In step (a), the integration of the first cosine with respect to theta gives zero because you are integrating over [-pi,pi]. The integration of the second cosine is taken with respect to theta but there is no theta inside the cosine. This will give us the constant 2pi, and the cosine term will stay.
great job with your explanation I understand the problem of autocorrelation of a random train pulses
Thank you professor, for uploading your lectures. You explain everything in such a structured way. Again, thank you!
brilliant!
sir for example 2,isnt the final answer should be in sin? since we r integratin the cos?
In step (a), the integration of the first cosine with respect to theta gives zero because you are integrating over [-pi,pi]. The integration of the second cosine is taken with respect to theta but there is no theta inside the cosine. This will give us the constant 2pi, and the cosine term will stay.
@@stanleychan2757 oh yaaa rightt..got it sir..thanks
Amazing Explanation