Question
Asked 19th Feb, 2022

# Do you know the reason for the unexpected increase in the frequency of this system?

I have a simulation code for a Horizontal Washing Machine.
The code solves the equations of motions of the system by Matlab ode45 and plots the vibration response of the system at the transient state of performance.
In this code, the frequency (omega) is an exponential function of time, as it's stated below (and its diagram is attached to 'the question'):
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omega= (1-exp((-0.5)*t))*omega_0+(1-exp((-0.5)*heaviside(t-t1).*(t-t1)))*(omega_1-omega_0);
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ode command:
-----------------------------------------------------------------------------
[T,Y]=ode45(@snowa1,tspan1,initial_vector1);
plot(T,Y(:,1)-mean(Y(:,1)))
-----------------------------------------------------------------------------
The resulting displacement response is attached to the question.
It is desired to :
First, increase the frequency to omega_0 by exponential1
Then, increase it to omega_1 by exponential2
But 'the problem' is that:
the displacement response shows an unexpected increase in frequency at the beginning of the second exponential increase (it becomes 20 Hz, which is much larger than the maximum frequency in the simulation- 10 Hz).
Do you know what could be the reason for this response?
Any help would be gratefully appreciated.

## Most recent answer

Frequency is related to stress. One of the forces is larger than expected, possibly from an imbalance.

