MaplePrimes Questions

Dear.All

 

I'm a beginner of Maple 13. 

 

In MATLAB, there is a command 'sigmaplot' which draw the 'SINGULAR VALUE OF SYSTEM H(jw)' over all frequency range. 

I want to obtain the function of the 'sigmaplot' graph about frequency variable 'w'.

 

so I defined matrix A,B,C,D in Maple13. 

 

and specify H like following

 

H:= Multiply(C,Multiply(Inversematrix(s*IdentityMatrix(8)-A),B))+D              

 

          --> It is to express H=C*((sI-A)^-1)*B+D that is the state-space matrix of frequency domain form.

 

and Maple 13 gave me a very long formular expressed by 's'.

 

Then I try to 'SingularValues' command 

 

but there came up 'Error Message : Error. (in content/polynom) general case of floats not handled'

 

I couldn't resolve this problem.... 

 

So I'm requesting your advice like this.

 

Could you give some advice ? 


Dear users,

In my attached file I have two PDES, (PDE1 and PDE2). PDE1 is a function of v(t) and w(x,t) and PDE2 is also a function of v(t) and w(x,t). I can solve PDE2 if I say v(t) is 1 for example and you can see the plot. But what if I put v(t) back in PDE2 and want to find v(t) and w(x,t) from PDE1 and PDE2 together? 

Many Thanks,

Baharm31

 

Define PDE Euler-Bernoulli Beam

 

NULL

restart:

Parametrs of piezoelectric and cantilever beam

 

``

Ys := 70*10^9: # Young's Modulus structure

Yp := 11.1*10^10: # Young's Modulus pieazo

ha := -0.00125: # Position

hb := 0.001: # Position

hc := 0.0015: # Position

d31 := -180*10^(-12): # Piezoelectric constant

b := 0.01: #Width of the beam

tb := 0.002:

epsilon33 := 15.92*10^(-9):

hp :=0.00025: # Position

hpc := 0.00125: # Position

YI := b*(Ys*(hb^3- ha^3)+Yp*(hc^3-hb^3))/3: # Bending stiffness of the composit cross section

cs := 0.564: # The equivqlent coefficient of strain rate damping

ca := 0: # Viscous air damping coefficient

Ibeam := (b * tb^3 )/12: # The equivalent moment of inertia

m := 0.101: # Mass of the structure

upsilon := - Yp*d31*b*(hc^2-hb^2)/(2*hp): # Coupling term

lb := 0.57:# Length of the structure (Cantilever Beam)

lp := 0.05:# Length of the Piezoelectric

R:= 10000: # Shunted resistor

Electrical circuit equation

 

PDE1:=(epsilon33 * b*lp / hp) * diff(v(t), t) + (v(t)/R)+ int(d31*Yp*hpc*b* diff(w(x, t),$(x, 2))*diff(w(x, t), t),x = 0..lp)=0;

0.3184000000e-7*(diff(v(t), t))+(1/10000)*v(t)+int(-0.2497500000e-3*(diff(diff(w(x, t), x), x))*(diff(w(x, t), t)), x = 0 .. 0.5e-1) = 0

(1.1.1.1)

``

 

PDE Equation

 

fn := 3.8:# Direct Excitation frequency;

wb(x,t) := 0.01*sin(fn*2*Pi*t):#Direct Excitation;

plot(wb(x,t),t = 0 .. 0.25*Pi,labels = [t,wb], labeldirections = ["horizontal", "vertical"], labelfont = ["HELVETICA", 15], linestyle = [longdash], axesfont = ["HELVETICA", "ROMAN", 10], legendstyle = [font = ["HELVETICA", 10], location = right],color = black);

 

 

FunctionAdvisor(definition, Dirac(n,x));

[Dirac(n, x) = (1/2)*(Int((I*_k1)^n*exp(I*_k1*x), _k1 = -infinity .. infinity))/Pi, `with no restrictions on `(n, x)]

(1.2.1)

 

PDE2 := YI*diff(w(x, t),$(x, 4))+ cs*Ibeam*diff(w(x, t),$(x, 4))*diff(w(x, t), t)+ ca* diff(w(x, t), t) + m * diff(w(x, t),$(t, 2))+ upsilon*v(t)*(Dirac(1,x) -Dirac(1,x-lp) ) =-m*diff(wb(x, t),$(t, 2))-ca*diff(wb(x, t), t);#PDE

