Continue to Site

Eng-Tips is the largest engineering community on the Internet

Intelligent Work Forums for Engineering Professionals

  • Congratulations waross on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!

Natural frequency extraction

Status
Not open for further replies.

gopi9

Electrical
Aug 3, 2012
14
Hi everyone.
I have the impulse response output of a 2nd order system.
What I did was got the frequency response from the transfer function(2nd order equation) and the peaks of the frequency response represents the natural frequency and I found that the natural frequency from these response are correct.
But the problem is that when I got the frequency response function(FRF) from the impulse response output. The values of frequency corresponding to peak FRF are not matching with the natural frequencies.
I am struck in here from 2 days. Please help me, I would really appreciate ur help.
Thanks.
 
Replies continue below

Recommended for you

I assume you are working with a computer simulation and not a real system?
(otherwise it would seem unlikely that you would have a transfer function).

The FRF should be fourier transform (or roughly fft) of the impulse response.

Perhaps you can share more details so we can understand what you've been doing and have a hope of figuring what might be wrong




=====================================
(2B)+(2B)' ?
 
Thanks for the reply.
yes i am working in matlab. I used an example and created 2nd order system. The transfer function is for that 2nd order system.
I got the impulse response for that system.
I used FFT to convert it in to FRF.
I have attached the file for the response that I got with the FRF. It is clear from that file that the frequency of corresponding peak magnitude is not equal to natural frequencies. The most left hand side of the figure has the natural frequencies.
 
 http://files.engineering.com/getfile.aspx?folder=747cd345-8a51-4bd9-96bc-57cd1c136296&file=Untitled.png
I did not get you.
What do u meant by bin number?
Do u mean the left most side of the figure?
The left most side of the figure represents the natural frequency.
I got those values by theoretical analysis. I have verified those values too, they are correct.
 
without running it

Y1_1=Y2:),1);
fft1=fft(Y1_1);
fft2=abs(fft1);
figure;
plot(fft2);
axis([0 1000 0 80e-03]);

you are plotting the array fft1 against the array index (ie the bin number), not the frequency

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376
 
Thank you so much Greg.
Should i take
f=1/t;
and plot(f,fft2).
Are you suggesting that this might help.



I tried this Greg. That did not help.
 
ok thanks.
I can do that.
But am i going in a correct way to get natural frequency from impulse response.
 
Thank you so much Greg. I figured out the solution for that. I got the values which matches with the required frequencies. The code is shown below
Time=6500;
Y1_1=Y2:),1);
N=length(Y1_1);
k=0:N-1;
T=N/Time;
freq=k/T;
X=fft(Y1_1);
plot(freq,abs(X));
The only problem is the rate that is 6500. I took this value with out any calculation. I guessed a value and took it. Can I know how can we get that rate.
 
Status
Not open for further replies.

Part and Inventory Search

Sponsor