for a second order differential equation,( mass-spring-damping) system, if we have harmonic force acting on the system, we can find the solution x(t), but how about if the frequency of harmonic force is the natural frequency? there will be resonant. we know it won't go to infinity because of damping. Are there any formula availabe for the response x(t) of the system under this kind of force? Thanks!
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harmonic forcing under natural frequency 1
- Thread starter hechengli
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electricpete
Electrical
Please don't attack my kid-on-a-swing example. When I say "intuition fails me"... it meant that this example fails to act like a resonant system. (I know it is easier to push at the top because physically you can apply the force longer while the child is moving slowly at that point.....unrelated to the resonance situation I am trying to model).
I have a modification of the yardstick example that I think makes it work. You are not holding the yardstick in your hand... the yardstick is rigidly clamped sticking out of the the ceiling or wall (cantilever). You apply a force (not displacement) very near the clamped end of the yardstick where it's not moving. If the yardstick is at rest and you start tapping it with constant force, you can get it to start swinging wildly... but not at first, only after repeated taps. You can also try to apply sinusoidal force with same result.
I have a modification of the yardstick example that I think makes it work. You are not holding the yardstick in your hand... the yardstick is rigidly clamped sticking out of the the ceiling or wall (cantilever). You apply a force (not displacement) very near the clamped end of the yardstick where it's not moving. If the yardstick is at rest and you start tapping it with constant force, you can get it to start swinging wildly... but not at first, only after repeated taps. You can also try to apply sinusoidal force with same result.
electricpete
Electrical
Greg - hope you don't mind me beating a dead horse on this one. I checked my calcs and I'm pretty sure you will come around to my way of thinking if you set up your differential equation the same as mine. And I'm also pretty sure you know a helluva lot more about resonant behavior than I do... but sometimes like all of us it takes awhile to get to the right answer.
I agree with you that on a dc step input there would be an overshoot of the final dc value. I can't exactly apply that to this suddenly applied ac input signal which I can't decompose into a sum of step dc and ac input signals. I can only think about it intuitively that the step response oscillates around a dc operating point which will be much lower than the peak level of the steady state response. I base this statement on the frequency response curve.... which will have the amplification much higher at the resonant frequency than the dc (zero frequency). So if there is any overshoot to the portion of our input that acts like a step-dc-like input... it is still much lower than the final ac value.
What do you think? Am I close or out in left field?
I agree with you that on a dc step input there would be an overshoot of the final dc value. I can't exactly apply that to this suddenly applied ac input signal which I can't decompose into a sum of step dc and ac input signals. I can only think about it intuitively that the step response oscillates around a dc operating point which will be much lower than the peak level of the steady state response. I base this statement on the frequency response curve.... which will have the amplification much higher at the resonant frequency than the dc (zero frequency). So if there is any overshoot to the portion of our input that acts like a step-dc-like input... it is still much lower than the final ac value.
What do you think? Am I close or out in left field?
electricpete
Electrical
I learned something new to me when exploring various cases with my computer model.
There are two classes of 2nd-order underdamped (less than critical) systems to consider, with the dividing line being the 45 degree angle from the imaginary axis in the complex plane. If a pole lies to-the-right of this line (closer to the imaginary axis=less damped), then the frequency plot will show a resonant peak. If a pole lies to-the-left of this line (closer to the real axis... more damped... closer to critical damping), then the frequency response will resemable a low-pass filter... with the only peak being at frequency=0 (no resonant peaks).
I stumbled into this by accident because my program would not work for the highly-damped cases. That is because the formula for wr breaks down in the area between critical damping and the 45-degree line. There is no maximum in the frequency spectrum (other than w=0).
For the resonant systems we are discussing, the system should be less damped than those systems on the 45 degree line.
There are two classes of 2nd-order underdamped (less than critical) systems to consider, with the dividing line being the 45 degree angle from the imaginary axis in the complex plane. If a pole lies to-the-right of this line (closer to the imaginary axis=less damped), then the frequency plot will show a resonant peak. If a pole lies to-the-left of this line (closer to the real axis... more damped... closer to critical damping), then the frequency response will resemable a low-pass filter... with the only peak being at frequency=0 (no resonant peaks).
I stumbled into this by accident because my program would not work for the highly-damped cases. That is because the formula for wr breaks down in the area between critical damping and the 45-degree line. There is no maximum in the frequency spectrum (other than w=0).
For the resonant systems we are discussing, the system should be less damped than those systems on the 45 degree line.
electricpete
Electrical
My 45-degree terminology may not be familiar to everyone. I know some others talk about "damping factors", but I am not familiar with that terminology.
An example system on the 45 degree line would be m=1, c=sqrt(2), k=1. The roots are -c/2m +/- I * sqrt(4ma-c^2)/2m = -sqrt(2)/2 +/- I * sqrt(4-2)/2 = -sqrt(2)/2 +/-I*sqrt(2)/2. The real parts and imaginary parts are equal, creating the 45 degree angle from origin in the complex plane.
