Fdtxjr

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 $e^jOmegathetaX[Omega]$ in the frequency domain. But how is $x[n]=sin(Omega_0n)$ transformed into the following?

$$X[Omega]=fracpijsum_k=-infty^infty {delta(Omega-Omega_0 -2pi k)-delta/Omega+Omega_0 -2pi k)$$

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

Thank you!

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
$$x_1[n]=e^iOmega_0n tag1.$$

Using the Fourier Transform
$$mathcal Fx_1[n] tag2=X_1(Omega)=sum_k=-infty^inftyx_1[k]e^-iOmega k$$

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

$$sum_k=-infty^inftye^iOmega_0 ke^-iOmega k=sum_k=-infty^inftye^i(Omega_0-Omega)k tag3$$

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

$$sum_k=-infty^infty1cdot e^-iOmega'k tag4$$

we cleary see that it's now just the DTFT of a constant $1$, i.e. $mathcal F 1 $, which is given by
$$mathcal F 1 =2pi delta(Omega') tag5$$
and $2pi delta(Omega-Omega_0)$ in our case. Now, why is this? Where does the $2pi$ come from?

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

$$mathcal F^-1 X_2(Omega) = frac12pi int_-pi^pi X_2(Omega)e^iOmega n dOmega tag6$$

and substitute $X_2(Omega)$ for our assumed $2pidelta(Omega)$ we get

$$frac12pi int_-pi^pi 2pidelta(Omega)e^iOmega n dOmega tag7$$

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

$$frac2pi2pi e^iOmega 0 = 1. tag8$$

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

If we now continue from equation $(4)$, we have that the resulting DTFT of an exponential $x_1[n]=e^iOmega_0n$ becomes

$$mathcal Fx_1[n] = sum_k=-infty^infty1cdot e^-iOmega'k = 2pidelta(Omega') = 2pidelta(Omega - Omega_0) tag9.$$

Secondly, show Discrete-Time Fourier Transform of a sine

We have the discrete-time function
$$y[n]=sin(Omega_0n+phi) tag10.$$

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

$$x[n]=sin(Omega_0n) tag11.$$

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

$$x[n]=frac12i left ( e^iOmega_0n - e^-iOmega_0n right ) tag12$$

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

$$mathcal F x[n] = sum_k=-infty^inftyfrac12i left ( e^iOmega_0k - e^-iOmega_0k right ) e^-iOmega k tag13$$

which can be rewritten as

$$frac12i left [ sum_k=-infty^inftye^-i(Omega-Omega_0)k - sum_k=-infty^inftye^-i(Omega+Omega_0)k right] tag14.$$

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

$$frac12i left [ 2pidelta(Omega-Omega_0) - 2pidelta(Omega+Omega_0 right] tag15$$

and

$$ipi left [ delta(Omega+Omega_0) - delta(Omega-Omega_0) right] tag16$$

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 $e^iOmegaphi$ and finally get

$$e^iOmegaphi cdot ipi left [ delta(Omega+Omega_0) - delta(Omega-Omega_0) right] tag17$$

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:

• https://www.youtube.com/watch?v=El1pSEUxg-c&


• https://www.youtube.com/watch?v=3USY_IHOfnk