Two systems H₁(z) and H₂(z) are connected in cascade as shown below. The overall output y(n) is the same as the input x(n) with a one unit delay. The transfer function of the second system H₂(z) is
$$x\left( n \right) \to \boxed{{H_1}\left( z \right) = \frac{{\left( {1 - 0.4{z^{ - 1}}} \right)}}{{\left( {1 - 0.6{z^{ - 1}}} \right)}}} \to \boxed{{H_2}\left( z \right)} \to y\left( n \right)$$

A. $$\frac{{\left( {1 - 0.6{z^{ - 1}}} \right)}}{{{z^{ - 1}}\left( {1 - 0.4{z^{ - 1}}} \right)}}$$

B. $$\frac{{{z^{ - 1}}\left( {1 - 0.6{z^{ - 1}}} \right)}}{{\left( {1 - 0.4{z^{ - 1}}} \right)}}$$

C. $$\frac{{{z^{ - 1}}\left( {1 - 0.4{z^{ - 1}}} \right)}}{{\left( {1 - 0.6{z^{ - 1}}} \right)}}$$

D. $$\frac{{\left( {1 - 0.4{z^{ - 1}}} \right)}}{{{z^{ - 1}}\left( {1 - 0.6{z^{ - 1}}} \right)}}$$

Answer: Option B