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Alasdair McAndrew? writes:

    I'd be grateful for a little help here!  (Then I'll see if I can use the
    z-transform to solve some difference equations.)

Ref:

On 29 May 2007 14:21:26 +0200 Martin Rubey wrote:

It seems that you hit a bug, but fortunately, there is an easy workaround. The problem is with rules of the form:

  rule ...a...b... | p(a,b) == ...

It seems that in this case, the predicate p is never tested, who knows why?

The workaround is to use the "suchThat" function.

fricas
Expr ==> Expression Integer
Type: Void
fricas
zt:=operator 'zt

\label{eq1}zt(1)
fricas
ztransrules :=  ruleset([                                            _
  suchThat(rule zt(a^n,n,z) == z/(z-a),                              _
                [a, n], l +-> freeOf?(l.1, l.2)),                    _
  suchThat(rule zt(cos(a*n),n,z) == z*(z-cos(a))/(1-2*z*cos(a)+z^2), _
                [a, n], l +-> freeOf?(l.1, l.2)),                    _
  suchThat(rule zt(sin(a*n),n,z) == z*sin(a)/(1-2*z*cos(a)+z^2),     _
                [a, n], l +-> freeOf?(l.1, l.2))                     _
])$Ruleset(Integer, Integer, Expr)

\label{eq2}\begin{array}{@{}l}
\displaystyle
\left\{{{zt \left({{{a}^{n}}, \: n , \: z}\right)}\mbox{\rm = =}{z \over{z - a}}}, \: \right.
\
\
\displaystyle
\left.{{zt \left({{\cos \left({a \  n}\right)}, \: n , \: z}\right)}\mbox{\rm = =}{{{z \ {\cos \left({a}\right)}}-{{z}^{2}}}\over{{2 \  z \ {\cos \left({a}\right)}}-{{z}^{2}}- 1}}}, \: \right.
\
\
\displaystyle
\left.{{zt \left({{\sin \left({a \  n}\right)}, \: n , \: z}\right)}\mbox{\rm = =}-{{z \ {\sin \left({a}\right)}}\over{{2 \  z \ {\cos \left({a}\right)}}-{{z}^{2}}- 1}}}\right\} 
(2)
Type: Ruleset(Integer,Integer,Expression(Integer))

We only need to use a rules to handle the case of a^n, sin and cos, but similar rules could be written for the rest or we can do the same thing by common recursive methods:

fricas
ztrans(f:Expr,n:Symbol,z:Symbol):Expr ==
  freeOf?(f,n) => f*z/(z-1)
  fs:= isPlus f; not (fs case "failed") =>
    reduce(+,map(x+->ztrans(x,n,z),fs::List Expr))
  fp:= isTimes f; not (fp case "failed") =>
    reduce(*,select(x+->freeOf?(x,n),fp::List Expr))* _
      ztrans(reduce(*,remove(x+->freeOf?(x,n),fp::List Expr)),n,z)
  fx:=isPower f; if not (fx case "failed") then
     fr:=fx::Record(val:Expr,exponent:Integer)
     k:=fr.exponent
     if fr.val=n and k>0 then
       return (-1)^k*limit(D(z/(z-exp(-x)),[x for i in 1..k]),x=0)::Expression Integer
  ztransrules zt(f,n,z)
Function declaration ztrans : (Expression(Integer),Symbol,Symbol) -> Expression(Integer) has been added to workspace.
Type: Void

On Tue, 29 May 2007 09:19:42 +1000 Alasdair McAndrew? wrote:

1) The commands:

fricas
ztrans(2+3^n,n,z)
fricas
Compiling function ztrans with type (Expression(Integer),Symbol,
      Symbol) -> Expression(Integer)

\label{eq3}{{3 \ {{z}^{2}}}-{7 \  z}}\over{{{z}^{2}}-{4 \  z}+ 3}(3)
Type: Expression(Integer)
fricas
--should return the result:
2*z/(z-1) + z/(z-3)

\label{eq4}{{3 \ {{z}^{2}}}-{7 \  z}}\over{{{z}^{2}}-{4 \  z}+ 3}(4)
Type: Fraction(Polynomial(Integer))

fricas
ztrans(0,n,z)

\label{eq5}0(5)
Type: Expression(Integer)
fricas
--should return the result:
0

\label{eq6}0(6)

fricas
ztrans((-1)^n,n,z)

\label{eq7}z \over{z + 1}(7)
Type: Expression(Integer)
fricas
--should return the result:
z/(z+1)

\label{eq8}z \over{z + 1}(8)
Type: Fraction(Polynomial(Integer))

fricas
ztrans(1,n,z)

\label{eq9}z \over{z - 1}(9)
Type: Expression(Integer)
fricas
--should return the result:
z/(z-1)

\label{eq10}z \over{z - 1}(10)
Type: Fraction(Polynomial(Integer))

fricas
ztrans(n,n,z)

\label{eq11}z \over{{{z}^{2}}-{2 \  z}+ 1}(11)
Type: Expression(Integer)
fricas
--should return the result:
z/(z-1)^2

\label{eq12}z \over{{{z}^{2}}-{2 \  z}+ 1}(12)
Type: Fraction(Polynomial(Integer))

fricas
ztrans(n^2,n,z)

\label{eq13}{{{z}^{2}}+ z}\over{{{z}^{3}}-{3 \ {{z}^{2}}}+{3 \  z}- 1}(13)
Type: Expression(Integer)
fricas
--should return the result:
z*(z+1)/(z-1)^3

