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I want to find the coefficient but I am stuck. The formula is:

$$\frac{1}{10}\sum_{n=0}^{9} e^{-jk\omega_0n}$$

In order to solve this I am using the following formula:

$$\sum_{n_1}^{n_2} \alpha^{n} = \frac{\alpha^{n_1}-\alpha^{n_2+1}}{1-\alpha}$$ I have tried everything but I am unable to reach the answer. The answer is: $$\frac{1}{10}(1-e^{-j\omega_08k})$$

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$\displaystyle S=\cfrac 1{10}\sum_{n=0}^9 e^{-jk\omega_0 n}=\cfrac 1{10}\cdot\cfrac{1-e^{-j10k\omega_0}}{1-e^{-jk\omega_0}}=\cfrac 1{10}\cdot \left(1+\cfrac{e^{-jk\omega_0}(1-e^{-j9k\omega_0})}{1-e^{-jk\omega_0}}\right)$

And $1-e^{j\theta}=1-\cos\theta-j\sin\theta=2\sin\cfrac{\theta}{2}\left(\sin\cfrac{\theta}{2}-j\cos\cfrac{\theta}{2}\right)=2\sin\cfrac{\theta}{2}e^{j(\theta-\pi)/2}$

Hence,

$$\cfrac{e^{-jk\omega_0}(1-e^{-j9k\omega_0})}{1-e^{-jk\omega_0}}= \cfrac {e^{-jk\omega_0}2\sin (-9k\omega_0/2) e^{-j4k\omega_0}}{2\sin(-k\omega_0/2)}$$

$$=\cfrac {e^{-j5k\omega_0}\sin (9k\omega_0/2) }{\sin(k\omega_0/2)}$$

$$S=\cfrac 1{10}\left(1+\cfrac {e^{-j5k\omega_0}\sin (9k\omega_0/2) }{\sin(k\omega_0/2)}\right)$$

It seems the definition of $\omega_0$ is needed if we want to go further.

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Edit: with integer $k$, generally we have

$\displaystyle \sum_{n=0}^{N-1}e^{-j2kn\pi/N}=\cfrac {1-e^{-j2k\pi}}{1-e^{-j2k\pi/N}}=0$

The correct answer you provided doesn't evaluate to $0$ with, say, $k=1$. So something must be wrong.

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  • $\begingroup$ $w_0$ is $(2\pi)/10$ $\endgroup$ – Ahmad Qayyum Nov 8 '18 at 17:50
  • $\begingroup$ can you complete the solution? $\endgroup$ – Ahmad Qayyum Nov 8 '18 at 18:26
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    $\begingroup$ @Ahmad Qayyum. If $k$ is integer, the sum apparently evalutes to $0$, as $\cfrac {1-e^{-j2\pi k}}{1-e^{-jk2\pi/10}}$ is $0$. Take $k=1$, your answer is $\cfrac {1-e^{-j8/5\pi}}{10}$ which is not $0$. Something still not right. $\endgroup$ – Lance Nov 8 '18 at 18:34

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