# Is the system stable ?

M

#### Miller

Please tell me if the system with following open loop transfer function is stable? Marginally stable? Or unstable?

<b>H(s) = 1/(s*(s+2))</b>

I think the system is marginally stable because of the pole at origin. But when I plot the step response in MATLAB I get a linear line which keeps on increasing that makes me think that system is unstable. Link for step response is given below:

http://postimg.org/image/rg9422r0t/

Any help would be appreciated.

#### PhilCorso

Miller... responding to your specific query... it is stable!

What response to you get when you plot the time-related function?

Regards,
Phil Corso

M

#### Miller

> What response to you get when you plot the time-related function?

Well, There is a pole at s= 0, does not this mean that system is marginally stable? How can you say that system is fully stable? Can you please elaborate? I shared the step response of the system earlier. Below is the link for inverse laplace transform of the transfer function:

http://postimg.org/image/verzlaq4x/

Thanks for the time.

D

#### d-

you can only consider "stability" in a traditional sense where you have feedback.

#### PhilCorso

Miller...

try the Inverse using The Partial Fraction Method!

Phil

M

#### Miller

> you can only consider "stability" in a traditional sense where you have feedback.

Inverse laplace transform using partial function is given as:

h(t) = 1/2- e^(-2*t)/2

How can I check now whether it is stable or not?

D

#### d-

"Please tell me if the system with following open loop transfer function is stable? "

at issue is that until you define the closed loop response, stability is not answerable other than the pole locations

again, you have not yet defined your system only a part of it, when you have done so then a discussion of stability is possible

pick up any text or website on controls to get started.

#### PhilCorso

Miller... by now you should have ploted the h(t) equation!

Note that the time-function meets the 'Stable' criterion, that is, the impulse-response decays to a steady-state value after a period of time!

The 'pole' sits on the negative-real axis, and its value is equal to the exponent of the exp term, i.e., "-2" !

Phil

G

#### Gerrit

If you are talking about stability, you have to know or define exactly what that means first. This should be defined in your lectures or study books. Alternatively, wikipedia is as good a place to start as any:

http://en.wikipedia.org/wiki/Control_theory#Stability

You have already noticed that you have one "non-repeated pole at complex plane origin". Therefore, using these definitions, your system is marginally stable, but not (asymptotically) stable.

Regarding the MATLAB step response you plotted: your system is marginally stable. This means roughly/intuitively that in the time domain the system output changes no faster than the input (time, for a step response), except for a constant gain (1/2 in your case). However, it does not approximate a final constant value either, which an asymptotically stable system would do.

In MATLAB, you can create the inverse Laplace transform you plotted (using Wolfram|Alpha?) by using the 'impulse' response plot function instead of 'step'.

Gerrit.

H

#### Hans H. Eder

There is no need to do any transformations etc.: H(s) = 1/(s*(s+2)) can be written as H(s) = (1/s) * (1/(s+2)) which means you have an integrator and a first order element in series. Thus you should see first a transient behavior followed by pure integration. However, the step response given at http://postimg.org/image/rg9422r0t/ does not show any transient part- why?

Because of the integrator such a system will not reach a new steady state – unless the initial change in the input is made undone.

J

#### John M. Correa

If the input is a step in time, then this input in Laplace domain is 1/s.

So the output (or response) of your system to the step input will be (in Laplace domain)

Y(s) = H(s)*U(s) = (1/s) * (1/(s*(s+2)))

The partial fractions expansion of the above is:

Y(s)= 1/(4s+2) - 1/(4s) + 1/(2s^2)

The output in time domain will be the sum of an exponential decaying term, plus a step, plus a linear term.

The third term is a function of time f(t)= K* t that increases (or decreases indefinitely with time). So the conclusion is that system H(s) subjected to bounded input (a step) will have an unbounded output (the k*t term).

PS. I have used an online partial fraction solver:
check:

http://www.quickmath.com/webMathematica3/quickmath/algebra/partialfractions/basic.jsp#c=partial_fractions_basicpartialfractions&v1=+(1/x^2)+*+(1/(x+2))

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