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Convert implicit linear relationship to explicit input-output relation

B = imp2exp(A,yidx,uidx)

`B = imp2exp(A,yidx,uidx)`

transforms a linear constraint between variables `Y`

and `U`

of
the form `A(:,[yidx;uidx])*[Y;U] = 0`

into an explicit
input/output relationship `Y = B*U`

. The vectors `yidx`

and `uidx`

refer
to the columns (inputs) of `A`

as referenced by the
explicit relationship for `B`

.

The constraint matrix `A`

can be a ```
double,
ss, tf, zpk
```

and `frd`

object as well as
an uncertain object, including `umat`

, `uss`

and `ufrd`

.
The result `B`

will be of the same class.

Consider the constraint `4y + 7u = 0`

. Solving
for `y`

gives `y = 1.75u`

. You form
the equation using `imp2exp`

:

A = [4 7]; Yidx = 1; Uidx = 2;

and then

B = imp2exp(A,Yidx,Uidx) B = -1.7500

yields `B`

equal to `-1.75`

.

Consider two motor/generator constraints among 4 variables `[V;I;T;W]`

,
namely `[1 -1 0 -2e-3;0 -2e-3 1 0]*[V;I;T;W] = 0`

.
You can find the 2-by-2 matrix `B`

so that ```
[V;T]
= B*[W;I]
```

using `imp2exp`

.

A = [1 -1 0 -2e-3;0 -2e-3 1 0]; Yidx = [1 3]; Uidx = [4 2]; B = imp2exp(A,Yidx,Uidx) B = 0.0020 1.0000 0 0.0020

You can find the 2-by-2 matrix `C`

so that ```
[I;W]
= C*[T;V]
```

Yidx = [2 4]; Uidx = [3 1]; C = imp2exp(A,Yidx,Uidx) C = 500 0 -250000 500

Consider two uncertain motor/generator constraints among 4 variables `[V;I;T;W]`

,
namely `[1 -R 0 -K;0 -K 1 0]*[V;I;T;W] = 0`

. You
can find the uncertain 2-by-2 matrix `B`

so that ```
[V;T]
= B*[W;I]
```

.

R = ureal('R',1,'Percentage',[-10 40]); K = ureal('K',2e-3,'Percentage',[-30 30]); A = [1 -R 0 -K;0 -K 1 0]; Yidx = [1 3]; Uidx = [4 2]; B = imp2exp(A,Yidx,Uidx) UMAT: 2 Rows, 2 Columns K: real, nominal = 0.002, variability = [-30 30]%, 2 occurrences R: real, nominal = 1, variability = [-10 40]%, 1 occurrence

Consider a standard single-loop feedback connection of controller `C`

and an uncertain plant `P`

, described by the equations `e = r-y; u = Ce; f = d+u; y = Pf`

.

P = tf([1],[1 0]); C = tf([2*.707*1 1^2],[1 0]); A = [1 -1 0 0 0 -1;0 -C 1 0 0 0;0 0 -1 -1 1 0;0 0 0 0 -P 1]; OutputIndex = [6;3;2;5]; % [y;u;e;f] InputIndex = [1;4]; % [r;d] Sys = imp2exp(A,OutputIndex,InputIndex); Sys.InputName = {'r';'d'}; Sys.OutputName = {'y';'u';'e';'f'};

pole(Sys)

ans = -0.7070 + 0.7072i -0.7070 - 0.7072i -0.7070 + 0.7072i -0.7070 - 0.7072i

stepplot(Sys)

The number of rows of `A`

must equal the length
of `yidx`

.

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