explanation of "damping"

Hello friends.
I wonder what damping of process instruments really is. I thpught it's time you define and after that process value is preserved on display for that time. For example if I choose damping 5 sec on my pressure transmitter I expect that one value is preserved on display for 5 sec, and then new value is displayed for the next 5 sec. Am I right? I ask this because on one particular transmitter valeu of pressure is not displayed for 5 sec If I choose damping 5 sec. What the damping really is?
Thanks

Sounds good to me, but your display may just have a refresh rate of 5sec,


Can you give any more detail, name brands and model numbers

"damping" can also be used to divert or stop something, a damper on in duct work would divert the air...damper on a cylinder would cushion the movement, so "damping" could be a ramp?

For field instruments, damping is another word for numerically filtering by averaging.

5 second damping amounts to a rolling average of the last 5 seconds worth of data conversions to produce an averaged value.

I've never seen damping implemented as "hold the value" for the damping period.

I did once find an exceptional use of 'damping' on a paperless recorder. Damping was defined there as the time delay value for alarm. The alarm state had to be held continuously for the damping value "X seconds" before the alarm output would initiate. I wondered why they didn't call it an alarm delay timer.

Dan

Damping and refresh rate are different. Damping is akin to filtering a signal.

Refresh rate defines how often a signal or display is updated. Each new update may be either a "snapshot" of the existing at that instant, or it could be a damped or filtered signal. This is like your turning your head every ten seconds to look at a pressure guage and turning away as soon as you have read it. All you know is the reading at that instant.

Damping is like putting an orifice or guage snubber ahead of the pressure guage. This is a restriction that minimizes the sudden changes in guage reading so the needle is steadier and easier to read. Filtering is the same thing - you electrically damp the analog signal or else average it over a number of readings to make it steadier.

They are not mutually exclusive. You can have a long refresh rate on an undamped signal, or you can have a short refresh rate on a higly damped signal.

I wonder what damping of process instruments really is.

Click to expand...

Damping is a filter applied to the instrumentation output. So that a step change in the actual process will produce a slower response from the instrumentation. Usually its done to help eliminate process noise.

Definition from Dictionary.com:
The capacity built into a mechanical or electrical device to prevent excessive correction and the resulting instability or oscillatory conditions.

The 5 sec. is probably the time constant for the filter that is creating the damping effect. The exact effect of the 5 sec. depends on what type of filter it is.

Example, for a simple filter it may be:
Pout=Pin(1-e-t/tc) that's e to the power of -t/tc, tc is the time constant, t is time elapsed. So for a step change in Pin, at t=tc the Pout=0.63 * Pin, for t=3*tc the Pout =0.95 x Pin.

There are many others on this site that can give a better answer than me, but here goes.

Damping is the that property that fights change. That was pupposely very general because it depends on the quantity being analysed.

Think of a shock absorber on a car. At least that is what we call them in the U.S. Apparently the Europenas correctly refer to them as, you guessed it, 'dampers'. Regardless of the name a shock absorber will fight the change of position of the car body relative to the wheels. The faster the car body tries to move the harder the shock absorber fights the motion. Dampers in general are used to 'damp out' oscillations. I usually think of dampers as mechanical component, but that's just a semantics thing. The function of a damper is just as likely to be an electronic/software function.

In your particular case you are most likely defining the time constant for a 1^st order filter. A first order response is the shape you get from a capacitor charging; fast change early and decreasing change as you get closer to the final value. A first order filter time constant defines the time required for the filter output to reach roughly 62.5% of the difference between the initial value and the new value given a step input. It is generally held that it will take 5 time constants for the output of a 1^st order filter to reach the value of a step input.

So by defining a damping time of 5 seconds you output should reach your step input within 25 seconds. The output wilkl change smoothly continuously during that time.

As an aside, the inverse of the time constant is the -3dB frequency of the filter in radians per second.

Hope this helps.
Keith

Actually instrument I have is Honeywell multivariable smart transmitter SMV3000.
I didn't know what damping means but I know now thatnks to you. Just to make sure
I understood correctly let me say it in another words.
Damping is basically a time constant of lowpass filter that is applied to the input
of instrument. Filter has the following transfer function: G(s) = 1/(Ts+1) and for
the step input (G=Y/R, R = 1/s) output from filter is Y(t) = invLaplace (1/s * 1/(Ts+1))
Y(t) = 1-e^(-t/T). So if input signal (i.e.) process variable suffer step change
(i.e. fast, instantaneous change), output variable (measured signal) will have slower,
"exponential" change.
At first I thought it's refresh rate, but I know now that it is not.
Am I right here? Can you confirm this?
On the other hand I'm not sure I understood part of time averaging.
Can you explain that a little further or all I need to know about damping is what I
wrote above?

