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: 4.506111145019531e-5
    50     7.515894e+02     7.099459e+02
 * time: 4.633913040161133
   100     7.370387e+02     9.003382e+01
 * time: 8.25938606262207
   150     7.319665e+02     6.576515e+01
 * time: 11.805922985076904
   200     7.300869e+02     1.292878e+01
 * time: 15.346028089523315
   250     7.297392e+02     1.681885e+01
 * time: 18.886148929595947
   300     7.296118e+02     1.322206e+01
 * time: 22.426326990127563
   350     7.295775e+02     5.851955e+00
 * time: 25.969537019729614
   400     7.295727e+02     2.074114e+00
 * time: 29.50413990020752
   450     7.295717e+02     1.132825e+00
 * time: 33.04593801498413

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}:
 0.057618688647415345
 0.15635501746727082
 0.01926955671261641

The authors got the following errors: [0.24, 0.39, 0.14]