Unexpected Results from my Transfer Function

Question

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

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:

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?

Answer 1

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.

Answer 2

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.)

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 $$ c=\frac{5 \pi \left(1+\sqrt{16000000000 \pi ^2-1599}\right)}{2-20000000 \pi ^2}$$ $$d=\frac{1+\sqrt{16000000000 \pi ^2-1599}}{4000000 \pi } $$ $$k=\frac{1}{1000} $$

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