An example system on the 45 degree line would be m=1, c=sqrt(2), k=1. The roots are -c/2m +/- I * sqrt(4ma-c^2)/2m = -sqrt(2)/2 +/- I * sqrt(4-2)/2 = -sqrt(2)/2 +/-I*sqrt(2)/2. The real parts and imaginary parts are equal, creating the 45 degree angle from origin in the complex plane.
GregLocock
Automotive
OK I ran it through mathcad and I found that the example I gave does not overshoot, according to their solver, ie my Excel model is faulty. So you were right.
That's fine at resonance, but would you believe that at c=.2, m=k=1, w=2, the mathcad solution does overshoot? Admittedly the amplitude is less, but it is still greater than the ss response.
Having got a model that does behave like you were predicting, I then had a look at how the amplitude builds up at resonance for the undamped case... and got quite a surprise (which was a bit less surprising on further thought, but is even more surprising on further further thought). The amplitude ramps up linearly, well ok, each cycle we are adding energy, so the PE at max amplitude is getting bigger each time. PE is proportional to x squared, F is constant, x is increasing, so we are adding more energy with each cycle, by F.x *some function. Makes sense, except that we are not adding energy since F and x are in quadrature. I shall mull that one over tonight.
i'm going to debug that excel thing now that I've got some better numbers to compare it with.
Incidentally all this malarkey in the time domain is probably a bit unnecessary for the original poster, his best bet is to get the frequency domain transfer function of his system (trivial, in all the books), FFT his drive signal, munge the two together and IFFT the result.
Cheers
Greg Locock
That's fine at resonance, but would you believe that at c=.2, m=k=1, w=2, the mathcad solution does overshoot? Admittedly the amplitude is less, but it is still greater than the ss response.
Having got a model that does behave like you were predicting, I then had a look at how the amplitude builds up at resonance for the undamped case... and got quite a surprise (which was a bit less surprising on further thought, but is even more surprising on further further thought). The amplitude ramps up linearly, well ok, each cycle we are adding energy, so the PE at max amplitude is getting bigger each time. PE is proportional to x squared, F is constant, x is increasing, so we are adding more energy with each cycle, by F.x *some function. Makes sense, except that we are not adding energy since F and x are in quadrature. I shall mull that one over tonight.
i'm going to debug that excel thing now that I've got some better numbers to compare it with.
Incidentally all this malarkey in the time domain is probably a bit unnecessary for the original poster, his best bet is to get the frequency domain transfer function of his system (trivial, in all the books), FFT his drive signal, munge the two together and IFFT the result.
Cheers
Greg Locock
electricpete
Electrical
Good point Greg, and some good thoughts.
I believe that we are adding energy each phase when F leads X by 90 degrees. Because V=d/dt(x) also leads X by 90 degrees and therefore will be in phase with F.
Power=F(t)v(t) will be always positive if F and v are in phase.
something like P=F0*V0*cos^2(wt)=F0V0(.5+.5cos(2w0t)
where F0 and V0 are peak amplitudes of F(t) and V(t)
I am not too surprised that there is an oversheet for your parameters with w=2*wresonant. In my mind the ballpark predictor of overshoot is the quantity H(w)/H(0) where w is the exciting frequency. That is based on previous discussion that the dc-step behavior produces overshoot and ac component does not. If the DC response exceeds the sinusoidal response, we might expect overshoot. For your parameters the dc response H(0)=1 and the ac response H(2)=0.33
I believe that we are adding energy each phase when F leads X by 90 degrees. Because V=d/dt(x) also leads X by 90 degrees and therefore will be in phase with F.
Power=F(t)v(t) will be always positive if F and v are in phase.
something like P=F0*V0*cos^2(wt)=F0V0(.5+.5cos(2w0t)
where F0 and V0 are peak amplitudes of F(t) and V(t)
I am not too surprised that there is an oversheet for your parameters with w=2*wresonant. In my mind the ballpark predictor of overshoot is the quantity H(w)/H(0) where w is the exciting frequency. That is based on previous discussion that the dc-step behavior produces overshoot and ac component does not. If the DC response exceeds the sinusoidal response, we might expect overshoot. For your parameters the dc response H(0)=1 and the ac response H(2)=0.33
electricpete
Electrical
Greg,
And since you have admitted that I was right, I had better return the favor and admit that you were right.
I was able to generate about 5% overshoot using the following parameters:
k=1,m=1,c=1.40,theta=2.8
That lies just to the right of the boundary of the resonant class of systems that I discussed above (boundary point is k=1,m=1,c=1.414=sqrt(2)). And it doesn't overshoot for any values of theta except very close to theta=2.8 (see my word file for definition of theta).
My explanation.. Moving toward the boundary (more damping) increase the dc response and decrease ratio of H(wr)/H(0), pushing towards overshoot.
And since you have admitted that I was right, I had better return the favor and admit that you were right.
I was able to generate about 5% overshoot using the following parameters:
k=1,m=1,c=1.40,theta=2.8
That lies just to the right of the boundary of the resonant class of systems that I discussed above (boundary point is k=1,m=1,c=1.414=sqrt(2)). And it doesn't overshoot for any values of theta except very close to theta=2.8 (see my word file for definition of theta).
My explanation.. Moving toward the boundary (more damping) increase the dc response and decrease ratio of H(wr)/H(0), pushing towards overshoot.
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