Lab Report 3_EELE3314
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Running head: EELE3314L LAB FOR CONTROL SYSTEMS
1
EELE3314L Lab for Control Systems
LAB #3 Frequency Response Characteristics
2
EELE3314L LAB FOR CONTROL SYSTEMS
Results Procedure 4.1
Measured Magnitude M at w = 1rad/s
(dB)
Calculated Kv =10^M/20
Calculated Kv = sim sG(s)
-20.1
0.0989
0.0999
Estimated Ess = 1/kv
Measured Ess from response
plot
10
9.9 (approximately 10)
Procedure 4.2
Required Ess
Calculated Kv =1/Ess
Calculated K
0.05
20
200.2
Measured Magnitude at w = 1rad/s (dB)
Calculated Kv =10^(M/20)
Measured Ess from
response plot
25.9
19.7
0.05
Procedure 4.3 Bode plot
Measured Mr (dB)
Measured Magnitude at wr
(rad/s)
Measured Mr linear
5.28
12.8
1.8365
Calculated ζ
Calculated wn (rad/s)
0.2839
13.9759
Calculated Mp (dB)
Calculated Ts (s)
39.4494%
1.0081
Procedure 4.3 Response plot
Measured Mp (dB)
Measured Ts (s)
39.4%
0.994
Procedure 4.3 Root Locus
Measured Mp (dB)
Calculated Ts (s) (using
measured values)
39.8%
1.017
Procedure 4.4
Measured Phase Margin
(degrees)
Measured Gain Margin
(Linear)
From Graph
89.4
From Matlab Command
89.3711
1101.1
Procedure and Analysis
3
EELE3314L LAB FOR CONTROL SYSTEMS
1. Unity Feedback System 1.1 Code for all of Part 1 1.2 Kv of Unity Feedback System
4
EELE3314L LAB FOR CONTROL SYSTEMS
From the Bode Diagram, the value for magnitude at 1rad/s was approximately -20.1dB.
Shown in the calculations below, the Kv that responds to this value of -20.1dB magnitude is
0.0989 (approximately 0.1). This value was verified using the calculation as well and found to be
nearly identical.
From the Bode Plot
Verify using Calculation
5
EELE3314L LAB FOR CONTROL SYSTEMS
1.3 Steady State Error of Unity Feedback System
6
EELE3314L LAB FOR CONTROL SYSTEMS
In this part of the lab, the steady state error was found mathematically using the value of Kv
and known equation for unit ramp response for a type 1 system, which is 1/Kv. Mathematically,
the steady state error was found to be 10. In order to confirm the correctness of this estimate, the
unit ramp response of the closed-loop system was plotted on the same graph as the input unit
ramp waveform using MatLab. At steady state, the difference in values between the input unit
ramp waveform and the unit ramp response of the closed-loop system will be equal to the steady
state error. From the MatLab plot, this error was found to be 9.9, i.e. approximately 10, as
mathematically expected.
Mathematically obtaining steady state error
Graphically obtaining steady state error in MatLab
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Related Questions
Given the following conditions.
Q = 0.05 m3/s
P1 = 12.5 kPa
P2 = 11.5 kPa
P3 = 10.3 kPa
Find HL (1-2)
P1
P2
(2)
- 50 mm
= 50 mm 10m
P3
arrow_forward
1
Study the below response which obtained from HVACexperiment to find the following:
Setpoint
Upper Limit
Lower Limit =
Digital Scopes
Chamber Temp 23.0
C
Setpoint
Ambient
21.4 22.5 C
Temperature (C)
MeasuredAV
23.5-
Signal Generator 2
23.4-
Signal Type
23.3-
23.2-
23.1-
Amplitude
0.00
23-
Frequency
0.0080
Hz
22.9-
Offset
0.50
22.8-
22.7-
Control Parameters
22.6-
Vh amp
2.50
V
22.5-
0.0
5.0
10.0
15.0
20.0 25.0
30.0
35.0
40.0
45.0
50.0
55.0
60.0
Vh off
2.50
V
ATh
0.25
arrow_forward
For the system shown below, find the %OS
if the input is T(t) and the output is 02(t)
|N=25
=7 kg m-HD =3 N-m-s/rad
Na =50
fh=15kg-m
N3 25
D;=5 N-m-s/rad
K=88 N-m/radr
N4-100|
D; = 4 N-m-s/rad
; = 8 ke-m0000
Select one:
O a. %O.S 39.9
O b. %O.S=21.8
O c. %O.S 13.5
O d. %O.S =8.06
O e. %0.S= 46.7
arrow_forward
QUESTION 3
Consider a system model given by
d²x (t)
dx (1)
dt²
dt
x (0) = 1
dx (t)
dt
3e
What is the impulse response function?
