I'm using an Endevco Model 2220D Accelerometer with an Endevco Model 2721B Charge Amplifier to get vibration readings from a motor. I was asked to get a reading in IPS (Inches per second) but the Charge Amplifier gives a normalized output in mV/g. How do I convert that to IPS? I've not been lucky finding any math on the google... And this stuff is pretty new to me. Thanks all!

Maybe the better way to ask is this: I see a dampened sine wave on my scope as the output of the charge amplifier. The datasheet for the amplifier says it's normalized to mV/g. I need my output converted to Inches pre Second (IPS). How do I go about that?

Update: The dampened sine wave, for clarification, was due to me pinging the motor with a hammer to get a test reading prior to running the system.

• Inches per second is a unit of velocity. Do you mean inches per second squared? Also, when you say mV/g, does g represent standard gravitational acceleration? Likely this question will have a difficult time lasting on this site if it is only about unit conversation. – davidmneedham Nov 29 '17 at 20:02
• @davidmneedham As I understand it, it's a gravitational unit. This was kinda placed on me last minute to figure out, and I've never encountered it before. Thus, asking the community. – Jedi Engineer Nov 29 '17 at 20:31
• Figure out the acceleration, integrate it over a period of time and accept the inherent error of the process. – Lior Bilia Nov 29 '17 at 20:59
• You need a unit of speed; the accelerometer gives you units of acceleration (g). As you know the waveform (sinusoidal) and its frequency, the rest is just arithmetic. – Brian Drummond Nov 29 '17 at 21:09
• I can imagine this is a common problem for embedded developers working with accelerometers, so I wouldn't dismiss the question as only a unit conversion question. As the answer as-of-now shows, it's not that trivial. – pipe Nov 30 '17 at 8:45

The equation for the accelerometer's output voltage at time 't' is,

$$y(t) = k\, a(t) \;\;\;\;\;\;\;\;\;\;(1)$$

where

y(t) := The accelerometer's output voltage at time 't'
k    := A conversion constant with units of V*s^2/m (see eq. 7 below)
a(t) := The acceleration at time 't' with SI units of m/s^2


You need velocity, not acceleration. We know that acceleration at time 't' is the first derivative of velocity at time 't' taken with respect to time 't':

$$a(t) = \frac{dv(t)}{dt} \;\;\;\;\;\;\;\;\;\;(2)$$

Substitute equation (2) into equation (1) and using separation of variables solve for velocity at time 't', v(t):

$$y(t) = k\, a(t)=k\,\frac{dv(t)}{dt} \;\;\;\;\;\;\;\;\;\;(3) \\[0.2in] \rightarrow dv(t)=\frac{1}{k}\,y(t)\,dt \;\;\;\;\;\;\;\;\;\;(4) \\[0.2in] \rightarrow \int_{v(t_0)}^{v(t)}dv(t)=\frac{1}{k}\int_{t_0}^{t}y(t)\,dt \;\;\;\;\;\;\;\;\;\;(5) \\[0.2in] \rightarrow v(t)=\frac{1}{k}\int_{t_0}^{t}y(t)\,dt+v(t_0) \;\;\;\;\;\;\;\;\;\;(6)$$

where the integral $\int y(t)dt$ has units of $Volts \cdot seconds$.

In other words, you'll need to integrate the accelerometer's output voltage $y(t)$ from time $t_0$ to $t$, divide that result by the conversion constant 'k', and then add in the accelerometer's initial velocity $v(t_0)$ at starting time $t_0$ to get the velocity at time 't', $v(t)$.

Solving for the value of the conversion constant 'k' is straightforward. For example, if the accelerometer's output voltage is $1\,mV$ for an applied acceleration of $1\,g$, then,

$$k=\frac{1\,mV}{9.807\,m/s^2} \bigg\rvert \frac{1\,m}{100\,cm} \bigg\rvert\frac{2.54\,cm}{1\,inch}=\frac{2.59\,\mu V}{in/s^2} \;\;\;\;\;\;\;\;\;\;(7)$$

I see a dampened sine wave on my scope as the output of the charge amplifier.

If the damped sinewave signal has exponential decay, then the formula for $y(t)$ in equation (6) is of the form shown in equation (8):

$$y(t)=A\,e^{-\lambda\,t}\,sin(2\pi f t + \phi) \;\;\;\;\;\;\;\;\;\;(8)$$

where

λ := Decay constant (1/s)
A := Sine wave's undamped amplitude
f := Sine wave's frequency (1/s)
φ := Sine wave's starting phase angle (radians)

• Thanks Jim, I should have clarified, the dampened sine wave was because I pinged the side of the motor with a hammer to make sure I could get an output from the charge amplifier. Otherwise, I'm expecting a sinusoidal wave at some frequency. – Jedi Engineer Nov 30 '17 at 12:31