1

I am awakard to deal with math problem

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I am trying to understand the first condition that is (k-r)=mN

I can understand when (k-r) is mN, the left formula is 1

But I don't know how to left formular to be right two conditions

I thought that

the left one is going to be (1/N) * (1-e^(j*2*pi*(k-r)))/(1-e^(j*2*pi*(k-r)/N))

  • I am curious about the upper formular is right or not.

so I substituted (mN) instead of (k-r) and

I got the result of 1/N, but it is not desirable.

please help me to understand this condition and formula

Thank you for advance

user2874612
  • 109

2 Answers2

1

Your calculation, if I read it correctly, is right. However, $$e^{(j)(2\pi)(k-r)}=1,$$ since $k-r$ is an integer.

It follows that the expression $1-e^{(j)(2\pi)(k-r)}$ that we get on top when summing the geometric series is equal to $0$. If $k-r$ is not of the shape $mN$, the result is $0$, since the denominator is non-zero.

If $k-r=mN$, the formula for summing the geometric series does not work, since the denominator is then $1-e^{jm}$, which is $0$. However, in that case the $N$ terms we are summing are all equal to $1$, so the sum is $N$, and when we multiply by the $\frac{1}{N}$ in front, we get $1$.

André Nicolas
  • 507,029
0

unless $k-r=mN$ the sum is, as you say: $$ (1/N)\frac{1-e^{2\pi j(k-r)}}{1-e^{2\pi j(k-r)/N}} = 0 $$ if $k-r=mN$ then this formula is indeterminate, but since each term in the sum is 1, the sum of $N$ terms, divided by $N$ is $1$

David Holden
  • 18,040