Computing derivative using "for" loop--why am I getting this error?

5 次查看（过去 30 天）
Kelsey
Kelsey 2014-6-5
评论： dpb 2014-6-6
Below is my code (note that in reality the vectors are much longer, but I simplified them to 5 entries of fake data...you get the picture). THIS code works, but when I use the actual data (each parameter is a 1x5036 matrix, as opposed to the 1x5 matrices that I present here for simplicity), I get a constant value of 0.75 for VehicleAccel for all t. Obviously, this is not the true value, as the data was for a real-world drive I did in a car around town (I have attached my actual m-file with the data). What is the problem?? Thanks in advance!
t = [1 2 3 4 5]; % sec
FLwhlRotation = [6.8 6.9 7.0 7.1 7.2]; % rad/s
FRwhlRotation = [6.8 6.9 7.0 7.1 7.2]; % rad/s
RLwhlRotation = [6.8 6.9 7.0 7.1 7.2]; % rad/s
RRwhlRotation = [6.8 6.9 7.0 7.1 7.2]; % rad/s
AvgWhlRotation = (FLwhlRotation + FRwhlRotation + RLwhlRotation + RRwhlRotation)./4; % rad/s
AvgWhlSpeed = AvgWhlRotation.*0.3683; % m/s (R = 0.3683 m)
VehicleSpeedMPH = AvgWhlSpeed.*2.2369; % mi/hr
VehicleSpeedKPH = AvgWhlSpeed.*3.6; % km/hr
% Instantaneous Acceleration (m/s^2)
VehicleAccel=0;
for i=1:4
VehicleAccel = (AvgWhlSpeed(i+1)-AvgWhlSpeed(i))./(t(i+1)-t(i));
end
%Plots
plot(t,AvgWhlSpeed,'b',t,VehicleAccel,'r')
legend('Speed (m/s)','Acceleration (m/s^2)')
xlabel('Time (s)')
  3 个评论
显示 1更早的评论
Kelsey
Kelsey 2014-6-6
When I run that, I get an error message saying that it couldn't plot because the vectors aren't the same lengths (i.e., VehicleAccel is 1x5035, while t is 1x5036). Any idea what the issue is?
dpb
dpb 2014-6-6
Of course...when you take the first differences there's one less value left than you start with. You don't show how you actually did the plot shown but your looping solution has the same issue; your initial code shows
for i=1:4
accel(i)=(spd(i+1)-spd(i))./(t(i+1)-t(i));
which leaves you with four accel values calculated from a time vector of length five. I presume you probably augmented the accel vector w/ a leading zero.
The vectorized solution would be
VehicleAccel = [0 diff(AvgWhlSpd)./diff(t)];

请先登录，再进行评论。

请先登录，再回答此问题。

回答（1 个）

Star Strider
Star Strider 2014-6-5
Change this line to create VehicleAccel as a vector and see if that improves things:
VehicleAccel(i) = (AvgWhlSpeed(i+1)-AvgWhlSpeed(i))./(t(i+1)-t(i));
Also, it’s best not to use ‘i’ and ‘j’ as loop indices. MATLAB uses them for its imaginary operators, and could cause confusion if you use them also as variables.
  2 个评论
Kelsey
Kelsey 2014-6-6
编辑：Kelsey 2014-6-6
Thanks!! That did it, I think...how does this plot look?
Would you say the acceleration appears to accurately represent the derivative of the speed?
Star Strider
Star Strider 2014-6-6
My pleasure!
I would. The derivative should be slightly lower than baseline for the areas where the speed is linearly decreasing, and it seems to be. I opened a larger version in a new tab in Firefox, and it looks good. The derivatives of the parabolas look appropriately linear.

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Loops and Conditional Statements 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by