The z-transform of a signal is given by $$C\left( z \right) = {1 \over 4}{{{z^{ – 1}}\left( {1 – {z^{ – 4}}} \right)} \over {{{\left( {1 – {z^{ – 1}}} \right)}^2}}}.$$ Its final value is

The correct answer is A.

The final value of a z-transform is the value of the z-transform as $z \to 1$. In this case, we have

$$C\left( z \right) = {1 \over 4}{{{z^{ – 1}}\left( {1 – {z^{ – 4}}} \right)} \over {{{\left( {1 – {z^{ – 1}}} \right)}^2}}} \to {1 \over 4}{{{z^{ – 1}}\left( {1 – 0} \right)} \over {{{\left( {1 – 0} \right)}^2}}} = {1 \over 4}$$

as $z \to 1$.

Option B is incorrect because the z-transform does not approach zero as $z \to 1$. Option C is incorrect because the z-transform does not approach 1 as $z \to 1$. Option D is incorrect because the z-transform is finite as $z \to 1$.