2
$\begingroup$

I need help with this exercise. I 've tried many times with pen and paper but I'm stuck. I don't want a full answer but some feedback . enter image description here

Here is my progess:

$$\frac{dx}{dt} = Ax(t) + Bu(t) \Rightarrow sX(S) = AX(S) + BU(S) \Rightarrow \\\Rightarrow X(s)=(sI-A)^{-1}BU(S) \ \ \ (1)$$
Now let's note as $$G(t) = \frac{x(t)}{u(t)} $$ So : $$G(s)= (sI-A)^{-1}B $$ Then we can write: $$\frac{U(s)}{R(s)} = \frac{1}{1+G(s)k} $$
Now , we need a zero - steady state: $$\lim_{t \rightarrow \infty}u(t) = \lim_{s \rightarrow 0}sU(s) =0$$ $$sU(s) = s\frac{1}{1+G(s)k}R(s)$$ but $$R(S) = \frac{1}{s}$$ so
$$sU(s) = \frac{1}{1+G(s)k}$$
This can happen , iff : $$\lim_{s \rightarrow 0}G(s)k =\infty$$
$$\lim_{s \rightarrow 0}((sI-A)^{-1}B)k)=\infty \rightarrow \lim_{s \rightarrow 0}(\frac{adj(sI-A)}{det(sI-A)}Bk)=\infty \rightarrow \lim_{s \rightarrow 0}adj(sI-A)Bk=\infty$$
But $$(sI-A)= \begin{bmatrix} s+7 & -10\\ 1 & s \end{bmatrix} \rightarrow adj(sI-A) = \begin{bmatrix} s & 10\\ -1 & s+7 \end{bmatrix}$$ and
$$\begin{bmatrix} s & 10\\ -1 & s+7 \end{bmatrix}\begin{bmatrix} 1\\ 0 \end{bmatrix}k=\infty$$ ... And here is my struggle...

$\endgroup$
2
$\begingroup$

Let's start by obtaining the state space form of the closed-loop system (closed loop means that you plug in the equations the expression of the controller). The controller of this specific system has the following form:

$$ u = -Kx+r $$ This is a full state feedback controller with feedforward gain of $1$ (feedforward is the gain by which the input signal is being multiplied). So, the state space equations are now:

$$ \dot{x} = (A-BK)x+r $$ $$ y = Cx $$ Since the system is a classic linear one, convert it into the corresponding $\text{s-domain}$ representation aka transfer function. The transformation from state space to transfer function is:

$$ \frac{Y(s)}{R(s)} = c(sI-(A-BK))^{-1}B $$ Plugging in the values of the matrices and doing the math (I leave them for you) but leaving the gain vector as $K =\begin{bmatrix} k_1 & k_2 \end{bmatrix}$ yields the following closed loop transfer function of the system:

$$T(s) = \frac{Y(s)}{R(s)} = \frac{1}{s^2+(7+k_1)s+(10+k_2)} $$ In order for the steady state error to be $e_{ss}=0$, we impose the requirement for the DC gain of the system to be equal to $1$. This is derived from the final value theorem:

$$ \lim_{t \to \infty}{y(t)} = \lim_{s \to 0}sY(s) $$

and in order to achieve satisfactory tracking we want :

$$ sY(s) \approx sR(s) \text{ as } s\rightarrow0 \Rightarrow \left. \frac{Y(s)}{R(s)}\right\vert_{s=0} = 1 $$

The DC gain of the system is the value by which the input of the system is multiplied (and produces the output) as $t \rightarrow \infty$ or in $\text{s-domain}$ notation as $s\rightarrow 0$. Inserting the value $s=0$ into the closed loop transfer function results to:

$$\left. \frac{Y(s)}{R(s)}\right\vert_{s=0} = \frac{1}{10+k_2} = \text{DC-Gain} $$ And as stated above, we want:

$$ \text{DC-Gain}=1 \Rightarrow \frac{1}{10+k_2} = 1 \Rightarrow k_2 = -9 $$ Obviously, only the gain $k_2$ influences the steady state error of the system. You can verify that by noticing that at the denominator of the closed loop transfer function the gain $k_1$ belongs to the expression which is multiplied by $s$ and as a result at steady state ($s\rightarrow 0$) this expression will vanish.

However, you can't conclude that any value of $k_1$ will be ok because you have to consider the stability of the system. In order for a second order system to be stable the coefficients of the characteristic polynomial (denominator of closed loop transfer function) have to be positive:

$$ 7+k_1 > 0 \Rightarrow k_1 > -7 $$ $$ 10 + k_2 > 0 \Rightarrow k_2 > -10 $$ Even though, you only have one possible choice where $k_2=-9$, it is always good to check everything that's essential for your system to function properly, Stability is, if not the most, among the most important characteristics.

To sum up, the final closed loop transfer function is:

$$ T(s) = \frac{1}{s^2+10s+1} $$

Notice that if you put $s=0$ at this equation you get:

$$\left. T(s)\right\vert_{s=0} = \frac{1}{1} = 1 = \text{DC-Gain} $$ which yields $e_{ss} = 0$ and a stable system with negative poles at $p=\begin{bmatrix}-9.899 & -0.1010\end{bmatrix}^T$. The step response of the system proves them also graphically:

enter image description here

$\endgroup$
4
  • $\begingroup$ thank you so much!, I have some questions : First, did I interprete wrong the question? you wrote that our goal is to prove that $t \rightarrow \infty$ $Y=R$ which means that eventually the input will be tranfered in the output, but I thought that we mean no error in the feedbackloop, and I wrote a different condition $\endgroup$ – brucebanner Nov 21 '20 at 8:37
  • $\begingroup$ so what I am basically asking is: I interpreted the question as $lim_{t \rightarrow\infty}e(t) = lim_{t \rightarrow\infty}u(t) =0$ while you did as : $lim_{t \rightarrow\infty}e(t) = lim_{t \rightarrow\infty}(y(t) - r(t)) =0 \rightarrow lim_{t \rightarrow\infty}\frac{y(t)}{r(t)}=1$ $\endgroup$ – brucebanner Nov 21 '20 at 9:53
  • 1
    $\begingroup$ The question states: “zero steady state tracking error for a step input”. This means: $e_{ss}=y_{ss}-r=0$. Your interpretation means that the control input at steady state will be equal to $0$. I can’t think where does this come from. $\endgroup$ – Teo Protoulis Nov 21 '20 at 11:56
  • $\begingroup$ okay I got a little bit confused , now it's crystal clear $\endgroup$ – brucebanner Nov 21 '20 at 16:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.