1 - 12 of 12 Posts

#### vasyachkin

·
##### Registered
Joined
·
2,136 Posts
Discussion Starter · ·
so Q of 1.0 = -0db = factor of 1 difference in energy

so Q of 0.7 = -3db = factor of 2 difference in energy

and Q of 0.5 = -6db = factor of 4 difference in energy

but wikipedia says that Q is a ratio of energies

this doesn't coupute because Q of 0.5 corresponds to 0.25% energy

#### mazurek

·
##### Registered
Joined
·
46 Posts
The factor Q is used by electrical engineers and is related to the inverse of the damping in a second order electrical system. A free resource for engineers would be the book, "op amps for everyone" hosted by Texas Instruments. Otherwise go to google/wikipedia/library/engineering library/classes for other explanations.

#### mazurek

·
##### Registered
Joined
·
46 Posts
Now I see the page you are confused by. Also look at this one:
http://en.wikipedia.org/wiki/Damping_ratio

There are many ways to define this value. The definition you found is not intuitive for electrical system. In mechanical systems you can understand those values by tracing the stress and strain over one cycle and look at the energy in the plot which is the area under the curve.

#### Rightbrained

·
##### Registered
Joined
·
810 Posts
Vas? Are you ok?

You are actually admitting you dont know everything and are asking for help understanding.

Bravo!!

#### djarchow

·
##### Registered
Joined
·
509 Posts
Vas, I thought you graduated with a EE degree?

#### DL86

·
##### Registered
Joined
·
537 Posts
Go study some advanced circuit analysis to really understand what Q is

#### vasyachkin

·
##### Registered
Joined
·
2,136 Posts
Discussion Starter · ·
come on i know what it is

i just need an explanation for why 6db corresponds to a factor of 2 difference in Q

6db is a factor of 2 difference in DISPLACEMENT but a factor of 4 difference in ENERGY

if this makes any sense to you please share ...

#### mazurek

·
##### Registered
Joined
·
46 Posts
Imagine that you stretch a perfect spring out to a given length. The total energy input into the system is work = force*displacement. If this is a perfect spring, this is the energy stored in the system, when you let go, the spring will bounce forever. Now imagine an imperfect spring, the spring's damping dissipates energy when you release it, and eventually after a certain number of cycles it will stop. The energy stored in the system is the initial work you put in, and the energy dissipated is the energy lost when the spring cycles. This ratio between them is damping.

There are many ways to define and measure damping. The wikipedia definition you are looking at is not very intuitive for electrical systems in my opinion, and I feel like you are extrapolating information.

A highly resonant second order system (low damping or high q) will not have any amplification of displacement for an impulsive signal. For a lightly damped system, with a steady state signal, energy can accumulate in the system over time resulting in amplification of the displacement. If there is no damping, the system will be unstable and blow up, otherwise due to losses there will be a finite amount of gain.

Consider taking a control systems class, or looking for lecture notes online. Often they start with a foundation of first and second order systems, and you can learn about all the ways to measure and characterize their performance (settling time, overshoot, step responses, impulse responses, bode plots, etc). Sometimes in speaker land we forget that just looking at frequency magnitude doesn't exactly describe what is a system is doing in time.

#### vasyachkin

·
##### Registered
Joined
·
2,136 Posts
Discussion Starter · ·

Quote:
Originally Posted by mazurek /forum/post/17006697

Imagine that you stretch a perfect spring out to a given length. The total energy input into the system is work = force*displacement

except the force equals K * displacement so energy = K * Displacement SQUARED

so 2X displacement = 4X energy = 6db. yet a filter that is down -6db at corner is considered critically damped. at the same time critically damped is supposed to be 1/2 energy.

ok i think i got it. 1/2 energy is lost PER CYCLE so the output at any time is 1/2 as much and the output lasts for 1/2 as long so total output is 1/4 = -6db. ! ! !

yess. i am awesome

#### whoaru99

·
##### Registered
Joined
·
6,946 Posts

Quote:
Originally Posted by vasyachkin /forum/post/17007045

yess. i am awesome

All I need are some tasty WAVs, a cool buzz, and I'm fine.

#### mazurek

·
##### Registered
Joined
·
46 Posts
You are right, I simplified it, it is really work = integral of force over displacement, so it would be 0.5*k*x^2

I'm still not sure about the other relationships you are claiming. I'd suggest that you graph a second order system in Excel, you should be able to find the equations for the transfer function magnitude and the impulse response in time, and play with each parameter to better understand the effects in the time and in the frequency domain. The other plot you should make is force vs distance plot and it should look like a bunch of loops.

#### vasyachkin

·
##### Registered
Joined
·
2,136 Posts
Discussion Starter · ·
no way. that's too complicated. no Excel. that would be work.

1 - 12 of 12 Posts