# Steady State Error of Control System

I have a question regarding a control loop system which is subject to disturbance, unfortunately I missed a couple of lectures on the subject and now am scratching my head as to how to do it. The loop is shown below. $$\D(s)=\frac{K}{s+10},\$$ $$\M_p(s)=\frac{M_o}{s},\$$ $$\G(s)=\frac{s+3}{s^2+4s+5},\$$ $$\I(s)=\frac{1}{s}\$$

$$\K_t\$$ isn't used for now so that can be ignored. $$\M_p\$$ is a step input disturbance.

Anyway, the question in hand is asking for the steady state error assuming the system to be stable, and I am simply struggling with getting the initial T.F. in order to get this.
It is given that I need to determine $$\\frac{\theta(s)}{M_p(s)}\$$ and furthermore am given that this is: $$\frac{(s+3)(s+10)}{s(s+10)(s^2+4s+5)+K(s+3)}$$

But I baffled as to how to get to this result, any ideas?
After this I know what to do but I just want to know how to get to this stage in the first place.

• Forward path is G(s)I(s), feedback path is D(s). Assume $\theta_{ref}=0$. To see this, re-draw the block diagram with the input signal of interest, $M_p$, on the left.
– Chu
Apr 28, 2016 at 9:31
• $\theta=G\cdot I(M_p+D\cdot (\theta_{ref}-\theta ))$, can you take it from here? Apr 28, 2016 at 9:55
• @VladimirCravero Yeah I can get that I struggle to get $\frac{\theta}{M_p}$ on its own though... Apr 28, 2016 at 10:10
• Hint 2: $\theta\cdot(1+D\cdot G\cdot I)=M_p\cdot G\cdot I$, using $\theta_{ref}=0$. If you can't get it from here you need to exercise on algebra, not control systems. Apr 28, 2016 at 10:20
• Then plug in all the other formula! Apr 28, 2016 at 10:24 You get Y = I.G(M - Y.D) which reduces to $\dfrac{Y}{M} = \dfrac{1}{\frac{1}{G.I} + D}$