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Recall the definition of the DTFT of a signal $x[n]$:$ X(\omega) ~=~ \sum_{n=-\infty}^{\infty} x[n]\, e^{-j \omega n} $Let's see what the DTFT of a complex exponential signal $e^{-j \omega_0 n}$: $X(\omega_0) ~=~ \sum_{n=-\infty}^{\infty} x[n]\, e^{-j \omega n} ~=~ \sum_{n=-\infty}^{\infty} e^{j \omega_0 n}\, e^{-j \omega n} ~=~ \sum_{n=-\infty}^{\infty} e^{-j (\omega-\omega_0) n}$ The nature of that sum is not immediately clear. When $\omega=\omega_0$, then the sum becomes unbounded: $\sum_{n=-\infty}^{\infty} 1=\infty$. But when $\omega\neq\omega_0$, what does an infinite sum for a sinusoid mean?
The dirac delta function
To find the DTFT of a sinusoid, it will to be necessary to consider another function, the dirac delta function. Suppose we have a function $d_\epsilon(\omega)$:
The dtft of a sinusoid
Having introduced the Dirac delta function, let's see what happens if we attempt to find its inverse DTFT: $\int_{-\pi}^\pi 2\pi \delta(\omega-\omega_0)\, e^{j\omega n} \, \frac{d\omega}{2\pi} ~=~ e^{j\omega_0 n}$ The inverse DTFT of a dirac delta function is a complex sinusoid, which means that the DTFT of a complex sinusoid is a dirac delta:$ e^{j\omega_0 n} ~\stackrel{\mathrm{DTFT}}{\longleftrightarrow}~ 2\pi\,\delta(\omega-\omega_0)$ Via Euler's formula, we use the above to also find the DTFT of discrete-time cosines and sines:$\cos(\omega_0 n) ~\stackrel{\mathrm{DTFT}}{\longleftrightarrow}~ \pi\,\delta(\omega-\omega_0) + \pi\,\delta(\omega+\omega_0)$ $\sin(\omega_0 n) ~\stackrel{\mathrm{DTFT}}{\longleftrightarrow}~ \frac{\pi}{j}\,\delta(\omega-\omega_0) + \frac{\pi}{j}\,\delta(\omega+\omega_0)$Read also:
OpenStax, Discrete-time signals and systems. OpenStax CNX. Oct 07, 2015 Download for free at https://legacy.cnx.org/content/col11868/1.2
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