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
(1) -> Expr ==> Expression Integer
Type: Void
fricas
zt:=operator 'zt
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\{{= = \left({{zt \left({{{a}^{n}}, \: n , \: z}\right)}, \:{\frac{z}{z - a}}}\right)}, \: \right. \ \ \displaystyle \left.{= = \left({{zt \left({{\cos \left({a \ n}\right)}, \: n , \: z}\right)}, \:{\frac{{z \ {\cos \left({a}\right)}}-{{z}^{2}}}{{2 \ z \ {\cos \left({a}\right)}}-{{z}^{2}}- 1}}}\right)}, \right. \ \ \displaystyle \left.\: \right. \ \ \displaystyle \left.{= = \left({{zt \left({{\sin \left({a \ n}\right)}, \: n , \: z}\right)}, \: -{\frac{z \ {\sin \left({a}\right)}}{{2 \ z \ {\cos \left({a}\right)}}-{{z}^{2}}- 1}}}\right)}\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 2007年5月29日 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}\frac{{3 \ {{z}^{2}}}-{7 \ z}}{{{z}^{2}}-{4 \ z}+ 3}
(3) Type: Expression(Integer)
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--should return the result:
2*z/(z-1) + z/(z-3)
\label{eq4}\frac{{3 \ {{z}^{2}}}-{7 \ z}}{{{z}^{2}}-{4 \ z}+ 3}
(4) Type: Fraction(Polynomial(Integer))
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ztrans(0,n,z)
Type: Expression(Integer)
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--should return the result:
0
Type: NonNegativeInteger
?
fricas
ztrans((-1)^n,n,z)
\label{eq7}\frac{z}{z + 1}
(7) Type: Expression(Integer)
fricas
--should return the result:
z/(z+1)
\label{eq8}\frac{z}{z + 1}
(8) Type: Fraction(Polynomial(Integer))
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ztrans(1,n,z)
\label{eq9}\frac{z}{z - 1}
(9) Type: Expression(Integer)
fricas
--should return the result:
z/(z-1)
\label{eq10}\frac{z}{z - 1}
(10) Type: Fraction(Polynomial(Integer))
fricas
ztrans(n,n,z)
\label{eq11}\frac{z}{{{z}^{2}}-{2 \ z}+ 1}
(11) Type: Expression(Integer)
fricas
--should return the result:
z/(z-1)^2
\label{eq12}\frac{z}{{{z}^{2}}-{2 \ z}+ 1}
(12) Type: Fraction(Polynomial(Integer))
fricas
ztrans(n^2,n,z)
\label{eq13}\frac{{{z}^{2}}+ z}{{{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}\frac{{{z}^{2}}+ z}{{{z}^{3}}-{3 \ {{z}^{2}}}+{3 \ z}- 1}
(14) Type: Fraction(Polynomial(Integer))
fricas
ztrans(n^3,n,z)
\label{eq15}\frac{{{z}^{3}}+{4 \ {{z}^{2}}}+ z}{{{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}\frac{{{z}^{3}}+{4 \ {{z}^{2}}}+ z}{{{z}^{4}}-{4 \ {{z}^{3}}}+{6 \ {{z}^{2}}}-{4 \ z}+ 1}
(16) Type: Fraction(Polynomial(Integer))
fricas
ztrans(b^n,n,z)
\label{eq17}\frac{z}{z - b}
(17) Type: Expression(Integer)
fricas
--should return the result:
z/(z-b)
\label{eq18}\frac{z}{z - b}
(18) Type: Fraction(Polynomial(Integer))
fricas
ztrans(cos(b*n),n,z)
\label{eq19}\frac{{z \ {\cos \left({b}\right)}}-{{z}^{2}}}{{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}\frac{{z \ {\cos \left({b}\right)}}-{{z}^{2}}}{{2 \ z \ {\cos \left({b}\right)}}-{{z}^{2}}- 1}
(20) Type: Expression(Integer)
fricas
ztrans(sin(b*n),n,z)
\label{eq21}-{\frac{z \ {\sin \left({b}\right)}}{{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}-{\frac{z \ {\sin \left({b}\right)}}{{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}\frac{z \ {\left({3 \ z}- 7 \right)}}{{\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