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Thanks to Joe and Jeremy!
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Dear all:
Desperately I'm trying to build a swept-sine displacement-history for
dynamic FE calculations. What I want to do is impose a sine-like
dispalcement in a frequency range from 10 to 50 Hz during one single
calcualtion. I found the following formular for swept-sine inputs:
y(a,b,c,d,x)=c*sin{PI/(b-a)*[((b-a)*x/d+a)^2-a^2]}
What do the parameters a,b and d represent? Can anybody enlighten me? At
least I know what c is for
Kind Regards,
Arno
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I haven't got a clue about that formular. Anyway, a swept sine:
Try:
y = a * sin (2*PI*t*f/s)
Where:
a = amplitude
t = sample index
f = frequency in Hz
s = samples per second
Then to sweep it, relate f to t and s, e.g:
f = 10.0 + 40.0 * t / (s * sweeptime)
This would sweep from 10 to 50Hz in sweeptime seconds
Hope this helps
Joe
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You have a rather complex formula for y. If you simplify the brackets
you will reduce to:
y = c sin { [ pi (b-a) x / d+2a] x}
which is of the form y = c sin {f(x) x} where the modulating
frequency
f(x) = pi (b-a) x / d+2a
is linear in x
If you want to simulate an instantaneous frequency F(x) then you have
to choose f(x) so that
F(x) = d [f(x) . x ] / dx
With a linear sweep this means that there is a factor of 2 in the
coefficient of the x term. So for a sweep from f0 to f1 in time
window x=0 to x=X you would want to generate
F(x) = f0 + (f1-f0) x / X
Then the parameters a, b and d are:
a = f0 / 2
b = f1 / 2
and
d = pi.X
So a and b are the bandwidth parameters and d is pi times the
duration. (There is some ambiguity between b and d but this is the
obvious way to write it.)
The formula you have is for a linear frequency sweep. You can model
other signals F(x) by integrating
F(x) = d [f(x) . x ] / dx
The constant of integration is equivalent to a phase shift in the
waveform: that might be important for your simulation or for the
response of any filters you are using, and you should not leave it
out. You could either use a complex exponential or write
y = c cos { [ pi (b-a) x / d+2a] x + 2.pi.e}
where 0
************************************************** **************************
*********
Dear all:
Desperately I'm trying to build a swept-sine displacement-history for
dynamic FE calculations. What I want to do is impose a sine-like
dispalcement in a frequency range from 10 to 50 Hz during one single
calcualtion. I found the following formular for swept-sine inputs:
y(a,b,c,d,x)=c*sin{PI/(b-a)*[((b-a)*x/d+a)^2-a^2]}
What do the parameters a,b and d represent? Can anybody enlighten me? At
least I know what c is for
Kind Regards,
Arno
************************************************** **************************
*********
I haven't got a clue about that formular. Anyway, a swept sine:
Try:
y = a * sin (2*PI*t*f/s)
Where:
a = amplitude
t = sample index
f = frequency in Hz
s = samples per second
Then to sweep it, relate f to t and s, e.g:
f = 10.0 + 40.0 * t / (s * sweeptime)
This would sweep from 10 to 50Hz in sweeptime seconds
Hope this helps
Joe
************************************************** **************************
*********
You have a rather complex formula for y. If you simplify the brackets
you will reduce to:
y = c sin { [ pi (b-a) x / d+2a] x}
which is of the form y = c sin {f(x) x} where the modulating
frequency
f(x) = pi (b-a) x / d+2a
is linear in x
If you want to simulate an instantaneous frequency F(x) then you have
to choose f(x) so that
F(x) = d [f(x) . x ] / dx
With a linear sweep this means that there is a factor of 2 in the
coefficient of the x term. So for a sweep from f0 to f1 in time
window x=0 to x=X you would want to generate
F(x) = f0 + (f1-f0) x / X
Then the parameters a, b and d are:
a = f0 / 2
b = f1 / 2
and
d = pi.X
So a and b are the bandwidth parameters and d is pi times the
duration. (There is some ambiguity between b and d but this is the
obvious way to write it.)
The formula you have is for a linear frequency sweep. You can model
other signals F(x) by integrating
F(x) = d [f(x) . x ] / dx
The constant of integration is equivalent to a phase shift in the
waveform: that might be important for your simulation or for the
response of any filters you are using, and you should not leave it
out. You could either use a complex exponential or write
y = c cos { [ pi (b-a) x / d+2a] x + 2.pi.e}
where 0