To find the Fourier Transform of the given signals, let's handle each one step by step.

a) $x(n) = \delta(n)$

The Fourier Transform of the discrete-time signal $\delta(n)$, where $\delta(n)$ is the Kronecker delta function, is a well-known result:

$X(\omega) = \mathcal{F}\{ \delta(n) \} = 1$

This is because $\delta(n)$ is the identity signal in the discrete-time domain, and its Fourier transform is constant (1) across all frequencies.

• Magnitude of $X(\omega)$: $|X(\omega)| = 1$ for all frequencies.
• Phase of $X(\omega)$: $\angle X(\omega) = 0$ for all frequencies.

b) $x(n) = a^n u(n)$

Here, $x(n) = a^n u(n)$, where $u(n)$ is the unit step function. The Fourier Transform of this signal can be computed using the standard formula for the discrete-time signal $a^n u(n)$, which is:

$X(\omega) = \mathcal{F}\{ a^n u(n) \} = \frac{1}{1 - a e^{-j\omega}} \quad \text{for} \ |a| < 1$

Where $X(\omega)$ is the Fourier Transform, and the magnitude and phase of this expression can be computed as follows:

• Magnitude of $X(\omega)$:

$|X(\omega)| = \frac{1}{|1 - a e^{-j\omega}|}$

• Phase of $X(\omega)$:

$\angle X(\omega) = \text{arg}\left( \frac{1}{1 - a e^{-j\omega}} \right)$

Now, let’s plot the magnitude and phase of these signals.

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Would you like me to go ahead and plot the graphs of the magnitude and phase for both of these functions (for a specific value of $a$ in case of the second signal), or would you prefer to focus on the theoretical details?