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
&
&
m1&&1 + ( b1 + b2 ) z1 + ( k1 + k2 ) z1 − b2z2 − k2z2 = 0
z Fc(t) &
&
m2&&2 + b2z2 + k2z2 − b2 z1 − k2 z1 = −Fc (t)
z 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
Si
• 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:
&
y + 5 y = 10u to a sinusoidal input:
– Forced response: u (t ) = sin(2t )
Y (s) = G (s) ⋅U (s) where G ( s ) = and ∴ Y (s) = Y ( s) = A1 U ( s ) = L [sin(2t ) ] = – PFE: + 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
−5
y (t ) = 1 24 t +
4 ⋅ e3 ⋅ sin(2t + φ )
1442443 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
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
&
y + 5y =10u and the input is general sinusoidal input: sin(
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:
residue
to
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.)
(Cont.) 10ω
52 + ω 2
1
10
1
1
A2 =
⋅
=
⋅ G ( jω ) =
⋅
2 j jω + 5 2 j
2j
−1
−1
−1
10
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
Calculate the magnitude and phase shift of the steady state response when the
system
system is excited by (i) sin(2t) and (ii) sin(10t). Compare your result with the
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
is
due to sinusoidal inputs at different frequencies.
In general, given a stable system:
general given stable system:
&
&
an 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 )
=
=
an s n + an −1s n −1 + L + a1s + a0 D( s ) an ( s − p1 )( s − p2 )L ( s − pn )
If the input is a sinusoidal signal with frequency ω , i.e.
i.e.
G(s) ≡ 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:
diff
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.
for G ( jω ) = G ( s ) s = jω ≡ bm ( jω ) m + bm −1 ( jω ) m −1 + L + b1 ( jω )+ b0
an ( jω ) n + an −1 ( jω ) n −1 + L + 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:
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
⎩
School of Mechanical Engineering
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
1st
The motion of a piston in a cylinder can be
modeled by 1st order system with force
modeled by a 1st order system with force
as
as input and piston velocity as output: (2) Calculate the steady state output of the
system when the input is Steady State Output v(t)
⎢G(jω)⎥ sin(ω t + φ )
sin(0t + sin(10t) f(t) Input f(t)
sin(ω t)
sin(0t) sin(10t + sin(20t) sin(20t + v
The EOM is:
&
Mv + Bv = f (t ) sin(30t) sin(30t + (1) Let M = 0.1 kg and B = 0.5 N/(m/s),
0.1
0.5
find the transfer function of the system: sin(40t) sin(40t + sin(50t) sin(50t + sin(60t) sin(60t + 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&&1 + B1 z1 + K1 z1 = f (t )
z 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). 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)
sin(11 z1 (m) 0.005 0 0.005
0 5 10 15 20 25 30 35 40 45 50 Time (sec)
School of Mechanical Engineering
Purdue University ME375 Frequency Response  21 Example  Vibration Absorber (II)
With vibration absorber:
z2
M2
K2 B2
z1
M1 K1 f(t) B1 TF (from f(t) to z1): EOM:
&
&
M 1&&1 + ( B1 + B2 ) z1 + ( K1 + K 2 ) z1 − B2 z2 − K 2 z2 = f (t )
z &
&
M 2 &&2 + B2 z2 + K 2 z2 − B2 z1 − K 2 z1 = 0
z
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
kg,
100
0.1
the steady state response of the system for f(t) =
(a) sin(8.5t) (b) sin(10t) (c) 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
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)
sin(11 0.02
z1 (m) 0.01
0
0.01
0.02
0 5 10 15 20 25 30 35 40 45 50 Time (sec)
School of Mechanical Engineering
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?
(Hint: when p = σ ± jω → the response is eσt e jωt →
the
) 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
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
Bode Plot
No absorber added Bode Plot
Absorber tuned at 10 rad/sec added
40 50 50 60 60 Phase (deg); Magnitude (dB) 30 40
Phase (deg); Magnitude (dB) 30 70
80
90
100
0
45
90 80
90
100
0
45
90 135 135 180
10 70 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 ...
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This note was uploaded on 12/22/2011 for the course ME 375 taught by Professor Meckle during the Fall '10 term at Purdue.
 Fall '10
 Meckle
 Mechanical Engineering

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