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I would like to know how to draw the step response given a transfer function.

For example, given \$G(s)=\frac{(1-\frac{s}{3})}{{(\frac{s}{0.1}+1)(\frac{s}{100}+1)^2}}\cdot 10\$ a transfer function, I calculated the initial value with the "Initial Value Theorem": \$g(0^+)=\lim_{s\rightarrow \infty}sG(s)=0\$, but other than that, I would also like to calculate the final value and the settling time.

I've been also given the Bode diagrams (I don't know if it helps):

enter image description here

Update: I figured out there's also a "Final Value Theorem"

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Well, a step response is the result you get when a Heaviside-step function is applied to a system. Mathematically speaking, the transfer function is gien by:

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

When a Heaviside-step function is applied to its input we get:

$$\text{Y}\left(\text{s}\right)=\mathcal{H}\left(\text{s}\right)\cdot\mathcal{L}_t\left[\theta\left(t\right)\right]_{\left(\text{s}\right)}=\mathcal{H}\left(\text{s}\right)\cdot\frac{1}{\text{s}}\tag2$$

In your case, you will get:

$$\text{Y}\left(\text{s}\right)=10\cdot\frac{1-\frac{\text{s}}{3}}{\left(1+10\text{s}\right)\left(1+\frac{\text{s}}{100}\right)^2}\cdot\frac{1}{\text{s}}\tag3$$

Transforming back to the time domain, gives:

$$\text{y}\left(t\right)=\frac{10\left(2994003- 100000\exp\left(-\frac{t}{10}\right)+\exp\left(-100t\right)\left(105997+10289700t\right)\right)}{2994003}\tag4$$

Plotting gives:

enter image description here

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  • \$\begingroup\$ Thank you, I'm actually asked to draw this manually, and to calculate the final value and the settling time (which would ultimately help me drawing the graph). Is there a more practical or intuitive way? \$\endgroup\$
    – Kevin
    Apr 6 '20 at 20:06
  • \$\begingroup\$ I was missing the Heaviside-step function, thanks! \$\endgroup\$
    – Kevin
    Apr 6 '20 at 20:36

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