sint
-21
- 4e
3e
+4-
=0
sint + e
-21
= 1.
-2t
-2t
sint + e
sint + e
cost
- 2t
+ 5x (t) =f(t)
sint - 2e
cost + 2
-2t
-21
cost + 2
cost
sint - 2cost + e-
arrow_forward
The 8kg body is moved to the right of the equilibrium position and released from rest at time t = 0. The viscous damping coefficient is 23NS/m
and the spring stiffness, K is 38N/m. Determine the damping factor (ratio) of the system.
Note: Give your answer to 3 decimal places.
Other Parameters:
Logarithmic Decrement (8):
Answer:
2ng
2T C
8 = In = In1 = 5 wnTd = 5wn
X2
Xn+1
1-5
wa 2m
k
Damping Ratio (5): 5=
2vkm
2mn
V(21)2+8
Frequency of damped vibration (wa): wa =
1-Wn
Undamped Forced Vibration:
Frequency Ratio (): r-
Xp-x sin w t
Next page
Hous page
Type here to search
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O %t. l a
O 11:70:£0
docs.google.com/forms
äbäi 30 For a damped one DOF system
where m, c, and k are shown
below: (a) k = 10 N/m, m = 1O
kg, c = 10 kg/s, (b) k = 2.5 N/m,
m = 10 kg, c = 10 kg/s, (c) k = 1
N/m, m = 10 kg, c = 10 kg/s.
Calculate for each case the
value of natural frequency in
rad/sec; time of period in sec3;
critical damping coefficient,
damping ratio; Are these
systems overdamped,
underdamped, or critically
* ?damped
اكتب المطلوب من السؤال فقط بدون ذکر
التفاصيل )النواتج المطلوبة فقط(. 30 درجة
إجابتك
صفحة 3 من 4
التالي
رجوع
عدم إرسال كلمات المرور عبر نماذج Go ogle مطلقا.
تم إنشاء هذا النموذج داخل STU. الإبلاغ عن إساءة الاستخدام
arrow_forward
Consider the closed loop system with response shown in following figure. What is the 10%-90% rise time of the response? Give your answer in seconds to the nearest third decimal place. Do not enter the units.
Amplitude
1.4
1.2
1
0.8
0.6
0.4
0.2
OF
0
A
H
| ¦¦
H
0.05
H
H
A
H
I
-14
LA
0.1
A
18
0.15
Time (seconds)
0.2
0.25
t
0.3
0.35
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H-W
Find Time response for
2S+3
Yes)=
s° (s+3)
arrow_forward
Question 1:
A pressure gauge which can be modeled as a Linear Time-Invariant (LTI) system has a time
response to a unit impulse input h(t) given by (Ge-2t +e-«") u(t). For a certain unknown
input, the output y(t) is observed to be (4e7t – 8e2")u(t). For this observed measurement,
derive the true pressure input x(t) to the gauge as a function of time. Assume that the input
signal is right-sided.
arrow_forward
Problem 7.65 - Enhanced - with Hints and Feedback
Figure
3 m
C
6 m
Part A
4 of 15
>
■Review
12 kN/m
1 of 1
Draw the shear diagram for the beam. Follow the sign convention. (Figure 1)
Click on "add vertical line off" to add discontinuity lines. Then click on "add segment" button to add functions between the
lines.
Note 1 - You should not draw an "extra" discontinuity line at the point where the curve passes the x-axis.
Note 2 - The curve you choose from the drop-down is only a pictorial representation of a real quadratic/cubic curve. The
equation of this curve is not mathematically equivalent to the correct answer. Consequently, slopes at discontinuities and
intercepts with the x-axis (if any) are not accurate.
