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Steady State Error Calculation to Input Step and Ramp Input

by spggodd
Tags: calculation, error, input, ramp, state, steady, step
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spggodd
#1
Mar8-13, 02:49 PM
P: 33
Hi all,

I am getting confused about how to calculate steady state error in a system.

My particular transfer function is:

G(s)= 4.992/(s^2+3s-1)

Firstly, with an input step am I right in saying that the Steady State error will always be zero?
If so can someone explain the reason behind this, is it because it has negative real parts?

Also, what happens when I apply a ramp input?
An example would be great if possible.

Thanks in advance
Steve
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milesyoung
#2
Mar9-13, 04:04 PM
P: 526
Since G(s) is unstable in open loop (it has poles in the right half-plane), I'm going to assume we're talking about using it in a feedback configuration.

Let the system error, e(t), be given as:
e(t) = r(t) - c(t)

where r(t) and c(t) are the system input and output, respectively.

For a unity feedback system, the Laplace transform of e(t), E(s), is then given as:
[tex]
E(s) = \frac{1}{1 + G(s)} R(s)
[/tex]

The system steady-state error, e_ss, is then given by the final value theorem as:
[tex]
e_{ss} = \lim_{s \rightarrow 0} s \frac{1}{1 + G(s)} R(s)
[/tex]

For a step input, R(s) = 1/s, we have:
[tex]
e_{ss} = \lim_{s \rightarrow 0} s \frac{1}{1 + \frac{4.992}{s^2 + 3s - 1}} \frac{1}{s} = \frac{1}{1 - 4.992} \approx -0.2505
[/tex]

For a ramp input, R(s) = 1/s^2, we have:
[tex]
e_{ss} = \lim_{s \rightarrow 0} s \frac{1}{1 + \frac{4.992}{s^2 + 3s - 1}} \frac{1}{s^2} = \lim_{s \rightarrow 0} \frac{1}{s + \frac{4.992s}{s^2 + 3s - 1}} = \infty
[/tex]

The steady-state error for a step input will thus be a constant and for a ramp input it will be unbounded. This is what you would expect for a type 0 system (no free integrators), if that makes sense to you.


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