Joe Riel

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These are answers submitted by Joe Riel

The DynamicSystems package has GainMargin and PhaseMargin commands to do precisely that.

If you want globals, a simple method is

Vector([cat(v__,1..3)]):

Note that you can use multiple ranges, or a mix of ranges and other stuff in cat
 

 cat(x, 1..3, `_`, 2..4);
                                  x1_2, x1_3, x1_4, x2_2, x2_3, x2_4, x3_2, x3_3, x3_4

Ranges do not have to be numeric
 

cat(x, "a" .. "b", `_`, 2..4);
                                           xa_2, xa_3, xa_4, xb_2, xb_3, xb_4

 

Have you looked at the ImageTools package?

One solution is

type(f(g(1),3,3,4),'And(specfunc(f), patfunc(specfunc(posint,g),posint))');
                             true

 

The annotation associated with each port defines its position.  Change the extents to

extent = {{-120, 40},{-100,60}}  // for first input
extent = {{-120, -60},{-100,-40}} // second input

Each extent has the form {{x1,y1},{x2,y2}}, the points define the diagonal of a rectangle.

Later

For a slightly nicer looking block (white background rather than see-through), add the following line, say after the model line

extends Modelica.Blocks.Icons.Block;

 

Another possiblity is to use ArrayTools:-Alias to allow access to a Matrix (or any rtable) created with a different offset.  For example

M := Matrix(3):  # 3x3 matrix
A := ArrayTools:-Alias(M, [0..2,0..2]):  # alias
A[0,0] := 23:
M[1,1];
                      23;

 

Here's how it can be solved using Syrup, which is available on the Maple Cloud:
 

with(Syrup):
ckt := [V(a+b*t+c*t^2),R,C]:
(deqs,rest) := Solve(ckt,'tran','returnall'):
dsol := dsolve(deqs):
subs(rest, dsol, v[C](t));

   -(-2*C^2*R*c+2*C*c*t+C*b+exp(-1/R/C*t)*(a/R+2*C^2*R*c-C*b))*R+c*t^2+b*t+a

 

Assuming this is a discrete system, here is one approach.

with(DynamicSystems):
K := [0.1, 0.2, 0.3, 0.4]:
# Assign the z-transform from the impulse response.
T := add(K[k]/z^(k-1), k=1..4);
sys := StateSpace(T,'discrete');
# Display the a, b, c, and d matrices
use sys in a,b,c,d; end use;
# Plot the impulse response
ImpulseResponsePlot(sys);

 

Currently there is no way to create matrix i/o using the CustomComponent template.  It wouldn't be hard to extend it to allow this.  In the meantime, you can do so by writing Modelica.  As an example, here is a block with a matrix output:

model foo
    import MBI = Modelica.Blocks.Interfaces;
    output MBI.RealOutput y[2,2] annotation (Placement(transformation(extent={{100,-10},{120,10}}))); 
equation
    y[1,1] = time;
    y[1,2] = 0;
    y[2,1] = 0;
    y[2,2] = sin(time);
end foo;

Currently the MapleSim GUI doesn't expect matrix signals; as such you won't be able to probe the output.  But you can connect it to another block that expects a Matrix input and it will work.  I've attached a simple model that does this; it uses the foo model shown above and connects it to a bar model.  MatrixIO.msim

Am not sure this is the best way to go.  It might make more sense to write a function (in the Modelica Code Editor) that returns a Matrix and then use that in a block. 

This is available via the Iterator:-Permute function.

P := Iterator:-Permute([2,1,1,3]):
for p in P do printf("%d\n", p); end do:
1 1 2 3
1 1 3 2
1 2 1 3
1 2 3 1
1 3 1 2
1 3 2 1
2 1 1 3
2 1 3 1
2 3 1 1
3 1 1 2
3 1 2 1

 

Using Kitonum's suggestion to answer the second question:

expand_power := proc(ex) 
local b,p; 
    (b,p) := op(ex); 
    if    p :: posint then `%*`(b $ p); 
    elif p :: negint then 1/`%*`(b$(-p)); 
    else ex; 
    end if; 
end proc:

y := x^2/z^3:
subsindets(y, `^`, expand_power);
      x %* x
     -----------
 z %* z %* z

 

Not quite sure what you mean by white space in this context; ReadFile returns a string with newline characters. You could do, for example,

str := FileTools:-Text:-ReadFile(somefile):
lines := StringTools:-Split(str, "\n");

to create a list of strings, one for each line in the original file.

The Syrup package, which is available on the Maple Cloud, has a ladder input notation that can be used to generate images of certain types of circuits.  For the example above you could do

ckt := [ (R1(4) &+ V1(3)) &// R2(3) &// (R3(1) &+ V2(4))]: 
Syrup:-Draw(ckt);

It's not the nicest looking rendition but is fast and easy to enter.

Differentiating the sliding average returns a delay-differential equation that dsolve can handle numerically.  That is,

deq := diff(xavg(t),t) = (x(t) - x(t-T))/T

You could then do, for the first case

dsol := dsolve(subs(T=1/2), [deq, xavg(0) = 0]), 'numeric'):
plots:=odeplot(dsol, 0..3);

The plot is shifted vertically because I set xavg(0) = 0.

You can enter the system using the ZeroPoleGain form:

with(DynamicSystems):
zpk := ZeroPoleGain([1],[1,2],1):

Alas, when converting to other forms the zero/pole will get cancelled.  Note that the cancellation option is normally false; it is used to actively cancel pole/zero pairs that are within a given distance, set by the relativeerror option.  So that won't help you.  What you can try, as a workaround, is to enter an offset as a symbolic parameter and give it a default of 0:

zpk := ZeroPoleGain([1+delta],[1,2],1, parameters=[delta=0]):

The parameter values are used with certain operations such as when generating plots, etc.  Here is the State Space representation.

ss := StateSpace(zpk):
use ss in a,b,c,d end use;
   Matrix(2, 2, [[0, 1], [-2, 3]]), Matrix(2, 1, [[0], [1]]), Matrix(1, 2, [[-1 - delta, 1]]), Matrix(1, 1, [[0]])
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