## All Answers (13)

I don't know the dynamics of the front-loading washer. And I don't know the purpose of your omega, though I can approximately recreate the signal from your 1st graph. So, please show the full mathematical model, which I believe it is in your snowa1 m-file. And if you are diligent, please attach the journal paper, we can find something to read about the Washer Dynamics and Control.
My initial guess is that your have triggered to double the value of omega (frequency) from t = 20 seconds. It takes about 5·(1/0.5) = 10 seconds from the triggering to arrive at the steady-state value.
20th Feb, 2022
Isfahan University of Technology
It's a 4DOF (2 translational and 2 rotational DOF) system. The 'omega' is one of the time-dependent input parameters of the system. This exponential profile is used to increase the rotational velocity of the system during the beginning of the spin-drying cycle of the washing machine.
The omega profile is usually 'linear' or 'exponential' in the papers and it's close to reality.
The two exponential steps profile has been used to check whether it makes the simulation more similar to the test results.
Here comes the ode function code:
function ds = snowa1(t,s)
NumberOfState=8; % 4 displacements/ 4 velocities (4 degrees of freedom)
NumberOfState_half=4;
global omega_0 omega_1
snowa_parameter %reads all needed parameters of the EOM for solving ode, from a file
omega= (1-exp((-0.5)*t))*omega_0+(1-exp((-0.5)*heaviside(t-t1).*(t-t1)))*(omega_1-omega_0);
EquationOfMotion % (M*ddx+H=0) reads matrices M and H which build equations of motion
MM=(M^(-1))*(-transpose(H)); %gives the 4 accelerations from the equations of motion
ds=zeros(NumberOfState,1); %first derivitive of the state vector (s). /velocities and accelerations
for i=1:NumberOfState
if i<NumberOfState_half+1 %i<5
ds(i)=s(i+NumberOfState_half); ٪the first 4 elements of ds, are attributed to the velocities
else
ds(i)=MM(i- NumberOfState_half); %the 2nd 4 elements of ds, are attributed to the accelerations ٪ %the final output .
end
end
end
And you can read the following articles to get more familiar with the dynamic modeling of horizontal washing machines:
1 Recommendation
20th Feb, 2022
Isfahan University of Technology
Dear
Sina S.
,
The system initial conditions are defined at the static equilibrium point.
I've set a time step of 0.001, but I guess Matlab ode would use finer steps by itself (depending on the solving requirements)
I've also checked the other ode solvers, but the results are alike.
Thanks for your reply and info. I think the general mathematical model (ODE) of the washer is given in Boyraz and Gündüz (2013) as Eqs. (1) and (2) in the first article that you recommended. Please correct me if I'm wrong.
But these two equations are like jigsaw puzzle to me because the components of the forces are detailed from Eq. (3) to Eq. (18). The forces are given in Eqs. (6), (7), (12a), (12b), (18a), (18b). However, the forces Fs2x, Fs2z, Fd2x, Fd2z, are not directly given. Also, though I can't understand your definition clearly, I think your omega (ω) is almost similar to the spin speed of the drum, β'd in Eq. (3).
Since you are the expert in washer, can you possibly assemble them into Eqs. (1) and (2)? It would certainly help me (and other interested scientists) to see & grasp the full equations.
By the way, are your simulated responses incorrect? If so, what do you expect, or how should the responses look like? Please enlighten.
21st Feb, 2022
Isfahan University of Technology
Yes. The 2D model of the system is given by the Newtonian method in Boyraz (2013). But I've used the Lagrangian method to extract the equations of motion(ODE) of the system, as it's been used in the second recommended article.
The damper and spring forces of the first article are assumed linear (Fs=k*x, Fd=-c*dx).
In the second article, all of the forces are specified and the process seems to be more clear and helpful.
The resulted equations of my 4DOF system are a little long.
And my omega is exactly the spin speed of the drum and is used according to the first article.
My response is similar to the test data, using one exponential function for omega. But during the test, there has been a period of constant spin speed before the speed rises again to reach the final speed. That's the reason I'm using 2 steps exponential increase of speed.
The maximum amount of displacement with one step exponential spin speed profile (up to 300 rpm) in the simulation is close to the maximum value in the experimental test. The final speed, of course, is 1200 rpm, which is obtained at the end of zone '3' in figure1 (the whole experimental test diagram). In zone 2, the speed reaches 600 rpm and remains constant for a while. Then (in zone 3) it increases to 1200. And in zone '4' the speed will remain 1200 rpm.
The whole diagram of the experimental test (Figure1) is attached to this answer, as well as the simulation diagram (with spin speed profile of: (1-exp((-1/30)*t))*omega_s).
Figure 2 shows zone 1 of test diagram, in which the speed reachs 300 rpm.
Thanks a lot for your attention and response.
22nd Feb, 2022
The American Institute of Aeronautics and Astronautics
I have not fully pondered your question and your explanations. However, at the first glance, it seems, you have made a mistake in using the Heaviside step function. Just after your question, where you have mentioned your MATLAB syntax as:
1-exp((-0.5)*heaviside(t-t1).*(t-t1))
In my opinion, in the third parenthesis from left, the value t1, is redundant. I have marked it as bold-faced letter in the syntax above. I think it should be amended as:
1-exp((-0.5)*heaviside(t-t1).*(t)).
It means you may simply remove, t1, from the third parenthesis.
WARNING: Using the Heaviside step-function to shift a function to a new position, differs from using the Heaviside step-function for cutting the function at a specific value.
You should reconsider how you have defined the Heaviside function with regard to your specific problem which seems to require to segment and patch the frequency values for specific time-intervals.
I understand that the 4DOF system for the washer can be quite lengthy. However, I'm unsure if your question is a control problem. If it is, and the 4DOF mathematical model can be expressed in such form:
x' = f(x) + g(xFc(ω)
where
1. f(x) = [f1(x), f2(x), f3(x), f4(x)]T and g(x) are the nonlinear terms in column vector forms that you derived from the Lagrangian method,
2. Fc(ω) is the control force that represents a function of ω in vector form, and
3. ω (omega) is the control input,
then I think it is possible to design the spin speed profile for the control input, ω, so that the desired responses of x can be achieved. If you want to design the profile, you need to at least understand the mathematical equation for Fc(ω). Do you want to regulate the spin speed at 300 rpm, 600 rpm, or 1200 rpm? Because I see only the signal oscillates within the dimensionless amplitudes ± 4×10–3.
I have plotted the signal according to your suggestion, and compared it with Mahdi's original signal. Note that if t1 > 5/τ, then exp(–τ·θ(t – t1)·t) ≈ 0 after t1, because exp(–τ·t) has decayed to almost zero after 5/τ.
2 Recommendations
22nd Feb, 2022
Haider Ali Bhatti
University of Engineering and Technology, Lahore
Unbalancing, vibrations, improper flooring,internal wearing of rings shafts, bearings,gears all the factors can increase the frequency of a system
1 Recommendation
23rd Feb, 2022
Isfahan University of Technology
According to the equations of motion I've extracted, I could put the model in this form:
M*x_dot = -H
in which 'M' and 'H' are both functions of 'omega' and 'x'.
And I consider 'omega' as a control input.
So it is a control problem.
I don't have it in a form in which a separated term of Fc(omega) exists. And those 2 matrices are full of 'omega' in their nonlinear terms. But I think this form doesn't differ from the form you mentioned, and the profile could be designed in a control system.
By the profile I've mentioned before, I want to:
First, regulate the speed at 300,
then, rise it again to regulate at 600,
and finally, regulate it at 1200.
One thing that I've noticed lately is that the 'exponential coefficient' ('alpha' in 'exp((-alpha )*t)' ) is effective on the response.
So if I use 'alpha=1/30' instead of '0.5' the aforementioned increase in frequency would disappear. But the values of 'displacement' don't seem to be desired according to the 'test result'.
So I checked it again with 'one exponential' profile:
omega = (1- exp((-1/30)*t))*omega_s;
(omega_s = 1200 rpm)
and I attained the diagram attached to this answer.
The amplitude has a dimension of 'meter' and is the value of displacement in the gravitational direction. And the values are close to the 'test result'.
Thanks for showing the model: M(x, ω)·x' = – H(x, ω).
I'm unsure of how M(x, ω) & H(x, ω) look like, but let xref be the reference of the desired responses, and the error is defined as e = xxref. If you can transform the model into
M(e, ω)·e' = – H(e, ω)
and design to satisfy the following
H(e, ω) = M(e, ω)·K·e,
such that K is a Hurwitz matrix (which can be a function of alpha, α), then xxref, as time t → ∞. The question now is whether or not you can solve the equation H(e, ω) = M(e, ω)·K·e, for ω.
1 Recommendation
26th Feb, 2022