Note 3 - Be sure to indicate the correct types of the functions between the lines, e.g. if in your answer the type of a function
is "linear increasing slope" for the function that actually has linear decreasing slope, the answer will be graded as
incorrect. Use the button "change…
arrow_forward
Q.11 - Consider the positions servomechanism is shown in figure below. Assume the following
numerical values for the system constant are.
Kp = gain of potentiometer detector = 7.64 v/ rad,
Kb= back e. m. f. constant =5.5*10-2 V/rad/sec
K; = motor torque constant = 6*10–5 Nm/A
C = 4*10-2 Nm/ (rad /sec).
Determine the natural frequency and damping ratio of the system.
Ra = armature winding resistant =2 2,
Ka=Gain of amplifier = 10
Jm = 1*10-5 Kg.m2. J =4.4*103 Kg.m2
n = gear ratio=1/10, Neglect Cm and la
Amp. %₂
ဦးအောင်မျိုာနစ် နှစ်
DH
0050
constant
N₁
arrow_forward
The following chart represents the Dynamic Amplification Factor (DLF) of pulses type Ramp.
Time from 0 to Maximum force: td
Natural Period of the structure: T
What is the maximum dynamic displacement of a structure with:
stiffness, k, of 150 kip/in
mass, m=50/g k-sec^2/in
g: gravity acceleration
The ramp force reach the maximum force F=100kip in a time td-0.15 sec
RAMP: DLF vs td/T
DLF
2500
1.500
1.000
0:000
O 0.81 in
O 1.22 in
O 0.67 in
O 1.50 in
td/T
15
arrow_forward
If characteristic equation is
minimum value for k to be in order to let the system stable
OK>-6
Ok>0
OK>-3
s³ + s² (K+6) + s(6K +9)+9K = 0
-3>K>-6
,what is the
5
arrow_forward
2
Find frequenciel
and mode shapes
Ic
of the two 9/ian
systems, assumiing
m=z ton/m
(b)
d=らm
hz 3m
E =210000 o kestlat
1 = 80Đ Cnt
ca)
arrow_forward
Question 3
Find the natural frequency (in rad/sec) of the system in In the figure below:
Use: m1= 5 kg, k1 = 53 N/m, k2 65 N/m and k3 18 N/m
Write your answer to FOUR significant figures. Don't write the units
X1
X2
k1
k2
k3
W ww
m1
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Theoretical value of
frequency
Wn =
L
(@n)? =
*thread length=28cm
Practical value of
frequency
No.
Tmain (sec)
= total time/5
time for (5) oscillations
1
T, =5.37s
2
T2 =5.64s
3
T3 =5.31s
4
T4
5.35s
5
T5 5.73s
Tmain
Tn =
1
fn =
Tn
Wn = 2 n fn
Compare (wn) the result of a
theoretical experiment with a
practical one
arrow_forward
Given a 2-DoF system below
ki
k.
m,
with parameters being m1 = 2kg and m2 = 3kg.
Determine the higher one of the two natural
frequencies wi and w2
in rad/s (w1
arrow_forward
the input and output impedances
amplifier configuration in Figure
*: below is
Aol = 175,000
Zin = 10 M
Zout = 75 N
%3D
%3D
Vin
R
560 k2
R
2.7 k2
Zi(NI) = 8.41 G0, Zout(NI)= 50 mQ O
Zi(NI) = 8.41 G0, Zout(NI)= 89.2 mQ O
%3D
Zi(NI) = 50 mQ, Zout(NI)= 8.41 GQ
%3D
ZI(NI)
= 89.2 m0, Zout(NI)= 8.41 GO
%3D
arrow_forward
2. The equation of motion for a damped multidegree of freedom system is
given by
Where [m]
=
[m]{x} + [c]{x} + [k]{x} = {f}
[100
=
0
0 0
10
0 10.
