Recent content by RandomUserName

Re: time-discrete odometry Yes, you are correct, I messed up the sign there => $$\theta' = -50°$$ Omg, the more complicated the task gets, the more simple stuff I seem to not think of :D edit: Ok, I figured out my mistake, was another sign error... But now I'm good. Thank you again!

Hello, I have another question regarding my cognitive robotics class. Here is my current task: And this is the slide from the lecture about this topic: Now just plugging in everything I know into the formulas, I get this: $$\boxed{3 = \sqrt{x'^{2} + y'^{2}}\\ -20 = atan2(y',x')\\ -30 = \theta'...

Alright :) Thanks for all your help!

I asked about this, and this was the reply: The noise has a shift of 1 (the mean), so if the real reading is 10, then the noisy reading (which will be the most probable) will be 11. Therefore, the difference between the expected reading at a certain x_i location (taking into account the noise...

Since you're asking like that, probably yeah :D. But I'm not 100% sure how yet... Can I just use this formula now? prob(z1|x2) = this prob(z2|x2) = this

Hey guys, I need some help. Here's my task: Here's what I came up with: I need to calculate prob(x1|z1), prob(x1|z2) and so on and then just write down which has the highest probability. To do that, I would use Bayes rule, but for that, I need p(z1|x1) and p(z1). How do I get these? I have...