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This example shows how to use a Padé approximant in control system theory to model time delays in the response of a first-order system.

Time delays arise in systems such as chemical and transport processes where there is a delay between the input and the system response. When these inputs are modeled, they are called dead-time inputs.

The Padé approximant of order `[m, n]`

approximates the function `f(x)`

around as

The Padé approximant is a rational function formed by a ratio of two power series. Because it is a rational function, it is more accurate than the Taylor series in approximating functions with poles. The Padé approximant is represented by the Symbolic Math Toolbox™ function `pade`

.

When a pole or zero exists at the expansion point , the accuracy of the Padé approximant decreases. To increase accuracy, use an alternative form of the Padé approximant which is

The `pade`

function returns the alternative form of the Padé approximant when you set the `OrderMode`

input argument to `Relative`

.

The behavior of a first-order system is described by this differential equation

Enter the differential equation in MATLAB®.

syms tau a x(t) y(t) xS(s) yS(s) H(s) tmp F = tau*diff(y)+y == a*x;

Find the Laplace transform of `F`

using `laplace`

.

F = laplace(F,t,s)

F =

Assume the response of the system at `t = 0`

is `0`

. Use `subs`

to substitute for `y(0) = 0`

.

F = subs(F,y(0),0)

F =

To collect common terms, use `simplify`

.

F = simplify(F)

F =

For readability, replace the Laplace transforms of `x(t)`

and `y(t)`

with `xS(s)`

and `yS(s)`

.

F = subs(F,[laplace(x(t),t,s) laplace(y(t),t,s)],[xS(s) yS(s)])

F =

The Laplace transform of the transfer function is `yS(s)/xS(s)`

. Divide both sides of the equation by `xS(s)`

and use subs to replace `yS(s)/xS(s)`

with `H(s)`

.

F = F/xS(s); F = subs(F,yS(s)/xS(s),H(s))

F =

Solve the equation for `H(s)`

. Substitute for `H(s)`

with a dummy variable, solve for the dummy variable using solve, and assign the solution to `Hsol(s)`

.

F = subs(F,H(s),tmp); Hsol(s) = solve(F,tmp)

Hsol(s) =

The input to the first-order system is a time-delayed step input. To represent a step input, use `heaviside`

. Delay the input by three time units. Find the Laplace transform using `laplace`

.

step = heaviside(t - 3); step = laplace(step)

step =

Find the response of the system, which is the product of the transfer function and the input.

y = Hsol(s)*step

y =

To allow plotting of the response, set parameters `a`

and `tau`

to specific values. For `a`

and `tau`

, choose values `1`

and `3`

, respectively.

y = subs(y,[a tau],[1 3]); y = ilaplace(y,s);

Find the Padé approximant of order `[2 2]`

of the step input using the Order input argument to `pade`

.

`stepPade22 = pade(step,'Order',[2 2])`

stepPade22 =

Find the response to the input by multiplying the transfer function and the Padé approximant of the input.

yPade22 = Hsol(s)*stepPade22

yPade22 =

Find the inverse Laplace transform of `yPade22`

using `ilaplace`

.

yPade22 = ilaplace(yPade22,s)

yPade22 =

To plot the response, set parameters `a`

and `tau`

to their values of `1`

and `3`

, respectively.

yPade22 = subs(yPade22,[a tau],[1 3])

yPade22 =

Plot the response of the system `y`

and the response calculated from the Padé approximant `yPade22`

.

fplot(y,[0 20]) hold on fplot(yPade22, [0 20]) grid on title 'Padé approximant for dead-time step input' legend('Response to dead-time step input', 'Padé approximant [2 2]',... 'Location', 'Best');

The `[2 2]`

Padé approximant does not represent the response well because a pole exists at the expansion point of `0`

. To increase the accuracy of `pade`

when there is a pole or zero at the expansion point, set the `OrderMode`

input argument to Relative and repeat the steps. For details, see `pade`

.

stepPade22Rel = pade(step,'Order',[2 2],'OrderMode','Relative')

stepPade22Rel =

yPade22Rel = Hsol(s)*stepPade22Rel

yPade22Rel =

yPade22Rel = ilaplace(yPade22Rel); yPade22Rel = subs(yPade22Rel,[a tau],[1 3])

yPade22Rel =

fplot(yPade22Rel, [0 20], 'DisplayName', 'Relative Padé approximant [2 2]')

You can increase the accuracy of the Padé approximant by increasing its order. Increase the order to `[4 5]`

and repeat the steps. The `[n-1 n]`

Padé approximant is better at approximating the response at `t = 0`

than the `[n n]`

Padé approximant.

`stepPade45 = pade(step,'Order',[4 5])`

stepPade45 =

yPade45 = Hsol(s)*stepPade45

yPade45 =

yPade45 = subs(yPade45,[a tau],[1 3])

yPade45 =

```
yPade45 = vpa(ilaplace(yPade45));
display('yPade45 ='); pretty(yPade45);
```

yPade45 = exp(-1.930807068546914778929595950184 t) cos(0.57815608595633583454598214328008 t) 3.2418384981662546679005910164486 - exp(-0.33333333333333333333333333333333 t) 2.7182817182817182817182817182817 - exp(-1.4025262647864185544037373831494 t) cos(1.7716120279045018112388813990878 t) 1.5235567798845363861823092981669 + exp(-1.930807068546914778929595950184 t) sin(0.57815608595633583454598214328008 t) 11.595342871672681856604670597166 - exp(-1.4025262647864185544037373831494 t) sin(1.7716120279045018112388813990878 t) 1.7803798379230333426855987436911 + 1.0

fplot(yPade45, [0 20], 'DisplayName', 'Padé approximant [4 5]')

The following points have been shown:

Padé approximants can model dead-time step inputs.

The accuracy of the Padé approximant increases with the increase in the order of the approximant.

When a pole or zero exists at the expansion point, the Padé approximant is inaccurate about the expansion point. To increase the accuracy of the approximant, set the

`OrderMode`

option to`Relative`

. You can also use increase the order of the denominator relative to the numerator.

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