Calculating response from FRF 2

[SrinivasAluri]

I have different FRFs for a 3DOF system. I want to calculate the response at 3rd DOF due to a sinusoidal input at the 2nd DOF. How do go about doing the calculation?

 

[GregLocock]

You need to know the principle of reciprocity, ie Hab=Hba, and then some cancellation.

You know

Haa
Hba
Hca

You want to know Hcb. I think it is Hcb= (Hca*Hba)/Haa

where all those operations are vector operations on the complex FRFs.




Cheers

Greg Locock

 

[SrinivasAluri]

I have H32 and a force of 100*cos(18t) at location 2, looking for a time response at location 2.

 

[SrinivasAluri]

Oops! I have H32 and a force of 100*cos(18t) at location 2, looking for a time response at location 3.

 

[GregLocock]

Good oh. You have all the information you need. Do you know how to read a Bode plot?



Cheers

Greg Locock

 

[SrinivasAluri]

Is Bode plot the FRF plotted in Mag vs Freq and Phase vs freq form? I don't know how to calculate the time response for say 2 seconds for the input I mentioned earlier using the plots.

 

[GregLocock]

Well, the magnitude of the FRF at 18 rad/s tells you the ratio of the response to the input, say it is 2.5, then you'll have an amplitude 200 sine wave. the phase at 18 rad/s tells you the phase delay with respect to the driving signal, say it is pi/2 then your response will be

250*cos(pi/2+18*t)





Cheers

Greg Locock

 

[SrinivasAluri]

One of the resonances for the system is close to 18 rad/sec, shouldn't the input force at 18 rad/sec cause the system response to increase considerably i.e., more than 2.5 times. I know damping of system keeps the response in check.

The reason I suspect this result is because I get a gradually increasing response using numerical integration.

 

[SrinivasAluri]

What I meant to say was, if I plot Mag*cos(18*t+phase)based on the FRF data, I am not getting same results as calculating the response directly using numerical integration, which is a increasing in amplitude for 0 to 2 secs. I am unable to reconcile the difference between the two solutions.

 

[GregLocock]

That is correct. You obviously need to revisit your understanding of Fourier analysis. Applying a steady state excitation of 100cos(18t) is not the same as switching on the same signal at t=0



Cheers

Greg Locock

 

[SrinivasAluri]

Greg, I am confused!! The excitation is steady state, its maintained at 100cos(18t) that's what I did using numerical integration, for calculating the response under steady state excitation.

 

[GregLocock]

Then you have made some other error, probably in your numerical integration.

Cheers

Greg Locock

 

[SrinivasAluri]

Are you saying I should get 250*cos(pi/2+18*t)(assuming 2.5 is gain and pi/2 is phase diff) from numerical integration, instead of a gradually increasing response that I got?

 

[GregLocock]

Steady state input implies steady state output in a Fourier context, because the timeline stretches from -infinity to +infinity.

In the real world that is not what will happen, but the fourier analysis of your excitation would not be a single frequency, and neither would the response.




Cheers

Greg Locock

 

[hacksaw]

what initial conditions are you using in the integration approach?

 

[SrinivasAluri]

Very small values equal to zero for all the initial displacements and initial velocities.

 

[hacksaw]

such initial conditions generally result in transients, how long are you waiting for "steady state"

 

[SrinivasAluri]

I actually evaluated the response for 16 secs, ofcourse the load is also applied for 16 secs.

 

[MikeyP]

Re your numerical integration results. If the response signal is still increasing after 16 seconds then (by definition!) you haven't reached steady state yet. I do this sort of thing by running the integration for one cycle at a time. After each cycle I subtract the current cycle from the previous one and calculate the error. When the error is small enough I call it steady state.

M

--
Dr Michael F Platten