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Suppose I have a signal in the frequency domain $x(\omega)$.

Suppose I know that its Nyquist rate is $\omega_0$ (and let $\omega_m$ be the its maximum frequency).

I need to get the Nyquist rate of

$$x(\omega)+x(\omega)e^{-j\omega}$$

The answers say it's still $\omega_0$.

Why is it not

$$\omega_m+\omega_me^{-j\omega}$$

and if we substitute $2\omega_m = \omega_0$

$$\frac{1}{2}\omega_0+\frac{1}{2}\omega_0e^{-j\omega}$$

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Let $Y(\omega)=X(\omega)+X(\omega)e^{-j\omega}$ with $X(\omega)$satisfying $X(\omega)=0$ for $|\omega|>\omega_m$. Then, by its definition, $Y(\omega)$ also satisfies $Y(\omega)=0$ for $|\omega|>\omega_m$, i.e., the maximum frequency remains unchanged, and, consequently, also the Nyquist frequency remains unchanged.

Note that if $x(t)$ is the signal corresponding to the spectrum $X(\omega)$, then $X(\omega)e^{-j\omega}$ corresponds to the signal $x(t-1)$, which clearly has the same maximum frequency as $x(t)$.

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