Electronic – How to find the system response given the transfer function without using Laplace transforms

controllaplace transformtransfer function

The given transfer function is:

$$\frac{6p+100}{p^2+12p+100}$$

and inputted into the system is a unit step response:

$$r(t) = \left\{\begin{array} f0 & t < 0\\1 & t > 0 \end{array}\right.$$

However I'm not supposed to use Laplace transforms to solve this. I believe it has something to do with finding the poles but I'm not exactly sure how to accomplish this.

Best Answer

I assume you are permitted to perform partial fractions, even if you aren't supposed to use \$\mathscr{L}^{-1}\$. The roots of your denominator are \$p_1=-6+j\:8\$ and \$p_2=-6-j\:8\$ and the root of the unit step function, \$\frac{1}{p}\$ is \$p_3=0\$. You have:

$$\begin{align*}\frac{1}{p}\cdot \frac{6p+100}{p^2+12p+100}&=\frac{6p+100}{p\cdot\left(p-p_1\right)\cdot\left(p-p_2\right)}\\\\&=\frac{A}{p-p_1}+\frac{B}{p-p_2}+\frac{C}{p-p_3}\\\\&=\frac{-0.5}{p-p_1}+\frac{-0.5}{p-p_2}+\frac{1}{p-p_3}\end{align*}$$

We know that the proposed solutions take the form of \$V_{\left(t\right)}=A\: e^{\:p\: t}\$. So it follows that:

$$\begin{align*} V_{\left(t\right)}&= A\: e^{\:p_1\: t}+B\: e^{\:p_2\: t}+C\: e^{\:p_3\: t}\\\\&=-\frac{1}{2}\: e^{\:p_1\: t}-\frac{1}{2}\: e^{\:p_2\: t}+ e^{\:p_3\: t}\\\\&=-\frac{1}{2}\: e^{-6\: t}\: e^{8j\: t}-\frac{1}{2}\: e^{-6\: t}\: e^{-8j\: t}+ 1\\\\&=1-\frac{1}{2}\: e^{-6\: t}\cdot\left(\left[\operatorname{cos}\left(8t\right)+i\operatorname{sin}\left(8t\right)\right]+\left[\operatorname{cos}\left(-8t\right)+i\operatorname{sin}\left(-8t\right)\right]\right)\\\\&=1-\frac{1}{2}\: e^{-6\: t}\cdot\left(\operatorname{cos}\left(8t\right)+i\operatorname{sin}\left(8t\right)+\operatorname{cos}\left(8t\right)-i\operatorname{sin}\left(8t\right)\right)\\\\&=1-\frac{1}{2}\: e^{-6\: t}\cdot\left(\operatorname{cos}\left(8t\right)+\operatorname{cos}\left(8t\right)\right)\\\\&=1- e^{-6\: t}\cdot\operatorname{cos}\left(8t\right) \end{align*}$$