Originally posted by Michael E Stanley of Freescale Semiconductor in The Embedded Beat on Mar 12, 2013
In Orientation Representations Part 1 and Part 2, we explore some of the mathematical ways to represent the orientation of an object. Now we’re going to apply that knowledge to build a virtual gyroscope using data from a 3axis accelerometer and 3axis magnetometer. Reasons you might want to do this include “cost” and “cost”. Cost #1 is financial. Gyros tend to be more expensive than the other two sensors. Eliminating them from the BOM is attractive for that reason. Cost #2 is power. The power consumed by a typical accel/mag pair is significantly less than that consumed by a MEMS gyro. The downside of a virtual gyro is that it is sensitive to linear acceleration and uncorrected magnetic interference. If either of those is present, you probably still want a physical gyro.
So how do we go from orientation to angular rates? It’s conceptually easy if you step back and consider the problem from a high level. Angular rate can be defined as change in orientation per unit time. We already know lots of ways to model orientation. Figure out how to take the derivative of the orientation and we’re there!
In our prior postings, we’ve discussed a number of ways to represent orientation. For this discussion, we will use the basic rotation matrix. Jack B. Kuipers has a nice derivation of the derivative of direction cosine matrices in his “Quaternions and Rotation Sequences” text – one of my most used textbooks. It makes a good starting point. Paraphrasing his math:
Let:
 v_{f }= some vector v measured in a fixed reference frame
 v_{b }= same vector measured in a moving body frame
 RM_{t} = rotation matrix which takes v_{f }into v_{b}
 ω = angular rate through the rotation
Then at any time t:
 v_{b}= RM_{t} v_{f}
Differentiate both sides (use the chain rule on the RHS):
 dv_{b}/dt = (dRM_{t}/dt) v_{f} + RM_{t}(dv_{f} /dt)
Our restrictions on no linear acceleration or magnetic interference imply that:
 dv_{f}/dt = 0
Then:
 dv_{b}/dt = (dRM_{t}/dt) v_{f}
We know that:
 v_{f} = RM_{t}^{1} v_{b}
Plugging this into (8) yields
 dv_{b}/dt = (dRM_{t}/dt) RM_{t}^{1} v_{b}
In a previous posting (Accelerometer placement – where and why) , we learned about the transport theorem, which describes the rate of change of a vector in a moving frame:
dv_{f}/dt = dv_{b}/dt – ω X v_{b}
Those who take the time to check will note that we have inverted the polarity of the ω in Equation 11 from that shown in the prior posting. In that case ω was the angular velocity of the body frame in the fixed reference frame. Here we want it from the opposite perspective (which would match gyro outputs).
And again,
 dv_{f}/dt = 0 so
 dv_{b}/dt = ω X v_{b}
Equating equations 10 and 13:
 ω X v_{b} = (dRM_{t}/dt) RM_{t}^{1}v_{b}
 ω X = (dRM_{t}/dt) RM_{t}^{1}
where:


0 
ω_{z} 
ω_{y} 
ω X = 
ω_{z} 
0 
ω_{x} 

ω_{y} 
ω_{x} 
0 
Going back to the fundamentals in our first calculus course and using a onesided approximation to the derivative:
 dRM_{t}/dt = (1/Δt)(RM_{t+1} – RM_{t})
where Δt = the time between orientation samples
 ω_{b} X = (1/Δt)(RM_{t+1} – RM_{t}) RM_{t}^{1}
Recall that for rotation matrices, the transpose is the same as the inverse:
 RM_{t}^{T} = RM_{t}^{1}
 ω_{b} X = (1/Δt)(RM_{t+1} – RM_{t}) RM_{t}^{T}
Equation 15 is a truly elegant equation. It shows that you can calculate angular rates based upon knowledge of only the last two orientations. That makes perfect intuitive sense, and I’m ashamed when I think how long it took me to arrive at it the first time.
An alternate form that is even more attractive can be had by carrying out the multiplications on the RHS:
 ω_{b} X = (1/Δt)(RM_{t+1} RM_{t}^{T} – RM_{t} RM_{t}^{T})
 ω_{b} X = (1/Δt)(RM_{t+1} RM_{t}^{T} – I_{3×3})
For the sake of being explicit, let’s expand the terms. A rotation matrix has dimensions 3×3. So both left and right hand sides of Eqn. 22 have dimensions 3×3.
 (1/Δt)(RM_{t+1} RM_{t}^{T} – I_{3×3}) = (1/Δt) W


