Thomas Dean

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14 years, 356 days

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These are replies submitted by Thomas Dean

@Joe Riel 

    |\^/|     Maple 2016 (X86 64 LINUX)
._|\|   |/|_. Copyright (c) Maplesoft, a division of Waterloo Maple Inc. 2016
 \  MAPLE  /  All rights reserved. Maple is a trademark of
 <____ ____>  Waterloo Maple Inc.
      |       Type ? for help.
> $define Time(x) restart; \                                     
syntax error, missing operator or `;`:
>                 proc() \
>                 local data := [seq(1..10^6)]; \
>                     CodeTools:-Usage(proc() to 10^4 do \
>                                                 evalb(x) \
>                                             end do; \
>                                      end proc()); \
>                 end proc()

 

Should

N:=<M|<2,0,2,0>>;

be

N:=<M | <Y,E,A,R>>;

And, the result be

YEARs := [2002, 2013, 2020, 2024, 2035, 2046, 2057, 2068, 2079]

F := [x^2+y+z-1, y^2+x+z-1, z^2+x+y-1]:

soln := solve(F):

lprint("Equations:  ",F);
for s in soln do
  if nops([allvalues(s)]) = 1 then
    lprint("substitute: ",s);
    lprint("   results: ",subs(s,F));
  else
    for t in allvalues(s) do
      lprint("substitute: ",t);
      lprint("   results: ",evala(subs(t,F)));
    end do;
  end if;
end do;

 

evalf(solve(...)) and fsolve return solutions in different orders.  One gives the first real solution and the other gives the second real solution.

Thanks,

@Joe Riel 

It looks like the call to Solve has some side effect that causes BodePlot to fail

restart;
    
with(Syrup);
Syrup:-Version;
with(DynamicSystems):

check_bode := proc()
    sys := TransferFunction( 1/(s-10) ):
    BodePlot(sys);
end proc:

################################
## This works
check_bode();

ckt := [V, Rsrc(50), C1(15e-9), L1(15e-6), C2(22e-9), L2(15e-6), C3(22e-9), L3(15e-6), C4(15e-9), 1, Rload(50)];
################################
## This works
check_bode();

(sol, other) := Solve(ckt, 'ac', 'returnall');
################################
## This fails
check_bode();

################################
## This works
restart;
with(DynamicSystems):
sys := TransferFunction( 1600000000000000000000000000000000000000000000000/(29403*s^7+78408000000*s^6+
670032000000000000*s^5+1298880000000000000000000*s^4+
4400400000000000000000000000000*s^3+5328000000000000000000000000000000000*s^2+
7360000000000000000000000000000000000000000*s+
3200000000000000000000000000000000000000000000000) ):
BodePlot(sys);

 

@Joe Riel 

Yes, I omitted some print statements.  In the cut-paste process, I omitted:

(sol, other) := Solve(ckt, 'ac', 'returnall');

I am using Maple 2018. My Syrup version is 0.1.16.

@Carl Love 

@tomleslie 

Reducing the function f to 1 reduced the problem to the volume of a sphere.

I work in Emacs, so I do a lot of cut/paste.  But, I do have a typo problem.

Thanks,

Tom Dean

@tomleslie 

Thanks for correcting my (silly) error.

Looking at my schematic, I numbered both the top ends of R5A and R5B then gave the output of U1A and U1B different numbers!

Sorry for the noise.

Thanks

@Joe Riel 

I tried this with ngspice-28.  The result was similar to using the IdealOpAmp in the above ckt.

I re-ran the analog.com filter design tool and got the same design.pdf.  I sent an email to analog.com.  No reply, yet.

@Joe Riel 

Here is the design file from analog.com Design.pdf

Look at the comments in my question.  This is a 4th order Butterworth.  Analog.com stated 10db gain and 3db down at 2kHz.  Looking at the component values, I questioned the performance.

Syrup gives very different results.

I get similar results to yours.

The netlist is from analog.com.  When I first saw the component values, I suspected the results, so I tried Syrup.  Syrup shows a much wider bandpass and 30db less gain.

@tomleslie  Thanks.  I just cut-paste and run,  Should read first.

@dharr 

limit(subs(theta2=theta1,soln[3]),theta1=0,left); ## or right

This is identical to one of the pdsolve examples, except the example uses

(D[2](u))(x, 0) = 0

as one of the initial conditions.

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