In this example, we will estimate a model for a four-stage evaporator to reduce the water content of a product, for example milk. The 3 inputs are feed flow, vapor flow to the first evaporator stage and cooling water flow. The three outputs are the dry matter content, the flow and the temperature of the outcoming product.
The example comes from STADIUS's Identification Database
Zhu Y., Van Overschee P., De Moor B., Ljung L., Comparison of three classes of identification methods. Proc. of SYSID '94,
using DelimitedFiles, Plots
using ControlSystemIdentification, ControlSystemsBase
url = "https://ftp.esat.kuleuven.be/pub/SISTA/data/process_industry/evaporator.dat.gz"
zipfilename = "/tmp/evaporator.dat.gz"
path = Base.download(url, zipfilename)
run(`gunzip -f $path`)
data = readdlm(path[1:end-3])
# Inputs:
# u1: feed flow to the first evaporator stage
# u2: vapor flow to the first evaporator stage
# u3: cooling water flow
# Outputs:
# y1: dry matter content
# y2: flow of the outcoming product
# y3: temperature of the outcoming product
u = data[:, 1:3]'
y = data[:, 4:6]'
d = iddata(y, u, 1)
InputOutput data of length 6305, 3 outputs, 3 inputs, Ts = 1
The input consists of two heating inputs and one cooling input, while there are 6 outputs from temperature sensors in a cross section of the furnace.
Before we estimate any model, we inspect the data
plot(d, layout=6)
We split the data in two, and use the first part for estimation and the second for validation. A model of order around 8 is reasonable (the paper uses 6-13). This system requires the option zeroD=false
to be able to capture a direct feedthrough, otherwise the fit will always be rather poor.
dtrain = d[1:3300] # first experiment ends after 3300 seconds
dval = d[3301:end]
model,_ = newpem(dtrain, 8, zeroD=false)
Iter Function value Gradient norm
0 7.796309e+02 2.166337e+02
* time: 5.078315734863281e-5
50 7.515894e+02 7.099459e+02
* time: 4.692425966262817
100 7.370387e+02 9.003382e+01
* time: 8.376680850982666
150 7.319665e+02 6.576515e+01
* time: 11.974751949310303
200 7.300869e+02 1.292878e+01
* time: 15.569308996200562
250 7.297392e+02 1.681885e+01
* time: 19.16288685798645
300 7.296118e+02 1.322206e+01
* time: 22.746511936187744
350 7.295775e+02 5.851955e+00
* time: 26.34854793548584
400 7.295727e+02 2.074114e+00
* time: 29.934436798095703
450 7.295717e+02 1.132825e+00
* time: 33.51741981506348
predplot(model, dval, h=1, layout=d.ny)
predplot!(model, dval, h=5, ploty=false)
The figures above show the result of predicting $h={1, 5}$ steps into the future.
We can visualize the estimated model in the frequency domain as well.
w = exp10.(LinRange(-2, log10(pi/d.Ts), 200))
sigmaplot(model.sys, w, lab="PEM", plotphase=false)
Let's compare prediction performance to the paper
ys = predict(model, dval, h=5)
ControlSystemIdentification.mse(dval.y-ys)
3×1 Matrix{Float64}:
0.05761868864741539
0.1563550174672708
0.019269556712616372
The authors got the following errors: [0.24, 0.39, 0.14]