0 
W_{1,2} 
W_{1,3} 
W = RM_{t+1} RM_{t}^{T} – I_{3X3 }= 
W_{2,1} 
0 
W_{2,3} 

W_{3,1} 
W_{3,2} 
0 
The zero value diagonal elements in W result from small angle approximations since the diagonal terms on RM_{t+1} RM_{t}^{T} will be close to one, which will be canceled by the subtraction of the identity matrix. Then:


0 
ω_{z} 
+ω_{y} 

0 
W_{1,2} 
W_{1,3} 
ω X = 
+ω_{z} 
0 
ω_{x} 
= (1/Δt) 
W_{2,1} 
0 
W_{2,3} 

ω_{y} 
+ω_{x} 
0 

W_{3,1} 
W_{3,2} 
0 
and we have:
 ω_{x}= (1/2Δt) (W_{3,2} – W_{2,3})
 ω_{y}= (1/2Δt) (W_{1,3 } W_{3,1})
 ω_{z}= (1/2Δt) (W_{2,1 } W_{1,2})
Once we have orientations, we’re in a position to compute corresponding angular rates with
 One 3×3 matrix multiply operation
 3 scalar subtractions
 3 scalar multiplications
at time each point. Sweet!
Some time ago, I ran a Matlab simulation to look at outputs of a gyro versus outputs from a “virtual gyro” based upon accelerometer/magnetometer readings. After adjusting for gyro offset and scale factors, I got pretty good correlation, as can be seen in the figure below.
You will notice that we started with an assumption that we already know how to calculate orientation given accelerometer/magnetometer readings. There are many ways to do this. I can think of three off the top of my head:
 Compute roll, pitch and yaw as described in Freescale AN4248. Use those values to compute rotation matrices as described in Orientation Representations: Part 1. This approach uses Euler angles, which I like to stay away from, but you could give it a go.
 Use the Android getRotationMatrix [4] to compute rotation matrices directly. This method uses a sequence of crossproducts to arrive at the current orientation.
 Use a solution to Wahba’s problem to compute the optimal rotation for each time point. This is my personal favorite, but I think I’ll save further explanation for a future posting.
Whichever technique you use to compute orientations, you need to pay attention to a few details:
 Remember that nonzero linear acceleration and/or uncorrected magnetic interference violate the physical assumptions behind the theory.
 The expressions shown generally rely on a small angle assumption. That is, the change in orientation from one time step to the next is relatively small. You can encourage this by using a short sampling interval. You should soon see an app note that my colleague Mark Pedley is working on that discards that assumption and deals with large angles directly. I like the form I’ve shown here because it is more intuitive.
 Noise in the accelerometer and magnetometer outputs will result in very visible noise in the virtual gyro output. You will want to low pass filter your outputs prior to using them. Mark will be providing an example implementation in his app note.
This is one of my favorite fusion problems. There’s a certain beauty in the way that nature provides different perspectives of angular motion. I hope you enjoy it also.
References
 Freescale Application Note Number AN4248: Implementing a TiltCompensated eCompass using Accelerometer and Magnetometer Sensors
 Orientation Representations: Part 1 blog posting on the Embedded Beat
 Orientation Representations: Part 2 blog posting on the Embedded Beat
 getRotationMatrix() function defined at http://developer.android.com/reference/android/hardware/SensorManager.htmlWikipedia entry for “Wahba’s problem”
 U.S. Patent Application 13/748381, SYSTEMS AND METHOD FOR GYROSCOPE CALIBRATION, Michael Stanley, Freescale Semiconductor