Thank you very much, with each thread I learn something.

This is just word play but, technically, 'damping' is something that counteracts change. Generally it is similar to the derivative function in a PID.

Having said that, I think Honeywell is using the term 'damping' a little loosely to describe a general purpose output filter. Also, I'm assuming that the filter is a simple first order filter. It could just as easily be a higher order linear filter, one of the non-linear forms of a filters or one of the averaging techniques described above. It's just that first order filters are pretty easy to implement digitally and are therefore pretty popular.

All details aside, 'dmaping' in your context is output filtering.

Keith

I don't know whether the cited math applies or not, but I do know for a certainty, having discusssed this with Honeywell's product specialist at noe time, that Honeywell's SMV3000 has a input conversion rate of 3 times per second. So the 'refresh rate' is 3 times per second.

If the damping is set of whatever value is close to the 1/3 second (I don't recall Honeywell's exact syntax) then there is no damping, and the calculation on the inputs is transferred immediately to the output.

If the damping is set at one second, then output is a rolling average of the last 3 calculations. If you have 2 second damping, the output is the rolling average of the last 6 calculations.

Damping is a rolling average in the SMV. What that amounts to mathematically, I'm not sure.

Dan

Thanks for replies, I tried to describe mathematically first order filter 1/(sT+1) where T is time constant I assumed to be damping in sec. An I simply found an output to step input (step process variable change) as shown in attachment. Regarding Honeywell and your informations I must say that it's another thing to average on time.

First of all I believe your transfer function is correct for a firstr order filter with time constant defined in seconds. And the attached PDF shows a first order response with a time constant of slightly more than 2 seconds.

However, the output of the Honeywell filter won't have this form. It will have a direct linear ramp (albeit very choppy)over the specified time given a step input change.

Keith

Damping:
The capacity built into a mechanical or electrical device to prevent excessive correction and the resulting instability or oscillatory conditions.

I like Randy's definiition.

See Gerry's Pout=Pin(1-exp(-t/tc). The tc is a time constant. Not damping. Gerry's equation can be rewritten Pout=Pin(1-exp(-a*t) where a is the corner frequency in radians per second. Neither
'a' or tc are damping. These equations are for first order systems and first order systems don't need to be damped.

Second order systems have a damping factor which is a unitless number. There is no time associated with the damping factor. Damping factors less than one resulut in an oscillitory or underdamped system.

I believe the term damping was misapplied in this case. Pandiani's 'damping' is really a time cosntant for a simple first order low pass filter as GerryM equation suggested.

Peter you're right. Damping constant (in control theory) cannot exist in first order system. In sistem of second order hooweverthere are damping factor and natural frequency but that is another story.

In my case damping coef. is in sec so in this case Damping should be understood different (not to be confuzed with damping factor of second order system).

I want to conclude this, so to have enough confidention to claim that I know what damping on Honeywell transmitter really is.
I'm a little confused now because at first I thought it's a time constant of first order filter. That maybe the case at other transmitters but as kamenges said that is not the case with Honeawell transmitter, so I give up my previous thoughts.
Now if I understood correctly Honeywell transmitter produce output from it's filter that is time average of input process variable.
So according to danw:
if resolution of transmitter is about 1/3 sec and damping is set to 2 sec. then I'll have six samples of the input. for example
X(0.3), X(0.6),..X(1.8)
and output from filter will be:
y = (X(0.3)+...+X(1.8))/6

So for the next 2 sec, another 6 samples are taknet and average value is present.

Can you confirm this?
If this the case that means that one value will be present on output for 2 sec (and that value is time average of samples)
But in this case how is possible that refresh rate and damping are different?
if filter is simple first order system, then output will be changed (refreshed) regardles of damping in sec. since it is just time constant.

Can you bring this a little colesr to me, maybe with little help of mathematics.

I don't have any particular experience with the Honeywell transmitter but, based on danw's post, the filter is a floating window.

You will get a new value every update period (every 1/3 second). This will be the time average of the previous samples in the time period. I'm going to ignore the time from power-up until the first time period expires. So given a 2 second filter time and a 1/3 second update rate you have:

Y_1.8 = (X₀ + X_0.3 + ... +X_1.8) / 6

The next output is produced 1/3 second later and is:

Y_2.1 = (X_0.3 + X_0.6 + ... +X_2.1) / 6

As a general form the equation would be:

Y₀ = X₀ + X_-1 + ... + X_{-(INT(Td/Ts)-1)} / INT(T_d/T_s)

where:
Y output
X input
T_d 'damping' time
T_s transmitter update time
There is a special case where the damping time is equal to or less than the update time.

I hope helps.

Keith