[moved] Can anyone please explain how we obtained equation from multiple equations

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n.emimal

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Can anyone please explain how we obtained equation (7.8) from (7.6)? in the attached

 

I do not understand which step is complicated. exp(ix)=cos(x)+ isin(x) // i is same as j

Same way, exp(2*pi*i*n)=cos(2*pi*n)+i*sin(2*pi*n) // if n is an integer, first term is 1 and the second term is zero

Now sum{n=0 to N-1} exp (2*pi*i*(n/N)*alpha) // sum the cos and sin terms separately
// 1+cos(2*pi*alpha/N)+cos(2*pi*2*alpha/N)+cos(2*pi*3*alpha/N) + ... // also add the sine terms

If alpha is multiple of N, then each term becomes 1 and we get N; sine terms all disappear
If alpha is not a multiple of N, we get N cos terms that will add to zero (clear??)

// same with sine
 

You should know

\[\displaystyle\sum_{n=0}^{N-1}{z^n}=\frac{1-z^N}{1-z} \; \text{if} \; z\ne 1\]


Then if \[z=e^{j \frac{2\pi}{N}(k-r)}\;\;\longrightarrow\;\;z^N=e^{j {2\pi}(k-r)}=1\]
because k,r are integers

Then
\[\displaystyle\sum_{n=0}^{N-1}{e^{j \frac{2\pi}{N}(k-r)n}}=\frac{1-1}{1-e^{j \frac{2\pi}{N}(k-r)}} = 0 \; \text{if} \; k \ne{r}\]

\[\; \text{if} \; k=r \;\;\longrightarrow\;\; \displaystyle\sum_{n=0}^{N-1}{e^{j \frac{2\pi}{N}(k-r)n}}=\displaystyle\sum_{n=0}^{N-1}{1}=N\]


At (7.6), all terms of the summation of k are multiplied by 0 except the term when k=r (by 1)
 
Thank you very much for your reply. I got this concept, however in the last step how do we get X(r) rather X(r+mN)

- - - Updated - - -

Thank you so much. what if the case k=r+mN and what will be the final answer?
 

k=r+mN is for the (7.7) identity.

in this demonstration 0<= k <= N-1 and 0<= r <= N-1 ==> m=0
 

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