 ωn 100 ωn Frequency (rad/sec) School of Mechanical Engineering
Purdue University ME375 Frequency Response  41 Example
• SecondOrder System
Second 40 Transfer Function: 2500
s2 + 10s + 2500
Plot the straight line approximation of
G(s)’s Bode diagram: 0 G (s) = Phase (deg); Magnitude (dB) 40 80 120 0 90 180
10 0 1 10 2 10 10 3 4 10 Frequency (rad/sec)
School of Mechanical Engineering
Purdue University ME375 Frequency Response  42 1 ME375 Handouts Bode Diagram Building Blocks
• 2nd Order Complex Zeros 80 Phase (deg); Magnitude (dB) Transfer Function:
2
s 2 + 2ζ ω n s + ω n
Gz 2 ( s ) =
, 1≥ ζ ≥ 0
2
ωn
Frequency Response:
ω2 ⎞
2ζ ω ⎛
⎟
G z 2 ( jω ) = j
+ ⎜1 −
⎜
ω n ⎝ ω n2 ⎟
⎠
G z 2 ( jω ) = 1 G p 2 ( jω )
4 ζ 2ω 2 = ωn 2 ⎛
ω2 ⎞
⎟
+ ⎜1 −
⎜
ω n2 ⎟
⎝
⎠ 2 60
40
20
0
20 180 90 ∠ G z 2 ( j ω ) = −∠ G p 2 ( j ω )
⎡ ζ
= tan −1 ⎢ 2ω ω
⎣ n ⎛1 −
⎜
⎝ ω2
ω n2 ⎞⎤
⎟⎥
⎠⎦ 0
0.01ωn 0.1 ωn School of Mechanical Engineering
Purdue University ωn
Frequency (rad/sec) 10 ωn 100 ωn ME375 Frequency Response  43 Bode Diagrams of Poles and Zeros R G ( jω ) = 1

G ( jω )
⇒S
 ∠ G ( jω ) = −∠ G ( jω )
T
R20 log e G ( jω ) j = − 20 log c G ( jω ) h

⇒S
 ∠ G ( jω ) = −∠ G ( jω )
T 20
0
20
40 p z p z 10 p 10 p z 180
Phase (deg) Let 1
G p (s) =
Gz ( s ) 40
Magnitude (dB) Bode Diagrams of stable complex zeros are
the mirror images of the Bode diagrams of
the identical stable complex poles w.r.t. the
0 dB line and the 0 deg line, respectively. 0 z 180 0.1ωn ωn 10ωn Frequency (rad/sec)
School of Mechanical Engineering
Purdue University ME375 Frequency Response  44 22 ME375 Handouts 2nd Order Bode Diagram Summary
• 2nd Order Complex Zeros • 2nd Order Complex Poles
ωn
, 1≥ ζ > 0
2
s + 2 ζω n s + ω n
– Break Frequency
ω b = ω n [ rad/s]
– Mag. Plot Approximation
2 G p 2 ( s) = ⇒ 0 dB from DC to ωn and a straight line with
40 dB/decade slope after ωn. – Phase Plot Approximation 1
2ζ 1 − ζ 0 deg from DC to ( 1 5 ) ω n . Between ( 1 5 )ζ ω n
and 5ζ ωn , a straight line from 0 deg to 180
deg (passing 90 deg at ωn). For frequency
higher than 5ζ ωn , straight line on 180 deg.
ζ 2 – Phase Plot Approximation
0 deg from DC to ( 1 5 ) ω n . Between ( 1 5 ) ω n
and 5ζ ωn , a straight line from 0 deg to −180
deg (passing −90 deg at ωn). For frequency
higher than 5ζ ωn , straight line on −180 deg.
ζ 1≥ ζ > 0 , – Mag. Plot Approximation n G p 2 ( jω r ) MAX = 2 – Break Frequency
ωb = ωn [ rad/s] 0 dB from DC to ωn and a straight line with
−40 dB/decade slope after ωn. Peak value
occurs at:
ω = ω 1 − 2ζ 2
r s 2 + 2ζωn s + ωn
2
ωn Gz 2 ( s ) = 2 ζ School of Mechanical Engineering
Purdue University ME375 Frequency Response  45 2nd Order System Frequency Response
A Closer Look: G p 2 ( s) = ωn 2
2
s 2 + 2ζω n s + ω n , 1≥ ζ ≥ 0 Frequency Response Function:
G p 2 ( jω ) = Phase: Magnitude:
G p 2 ( jω ) = ωn 2
1
=
2
( jω ) + 2ζω n ( jω ) + ω n
ω2 ⎞
2 ζω ⎛
j
+ ⎜1 − 2 ⎟
ωn ⎝ ωn ⎠
2 2 ⎛ 2 ζω ⎞ ⎛ ω ⎞
⎜ ω ⎟ + ⎜1 − 2 ⎟
⎝ n ⎠ ⎝ ωn ⎠
2 ( ) ζ
ω
∠G p 2 ( jω ) = −∠ ⎡ 1 − ω 2 + j 2ωnω ⎤
⎢
⎥
n
⎣
⎦ 1
2 2 ( ) ζ
ω2
= − atan2 ⎡ 2ωnω , 1 − ω 2 ⎤
⎢
⎥
n
⎣
⎦ The maximum value of G(jω) occurs at the Peak (Resonant) Frequency ωr :
G
ω r = ω n 1 − 2ζ 2 and G p 2 ( jω r ) = School of Mechanical Engineering
Purdue University 1
2ζ 1 − ζ 2
ME375 Frequency Response  46 23 ME375 Handouts 2nd Order System Frequency Response
40
20 Phase (deg); Magnitude (dB) 0
20
40
60
0
45
90
135
180
0.1ωn ωn 10ωn Frequency (rad/sec)
School of Mechanical Engineering
Purdue University ME375 Frequency Response  47 2nd Order System Frequency Response
A Few Observations:
• Three different characteristic frequencies:
– Natural Frequency (ωn)
– Damped Natural Frequency (ωd): ω d = ω n 1 − ζ 2
– Resonant (Peak) Frequency (ωr): ω r = ω n 1 − 2ζ 2 ωr ≤ωd ≤ωn
• When the damping ratio ζ > 0.707, there is no peak in the Bode magnitude plot.
0.707,
DO NOT confuse this with the condition for overdamped and underdamped
overundersystems: when ζ < 1 the system is underdamped (has overshoot) and when ζ > 1
underthe system is overdamped (no overshoot).
over• As ζ → 0 , ωr → ωn and ⏐G(jω)⏐ΜΑΧ increases; also the phase transition from 0
deg to −180 deg becomes sharper.
School of Mechanical Engineering
Purdue University ME375 Frequency Response  48 24 ME375 Handouts Example
• Combination of Systems
Transfer Function: G (s) = 2000(s2 + s + 25)
s( s + 2 0 0 )( s 2 + 1 0 s + 2 5 0 0 ) Plot the straight line approximation of G(s)’s Bode diagram: School of Mechanical Engineering
Purdue University ME375 Frequency Response  49 Example
40
Magnitude (dB) 20
0
20
40
60
80 180 Phase (deg) 90
0
90
180
270
1
10 10 0 1 10
Frequency (rad/sec)
School of Mechanical Engineering
Purdue University 10 2 10 3 ME375 Frequency Response  50 25 ...
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Full Document
 Spring '10
 Meckle
 Mechanical Engineering, Purdue University, Nyquist plot, School of Mechanical Engineering

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