# How do you show the Discrete Time Fourier Transform of $x[n]=\cos(2\pi f_0n)$ is $\frac{1}{2} \delta(f+f_0) + \frac{1}{2}\delta(f-f_0)$? [duplicate]

How do you show the Discrete Time Fourier Transform of $x[n]=\cos(2\pi f_0n)$ is $\frac{1}{2} \delta(f+f_0) + \frac{1}{2}\delta(f-f_0)$?

Here is my thought process:

The definition of DTFT in my textbook is: $$X(f) = \sum_{n=-\infty}^{\infty} x[n] e^{-j2\pi fn}\quad\text{for}\quad \frac{1}{2} \leq f \leq \frac{1}{2}$$

Therefore for this specific problem we have: $$X(f) = \sum_{n=-\infty}^{\infty} \cos(2\pi f_0n) e^{-j2\pi fn}$$

I know from Euler's identity that $$\frac { e^{i\theta} + e^{-i\theta}}{2} = \cos\theta$$

So we can convert the cosine above into the complex exponential like so: \begin{align} X(f) &= \sum_{n=-\infty}^{\infty} \frac { e^{j2\pi f_0n} + e^{-j2\pi f_0n}}{2} e^{-j2\pi fn}\\ &=\sum_{n=-\infty}^{\infty} e^{-j2\pi n(f-f_0)} + e^{-j2\pi n(f+f_0)} \end{align}

This looks very close to the $\delta(f+f_0) + \frac{1}{2}\delta(f-f_0)$ I'm looking for - but I'm not sure how to get there.

• Here is a useful answer related to your question. – Gilles Nov 29 '17 at 21:12

The sequence $$~x[n] = \cos(\omega_0 n) ~$$ , $$-\infty < n < \infty$$, is neither absolutely nor square summable, therefore its DTFT formally does not exist; i.e., the DTFT sum does not converge to a finite number, but diverges to infinity.

Because of the extreme importance of that signal in the context of both the theory and the practice of signal processing, we would like to express its DTFT in an adequate and consistent manner under all operations of signal processing. One such possible representation makes use of the Dirac impulse functions with the following observation:

Let $$x[n] = 1$$, for all $$n$$, be a discrete-time (unstable) signal whose DTFT $$X(\omega)$$ given by the sum: $$X(\omega) = \sum_{n=-\infty}^{\infty} 1 e^{-j \omega n} \tag{1}$$

This sum, resembles a CTFS (continous time Fourier series) synthesis relation, remembering the fact that the DTFT $$X(\omega)$$ is a periodic function of continuous -frequency $$\omega$$. Then the above sum in Eq.1 is re-interpreted as a CTFS synthesis as:

$$X(\omega) = \sum_{n=-\infty}^{\infty} a_n e^{-j \frac{2\pi}{W} n \omega } \tag{2}$$ where $$\frac{2\pi}{W}$$ is the fundamental frequency with the fundamental period being $$W = 2\pi$$, and the CTFS coefficients are all $$a_n = 1$$. From CTFS pairs, we find the signal $$X(\omega)$$ whose CTFS coefficents are all $$1$$, as the Dirac impulse train :

$$X(\omega) = 2\pi \sum_{n=-\infty}^{\infty} \delta(\omega - 2\pi n) \tag{3}$$

and using the CTFS analysis equation, we have : \begin{align} a_n &= \frac{1}{2\pi} \int_{-\pi}^{\pi} \left( 2\pi \sum_{n=-\infty}^{\infty} \delta(\omega - 2\pi n) \right) e^{-j \frac{2 \pi}{2\pi} \omega n } d\omega \\\\ &= \frac{1}{2\pi} \int_{-\pi}^{\pi} 2\pi \delta(\omega) e^{-j \omega n } d\omega = 1 \tag{4} \end{align} The integral in Eq.4 is identical to an inverse DTFT of the impulse function $$2 \pi \delta(\omega)$$, hence we deduce that the DTFT $$X(\omega)$$ of $$x[n]=1$$ is $$x[n] = 1 \implies X(\omega) = 2\pi \delta(\omega) \tag{5}$$

Having found the DTFT of the sequence $$x[n]=1$$, we apply an estabslihed property of the DTFT :

$$x[n] \leftrightarrow X(\omega) \implies x[n]e^{j\omega_0 n} \leftrightarrow X(\omega - \omega_0) \tag{6}$$

$$1 \leftrightarrow 2 \pi \delta(\omega) \implies 1 \cdot e^{j\omega_0 n} \leftrightarrow 2\pi \delta(\omega - \omega_0) \tag{7} \\\\$$

to the seqeunce $$x[n] = \cos(\omega_0 n)$$ after decomposing it into complex exponentials:

\begin{align} x[n] &= \cos(\omega_0 n) = 0.5 \left( e^{j \omega_0 n} + e^{-j \omega_0 n} \right) \\ \\ X(\omega) &= \pi \delta( \omega - \omega_0) + \pi \delta(\omega + \omega_0) \tag{8} \\ \\ \end{align}

Eq.8 gives us the requested DTFT of the sequence $$x[n] = \cos(\omega_0 n)$$.

• I hope you teach somewhere so that your explanation skills reach the masses. atleast the DSP masses !!!!! – mark leeds Nov 30 '17 at 3:25
• @markleeds Thanks Mark! I'm glad if I was helpful... – Fat32 Nov 30 '17 at 9:58
• incredibly helpful. – mark leeds Dec 1 '17 at 19:05