Consider the following periodic and time-discrete signal:

$$y[n] = \sin (\Omega_0 n+\phi)$$

How do I determine the discrete-time Fourier transform for $-\pi<\Omega<\pi$ for that signal?

I know that the $\theta$ causes a time shift, equal to $e^{j\Omega\theta}X[\Omega]$ in the frequency domain. But how is $x[n]=\sin(\Omega_0n)$ transformed into the following?

$$X[\Omega]=\frac{\pi}{j}\sum_{k=-\infty}^{\infty} \{\delta(\Omega-\Omega_0 -2\pi k)-\delta/\Omega+\Omega_0 -2\pi k)$$

Where does $\pi$ come from and the delta function?

Thank you!


1 Answer 1


First, an example with simple exponential

We start off by a simplified example with the determining the Discrete-Time Fourier Transform (DTFT) of the exponential function given by $$x_1[n]=e^{i\Omega_0n} \tag{1}.$$

Using the Fourier Transform $$\mathcal F\{x_1[n] \} \tag{2}=X_1(\Omega)=\sum_{k=-\infty}^{\infty}x_1[k]e^{-i\Omega k}$$

we can calculate $\mathcal F \{ x_1[n]\}$ by substituting $x_1[n]$ and simplify the expression as follows:

$$\sum_{k=-\infty}^{\infty}e^{i\Omega_0 k}e^{-i\Omega k}=\sum_{k=-\infty}^{\infty}e^{i(\Omega_0-\Omega)k} \tag{3}$$

Then, by defining $\Omega'=\Omega - \Omega_0$ (observe the changed order) so that

$$\sum_{k=-\infty}^{\infty}1\cdot e^{-i\Omega'k} \tag{4}$$

we cleary see that it's now just the DTFT of a constant $1$, i.e. $\mathcal F \{1 \}$, which is given by $$\mathcal F \{1 \}=2\pi \delta(\Omega') \tag{5}$$ and $2\pi \delta(\Omega-\Omega_0)$ in our case. Now, why is this? Where does the $2\pi$ come from?

So we have assumed that $\mathcal F \{1 \} = 2\pi\delta(\Omega)$, which defines the constant $1$ as a function $x_2[n]=1$. If we use the Inverse Discrete-Time Fourier Transform (IDTFT)

$$\mathcal F^{-1} \{ X_2(\Omega) \} = \frac{1}{2\pi} \int_{-\pi}^{\pi} X_2(\Omega)e^{i\Omega n} d\Omega \tag{6}$$

and substitute $X_2(\Omega)$ for our assumed $2\pi\delta(\Omega)$ we get

$$\frac{1}{2\pi} \int_{-\pi}^{\pi} 2\pi\delta(\Omega)e^{i\Omega n} d\Omega \tag{7}$$

which, since $\delta(\Omega)$ is "on" or $1$ only for $\Omega=0$, evaulates to

$$\frac{2\pi}{2\pi} e^{i\Omega 0} = 1. \tag{8}$$

We have thereby shown that the DTFT of a constant $1$ is equal to $2\pi\delta(\Omega)$.

If we now continue from equation $(4)$, we have that the resulting DTFT of an exponential $x_1[n]=e^{i\Omega_0n}$ becomes

$$\mathcal F\{x_1[n] \} = \sum_{k=-\infty}^{\infty}1\cdot e^{-i\Omega'k} = 2\pi\delta(\Omega') = 2\pi\delta(\Omega - \Omega_0) \tag{9}.$$

Secondly, show Discrete-Time Fourier Transform of a sine

We have the discrete-time function $$y[n]=\sin(\Omega_0n+\phi) \tag{10}.$$

Observe that the $\phi$ causes a time-shift and can be handled separately. Therefore consider, for now, only the function

$$x[n]=\sin(\Omega_0n) \tag{11}.$$

Using Euler's formula we can write the sine as exponentials in the form

$$x[n]=\frac{1}{2i} \left ( e^{i\Omega_0n} - e^{-i\Omega_0n} \right ) \tag{12}$$

Using the DTFT of $x[n]$ we get

$$\mathcal F\{ x[n] \} = \sum_{k=-\infty}^{\infty}\frac{1}{2i} \left ( e^{i\Omega_0k} - e^{-i\Omega_0k} \right ) e^{-i\Omega k} \tag{13}$$

which can be rewritten as

$$\frac{1}{2i} \left [ \sum_{k=-\infty}^{\infty}e^{-i(\Omega-\Omega_0)k} - \sum_{k=-\infty}^{\infty}e^{-i(\Omega+\Omega_0)k} \right] \tag{14}.$$

We can now see that each term in equation $(14)$ is on the form in equation $(9)$ and therefore we get

$$\frac{1}{2i} \left [ 2\pi\delta(\Omega-\Omega_0) - 2\pi\delta(\Omega+\Omega_0 \right] \tag{15}$$


$$i\pi \left [ \delta(\Omega+\Omega_0) - \delta(\Omega-\Omega_0) \right] \tag{16}$$

if we shift the order of the terms to compensate the negative sign when $i$ jumps up from the denominator.

To consider the time-shift $\phi$ we mutltiply our results with $e^{i\Omega\phi}$ and finally get

$$e^{i\Omega\phi} \cdot i\pi \left [ \delta(\Omega+\Omega_0) - \delta(\Omega-\Omega_0) \right] \tag{17}$$

I'm not a hundred percent certain of my answer and I don't fully understand how to get the general form in the questions asked. If someone has edits to suggest, please do so.

The solutions are inspired and based on the following YouTube videos:


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .