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     7.896730e+02     2.039439e+02
 * time: 4.482269287109375e-5
    50     7.576979e+02     4.949210e+02
 * time: 4.682520866394043
   100     7.371685e+02     7.425408e+01
 * time: 8.307886838912964
   150     7.314176e+02     9.003250e+01
 * time: 11.87750792503357
   200     7.298205e+02     2.482114e+01
 * time: 15.450646877288818
   250     7.295992e+02     5.189602e+00
 * time: 18.98344397544861
   300     7.295811e+02     1.004392e+01
 * time: 22.513665914535522
   350     7.295730e+02     5.093855e+00
 * time: 26.046955823898315
   400     7.295718e+02     8.614356e-01
 * time: 29.578061819076538

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}:
 0.057618688637127706
 0.15635501746397473
 0.01926955672719852

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