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For the system to be invertible, we should have different outputs for different inputs.

In terms of constant functions say,

$$X_1[n]=3 \quad \forall n \in \mathbb{Z}$$ and $$X_2[n]=4 \quad \forall n \in \mathbb{Z}$$

It is perfectly clear that the above system gives 0 for both the inputs hence can be concluded as non invertible. But how can this be explained in terms of transform method (Bilateral z transform or unilateral z transform )?

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The impulse response of your system is $h[n] = \delta[n] - \delta[n-1]$ and its Z-transform is accordingly

$$H(z)=1-z^{-1}$$

Now, you can evaluate the frequency response of this system:

$$ |H(z)|_{z=\exp(j\omega)}=|1-\exp(j\omega)| $$

you see, that $H(z)|_{z=\exp(j0)}=0$ (i.e. it vanishes for $\omega=0$). Now, in order for a filter to be invertible, its Z-Transform on the unit circle (i.e. when setting $z=\exp(j\omega)$ must not vanish. If it does not vanish, the inverse is given by $\frac{1}{H(z)}$. If it vanishes, the inverse does not exist.

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  • $\begingroup$ to put it a little differently, the original filter $$ H(z) = 1 - z^{-1}$$ completely destroys DC. it is a "DC blocker" and it blocks DC by multiplying DC by zero. an inverse filter would be able to undo what the original filter does to a signal, but it can't for DC. it cannot divide by zero and it cannot recover information that is totally lost. $\endgroup$ – robert bristow-johnson Jan 9 '17 at 22:25
  • $\begingroup$ and, putting it another way, the inverse filter turns poles into zeros and zeros into poles. that swaps the numerator and denominator which causes the zeros in the original to cancel the poles in the inverse filter. when the zero on the unit circle becomes a pole, the inverse filter is unstable (or, at best, marginally stable). $\endgroup$ – robert bristow-johnson Jan 9 '17 at 22:31
  • $\begingroup$ Clearly, you meant $|H(z)|_{z=\exp(j\omega)}=|1-\exp(-j\omega)|$, i.e. with the minus sign on the exponent, right? $\endgroup$ – Jason Jan 15 '18 at 21:20

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