[amp_mcq option1=”$$\frac{{\left( {1 – 0.6{z^{ – 1}}} \right)}}{{{z^{ – 1}}\left( {1 – 0.4{z^{ – 1}}} \right)}}$$” option2=”$$\frac{{{z^{ – 1}}\left( {1 – 0.6{z^{ – 1}}} \right)}}{{\left( {1 – 0.4{z^{ – 1}}} \right)}}$$” option3=”$$\frac{{{z^{ – 1}}\left( {1 – 0.4{z^{ – 1}}} \right)}}{{\left( {1 – 0.6{z^{ – 1}}} \right)}}$$” option4=”$$\frac{{\left( {1 – 0.4{z^{ – 1}}} \right)}}{{{z^{ – 1}}\left( {1 – 0.6{z^{ – 1}}} \right)}}$$” correct=”option3″]
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.