Amplitude Spectrum, Nyquist Frequency, mixed/min/max wavelets The problem is here. Now I know the definition of mixed/max/min phase wavelets, whether the roots lie within the unit circle or not. Starting from n = 1, let
$$ x_t = ( 5, 6) $$ 
$$ X(z) = 5 + 6z $$
This would have root -5/6, making it a max phase wavelet (since it's within unit circle). If we switch 5 and 6 around in xt, we would get root -6/5, which would make it a min phase wavelet.

I don't know how I would show they have same amplitude spectrum. I know |X(f)| amplitude spectrum would lie between -fc and fc (Nyquist frequency) and fc = 1/2*delta t. Not sure how I would go about proving it.
Note that moving between minimum phase, maximum phase, and mixed phase involves basically two operation in the time domain:


shifting
time-reversal


If $X(e^{j\omega})$ is the Fourier transform of the sequence $x(n)$ then the transforms of the shifted and the time-reversed sequences are given by

$$x(n-k)\Longleftrightarrow e^{-jk\omega}X(e^{j\omega})\\
x(-n)\Longleftrightarrow X(e^{-j\omega})$$

Since $|e^{-jk\omega}|=1$, the magnitudes of the spectra of $x(n)$ and $x(n-k)$ are identical. Furthermore, if $x(n)$ is real-valued then $X(e^{-j\omega})=X^*(e^{j\omega})$ and consequently the spectra of $x(n)$ and $x(-n)$ also have the same magnitudes.