Let X(Ω) is the DTFT of the signal x[n], recall X(Ω) is periodic with 2π hence, X(Ω+2π) = X(Ω),

Step 1: State the inverse DTFT and periodicity. X(Ω) is 2π-periodic: X(Ω+2π)=X(Ω) for all Ω. The

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Let $X(\Omega)$ is the DTFT of the signal $x[n]$, recall $X(\Omega)$ is periodic with $2\pi$ hence, $X(\Omega+2\pi) = X(\Omega)$, $\forall \Omega$. Show that for $0 < \beta_o < 2\pi$, following holds: $x[n] = \frac{1}{2\pi} \int_{-\pi}^{\pi} X(\Omega)e^{j\Omega n}d\Omega = \frac{1}{2\pi} \int_{-\pi + \beta_o}^{\pi + \beta_o} X(\Omega)e^{j\Omega n}d\Omega$ This is true for any $\beta_o \in \mathbb{R}$ but it is enough to show for $0 < \beta_o < 2\pi$ for this exercise. This means for the inverse DTFT, we can start integrating from anywhere as long as we cover $2\pi$.

          Let $X(\Omega)$ is the DTFT of the signal $x[n]$, recall $X(\Omega)$ is periodic with $2\pi$ hence, $X(\Omega+2\pi) = X(\Omega)$, $\forall \Omega$. Show that for $0 < \beta_o < 2\pi$, following holds:

$x[n] = \frac{1}{2\pi} \int_{-\pi}^{\pi} X(\Omega)e^{j\Omega n}d\Omega = \frac{1}{2\pi} \int_{-\pi + \beta_o}^{\pi + \beta_o} X(\Omega)e^{j\Omega n}d\Omega$

This is true for any $\beta_o \in \mathbb{R}$ but it is enough to show for $0 < \beta_o < 2\pi$ for this exercise. This means for the inverse DTFT, we can start integrating from anywhere as long as we cover $2\pi$.

Let X(Ω) is the DTFT of the signal x[n], recall X(Ω) is periodic with 2π hence, X(Ω+2π) = X(Ω), ∀Ω. Show that for 0 < < 2π, following holds:

x[n] = (1)/(2π)∫-π^π X(Ω)e^jΩ ndΩ = (1)/(2π)∫-π + ^π + X(Ω)e^jΩ ndΩ

This is true for any ∈ℝ but it is enough to show for 0 < < 2π for this exercise. This means for the inverse DTFT, we can start integrating from anywhere as long as we cover 2π.

Added by Erica H.

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