pkg load symbolic f0=20; t0=0; f1=200; t1=2; syms alpha A B sol = solve( f0==A+B*(1-exp(-alpha*t0)), f1==A+B*(1-exp(-alpha*t1)),alpha, A, B) aa=double(sol.A) bb=double(sol.B) cc=double(sol.alpha) I get and error "error: structure has no member 'alpha'" and I'm not sure why. I'm working a problem to see if I can solve it using Octave the problem I'm working is below: The general formula for the exponential rise in frequency is given by f = A + B(1 − exp(−αt)) We need to solve for the values of A , B and α . Let's assume you want an initial frequency of 20 at time t = 0 , and an asymptotic frequency of 200 Hz at time t = ∞ . Substituting t = 0 and f = 20 into the above equation we find A = 20. If we substitute t = ∞ and f = 200 we find that B = 180 . For α you really have a free choice - it controls how fast the frequency rises. If you have another control point say at time t = 2 we want the frequency to be at 95 of the final frequency i.e. f = .95 × 200 = 190 then we can substitute these into the above equation (Since we know what A , and B are). 190 = 20 + 180(1 − exp(−2α)) Solving gives α = −0.5 ln(10/180) choose a larger value of α . If you want a faster rise then you'll need to = −1.445 To find the expression for phase we simply integrate with respect to time to give φ = 2π (At + Bt +(B/α)*exp(−αt)) Thanks _______________________________________________ Help-octave mailing list [hidden email] https://lists.gnu.org/mailman/listinfo/help-octave |
On Sun, Jul 23, 2017 at 2:12 PM, RT <[hidden email]> wrote:
as you say for alpha you have a free choice so the symbolic pkg leaves it as a free choice B = (sym) 2⋅α 180⋅ℯ ──────── 2⋅α ℯ - 1 therefore you can't print this out as a double. _______________________________________________ Help-octave mailing list [hidden email] https://lists.gnu.org/mailman/listinfo/help-octave |
How does one set t = ∞ by setting "t = ∞ and f = 200 we find that B = 180" Is this possible? On Sun, Jul 23, 2017 at 2:39 PM, Doug Stewart <[hidden email]> wrote:
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