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 $$x\left( n \right) \to \boxed{{H_1}\left( z \right) = \frac{{\left( {1 – 0.4{z^{ – 1}}} \right)}}{{\left( {1 – 0.6{z^{ – 1}}} \right)}}} \to \boxed{{H_2}\left( z \right)} \to y\left( n \right)$$

The correct answer is $\boxed{{H_2}\left( z \right) = \frac{{z^{ – 1}}\left( {1 – 0.4{z^{ – 1}}} \right)}}{{\left( {1 – 0.6{z^{ – 1}}} \right)}}}$.

The transfer function of a system is the ratio of its output to its input, evaluated at $z=e^{j\omega}$. The transfer function of a cascade of two systems is the product of the transfer functions of the individual systems.

In this case, the first system is a first-order low-pass filter with a transfer function of $\frac{{\left( {1 – 0.4{z^{ – 1}}} \right)}}{{\left( {1 – 0.6{z^{ – 1}}} \right)}}$. The second system is a first-order delay with a transfer function of $z^{-1}$.

The transfer function of the cascade of these two systems is therefore:

$$H_2(z) = H_1(z)H_2(z) = \frac{{\left( {1 – 0.4{z^{ – 1}}} \right)}}{{\left( {1 – 0.6{z^{ – 1}}} \right)}}\cdot z^{-1} = \frac{{z^{ – 1}}\left( {1 – 0.4{z^{ – 1}}} \right)}}{{\left( {1 – 0.6{z^{ – 1}}} \right)}}$$

This transfer function has a pole at $z=0.6$ and a zero at $z=0.4$. The pole is closer to the unit circle than the zero, so the system is stable. The system has a time delay of one unit, as desired.