State space may not contain all states when there are zeros in the matrix entries

[baggepinnen]

In the following example, there are three inputs passed along to io_preprocessing which in turn calls structural_simplify. All three inputs are present as states in the system upon calling io_preprocessing. Before markio! is called within structural_simplify, one of the inputs, (model₊input₊u(t))[2] has been removed from fullvars, leading to the error below:

ERROR: Some specified inputs were not found in system. The following Dict indicates the found variables
Dict{Num, Bool}(
    u(t) => 1,
    (model₊input₊u(t))[2] => 0,
    d(t) => 1
)
Stacktrace:
 [1] error(::String, ::Dict{Num, Bool})
   @ Base ./error.jl:44
 [2] markio!(state::TearingState{ODESystem}, inputs::Vector{Num}, outputs::Vector{Any}; check::Bool)

using ModelingToolkitStandardLibrary.Blocks
A,C = [randn(2,2) for i in 1:2]
B = [1.0 0; 0 0]
@named model = Blocks.StateSpace(A,B,C)
@named integrator = Blocks.StateSpace( [-0.001;;], [1.0;;], [1.0;;], [0.0;;])
ins = collect(model.input.u)
disturbed_input = ins[1]
@named dist_integ = ModelingToolkit.DisturbanceModel(disturbed_input, integrator)
function add_input_disturbance_(sys, dist::DisturbanceModel, inputs)
    t = ModelingToolkit.get_iv(sys)
    @variables d(t)=0 [disturbance = true]
    @variables u(t)=0 [input = true] # Potential new system input
    dsys = ModelingToolkit.get_disturbance_system(dist)
    all_inputs = inputs
    all_inputs = Base.setindex(all_inputs, u, findfirst(isequal(dist.input), inputs)) # The input where the disturbance acts is no longer an input, the new 
    eqs = [dsys.input.u[1] ~ d
           dist.input ~ u + dsys.output.u[1]]
    augmented_sys = ODESystem(eqs, t, systems = [sys, dsys], name = gensym(:outer))

    ModelingToolkit.generate_control_function(augmented_sys, all_inputs, [d])
end

add_input_disturbance_(model, dist_integ, ins)

[YingboMa]

This doesn't have anything to do with tearing solving inputs. It's because B[2,2]==0 so we get

julia> equations(model)
4-element Vector{Equation}:
 Differential(t)((x(t))[1]) ~ 1.4127911057946012(x(t))[1] + 1.6442707924892608(x(t))[2] + (input₊u(t))[1]
 Differential(t)((x(t))[2]) ~ 0.9732693843806025(x(t))[1] + 1.8506489974903615(x(t))[2]
 (output₊u(t))[1] ~ -0.27319983060901565(x(t))[1] - 0.028715875023662633(x(t))[2]
 (output₊u(t))[2] ~ 0.46630447252950014(x(t))[1] - 0.4643074023821421(x(t))[2]

and note that (input₊u(t))[2] appear in the system. If you need to assume all states still exist, then I suggest to keep A, B, C, and D parameters making all variables structurally visible. But then, MTK could accidentally divide by 0. Or, just simply remove the check in IO.

[YingboMa]

Alternatively, you need to filter ins = collect(model.input.u) based on B