Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
Drexel University or The Math Forum.


stainburg
Posts:
1
From:
china
Registered:
10/18/10


sum of csc
Posted:
Oct 18, 2010 10:42 PM


sum{csc[pi*(n+.5)/N]} with n where k+1<=n<=Nk2 0<=k<=N1
n,k,N are natural numbers
[Difficulty] I doubt if it's concerned about the gamma function. gamma(x)*gamma(1x)=pi*csc(pi*x)
[Thoughts] This question is from the problem X(k+1/2)=DFT[cot(pi*(n+.5)/N)*exp(1j*pi*n/N)],where DFT denotes the discrete fourier transform.
while x(n) denotes a time sequence,X(k) denotes its DFT sequence and we have X(k)=DFT[x(n)]=sum[x(n)*exp(1j*2*pi*n*k/N)],(0<=n<=N1),where N is the length of x(n),k is the sampling frequency,n is the sampling time. X(k+1/2)=DFT[x(n)*exp(1j*pi*n/N)] =(1/N)*{sum[X(g)]+sum[X(g)*cot[pi*(kg+1/2)/N]]},with 0<=g<=N1. and sum[X(g)*cot[pi*(kg+1/2)/N]] denotes a circle convolution with X(k)and cot[pi*(k+1/2)/N]. I have known that X(k)=DFT[cot(pi*(n+.5)/N)]=T(k)N*exp(1j*pi*k/N), where T(k) is a sampling function, which is equal to [N,0,0,0,0...]. so X(k+1/2)=DFT[cot(pi*(n+1/2)/N)*exp(1j*pi*n/N)] can be expressed by X(k+1/2)=cot(pi*(k+1/2)/N)+1j exp(1j*pi*(k+1/2)/N)*{sum{csc[pi*(g1+1/2)/N]}sum{csc[pi*(g2+1/2)/N]}} with, 0<=g1<=k,k+1<=g2<=N1,0<=k<=N1 and there comes my question. thanks for reading!
Message was edited by: stainburg



