Design a controller to stabilize plant $\frac{1}{s-\alpha}$, but no zeros in closed loop.

Question

I want to control an unstable plant system with transfer function $\frac{1}{s-\alpha}$ using controller in the feedforward loop and unity feedback. However, my requirement is my closed loop system has i) No zeros in the complex plane ii) Settling time Ts (2%). No pole zero cancellation allowed.

Using proportional controller with control gain $K_p$, my closed loop transfer becomes $\frac{C(s)}{R(s)}=\frac{K_p}{s+(K_p-\alpha)}$. If $K_p>\alpha$, my system becomes stable. No zeros in the closed loop transfer function and I can choose $K_p$ by equating $T_s=4*T=\frac{4}{K_p-1}$.

Apart from proportional control, is there any way I can stabilize the system and achieve my other constraints?

Answer 1

Well, when we have a transfer function that looks like:

$$\mathcal{T}:=\frac{\text{Y}\left(\text{s}\right)}{\text{X}\left(\text{s}\right)}=\frac{\text{G}\left(\text{s}\right)}{1+\text{G}\left(\text{s}\right)\text{H}\left(\text{s}\right)}\tag1$$

Where $\text{H}\left(\text{s}\right):=1$ because of the unity feedback and $\text{G}\left(\text{s}\right):=\text{K}\cdot\frac{1}{\text{s}-\alpha}$

We get a complete transfer function that looks like:

$$\mathcal{T}=\frac{\text{Y}\left(\text{s}\right)}{\text{X}\left(\text{s}\right)}=\frac{\text{K}\cdot\frac{1}{\text{s}-\alpha}}{1+\text{K}\cdot\frac{1}{\text{s}-\alpha}}\tag2$$

Now, for the poles and zeros:

• Poles: $$1+\text{K}\cdot\frac{1}{\text{s}-\alpha}=0\space\Longleftrightarrow\space\text{s}=\alpha-\text{K}\tag3$$
• Zeros: $$\text{K}\cdot\frac{1}{\text{s}-\alpha}=0\space\Longleftrightarrow\space\text{K}=0\tag4$$

For the settling-time we have that:

$$t\space_{2\text{%}}=-\frac{\ln\left(50\right)}{\lambda}\space\Longleftrightarrow\space\lambda=-\frac{\ln\left(50\right)}{t\space_{2\text{%}}}\tag5$$

We need to use:

$$\text{s}=\lambda+\omega\text{j}\tag6$$

Where $\lambda\space\wedge\space\omega\in\mathbb{R}$ and $\text{j}^2=-1$

Now, we need to solve for $\text{K}$ in the pole equation (assuming $\text{K}\ne0$):

$$1+\text{K}\cdot\frac{1}{\left(-\frac{\ln\left(50\right)}{t\space_{2\text{%}}}+\omega\text{j}\right)-\alpha}=0+0\text{j}\space\Longleftrightarrow\space\text{K}=\alpha+\frac{\ln\left(50\right)}{t\space_{2\text{%}}}\space\space\space\wedge\space\space\space\omega=0\tag7$$

Conclusion, we get a transfer function that looks like:

$$\color{red}{\mathcal{T}=\frac{\text{Y}\left(\text{s}\right)}{\text{X}\left(\text{s}\right)}=\frac{\alpha\cdot t\space_{2\text{%}}+\ln\left(50\right)}{\text{s}\cdot t\space_{2\text{%}}+\ln\left(50\right)}}\tag8$$