# First order system ramp input

Discussion in 'Homework Help' started by kuannygohcheetatt, Oct 3, 2014.

1. ### kuannygohcheetatt Thread Starter Member

Oct 31, 2013
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Hi guys , my transfer function is 1/(s+alpha) , by varying the value of alpha, i plotted the graph below. The graph below shows the error by using different value of alpha . It is attached below

I am in doubt, based on my understanding in control engineering, ramp input should cause a constant error to first order system, why is the follownig shows all infinity error except when alpha is one, the graph is certainly not wrong, i just dun understand

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
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I'm guessing the transfer function 1/(s+alpha) is the closed loop (unity feedback) form.
This means the open loop function has the form G(s)=1/(s+a-1).
The error relationship is given by e(s)=u(s)/(1+G(s)) where u(s) is the input.
So:
$e(s)=\frac{u(s)}{1+\frac{1}{$$s+a-1$$}} =\frac{u(s) $$s+a-1$$}{$$s+a$$}$
For u(s) being a unit ramp
$u(s)=\frac{1}{s^2}$
So:
$e(s)=\frac{\frac{1}{s^2} $$s+a-1$$}{$$s+a$$}=\frac{$$s+a-1$$}{s^2 $$s+a$$}$
And the steady state error will be

$lim_{s->0} \{ s \times e(s)\}=lim_{s->0} \{ s \times \frac{$$s+a-1$$}{s^2 $$s+a$$}\}=lim_{s->0} \{\frac{$$s+a-1$$}{s $$s+a$$} \}$

A little thought will show this limit will always tend to infinity unless a=1. Only when a=1 does the numerator become a pure 's' which cancels the pure 's' in the denominator and one is left with 1/(s+1), which in the limit tends to unity as s tends to zero.

So you will hopefully also see that the open loop transfer function is a pure integrator with a=1.

Last edited: Oct 3, 2014