## Main.DynamicEstimation History

January 15, 2021, at 04:00 AM by 10.35.117.248 -

plt.show() (:sourceend:) (:divend:)

The approach can also be extended to multiple data sets and when the experimental values are at different times.

(:toggle hide solution3b button show="Show GEKKO Solution":) (:div id=solution3b:)

(:source lang=python:) from gekko import GEKKO import numpy as np import pandas as pd import matplotlib.pyplot as plt

1. data set 1

t_data1 = [0.0, 0.1, 0.2, 0.4, 0.8, 1.00] x_data1 = [2.0, 1.6, 1.2, 0.7, 0.3, 0.15]

1. data set 2

t_data2 = [0.0, 0.15, 0.25, 0.45, 0.85, 0.95] x_data2 = [3.6, 2.25, 1.75, 1.00, 0.35, 0.20]

1. combine with dataframe join

data1 = pd.DataFrame({'Time':t_data1,'x1':x_data1}) data2 = pd.DataFrame({'Time':t_data2,'x2':x_data2}) data1.set_index('Time', inplace=True) data2.set_index('Time', inplace=True) data = data1.join(data2,how='outer') print(data.head())

1. indicate which points are measured

z1 = (data['x1']==data['x1']).astype(int) # 0 if NaN z2 = (data['x2']==data['x2']).astype(int) # 1 if number

1. replace NaN with any number (0)

data.fillna(0,inplace=True)

m = GEKKO(remote=False)

1. measurements

xm = m.Array(m.Param,2) xm[0].value = data['x1'].values xm[1].value = data['x2'].values

1. index for objective (0=not measured, 1=measured)

zm = m.Array(m.Param,2) zm[0].value=z1 zm[1].value=z2

m.time = data.index x = m.Array(m.Var,2) # fit to measurement x[0].value=x_data1[0]; x[1].value=x_data2[0]

k = m.FV(); k.STATUS = 1 # adjustable parameter for i in range(2):

    m.free_initial(x[i])               # calculate initial condition
m.Equation(x[i].dt()== -k * x[i])  # differential equations
m.Minimize(zm[i]*(x[i]-xm[i])**2)  # objectives


m.options.IMODE = 5 # dynamic estimation m.options.NODES = 2 # collocation nodes m.solve(disp=True) # solve k = k.value[0] print('k = '+str(k))

1. plot solution

plt.plot(m.time,x[0].value,'b.--',label='Predicted 1') plt.plot(m.time,x[1].value,'r.--',label='Predicted 2') plt.plot(t_data1,x_data1,'bx',label='Measured 1') plt.plot(t_data2,x_data2,'rx',label='Measured 2') plt.legend(); plt.xlabel('Time'); plt.ylabel('Value') plt.xlabel('Time');

June 06, 2020, at 02:02 PM by 136.36.211.159 -
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x_data = [2.0,1.6,1.2,0.7,0.3,0.15,0.1, 0.05,0.03,0.02,0.015,0.01]

to:

x_data = [2.0,1.6,1.2,0.7,0.3,0.15,0.1,0.05,0.03,0.02,0.015,0.01]

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Use an initial condition of x=2 that matches the data.

to:

Use an initial condition of x=2 that matches the data. Create new states y=dx/dt and z=dy/dt for the higher order derivative terms.

$$\frac{dx}{dt} = y$$ $$\frac{dy}{dt} = z$$ $$\frac{dz}{dt} = az+by+cx+d$$

June 06, 2020, at 01:57 PM by 136.36.211.159 -

plt.legend() plt.xlabel('Time'), plt.ylabel('Value') plt.show() (:sourceend:) (:divend:)

#### Dynamic Parameter Estimation Example 4

Estimate the parameter a,b,c,d in the differential equation:

$$\frac{d^3x}{dt^3} = a\frac{d^2x}{dt^2}+b\frac{dx}{dt}+c x+d$$

by minimizing the error between the predicted and measured x values. The x values are measured at the following time intervals.

(:source lang=python:) t_data = [0,0.1,0.2,0.4,0.8,1,1.5,2,2.5,3,3.5,4] x_data = [2.0,1.6,1.2,0.7,0.3,0.15,0.1, 0.05,0.03,0.02,0.015,0.01] (:sourceend:)

Use an initial condition of x=2 that matches the data.

(:toggle hide solution4 button show="Show GEKKO Solution":) (:div id=solution4:)

(:source lang=python:) from gekko import GEKKO

t_data = [0,0.1,0.2,0.4,0.8,1,1.5,2,2.5,3,3.5,4] x_data = [2.0,1.6,1.2,0.7,0.3,0.15,0.1, 0.05,0.03,0.02,0.015,0.01]

m = GEKKO() m.time = t_data

1. states

x = m.CV(value=x_data); x.FSTATUS = 1 # fit to measurement y,z = m.Array(m.Var,2,value=0)

a,b,c,d = m.Array(m.FV,4) a.STATUS=1; b.STATUS=1; c.STATUS=1; d.STATUS=1

1. differential equation
2. Original: x' = a*x + b x' + c x + d
3. Transform: y = x'
4. z = y'
5. z' = a*z + b*y + c*x + d

m.Equations([y==x.dt(),z==y.dt()]) m.Equation(z.dt()==a*z+b*y+c*x+d) # differential equation

m.options.IMODE = 5 # dynamic estimation m.options.NODES = 3 # collocation nodes m.solve(disp=False) # display solver output print(a.value[0],b.value[0],c.value[0],d.value[0])

import matplotlib.pyplot as plt # plot solution plt.plot(m.time,x.value,'bo',label='Predicted') plt.plot(m.time,x_data,'rx',label='Measured')

