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

 

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.