1.567812500*(diff(diff(diff(diff(w(x, t), x), x), x), x))+0.3760000000e-11*(diff(diff(diff(diff(w(x, t), x), x), x), x))*(diff(w(x, t), t))+.101*(diff(diff(w(x, t), t), t))+0.4995000000e-3*Dirac(1, x)-0.4995000000e-3*Dirac(1, x-0.5e-1) = 0.583376e-1*sin(7.6*Pi*t)*Pi^2

(1.2.2)

tmax := 0.3:

xmin := 0:

xmax := lb:

N := 20:#NUMBER OF NODE POINT

bc1 := dw(xmin, t) = 0:

bc2 := dw(xmax, t) = 0:

bc3 := w(xmin, t) = 0:

ic1 := wl(x, 0) = 0:

Maple's pdsolve command

 

 

 

bcs := { w(x,0)=0 , D[2](w)(x,0)=0 , w(0, t) = rhs(bc1), D[1](w)(0, t)= rhs(bc1), D[1,1](w)(lb,t) = rhs(bc2), D[1,1,1](w)(lb,t) = rhs(bc2)}; # Boundary conditions for PDE2.

{w(0, t) = 0, w(x, 0) = 0, (D[1](w))(0, t) = 0, (D[2](w))(x, 0) = 0, (D[1, 1](w))(.57, t) = 0, (D[1, 1, 1](w))(.57, t) = 0}

(2.1)

PDES := pdsolve(PDE2, bcs, numeric, time = t, range = 0 .. xmax, indepvars = [x, t], spacestep = (1/1000)*xmax, timestep = (1/1000)*tmax);

 

module () local INFO; export plot, plot3d, animate, value, settings; option `Copyright (c) 2001 by Waterloo Maple Inc. All rights reserved.`; end module

(2.2)

PDES:-plot3d(t = 0 .. tmax, x = 0 .. xmax, axes = boxed, orientation = [-120, 40], shading = zhue, transparency = 0.3);

 

 

NULL


Download Euler-Bernoulli_Beam-last_version.mw

 

I wnat to print the polar that contain a part of loops,but it always contains all loops.

how do I solve this preblem?

Any one can help me,please?

 

 

 

 

Dear Maple users

 

I have a question about applying pdsolve MAPLE for solving two dimensional heat equations:

My codes have been provided but it shows to me this error:

Error, (in pdsolve/numeric/process_PDEs) can only numerically solve PDE with two independent variables, got {t, x, y}

If kindly is possible, please help me in this case.

 

With kind regards,

Emran Tohidi.

 

> restart;
> with(plots);
print(??); # input placeholder
> with(PDEtools);
print(??); # input placeholder
> declare(u(x, y, t));
print(`output redirected...`); # input placeholder
                    u(x, y, t) will now be displayed as u
> S := 1/100; tR := 0 .. 1; xR := 0 .. 1; yR := 0 .. 1; NF := 30; NP := 100;
print(??); # input placeholder
> N := 3; L1 := [red, blue, green]; L2 := [0, 1/2, 1]; Ops := spacestep = S, timestep = S;
print(??); # input placeholder
> Op1 := frames = NF, numpoints = NP;
print(??); # input placeholder
> PDE1 := diff(u(x, y, t), t)-(diff(u(x, y, t), `$`(x, 2)))-(diff(u(x, y, t), `$`(y, 2))) = 0;
print(??); # input placeholder
> IC := {u(x, y, 0) = exp(x+y)}; BC := {u(0, y, t) = exp(2*t+y), u(1, y, t) = exp(2*t+y+1), u(x, 0, t) = exp(2*t+x), u(x, 1, t) = exp(2*t+x+1)};
print(??); # input placeholder
> Sol := pdsolve(PDE1, `union`(IC, BC), numeric, u(x, t), Ops);
Error, (in pdsolve/numeric/process_PDEs) can only numerically solve PDE with two independent variables, got {t, x, y}


A linear system with three variables has augmented matrix that is row-equivalent to the 
following matrix: 

k + 3 2 k − 4 = 3 
0 2 −9 = 5 
0 0 k^2 + k − 2 = k − 1 
Determine the values of k for which the system has: 
(a) exactly one solution, 
(b) infinitely many solutions, 
(c) no solutions.

(a) Write down the augmented matrix corresponding to the linear system: 
4x1 + 2x2 − 7x3 − 11x4 = 5 
2x1 + 1x2 − 2x3 − 4x4 = 2 
4x1 + 2x2 − 10x3 − 14x4 = 6 

(b) Use Gauss-Jordan elimination to reduce this matrix to reduced row-echelon form. 