1000-4
{f}
[c]
8 –4 0
8
– 4
-4
4
0
8
4
= 100 2
0
=
Focos(wt)
-2
-2
The value of Fo
50N and w
50 rad/sec. Assuming the initial
conditions to be zero and using the modal coordinate approach, find out the
steady state solution of the system in the modal coordinate for first mode. Plot
the steady state solution using the computational tools.
0
−2], [k] =
=
2
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5 Problem
Match the following two frequency response diagrams:
-0
0.4
|G(i w)|
0.2
10
|G(i w)|
5
Response A
2
3
4
5
6
7
8
Frequency, (rad/s)
Response B
5
6
7
8
Frequency, w(rad/s)
1
2
3
4
6
6
10
10
10
to two of the following ODES
1. +x+100x = u(t)
5.
+100x = u(t)
9. +100x = u(t)
3. x+x+5x= u(t)
4. x+x+x= u(t)
2. x+x+25x = u(t)
:
6.
x +
25xu(t)
10. +25x = u(t)
7. x+5x= u(t)
8. u(t)
11.
+5x= u(t)
=
12.
xu(t)
arrow_forward
An experiment was carried out on a SDOF system to estimate the natural frequency and damping. The time history plotted below has the response in centimetres and the time in
seconds, as shown in Figure Q10 below. Estimate values for the following and choose the nearest values from the list given below:
the damped natural frequency in Hz;
the damping ratio using the logarithmic decrement method; and
the natural frequency in rad/s.
Ju(t)
2.00
AMA
1.00
0.00
0.00
Figure Q10
0.80
1.60
Select one or more:
O a. 4.92
b. 0.781
c. 0.625
O d. 0.330
O e.
0.052
2.40
3.20
4.00
t
arrow_forward
Match the transfer function with correct Bode amplitude plots.
G(s)=
S
s²+7s+10
S
G(s) =
s² + 10s+ 100
s+1
G(s) =
s²+10x+100
S+ 10
G(s) =
G(s) =
s²+s+10
s+ 10
s²+101s+ 100
A.
20 dB/dec
-20 dB/dec
B.
-20 dB/dec
-20 dB/dec
C.
20 dB/dec
20 dB/dec
E.
20 dB/dec
-20 dB/dec
-40 dB/dec
-20 dB/dec
arrow_forward
H.w 2: The open loop response, that is, the speed of the motor to a voltage input of 20V, assuming a
system without damping is
dw
20 = (0.02) + (0.06)w.
dt
If the initial speed is zero (w(0) = 0) ,and using the Runge-Kutta 4th order method, what is the speed at
t = 0.8s? Assume a step size of h = 0.4s.
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Q) solve exersice
The response of a system is given by
x() = 0.003 cos (30) + 0.004 sin(30) m
(a)
Determine (a) the amplitude of motion, (b) the period of motion, (c) the frequency in
Hz, (d) the frequency in rad/s, (e) the frequency in rpm, (f) the phase angle, and (g) the
response in the form of Equation (1.12)
SOUTION
arrow_forward
Given: A single degree of freedom system as shown, at t=0, y=2, and v=3, solve problems
9-10.
K
m
Q9) The response of undamped single degree of freedom system is:
O a. 2*cos(wt) +(3/w)"sin(wt)
O b. 3*cos(wt)+(2/w)*sin (wt)
O. 2°sin(wt) +(3/w)*cos(wt)
O d. 2*cos(wt)+(2/w)*sin(wt)
O e. cos(wt)+(3/w)*sin(wt)
vilian teare
Q10) The physical natural frequency, f is:
a. sąrt(m/k)/(2*T)
O b. sąrt(k/m)/(2*TT)
O. (k/m)/(2*rt)
Od. (m/k)/(2*T)
O e. (2*T)/sqrt(k/m)
arrow_forward
Question (2).
The following system has a block of a mass m = xykg(from your registration mUrnber),
stiffness k= 12000 N/m, damping coefficient c = xy• 10 kg/s, base motion y(t)=
0.01 sin(3t), calculate the following:
%3D
1 Natural frequency.
2 Base frequency.
3. Steady-state part of the response
X(t)
k
y(1)
base
wW
arrow_forward
Question (2).
