First Order System's Time Constant

[yanaibarr]

Hello,
I have a question on a the units of a first order system's time constant.
If i have a first order system the basic transfer function will be:
K/(tau*s+1)
where K is the Gain, and tau is the system's time constant.
tau's units, according to what I've learned, are [sec].
but aren't the s plane's units in [rad/sec] (s=jw+sigma)?
That means that tau should be given in [sec/rad] to match the "1"-'s units in the transfer function.
I know that rad can be considered "unitless" but when dealing with actual numbers it matters if the system's time constant is 1 [sec] or 1[sec/rad]= 2*pi [sec].

My question is specifically about the units of tau in the transfer function,
not when it is used in the decay rate of e (e^(-t/tau)), there it has to be sec.

I'll appreciate a clarification.

Thanks

 

[tiny-tim]

welcome to pf!

hello yanaibarr! welcome to pf!

tau is always in seconds …

the difference between radians and (eg) degrees is absorbed into the k

 

[MisterX]

tau is always in seconds …

No, one may use any unit for tau. For exponential decay, Ae^(-t/tau), the exponent (-t/tau) should be unit-less.

 

[yanaibarr]

tau is always in seconds …


Thanks for he replay.
One more question about it,
if tau's units should be seconds, then the s-plane units should be Hz [1/s].
According to what I've learned, the s-plane's units are [rad/sec] (s=jw+sigma).
Can i take the s-plane's units as Hz?

I tried working with an actual differential equation, and according to it the s-plane's units will always be [1/sec], because the s represents the derivative.
If it's so, when do i use the [rad/sec] units and when [Hz] in the s-plane?

Thanks,

Yanai barr

 

[tiny-tim]

sorry, i don't know, i haven't come across the s-plane

 

[viscousflow]

tau is always in seconds …Thanks for he replay.
One more question about it,
if tau's units should be seconds, then the s-plane units should be Hz [1/s].
According to what I've learned, the s-plane's units are [rad/sec] (s=j\omega+\sigma).
Can i take the s-plane's units as Hz?

I tried working with an actual differential equation, and according to it the s-plane's units will always be [1/sec], because the s represents the derivative.
If it's so, when do i use the [rad/sec] units and when [Hz] in the s-plane?

Thanks,

Yanai barr

\omega has units of \frac{rad}{sec} (s = jw+sigma) , Hz has units of \frac{1}{s} so the connection you made between the derivative, 1/s and, Hz for the s domain is correct.

 

[yanaibarr]

sorry, i don't know, i haven't come across the s-plane

The s-plane is what u get after using the Laplace Transform.

 

[yanaibarr]

\omega has units of \frac{rad}{sec} (s = jw+sigma) , Hz has units of \frac{1}{s} so the connection you made between the derivative, 1/s and, Hz for the s domain is correct.

Thank u for the reply ,
but Hz [1/s] and omega's units [rad/s] are not the same, u should divide\multiply it by 2*pi.
This is exactly my question, the units don't match (according to the theory I've learned).
In theoretical problems it doesn't matter, but when i use actual numbers i need to decide how to use the data, and how to convert the units accordingly.

Yanai Barr

 

[valjok]

I've stumbled at the same problem. All learning materials seem to expose the concept but none gives example with exact units.

So, if I want a frequency break at 1 Hz, should I write 1/(s+1) or 1/(s+2Pi)? Second seems more plausible. However, when Laplace-transfromed, it gives e^{-2pi t} meaning that time constant is T = 1/2pi. Yet, I'm customed that periods are measured in seconds rather than seconds per radian. I mean that 2pi is not usually a part of period. But, wikipedia article on time constant does not clarify what are the units.

https://www.physicsforums.com/showthread.php?t=516891"

 

[valjok]

In other words, is it right that time constant 2Pi corresponds to frequency of 1 Hz?
http://www.google.ee/url?sa=t&rct=j...g=AFQjCNFCHWRk_HC4jpb0BTtqc2jeeXzz6g&cad=rja" guide saying that a pole of 1, H(s)=1/(s+1), corresponds to time constant of 1 sec, seems to agree with this.