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I have the following diagram of a system's step response:
I'm having trouble understanding how to calculate the system's transfer function, given this diagram. Specifically, I don't understand how exactly I can calculate the natural frequency and damping ratio.
Nothing I've read on this has helped me get a clear picture of what I should do. Can anyone help me understand step-by-step how to think about this problem?
Write the TF as \$ \small G(s)=\large\frac{ab}{(s+a)(s+b)}\$, where \$a\$ and \$b\$ are real and distinct.
This factorises to: \$\small G(s)=\large\frac{ab}{(b-a)} \large(\frac{1}{s+a}-\frac{1}{s+b}\large)\$
From which the inverse LT, i.e. the unit impulse response, is:
\$g(t)= \large\frac{ab}{(b-a)} \normalsize (e^{-at}-e^{-bt})\$
Differentiating and setting to zero determines the time to peak value, \$t_P\$:
\$t_P =\large\frac{ln(\frac{b}{a})}{(b-a)}\$, with corresponding peak value: \$g(t_P)=be^{-b t_P}\$
Hence, measuring \$t_P\$ and \$g(t_P)\$ will yield the value of \$b\$; and substituting back into the expression for \$t_P\$ will yield \$a\$
Note that the unit impulse response can be obtained by differentiating the unit step response, but the differentiator must be carefully designed to avoid introducing excessive noise.
Typical overdamped step and impulse responses are shown below.
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:
$$\begin{align*} V_{\left(t\right)}&= A\: e^{\:p_1\: t}+B\: e^{\:p_2\: t}+C\: e^{\:p_3\: t}\\\\&=-\frac{1}{2}\: e^{\:p_1\: t}-\frac{1}{2}\: e^{\:p_2\: t}+ e^{\:p_3\: t}\\\\&=-\frac{1}{2}\: e^{-6\: t}\: e^{8j\: t}-\frac{1}{2}\: e^{-6\: t}\: e^{-8j\: t}+ 1\\\\&=1-\frac{1}{2}\: e^{-6\: t}\cdot\left(\left[\operatorname{cos}\left(8t\right)+i\operatorname{sin}\left(8t\right)\right]+\left[\operatorname{cos}\left(-8t\right)+i\operatorname{sin}\left(-8t\right)\right]\right)\\\\&=1-\frac{1}{2}\: e^{-6\: t}\cdot\left(\operatorname{cos}\left(8t\right)+i\operatorname{sin}\left(8t\right)+\operatorname{cos}\left(8t\right)-i\operatorname{sin}\left(8t\right)\right)\\\\&=1-\frac{1}{2}\: e^{-6\: t}\cdot\left(\operatorname{cos}\left(8t\right)+\operatorname{cos}\left(8t\right)\right)\\\\&=1- e^{-6\: t}\cdot\operatorname{cos}\left(8t\right) \end{align*}$$
Best Answer
From the step response plot, the peak overshoot, defined as
$$M_p = \frac{y_{peak}-y_{steady-state}}{y_{steady-state}}\approx\frac{1.25-0.92}{0.92}=0.3587$$ Also, the relationship between \$M_p\$ and damping ratio \$\zeta\$ (\$0\leq\zeta<1\$) is given by:
$$M_p=e^\frac{-\pi\zeta}{\sqrt{1-\zeta^2}}$$
Or, in terms of \$\zeta\$:
$$\zeta=\sqrt{\frac{\ln^2M_p}{\ln^2M_p+\pi^2}}$$ So, replacing that estimated \$M_p\$: $$\zeta\approx0.31$$ Also, from the step response plot, the damped natural frequency is aprox. 0.5 Hz or \$\pi\$ rad/s. The relationship with the undamped natural frequency is: $$\omega_n=\frac{\omega_d}{\sqrt{1-\zeta^2}}\approx3.3 rad/s$$ Finally, the gain \$G_{DC}=y_{steady-state}\approx0.92\$
A standard second order transfer function has the form: $$H(s)=G_{DC}\frac{\omega_n^2}{s^2+2\zeta\omega_ns+\omega_n^2}$$ Putting the obtained values: $$H(s)\approx\frac{10}{s^2+2s+11}$$ Compare the step response below with that supplied by you: