First order system ramp input

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

  1. kuannygohcheetatt

    Thread Starter Member

    Oct 31, 2013
    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
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    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.
    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
    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