How to calculate frequency at given time using sweep rate? 3

[jcm61]

We recently subjected one of our devices to a sinusoidal vibration test (per MIL STD 810F, method 514C-18) at a local test house that used a sweep-rate of 0.2215 octaves per minute. The 810F test is performed in 3 axes, consisting of six (6), 30 minute sweeps per axis, ranging from 5 Hz to 500 Hz (acceleration .2 to 5.0 G's).
Two incidents occured at elapsed times of 00:37:53 and 01:37:50 {hr:min:sec}. I need to determine the respective frequencies that correspond with these times. I do know that a frequency doubles after each succesive octave (7 in this case*) but I am not sure how to use this towards solving for the desired frequencies.

I have spent much of the day searching the web and am more frustrated after encountering many rigorous, seemingly endless mathematical equations on the subject (Fourier FFT, etc)but am no closer to a solution.

I appreciate your help.

* 810F test Octaves @ frequencies (Hz):
5-10, 10-20, 20-40, 40-80, 80-160, 160-320, 320-640.

 

[IRstuff]

Your information seems to still be incomplete. Is each sweep immediately after the previous? i.e., does 00:37:53 represent 7:53 into the second sweep?

If so, then the frequency should be

5 Hz * 2^(7.8833) = 1180.5647 Hz

TTFN

FAQ731-376

 

[jcm61]

THANKS FOR YOUR RESPONSE

[ANSWER TO YOUR QUESTION AND MY COMMENTS ARE BRACKETED]
Your information seems to still be incomplete.
Is each sweep immediately after the previous? [YES]
i.e., does 00:37:53 represent 7:53 into the second sweep?
[YES]

If so, then the frequency should be
5 Hz * 2^(7.8833) = 1180.5647

[THIS IS INCORRECT SINCE THE TEST FREQUENCY RANGES BETWEEN 5 TO 500 Hz. WHERE DID THE 7.8833 COME FROM & WHAT EQUATION ARE YOU USING? I DONT SEE WHERE THE GIVEN SWEEP RATE OF .2215 OCTAVES PER MIN IS USED IN THIS SOLUTION].

Please see attachment 810F profile for clarification.

Thanks,
JCM

 

[btrueblood]

Hmm.

IRstuff's equation should be

freq = (5 Hz) * 2^(t/.2215) for 0<t<t1

and freq = 500*2^(-t/.2215) for t1<t<2t1

and repeating thereafter


t1 occurs when f = 500 Hz, or

2^(t1/.2215) = 100
(t1/.2215)ln(2) = ln(100)
t1 = 1.4716 minutes or 01:28.296

At 7:53 into run #2, you are in period #5, so use eqtn 1 with t = .5253 minutes, which gives freq = 25.9 Hz

Or at least, that's what I get...

 

[IRstuff]

Sorry, start over again

7.8833 min is 7:53 in decimal minutes. I forgot to multiply by .2215 oct/min

5Hz*2^(7.8833min*.2215oct/min) = 16.77 Hz

The period is 30 minutes. The .221462 oct/min puts the frequency at 500 Hz at the 30 minute elapsed time: 5Hz*2^(30min*.2215oct/min) = 500.3966 Hz

There is only one equation:

5Hz*2^(modulo(time,30min)*.2215oct/min)

This is cautionary tale of why you should ALWAYS keep units in the calculations, since it would immediately be obvious that the construct:
t1*min / .2215oct/min should have been t1*min * .2215oct/min

TTFN

FAQ731-376

 

[jcm61]

THANKS FOR YOUR RESPONSE btrueblood!
PLEASE ANSWER MY QUESTIONS BELOW [BRACKETED]:

freq = (5 Hz) * 2^(t/.2215) for 0<t<t1

[I ASSUME THAT THIS IS "eqtn 1" you are referring to below?
CAN YOU TYPE OUT THIS EQUATION OR DIRECT ME TO A WEB SOURCE WHERE I CAN SEE THE DERIVATION OF THIS EQUATION?
ALSO, WHY IS 5 Hz PART OF EQUATION 1 SINCE FREQUENCY OF 5 Hz OCCURS ONLY AT THE BEGINNING OF SWEEPS #1, 3, & 5 AND AT THE END OF SWEEPS #2 , 4 , & 6? I AM TRYING TO SOLVE FOR A FREQUENCY THAT OCCURED AFTER 5 Hz (i.e., at ELAPSED TIMES OF 37.8 MINUTES {OR 7.8 MINUTES} INTO SWEEP #2 AND 96.8 MIN {OR 6.8 MINUTES} INTO SWEEP #4]

and freq = 500*2^(-t/.2215) for t1<t<2t1

[LIKEWISE, WHY IS 500 Hz PART OF THIS EQUATION AND WHY IS A MINUS t (-t) USED IN THIS EQUATION?]
and repeating thereafter

t1 occurs when f = 500 Hz, [WHY DOES t1 OCCUR AT 500 HZ AND NOT AT f TO BE DETERMINED AT 7.8 MINUTES?]

or 2^(t1/.2215) = 100 [WHAT IS AND HOW DID YOU GET 100?]
(t1/.2215)ln(2) = ln(100)
t1 = 1.4716 minutes or 01:28.296

At 7:53 into run #2, you are in period #5
[IT'S UNCLEAR WHAT YOU MEAN BY "period #5"? DO YOU MEAN A PERIOD OF TIME OR AN OCTAVE? OCTAVE #5, e.g., STARTS AT 80 AND ENDS AT 160 Hz.],
so use eqtn 1 with t = .5253 minutes, which gives freq = 25.9 Hz
[25.9 Hz SOUNDS RIGHT BUT NOT SURE HOW YOU ARRIVED THERE. HOW DID YOU FIND t=.5235?]