The following system has a block of a mass m = xykg(from your registration mumber),
stiffness k= 12000 N/m, damping coefficient c= xy• 10 kg/s, base motion y(t) =
0.01 sin(3t), calculate the following:
%3D
%3D
1. Natural frequency.
2 Base frequency.
3. Steady-state part of the response
4. Isolation range.
X(t)
m
k
Ay()
base
arrow_forward
15
The solution to free vibration of a damped single degree of freedom system is
a. x(t) = Ae"Sin(o,t+n)
b. x(t) = Ae"Sin(o,t+2)
c. x(t) = ASin(@,t+o)
d. x(t) = Ae"Sin(o,t+q)
%3D
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For the system shown in Figure 2, u (t) and y (t) denote the absolute displacements of Building A and
Building B, respectively. The two buildings are connected using a linear viscous damper with damping
coefficient c. Due to construction activity, the floor mass of Building B was estimated that vibrates with
harmonic displacement that is described by the following function: y(t) = y,cos (wt).
k/2
Building A
H
k/2
M(t)
C
73(1)
Building B
Equivalent
(Building A)
00000000
Figure 2: Single-degree-of-freedom system in Problem 2.
111
uft)
Please compute the following related to Building A:
(a) Derive the equation of motion of the mass m.
(b) Find the expression of the amplitude of the steady-state displacement of the mass m.
(1)
(Building B)
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a) Find the total response of a single-degree-of-freedom system with m = 10 kg,
c = 20 N-s/m, k = 4000 N/m, xo = 0.01 m, and x'o=0 under the following
conditions:
a. An external force F (t) = Fo cos wt acts on the system with Fo = 100 N and
o= 10 rad/s.
b. Free vibration with F(t) = 0.
arrow_forward
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Related Questions
- Given the following conditions. Q = 0.05 m3/s P1 = 12.5 kPa P2 = 11.5 kPa P3 = 10.3 kPa Find HL (1-2) P1 P2 (2) - 50 mm = 50 mm 10m P3arrow_forward1 Study the below response which obtained from HVACexperiment to find the following: Setpoint Upper Limit Lower Limit = Digital Scopes Chamber Temp 23.0 C Setpoint Ambient 21.4 22.5 C Temperature (C) MeasuredAV 23.5- Signal Generator 2 23.4- Signal Type 23.3- 23.2- 23.1- Amplitude 0.00 23- Frequency 0.0080 Hz 22.9- Offset 0.50 22.8- 22.7- Control Parameters 22.6- Vh amp 2.50 V 22.5- 0.0 5.0 10.0 15.0 20.0 25.0 30.0 35.0 40.0 45.0 50.0 55.0 60.0 Vh off 2.50 V ATh 0.25arrow_forwardFor the system shown below, find the %OS if the input is T(t) and the output is 02(t) |N=25 =7 kg m-HD =3 N-m-s/rad Na =50 fh=15kg-m N3 25 D;=5 N-m-s/rad K=88 N-m/radr N4-100| D; = 4 N-m-s/rad ; = 8 ke-m0000 Select one: O a. %O.S 39.9 O b. %O.S=21.8 O c. %O.S 13.5 O d. %O.S =8.06 O e. %0.S= 46.7arrow_forward
- QUESTION 3 Consider a system model given by d²x (t) dx (1) dt² dt x (0) = 1 dx (t) dt 3e What is the impulse response function? sint -21 - 4e 3e +4- =0 sint + e -21 = 1. -2t -2t sint + e sint + e cost - 2t + 5x (t) =f(t) sint - 2e cost + 2 -2t -21 cost + 2 cost sint - 2cost + e-arrow_forwardThe 8kg body is moved to the right of the equilibrium position and released from rest at time t = 0. The viscous damping coefficient is 23NS/m and the spring stiffness, K is 38N/m. Determine the damping factor (ratio) of the system. Note: Give your answer to 3 decimal places. Other Parameters: Logarithmic Decrement (8): Answer: 2ng 2T C 8 = In = In1 = 5 wnTd = 5wn X2 Xn+1 1-5 wa 2m k Damping Ratio (5): 5= 2vkm 2mn V(21)2+8 Frequency of damped vibration (wa): wa = 1-Wn Undamped Forced Vibration: Frequency Ratio (): r- Xp-x sin w t Next page Hous page Type here to searcharrow_forwardO %t. l a O 11:70:£0 docs.google.com/forms äbäi 30 For a damped one DOF system where m, c, and k are shown below: (a) k = 10 N/m, m = 1O kg, c = 10 kg/s, (b) k = 2.5 N/m, m = 10 kg, c = 10 kg/s, (c) k = 1 N/m, m = 10 kg, c = 10 kg/s. Calculate for each case the value of natural frequency in rad/sec; time of period in sec3; critical damping coefficient, damping ratio; Are these systems overdamped, underdamped, or critically * ?damped اكتب المطلوب من السؤال فقط بدون ذکر التفاصيل )النواتج المطلوبة فقط(. 