(c) Solve the linear system.

I am having problems with the curly brackets in math mode. I am using a Danish keyboard, and since changing to Maple 17, the inline evaluation (usually [CTRL]+=) has been placed as [CTRL]+[ALT]+0 which - by Maple - is interpreted the same way as [ALT-GR]+0. Unfortunately [ALT-GR]+0 is the way to get the end-bracket for curly brackets, i.e. }

Therefore, I am not able to write } in math mode. Maple interprets my keystrokes as a wish to do inline evaluation when trying to write }.

Has anyone experienced this issue - and if so: Is there a way to fix it?

Is there a way to change the shortcut keys in Maple (e.g. make an alternative shortcut for inline evaluations)?

 

Hi.

I am new in Maple and I'm trying to get functions from system of equations.

Constants are defined in line 4 and equations are:

eq1 := E2 = fE2(1+(KaE2+Ca)/(1+KaE2*fE2+KaT*fT+KaDHT*fDHT)+KsE2*Cshbg/(1+KsE2*fE2+KsT*fT+KsDHT*fDHT))

eq2 := T = fT(1+KaT*Ca/(1+KaE2*fE2+KaT*fT+KaDHT*fDHT)+KsT*Cshbg/(1+KsE2*fE2+KsT*fT+KsDHT*fDHT))

eq3 := DHT = fDHT(1+KaDHT*Ca/(1+KaE2*fE2+KaT*fT+KaDHT*fDHT)+KsDHT*Cshbg/(1+KsE2*fE2+KsT*fT+KsDHT*fDHT))

KsT = 0.10e11; KaT = 4.6*0.10e6; KsE2 = 3.14*0.10e10; KaE2 = 4.21*0.10e6; KsDHT = 3*0.10e6; KaDHT = 3.5*0.10e6;

fT, fE2 and fDHT are variables, not functions (i.e. fT is not f(T) ) and I am trying to get fT=f(E2,T,DHT,Ca,Cshbg), fE2=f(E2,T,DHT,Ca,Cshbg) and fDHT=f(E2,T,DHT,Ca,Cshbg).

When I type:

eliminate({eq1, eq2, eq3}, {fE2, fT, fDHT})

Maple gives me a blank field. No error, no other comment.

I have no idea where I'm making mistakes.

Any suggestion is appreciated.

 

Thanks in advance.

Could you tell me what the maple player is ?

How can I calculate the residue of a function with complex variable in maple ?

Ok so i have a matrix M filled with complex roots of unity. I want to execute the following code but it seems to be failing to do so correctly

t:=exp(2 Pi I/11);

m := (i, j) -> M[(i mod 11)+1, (j mod 11)+1] ;  

mu :=(i,j)->(add(add(add(a[k]*a[m]*a[n]*t^(m)*m((i+k-m),(j+n-m))),n=0..10),m=0..10),k=0..10));

 

Something is wrong with my indexing of the matrix . or when the mod procs.

restart:with(plots):with(PDEtools):

 pde:=1/r*diff(r*diff(U(r,z),r),r)+1/b^2*diff(U(r,z),z,z)=0;

ans := pdsolve(pde);

ics:=(D[1](U))(0, z) = 0,(D[2](U))(r, 0)-B*U(r,0) = 0;

bcs:=(D[2](U))(r, 1)+B*U(r,1) = B,(D[1](U))(1, z)+B*U(1,z) = 0;

B:=1:b:=1:

S:= pdsolve(pde, {bcs, ics}, numeric);

Error, (in pdsolve/numeric) unable to handle elliptic PDEs

anyway around this?

 

RIZPDE.mw

I have a matrix with repeated columns

with(LinearAlgebra):

Q := Matrix([1,1,1,0,0,0,-2,-1],[0,0,0,1,1,1,0,-3]);

I'd like to write a loop that defines u[1],...,u[n] to be the unique columns of Q. Is there a way to do this?

I'm writing some code that computes a positive integer N and I'd like Maple to define a function of N variables. I've tried writing 

f := (seq(t[i],i=1..N))-> MY EXPRESSION;

but I receive the error

Error, invalid parameter; functional operators require their parameters to be of type symbol or (symbol::type)

Is there anything I can do here?

123.mw

i make two component that the left named MathContainer0 and right named MathContainer1

which can enter any numers on it

and i want to solve the eqution such as  

x*%MathContainer0+y*%MathContainer1=0 ,and y*%MathContainer0+x*%MathContainer1=0

but it still not shows the answers i want

anyone can help thx

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