I'm a noob at basically everything.
What an excellent blog post. Now, can you help me show that the sum from n = -(N-1)/2 to (N-1)/2 of e^(j*n*k*d*sin(theta)) is equal to 1/N * sin(1/N*k*d*sin(theta))/sin(1/(2*k*d*sin(theta)) ? awesome! thanks.
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1 comment:
What an excellent blog post.
Now, can you help me show that the sum from n = -(N-1)/2 to (N-1)/2 of e^(j*n*k*d*sin(theta)) is equal to 1/N * sin(1/N*k*d*sin(theta))/sin(1/(2*k*d*sin(theta)) ? awesome! thanks.
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