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 $d_\epsilon(\omega)$:
Note how the non-zero portion of this function has a width of $\epsilon$ and a height of $\frac{1}{\epsilon}$. The area underneath the function is therefore always $1$, no matter what $\epsilon$ may be.
We will let the value of $\epsilon$ get smaller and smaller. As it approaches zero, the function will become taller and narrower, until at the limiting case it is "infinitely" tall and "infinitesimally" narrow. We will denote the function in this limiting case $\delta(\omega)$. But it will still have an area of $1$, thus$\int \delta(\omega)d\omega=1$
Suppose we scale the $\delta(\omega)$ by another function $X(\omega)$. Since $\delta(\omega)$ is non-zero at all values other than at $\omega=0$, scaling $\delta(\omega)$ by $X(\omega)$ within the integral will yield the following:$\int X(\omega)\delta(\omega)d\omega=X(0)$
What if the integrand has $\delta(\omega-\omega_0)$ and is scaled by $X(\omega)$. In this case the integrand is only nonzero at $\omega=\omega_0$, thus:$\int X(\omega)\delta(\omega-\omega_0)d\omega=X(\omega_0)$
The $\delta(\omega)$ "function" we have derived is known as the
Dirac delta function . It is not technically a function (it is a generalized function, or distribution), but for our purposes we can treat it as a function with infinite height, infinitesimal narrowness, and an area of 1 at $\omega=0$.
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:$\cos(\omega_0 n) ~\stackrel{\mathrm{DTFT}}{\longleftrightarrow}~
\pi\,\delta(\omega-\omega_0) + \pi\,\delta(\omega+\omega_0)$
$\sin(\omega_0 n) ~\stackrel{\mathrm{DTFT}}{\longleftrightarrow}~
\frac{\pi}{j}\,\delta(\omega-\omega_0) + \frac{\pi}{j}\,\delta(\omega+\omega_0)$
Questions & Answers
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