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In the lecture slides from my university it says that:

"The response of a stable second-order transfer function to a unit sine wave input is:"

$$Y(s)=\frac{1}{s^2+2\zeta\omega_n+\omega_n^2}*\frac{\omega}{s^2+\omega^2}$$

Isn't this missing an \$ω_n^2\$ in the numerator? Since the standard form of a second order transfer function is:

$$H(s) = \frac{\omega_n^2}{s^2+2 \zeta \omega_n s + \omega_n^2}$$

and the laplace transform of the sine wave input is:

$$\frac{\omega}{s^2+\omega^2}$$

Is this a mistake in the lecture slides?

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2 Answers 2

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You are probably right, but it depends on how you define the transfer function. The sine part is right, while as you can see your \$H(s)\$ is not adimensional, it's something like \$s^2\$, that is pretty strange for a transfer function[^seconds].

You are safe assuming that's a slide mistake. For the future keep in mind that checking the measurement units is always a great idea.

[^seconds]: the s is for seconds, not for the s variable.

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  • \$\begingroup\$ The symbol "s" stands for the complex frequency variable - not for "seconds". Units do not appear in such functions. However, the first function is not correct - there is a missing "s" in the middle term of the denominator (the second expression for H(s) looks good). More than that, the 2nd part of the first equqation looks strange to me. Both parts together are not equivalent to a second-order function (it´s 4th order). \$\endgroup\$
    – LvW
    Commented Aug 2, 2014 at 8:01
  • \$\begingroup\$ @LvW I didn't see the missing s, I bet that's an OP error, what I meant is that the first functions measurement units is s^2 while usually a transfer function is dimensionless. \$\endgroup\$ Commented Aug 2, 2014 at 8:12
  • \$\begingroup\$ Ahhh OK - I see what you mean: Because the numerator is only "1", right? \$\endgroup\$
    – LvW
    Commented Aug 2, 2014 at 8:18
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    \$\begingroup\$ Blue7 and V. Cravero: I have to correct my first comment. Only now I have realized that the given function is NOT the transfer function H(s) but the OUTPUT Y(s). In this case, the second part of the equation is the Laplace transform for the input X(s) which is a sinus signal (frequency w) switched on at t=0. Hence, the equation is correct (except the missing "s" in the middle of the denominator and the missing wn^2 in the numerator). \$\endgroup\$
    – LvW
    Commented Aug 2, 2014 at 8:36
  • \$\begingroup\$ @LvW now we have an agreement ;) \$\endgroup\$ Commented Aug 2, 2014 at 8:45
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If the DC gain (= steady state gain) of your transfer function should be unity, then yes, there should be a wn^2 in the numerator of the TF. But there's nothing essentially wrong with the TF as it stands, it could legitimately have a DC gain of 1/wn^2 (but include the missing 's' in the denominator!).

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