Units of constants in transfer functions?

[Excom]

Hi All

Probably a very basic question.

What are the units of the constants in transfer functions?

It we take a look at the transfer function of a second order system we then have:

H(s) = ω₀²/(s²+2ζω₀s+ω₀²)

ω₀ is the natural resonance frequency and has a unit of rad/sec. ζ is the damping and has no unit. Hence everything ends up being without an unit if s also has the unit rad/sec. So for so good.

If we now look at the transfer function for a first order response:

H(s) = a/(s+a)

where a = 1/τ, where τ is the time constant. The unit of τ is sec. Assuming again that the unit of s is rad/sec, a/(s+a) is not without a unit.

When looking at more complex transfer functions the units become more confusion to me.

I am most likely missing a small detail. However, I will be very grateful if someone can tell me what I am missing.

Thanks.

 

[MisterX]

Remember the definition of the transfer function
$$ H(s) = \frac{Y(s)}{X(s)} $$
It will always have units
$$ \frac{\text{units of }Y(s)}{\text{units of }X(s)} $$
If $X(s)$ and $Y(s)$ (or $x(t)$ and $y(t)$ ) have the same units, the transfer function is unit-less.
I am not sure exactly where your confusion lies.
$$Y(s) \propto \int y(t) e^{-st}\,dt$$
$Y(s)$ should then have units $$\text{units of }Y(s) = \text{units of }y(t) * \text{ time}. $$

 

[Excom]

Hi MisterX

Thanks for your reply.

I do understand that the input and output of a system often have units. However, from your definition of the transfer function above you can rewrite like this: H(s)=K⋅G(s) Where K is the gain and has some units that are specifik to the system of interest.

We are then back to the situation where the transfer function, without the gain, of a first order system is: G(s) = a/(s+a) And we have a = 1/τ, where τ is the time constant. The unit of τ is sec. Assuming that the unit of s is rad/sec, we have (1/sec)/(rad/sec+1/sec) which does't make any sense to me.

 

[MisterX]

Hi MisterX

Thanks for your reply.

I do understand that the input and output of a system often have units. However, from your definition of the transfer function above you can rewrite like this: H(s)=K⋅G(s) Where K is the gain and has some units that are specifik to the system of interest.

We are then back to the situation where the transfer function, without the gain, of a first order system is: G(s) = a/(s+a) And we have a = 1/τ, where τ is the time constant. The unit of τ is sec. Assuming that the unit of s is rad/sec, we have (1/sec)/(rad/sec+1/sec) which does't make any sense to me.

Yes your confusion comes from the fact that radians aren't a unit. Consider the relation between arc length and radius
$$S = \theta R $$
Both arc length $S$ and radius $R$ are lengths, measured in meters for example, and so $\theta$ in radians is unitless. Radians/second is the same unit as $1/second$