30 درجة إجابتك صفحة 3 من 4 التالي رجوع عدم إرسال كلمات المرور عبر نماذج Go ogle مطلقا. تم إنشاء هذا النموذج داخل STU. الإبلاغ عن إساءة الاستخدامarrow_forward
- Consider the closed loop system with response shown in following figure. What is the 10%-90% rise time of the response? Give your answer in seconds to the nearest third decimal place. Do not enter the units. Amplitude 1.4 1.2 1 0.8 0.6 0.4 0.2 OF 0 A H | ¦¦ H 0.05 H H A H I -14 LA 0.1 A 18 0.15 Time (seconds) 0.2 0.25 t 0.3 0.35arrow_forwardH-W Find Time response for 2S+3 Yes)= s° (s+3)arrow_forwardQuestion 1: A pressure gauge which can be modeled as a Linear Time-Invariant (LTI) system has a time response to a unit impulse input h(t) given by (Ge-2t +e-«") u(t). For a certain unknown input, the output y(t) is observed to be (4e7t – 8e2")u(t). For this observed measurement, derive the true pressure input x(t) to the gauge as a function of time. Assume that the input signal is right-sided.arrow_forward
- Problem 7.65 - Enhanced - with Hints and Feedback Figure 3 m C 6 m Part A 4 of 15 > ■Review 12 kN/m 1 of 1 Draw the shear diagram for the beam. Follow the sign convention. (Figure 1) Click on "add vertical line off" to add discontinuity lines. Then click on "add segment" button to add functions between the lines. Note 1 - You should not draw an "extra" discontinuity line at the point where the curve passes the x-axis. Note 2 - The curve you choose from the drop-down is only a pictorial representation of a real quadratic/cubic curve. The equation of this curve is not mathematically equivalent to the correct answer. Consequently, slopes at discontinuities and intercepts with the x-axis (if any) are not accurate. Note 3 - Be sure to indicate the correct types of the functions between the lines, e.g. if in your answer the type of a function is "linear increasing slope" for the function that actually has linear decreasing slope, the answer will be graded as incorrect. Use the button "change…arrow_forwardQ.11 - Consider the positions servomechanism is shown in figure below. Assume the following numerical values for the system constant are. Kp = gain of potentiometer detector = 7.64 v/ rad, Kb= back e. m. f. constant =5.5*10-2 V/rad/sec K; = motor torque constant = 6*10–5 Nm/A C = 4*10-2 Nm/ (rad /sec). Determine the natural frequency and damping ratio of the system. Ra = armature winding resistant =2 2, Ka=Gain of amplifier = 10 Jm = 1*10-5 Kg.m2. J =4.4*103 Kg.m2 n = gear ratio=1/10, Neglect Cm and la Amp. %₂ ဦးအောင်မျိုာနစ် နှစ် DH 0050 constant N₁arrow_forwardThe following chart represents the Dynamic Amplification Factor (DLF) of pulses type Ramp. Time from 0 to Maximum force: td Natural Period of the structure: T What is the maximum dynamic displacement of a structure with: stiffness, k, of 150 kip/in mass, m=50/g k-sec^2/in g: gravity acceleration The ramp force reach the maximum force F=100kip in a time td-0.15 sec RAMP: DLF vs td/T DLF 2500 1.500 1.000 0:000 O 0.81 in O 1.22 in O 0.67 in O 1.50 in td/T 15arrow_forward
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