- For your circle question, yes you can simply use 2*pi*rand (N,1) to get a uniform distribution of angles in the [0,2pi] range. You can get the same result by using P1=0.25 in the code I supplied. As far as your "densely placed" observation goes, it is likely just due to not many samples being generated. Try it with a large
**number**of samples. - Answer: x=
**0**: pi/100:10*pi Here this represents about x i.e what are values of x, from which point to which. So here x varies from**0**to 10*pi. x takes these values only ; only this interval. - Use the reference
**MatLab**code to generate the test vectors and design your receiver. ... (1,1))%2*pi*rand() provides**random****number****between****0****and 2pi**with uniform pdf ... - Search: Quaternion Kalman Filter Python. Applying the Kalman Filter to a Pair of ETFs Given a sequence of noisy measurements, the Kalman Filter is able to recover the "true state" of the underling object being tracked This is code implements the example given in pages 11-15 of An Introduction to the Kalman Filter by Greg Welch and Gary Bishop, University of North Carolina
**MATLAB**is a powerful programming language, that can be used to draw various plots used in machine learning, deep learning, computer vision, and big data programming. Let us start with coding for plots in**MATLAB**. Example 1: Let us first understand the simple plot: % angle from**0**to**2pi**. theta =**0**:**0**.01 : 2 * pi; % sin function works on an array.