enter image description here [from Discrete-time Signal Processing by Oppenheim and Schafer, 3rd ed.]

I was not able to prove this:

If the analysis and synthesis filters are ideal so that they exactly split the band 0 ≤ |ω| ≤ π into two equal segments without overlapping, then it is straightforward to verify that Y(ejω) = X(ejω); i.e., the synthesis filter bank reconstructs the input signal exactly.


$G_0(e^{j\omega})$ and $H_0(e^{j\omega})$ are ideal low pass filters with passband $[0,\pi/2]$ and stopband $[\pi/2,\pi]$. $G_1(e^{j\omega})$ and $H_1(e^{j\omega})$ are ideal high pass filters that are complementary to the lowpass filters, i.e., their passbands coincide with the lowpass filters' stopbands and vice versa.

Note that shifting the frequency axis by $\pi$ transforms a lowpass filter to a highpass filter and vice versa. I.e., $H_0(e^{j(\omega-\pi)})$ is a highpass filter, and $H_1(e^{j(\omega-\pi)})$ is a lowpass filter. Consequently, $G_0(e^{j\omega})H_0(e^{j(\omega-\pi)})=0$ and $G_1(e^{j\omega})H_1(e^{j(\omega-\pi)})=0$. So the second term $(111b)$ corresponding to the aliased input spectrum vanishes. We're left with the first term, which is constant because we add an ideal lowpass spectrum and its complementary highpass spectrum.


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