March 10, 2020, at 01:30 PM by 136.36.211.159 -
March 10, 2020, at 01:29 PM by 136.36.211.159 -
Changed lines 37-38 from:

$$\frac{x}{dt} = -k\,x$$

to:

$$\frac{dx}{dt} = -k\,x$$

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Use an initial condition of x=2 that matches the data. Verify the solution of x with the analytic expression x(t)=2 exp(-k\,t).

to:

Use an initial condition of x=2 that matches the data. Verify the solution of x with the analytic expression x(t)=2 exp(-k t).

March 10, 2020, at 01:28 PM by 136.36.211.159 -
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by matching the predicted x to the measured x data:

to:

by minimizing the error between the predicted and measured x values. The x values are measured at the following time intervals.

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time = [0, 0.1, 0.2, 0.4, 0.8, 1] x_data = [2.0081, 1.5512, 1.1903, 0.7160, 0.2562, 0.1495]

to:

t_data = [0, 0.1, 0.2, 0.4, 0.8, 1] x_data = [2.0, 1.6, 1.2, 0.7, 0.3, 0.15]

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Verify the solution of x with the analytic expression x(t)=exp(-k\,t).

to:

Use an initial condition of x=2 that matches the data. Verify the solution of x with the analytic expression x(t)=2 exp(-k\,t).

(:toggle hide solution3 button show="Show GEKKO Solution":) (:div id=solution3:)

(:source lang=python:) from gekko import GEKKO

t_data = [0, 0.1, 0.2, 0.4, 0.8, 1] x_data = [2.0, 1.6, 1.2, 0.7, 0.3, 0.15]

m = GEKKO(remote=False) m.time = t_data x = m.CV(value=x_data); x.FSTATUS = 1 # fit to measurement k = m.FV(); k.STATUS = 1 # adjustable parameter m.Equation(x.dt()== -k * x) # differential equation

m.options.IMODE = 5 # dynamic estimation m.options.NODES = 5 # collocation nodes m.solve(disp=False) # display solver output k = k.value[0]

import numpy as np import matplotlib.pyplot as plt # plot solution plt.plot(m.time,x.value,'bo', label='Predicted (k=str(np.round(k,2)))') plt.plot(m.time,x_data,'rx',label='Measured')

1. plot exact solution

t = np.linspace(0,1); xe = 2*np.exp(-k*t) plt.plot(t,xe,'k:',label='Exact Solution') plt.legend() plt.xlabel('Time'), plt.ylabel('Value') plt.show() (:sourceend:) (:divend:)

March 10, 2020, at 01:12 PM by 136.36.211.159 -

#### Dynamic Parameter Estimation Example 3

Estimate the parameter k in the exponential decay equation:

$$\frac{x}{dt} = -k\,x$$

by matching the predicted x to the measured x data:

(:source lang=python:) time = [0, 0.1, 0.2, 0.4, 0.8, 1] x_data = [2.0081, 1.5512, 1.1903, 0.7160, 0.2562, 0.1495] (:sourceend:)

Verify the solution of x with the analytic expression x(t)=exp(-k\,t).

January 04, 2017, at 05:57 PM by 10.10.144.119 -
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#### Dynamic Parameter Estimation Example 1

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#### Dynamic Parameter Estimation Example 2

April 04, 2015, at 02:03 PM by 45.56.12.124 -
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April 03, 2015, at 01:39 AM by 10.24.17.95 -
April 02, 2015, at 04:03 PM by 45.56.12.124 -
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April 02, 2015, at 03:37 PM by 45.56.12.124 -
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#### Dynamic Parameter Estimation Example

April 02, 2015, at 03:37 PM by 45.56.12.124 -
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#### Numerical Solution Tutorial for Dynamic Parameter Estimation

April 02, 2015, at 03:30 PM by 45.56.12.124 -

#### Numerical Estimation Introduction

A method to solve dynamic estimation is by numerically integrating the dynamic model at discrete time intervals, much like measuring a physical system at particular time points. The numerical solution is compared to measured values and the difference is minimized by adjusting parameters in the model. Excel, MATLAB, Python, and Simulink are used in the following example to both solve the differential equations that describe the velocity of a vehicle as well as minimize an objective function.

(:html:) <iframe width="560" height="315" src="https://www.youtube.com/embed/y0ERNz5Kms8?rel=0" frameborder="0" allowfullscreen></iframe> (:htmlend:)

April 02, 2015, at 06:09 AM by 45.56.12.124 -

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(:title Estimation Introduction:)

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(:title Dynamic Estimation Introduction:)