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## Re: plotting transfer function in octave 5.2: How to fix error: set: "da

 From: Doug Stewart Subject: Re: plotting transfer function in octave 5.2: How to fix error: set: "dataaspectratio' must be finite Date: Fri, 31 Jul 2020 16:59:29 -0400

On Fri, Jul 31, 2020 at 4:49 PM shall689 <sahallacy@hotmail.com> wrote:
Hello Doug,

Won't s^8 counteract that small value.  What if the frequency is 500Hz, then
w=2pi*500= 1000pi and (1000pi)^8*6.25e-18 = 59303318850.

I would have to solve the denominator equation to determine which ones to
exclude?

Stephen

Here is what I tried. I am not saying this is the best, but it is a starting point.

I do not know what you expect so just use it as a starting point.

s=tf('s')
n=4.63e-12*s^7+1.346e-07*s^6-0.02167*s^5-632.2*s^4+1.04e8*s^3+3.035e12*s^2+1.284e16*s+3.225e20
d=6.25e-5*s^8+6.136e-12*s^7+2.875e-7*s^6-0.01399*s^5-448.9*s^4+1.057e8*s^3+3.068e12*s^2+1.293e16*s+3.241e20
#Note ^^^^^^^^ I changed this number.
sys =n/d
#step(sys)

[nn,dd]=tfdata(sys,'v')

rn=roots(nn)
n1=rn(5:7)
n2=poly(n1)

rd=roots(dd)
[sr,idx]=sort(real(rd))
idx1=idx(1:4)
d1=rd(idx1)

d2=poly(d1)

sys2=tf(n2,d2)
step(sys2)

--
DAS 