6
$\begingroup$

I have a system that can be modeled with the following picture:

enter image description here

There is a mass $m$ connected to a spring $k$ and a dashpot $d$. These are both connected to another dashpot $c$. A force $F(t)$ is applied at the junction.

After some effort tackling a system of linear ODEs, I've found a transfer function that describes this system:

$$\frac{X(s)}{F(s)} = \frac{ds + k}{(mc + md)s^3 + (mk + cd)s^2 + (ck)s}$$

I'm almost certain this is correct considering I've checked over the math multiple times (and the units check out, which is always a plus).

The system has certain parameters: $m = 1$, $Q = 20$, $\omega = 50$ Hz.

From this we can calculate the following parameters: $k = m(2\pi\omega) = 98696$ N/m, $d = \frac{sqrt(mk)}{Q} = 15.7$ Ns/m.

I estimate $c = 4$ Ns/m.

Given these parameters, I expect the following:

  • Impulse response that asymptotically approaches a value.
  • Step response that grows without bound.
  • Bode plot that has a peak at around 50 Hz.

The first two are true, but this is what my Bode plot actually looks like:

enter image description here

The reason I expect a peak at 50 Hz is because when $F(t)$ has a frequency near $\omega$, the amplitude of motion of the mass should rapidly increase.

Is there something that I'm modeling incorrectly here? Or is there some major misconception about how I think this system operates?

$\endgroup$
  • $\begingroup$ I suggest you make a Bode plot for $Y$, and also compare the force in damper $c$ with the driving force $F$. I haven't done any math on your system, but my instinct says that $F$ is going down the "wrong" load path (through $c$) which will kill off the resonance you expected to see. Even if that guess is wrong, making those plots should help understanding what's going on. $\endgroup$ – alephzero Mar 1 '16 at 2:12
  • $\begingroup$ When I think about it in terms of "circuitry," I think of the spring as an inductor and the damper as a resistor, so the damper has constant complex impedance (and no frequency dependence), while the spring has complex impedance proportional to $\omega$. I don't know if this frequency dependence just doesn't show up in the Bode plot (but might show up in an actual simulation). But I'll go ahead and make a Bode plot for $Y$ and see where it leads me. Thanks! ^^ $\endgroup$ – anonymouse Mar 1 '16 at 14:46
2
$\begingroup$

Checking the units is an excellent way to double check your work; kudos for doing so. However, the next step in checking to see if your results make sense is to check limits. In your case, you can use physical intuition to identify how the system would act at very low frequencies, and how it would act without any damping.

At very low frequencies ($s\rightarrow0$) the responsiveness of the system should be to move in phase with an amplitude transfer function that goes flat at DC ($\frac{X(s)}{Y(s)}\rightarrow1$). Right now you have an infinite responsiveness at DC; applying a slight force moves the mass by an infinite amount. Another way to see this is by considering the system without any damping ($c\rightarrow0$, $d\rightarrow0$) in which case your TF would become $$ \frac{X(s)}{F(s)}\rightarrow\frac{1}{ms^2}, $$ when it should look more like the undamped harmonic oscillator TF which is given by $$ \frac{k}{ms^2 + k}. $$ So, I don't know exactly where you've gone wrong, but you are missing a constant term in the denominator of your TF.

$\endgroup$
  • $\begingroup$ Thanks for the swift answer! Two things. One, what makes you say that I have infinite responsiveness at DC? Two, I'm not sure that letting $c$ and $d$ go to zero will result in an undamped harmonic oscillator. Here, the force is being applied to the back side of the spring and not to the mass, so intuitively (and my intuition might completely be wrong) I think that $\frac{1}{ms^2}$ might make sense. Any thoughts? $\endgroup$ – anonymouse Mar 1 '16 at 14:42
  • $\begingroup$ @SSS Responses: 1) Set $s$ equal to zero in your TF and recall that $\lim_{s\rightarrow0}\frac{1}{s}=\infty$. 2) You are correct that the TF will not be exactly the same as the harmonic oscillator, but it will still have a resonance peak which $\frac{1}{ms^2}$ does not have. $\endgroup$ – Chris Mueller Mar 1 '16 at 16:23
1
$\begingroup$

The transfer function appears correct, except that the x and y deviations from the unstreched locations and not as depicted in the figure. Otherwise, when $y=x$ then there is no space for the spring $k$ and damper $d$.

