Control Systems-Design controller to achieve the following

I'm given this system $$G(s)=\frac{10}{s(s+1)}$$ I am asked to design the least complex controller C(s) so I can achieve the following 2 conditions at the same time :

1) steady state value( y(t) for t going to infinity) is 1 for a unit step input u(t).

2)The dominant poles of the closed loop system are between -1 and -0.5. $$\require{cancel}$$ I have many ideas but that part about the least complex controller troubles me a bit. Regarding the steady state value: $$C(s)=N_c(s)/D_c(s)\\T(s)=\frac{N_c(s)(10)}{s(s+1)D_c(s)+10N_c(s)}$$ Nc is the numerator of my controller and Dc is the denominator. Using final value theorem I will get : $$\lim_{t->\infty}y(t)=sY(s)=sT(s)2U(s)=\cancel sT(s)2\frac{1}{\cancel s}=\frac{20\cancel {N_c(s)}}{10\cancel {N_c(s)}}=2$$

This can't be 1 no matter what C(s) is. Maybe I could cancel the origin pole with a zero at the origin. That way my denominator in the limit will be left with more terms without an s.

• Doesn't say unit step input does it?
– Chu
Aug 30 '17 at 19:47
• It is unit Chu. We just don't use it in my language, my bad. I figured this one out I think. I used C(s)=sk/(s+a) and chose some acceptable values for k and a to meet the conditons. Aug 30 '17 at 19:50

Let's write in Laplace domain for the closed loop Y=2RCG/(1+CG) (sorry for missing math typesetting)

If we write the final value of y(t)=1, we must write in Laplace domain that the limit of sY is 1 when s approaches zero.

By applying this to the first formula we get CG=1. This is not the whole truth, because CG=1 must be true as the limit when we let s to approach zero. Thus we can add anything that approaches zero or multiply with anything that approaches 1 when s approaches zero.

To make it simple let's select CG=A where A is one when s is zero.

To check the second condition we write for the closed loop Y=2RA/(1+A)

If we select A=1+2s we get Y=R(1+2s)/(1+s) whis has a pole at -1. That's ok.

the controller: C = (1+2s)/G = s(1+s)(1+2s)/10

I wrote that something can be added to CG if that something is zero when s=0. A short calculation shows that it does not add new possiblities to simplify

One should note that altough C seems to compensate G's poles, that's not true in practice because in practical circuits the poles and zeros are not exact. This can kill the solution depending on the given declarations.

• Where is the mistake exactly ? By the way, I'm studying for an exam and this is from a previous exam sheet so no homework. Aug 30 '17 at 20:40
• @JohnKatsantas actually I must check my own calculations. I apologize and delete this answer soon. I'll return if I found a proper one. Aug 30 '17 at 20:51
• @JohnKatsantas I wrote a new text. Aug 30 '17 at 22:20
• That's a nice way to look at the problem. It will definitely help me with other ones as well, thanks. What I tried was a controller like C(s)=sk/(s+a). The s cancels out with the s pole. I calculated my closed loop transfer function including a and k and picked some values for them so I can meet the conditions. Is this correct? I've always 'worried' in these problems because with the proper controller we could cancel out the initial G function completely and I'm not sure we can 'manipulate' it that much. We create a completely different system. Here I just move the pole at the origin though. Aug 30 '17 at 22:32
• @JohnKatsantas actually your system leads to lower order C which can be considered the right thing. Proper a and k seem to be possible but you must solve some equations. No quarantee, but try k=0,01 and a=0,1 Aug 30 '17 at 22:59

A PD controller, just guessing. The steady state error is eliminated by the integral part of the system, then adding a P controller you get a similar response like I controller. Adding an extra D component you get PI controller like response. Where P gain adjusts the integration time, while D component adjust the proportional gain with your system I + FOD (integrator and First Order System)