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Desinging a very basic 1st order integrator with the continous transfer function:

$H(s) = -1 / (0.001 s)$ [No zeros, one pole at origin]

I was expecting the unit step response to be positive-slope. However in my Python code I get a negative slope.

num = (0 , -1)
den = (0.001 , 0)
Hs = sig.lti(num,den)
t, s = sig.step(Hs)
plt.plot(t,s)

Provides:

A negative slope of Output Vs. time

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    $\begingroup$ $H(s)$ has a negative sign, that's why. $\endgroup$ – Matt L. Dec 4 '18 at 12:40
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    $\begingroup$ That's the sign for more coffee needed. $\endgroup$ – Malcolm Rest Dec 4 '18 at 12:49
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    $\begingroup$ I'll drink to that! :-) $\endgroup$ – Peter K. Dec 4 '18 at 13:01
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Your integrator has a negative sign:

$$H(s)=-\frac{1000}{s}\tag{1}$$

That's why its step response also has a negative sign:

$$y(t)=-1000\int_{-\infty}^tu(\tau)d\tau=-1000\cdot t\cdot u(t)\tag{2}$$

where $u(t)$ is the unit step.

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  • $\begingroup$ Man! Do we really need to post answer for a single negative sign ;-))) $\endgroup$ – Fat32 Dec 4 '18 at 18:25
  • $\begingroup$ @Fat32: One yes, otherwise it keeps popping up as unanswered, two or three, probably not :) $\endgroup$ – Matt L. Dec 4 '18 at 20:37
  • $\begingroup$ that's a reason i bitched on the meta about SE always regurgitating unanswered questions. sometimes they're unanswered because the question is useless. hopefully the OP will check-mark Matt's answer and then SE will be satisfied that it's answered and we all can move on. $\endgroup$ – robert bristow-johnson Dec 4 '18 at 22:44
  • $\begingroup$ @robertbristow-johnson: Yes, what we can do about it is suggesting to close bad questions, and answer all others. $\endgroup$ – Matt L. Dec 5 '18 at 8:27

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