Discrete Harmonics Concepts

If we add $$2 \pi$$ to the discrete frequency $$e^{j(\omega_0+2\pi k)n}$$ it gives the next frequency, why is this true?

For continuous domains $$e^{jk\Omega_0n}$$ gives a complete set of orthogonal harmonics for fourier decomposition but For discrete domain $$e^{jk\omega_0n}$$ does not form a complete set orthogonal basis set to decompose a signal because... ?

• No Answer yet on 14 Nov 2018, if you are able to know - please share knowledge. I have edited this post and shortened it to make it clear what my questions are.
• where have read your definitions? – Stanley Pawlukiewicz Nov 8 at 13:18
• This book - complextoreal.com/fftguide and chapter 3 which is free to view here...complextoreal.com/tutorials/… its a portion of the chapter from pg 86. Is my approach above wrong? – Natalie Johnson Nov 8 at 13:21
• i would reverse the roles of your symbols $\Omega_0$ and $\omega_0$ to be consistent with the convention we see in DSP books. i know that in the old analog days we use $\omega=2\pi f$ for continuous-time signals (like $s=j\omega$) but in DSP we now say $z=e^{j\omega}$ and when we need to reference an analog context, we say $s=j\Omega$. – robert bristow-johnson Nov 8 at 22:35
• @NatalieJohnson: The set $\phi_k[n]=e^{j2\pi nk/N}$ is an orthogonal set, and that's what you use to find the Fourier series of a discrete-time signal. – Matt L. Nov 9 at 10:43
• I know that. But I am confused how we get there - I edited my post to make it crystal clear what I am confused about. – Natalie Johnson 15 hours ago

ok. Let $$n$$ nd $$k$$ be in integers,
Since by Euler identity we have : $$e^{j \omega_0 n } = \cos(\omega_0 n ) + j \sin(\omega_0 n)$$
Then we can see that \begin{align} e^{j (\omega_0 + 2\pi k) n } &= \cos((\omega_0 + 2\pi k) n ) + j \sin((\omega_0 + 2\pi k) n) \\ &= \cos( \omega_0 n + 2\pi k n ) + j \sin(\omega_0 n + 2\pi k n) \\ &= \cos( \omega_0 n ) + j \sin(\omega_0 n) \\ &= e^{j \omega_0 n } \end{align}
it's ok now. They are the same. $$\omega_0$$ and $$\omega_0 + 2 \pi k$$ are the same in discrete time case when $$n$$ and $$k$$ are integers.