P.S.SORRY TO ALL FOR THE LONG STRING.
Thanks,
JCM

 

[electricpete]

You want the derivation for: freq = (5 Hz) * 2^(t/.2215)

If you had a linear sweep of frequency vs time, the units would be hertz per minutes.

But you have not specified a linear sweep, but a sweep in the form of octaves per minute. A little thought should convince you that requires some form of exponential. This is the exponential equation which fits your data.

If that is not obvious, at least consider that it amounts to a straight line on a curve of ln(f) vs time. (agreed?)

Find the equation of the line:
ln(f) = m * t + b
that satisfies:

ln(5) = m *(0) + b
and
ln (2*5) = m * 0.2215 + b

Solution of two unknowns m and b from two equations gives:
b = ln(5)
m = [ln(10) - ln(5)]/0.2215

So the equation is:
ln(f) = ln(5) + {[ln(10) - ln(5)]/0.2215} * t
ln(f) = ln(5) + {[ln(10/5)]/0.2215} * t
ln(f) = ln(5) + {[ln(2)]/0.2215} * t
ln(f) - ln(5) = {[ln(2)]/0.2215} * t
ln(f/5) = {[ln(2)]/0.2215} * t
ln(f/5) = [ln(2)] * (t/0.2215)
take exp ( ) of both sides
f / 5 = 2 ^(t/0.2215)
f = 5 * {2 ^(t/0.2215)}
where f in hz, t in minutes

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[electricpete]

One correction:
"This is the exponential equation which fits your data."
should have been:
"This is the exponential equation which fits your specification (0.2215 octaves per minutes, starting at f=5 when t = 0)."

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[IRstuff]

Your resulting equation does not result in 500 Hz at 30 minutes, which is also part of the requirement.

The derivation simply comes from

f(30minutes) = 500 Hz = 5 Hz * 2^(alpha * 30 minutes)

ln(100)/ln(2) = alpha * 30minutes

alpha = 0.2215 octaves /min.

Therefore, the correct equation is

freq = 5 Hz * 2^(time * 0.2215/minute)



TTFN

FAQ731-376

 

[electricpete]

You're right, mine would be 0.2215 minutes per octave.

The derivation is the same.

Change
ln (2*5) = m * 0.2215 + b
to
ln (2*5) = m * 1/0.2215 + b

And invert 0.2215 everywhere it appears after that.



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[electricpete]

And I agree whole-heartedly with your derivation. But the OP did not seem satisfied with the equation, so apparently wanted some justification for the form of the equation. My derivation shows that it arises from a linear change of ln(f) vs t

Either derivation is fine imo.

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[IRstuff]

Sure, the check point is that every 4.5147 minutes, the frequency is doubled.


Since 7:53 is less than twice that value, the frequency must be less than 20 Hz for the "event."

TTFN

FAQ731-376

 

[jcm61]

That's what I was looking for...
Thanks to IRstuff and electricpete for your collective input.
JCM

 

[btrueblood]

Sorry not to get back sooner, jcm, but the network server at work went out after a power outage. Irstuff and pete got things squared away well enough, though; 16.7 Hz is my final answer also. Amazing that we all three could start out tripping over our...er, selves...at first go.

 

[jcm61]

I was curious what frequency IRstuff and electricpete calcuated for the second incident, at t= 6.783 minutes, into the 4th, 30 minute sweep which begins at 500Hz and A= 5 g's and descends to 5 Hz and A=.23g as shown in the attached 810F figure 514C-5C test sweep profile (each sweep is a mirror image of the previous sweep, e.g., sweep 4 is a mirror image of sweep 3, etc..)

Using the previously derived equation to calculate frequency f = 5 Hz * 2^(t * .2215/min) at t = 6.783 minutes yields
f= 5* 2^(6.783 * .2215)= 14.36 Hz?? This is incorrect since I know at 14 Hz, A = 2 g's per the 810F sweep file attached, and the frequency must occur at A = 5 g's.

Since we are starting the 4th sweep at 500 Hz and heading back towards 5 Hz, I substituted t = 30 minutes - 6.783 = 23.217 minutes which yields a more reasonable frequency of f = 5* 2^(23.217*.2215)= 176.6 Hz when compared to figure 514C. However, following this approach in conjunction with the derived equation for f, I question why t = 6.783 is not substituted into the equation accordingly and likewise
500 Hz for 5 Hz...that is until realizing that calculated f would exceed 500 Hz???

JCM

 

[IRstuff]

If it's a triangle and not a sawtooth, then the equation is simply reversed:

f = 50Hz * 2(-t * .2215 oct/min) = 176.48 Hz.

As a check, t = 0 min results in the starting frequency of 500 Hz, and t = 30 min results in 4.996 Hz.

TTFN

FAQ731-376

 

[tomirvine]

Here is a reference paper:

http://www.vibrationdata.com/tutorials2/sweep.pdf

Tom Irvine