\label{eq14}{{{z}^{2}}+ z}\over{{{z}^{3}}-{3 \ {{z}^{2}}}+{3 \  z}- 1}(14)
Type: Fraction(Polynomial(Integer))

fricas
ztrans(n^3,n,z)

\label{eq15}{{{z}^{3}}+{4 \ {{z}^{2}}}+ z}\over{{{z}^{4}}-{4 \ {{z}^{3}}}+{6 \ {{z}^{2}}}-{4 \  z}+ 1}(15)
Type: Expression(Integer)
fricas
--should return the result:
z*(z^2+4*z+1)/(z-1)^4

\label{eq16}{{{z}^{3}}+{4 \ {{z}^{2}}}+ z}\over{{{z}^{4}}-{4 \ {{z}^{3}}}+{6 \ {{z}^{2}}}-{4 \  z}+ 1}(16)
Type: Fraction(Polynomial(Integer))

fricas
ztrans(b^n,n,z)

\label{eq17}z \over{z - b}(17)
Type: Expression(Integer)
fricas
--should return the result:
z/(z-b)

\label{eq18}z \over{z - b}(18)
Type: Fraction(Polynomial(Integer))

fricas
ztrans(cos(b*n),n,z)

\label{eq19}{{z \ {\cos \left({b}\right)}}-{{z}^{2}}}\over{{2 \  z \ {\cos \left({b}\right)}}-{{z}^{2}}- 1}(19)
Type: Expression(Integer)
fricas
--should return the result:
z*(z-cos(b))/(1-2*z*cos(b)+z^2)

\label{eq20}{{z \ {\cos \left({b}\right)}}-{{z}^{2}}}\over{{2 \  z \ {\cos \left({b}\right)}}-{{z}^{2}}- 1}(20)
Type: Expression(Integer)

fricas
ztrans(sin(b*n),n,z)

\label{eq21}-{{z \ {\sin \left({b}\right)}}\over{{2 \  z \ {\cos \left({b}\right)}}-{{z}^{2}}- 1}}(21)
Type: Expression(Integer)
fricas
--should return the result:
z*sin(b)/(1-2*z*cos(b)+z^2)

\label{eq22}-{{z \ {\sin \left({b}\right)}}\over{{2 \  z \ {\cos \left({b}\right)}}-{{z}^{2}}- 1}}(22)
Type: Expression(Integer)

2) How do I force answers to be returned in factored form?

fricas
( x+->factor(numer(x))/factor(denom(x)) ) ztrans(2+3^n,n,z)

\label{eq23}{z \ {\left({3 \  z}- 7 \right)}}\over{{\left(z - 3 \right)}\ {\left(z - 1 \right)}}(23)
Type: Fraction(Factored(SparseMultivariatePolynomial?(Integer,Kernel(Expression(Integer)))))

Bill Page wrote:

Reduce does z-transforms

http://www.uni-koeln.de/REDUCE/3.6/doc/ztrans/ztrans.html

load_package ztrans;
*** binomial already defined as operator
*** ~f already defined as operator
reduce

ztrans((-1)^n*n^2,n,z);
reduce
\displaylines{\qdd
\frac{z\cdot 
      \(-z
        +1
       
ztrans(cos(n*omega*t),n,z);
reduce
\displaylines{\qdd
\frac{z\cdot 
      \(\cos 
        \(\omega \cdot t
         
ztrans(cos(b*(n+2))/(n+2),n,z);
reduce
\displaylines{\qdd
z\cdot 
\(-\cos 
  \(b
   
ztrans(n*cos(b*n)/factorial(n),n,z);
reduce
\displaylines{\qdd
\frac{e^{\frac{\cos 
               \(b
                
ztrans(sum(1/factorial(k),k,0,n),n,z);
reduce
\displaylines{\qdd
\frac{e^{\frac{1}{
               z}}\cdot z}{
      z
      -1}
\cr}
 
operator f$
ztrans((1+n)^2*f(n),n,z);
reduce
\displaylines{\qdd
\frac{\partial ^{2}ztrans
      \(f
        \(n
         

Inverse z-transforms

invztrans((z^2-2*z)/(z^2-4*z+1),z,n);
reduce
\displaylines{\qdd
\frac{\(-
        \sqrt{3}
        +2
       
invztrans(z/((z-a)*(z-b)),z,n);
reduce
\displaylines{\qdd
\frac{a^{n}
      -b^{n}}{
      a
      -b}
\cr}
 
invztrans(z/((z-a)*(z-b)*(z-c)),z,n);
reduce
\displaylines{\qdd
\frac{a^{n}\cdot b
      -a^{n}\cdot c
      -b^{n}\cdot a
      +b^{n}\cdot c
      +c^{n}\cdot a
      -c^{n}\cdot b}{
      a^{2}\cdot b
      -a^{2}\cdot c
      -a\cdot b^{2}
      +a\cdot c^{2}
      +b^{2}\cdot c
      -b\cdot c^{2}}
\cr}
 
invztrans(z*log(z/(z-a)),z,n);
reduce
\displaylines{\qdd
\frac{a^{n}\cdot 
      \(-1
       
invztrans(e^(1/(a*z)),z,n);
reduce
\displaylines{\qdd
\frac{1}{
      a^{n}\cdot factorial
      \(n
       
invztrans(z*(z-cosh(a))/(z^2-2*z*cosh(a)+1),z,n);
reduce
\displaylines{\qdd
cosh
\(a\cdot n
 




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