This needs to be equated to

$$ \frac{1}{c m+d m}\frac{d s+k}{s}\frac{1}{s^2+\frac{s w}{Q}+w^2} $$

Comparing terms we get the following two equations

$$ w^2=\frac{c k}{c m+d m} $$ $$\frac{w}{Q}=\frac{c d+k m}{c m+d m}$$

There are 6 variables ($m$, $k$, $d$, $c$, $Q$, $w$) of which 3 are known ($m$, $Q$, $w$). This leaves us with 2 equations and three unknowns, and thus there are multiple solutions.

One thing I noticed is that $c$ and $d$ have opposite signs. This is a bit odd because it means that $c$ or $d$ must have a negative sign! (The following is a screen shot of the computations in Mathematica.)

enter image description here

If $k << d$ then the zero will almost cancel out the pole at the origin. This was the additional condition I used to arrive at the following values $$ c=\frac{5 \pi \left(1+\sqrt{16000000000 \pi ^2-1599}\right)}{2-20000000 \pi ^2}$$ $$d=\frac{1+\sqrt{16000000000 \pi ^2-1599}}{4000000 \pi } $$ $$k=\frac{1}{1000} $$

enter image description here

Then this is the Bode plot with the peak as expected at ~$314 rad/s$ or $50 Hz$.

enter image description here

$\endgroup$
  • $\begingroup$ Thanks for the answer. I have two questions. One, does it matter that the spring and damper have zero length when $y = x$? Is it not okay to assume that the unstretched length of the spring is much smaller than the characteristic lengths of $x(t)$ and $y(t)$? Second, I followed your analysis and it seems to make sense, but $c$ is negative. While this definitely produces the Bode plot I want, it seems really nonphysical - the $c$ damper will provide a force to increase $\dot{y}$. I'll try the values in my simulation to see what the impulse and step responses are, though. $\endgroup$ – anonymouse Mar 1 '16 at 17:25
  • $\begingroup$ SSS, It is just not possible for a physical spring or damper to have zero length. This is a transfer function model, there is nothing preventing x and y from being equal, so we just got to make sure what it physically means in this context. I can add the impulse and step response to the answer if you want me to do so. The impulse response converges to zero and the step response grows without bound. $\endgroup$ – Suba Thomas Mar 1 '16 at 17:48
  • $\begingroup$ Hm, I'm not sure if that impulse response is what I expect from this system. If I give an impulse force $F = \delta$, I expect $y$ to grow and eventually be damped by $c$, and I expect the quantity $x - y$ to perhaps experience some oscillations because of the spring. In total, I think the impulse response should converge to some nonzero value (and the step response should grow without bound). In fact, this is the system I'm attempting to characterize, so do you think I need a change in my actual model to reflect this? $\endgroup$ – anonymouse Mar 1 '16 at 18:09
  • $\begingroup$ To elucidate this further, the type of system I'm attempting to characterize is one that, on a macro-scale, moves like a mass-damper system where an impulse force will move it some distance but it eventually has to stop. On a micro-scale, there is some resonant frequency $f$ such that, if I attempt to force the system to move with this frequency, I'll experience wild high-amplitude oscillations. This is why I've set up the system as a mass-spring-damper to the right, but with a "higher-level" damper to the left. $\endgroup$ – anonymouse Mar 1 '16 at 18:11
  • $\begingroup$ My bad, the impulse response at steady-state is actually a nonzero negative value. $\endgroup$ – Suba Thomas Mar 1 '16 at 18:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.