Question

Consider a causal and stable LTI system with rational transfer function H(z), whose corresponding impulse response begins at n = 0. Furthermore, \(H\left( 1 \right) = \frac{5}{4}\). The poles of H(z) are \({p_k} = \frac{1}{{\sqrt 2 }}\exp \left( {j\frac{{\left( {2k - 1} \right)}}{4}\pi } \right)\) for k = 1, 2, 3, 4. The zeros of H(z) are all at z = 0. Let g[n] = jⁿh[n]. The value of g[8] equals __________. (Give the answer up to three decimal places.)

Answer 1

Poles of H(z): \({P_k} = \frac{1}{2}\exp \left( {\frac{{j\;\left( {2k - 1} \right)\pi }}{4}} \right)for\;k = 1,2,3,4\)

\({Z_1} = {P_1} = \frac{1}{2}\exp \left( {j\frac{\pi }{4}} \right) = \frac{1}{2} + \frac{j}{2}\)

\({Z_2} = {P_2} = \frac{1}{2}\exp \left( {j\frac{{3\pi }}{4}} \right) = - \frac{1}{2} + \frac{j}{2}\)

\({Z_3} = {P_3} = \frac{1}{2}\exp \left( {j\frac{{5\pi }}{4}} \right) = \frac{{ - 1}}{2} - \frac{j}{2}\)

\({Z_4} = {P_4} = \frac{1}{2}\exp \left( {j\frac{{7\pi }}{4}} \right) = \frac{1}{2} - \frac{j}{2}\)

Z₁ = -Z₃ and Z₂ = -Z­₄

Since, h(n) is causal and it start from n = 0 , so numerator will have same order as denominator.

Hence, transfer function is given by,

\(H\left( Z \right) = \frac{{k{Z^4}}}{{\left( {Z - {Z_1}} \right)\left( {Z - {Z_2}} \right)\left( {Z - {Z_3}} \right)\left( {Z - {Z_4}} \right)}}\)

\(= \frac{{k{Z^4}}}{{\left( {Z - {Z_1}} \right)\left( {Z - {Z_2}} \right)\left( {Z + {Z_1}} \right)\left( {Z + {Z_2}} \right)}}\)

\(= \frac{{k{Z^4}}}{{\left( {{Z^2} - Z_1^2} \right)\left( {{Z^2} - Z_2^2} \right)}}\)

\(= \frac{{k{Z^4}}}{{\left( {{Z^2} - \frac{1}{2}j} \right)\left( {{Z^2} + \frac{1}{2}j} \right)}}\)

\(H\left( Z \right) = \frac{{k{Z^4}}}{{{Z^4} + \frac{1}{4}}}\)

\(H\left( 1 \right) = \frac{5}{4}\)

\(\Rightarrow \frac{5}{4} = \frac{{k{{\left( 1 \right)}^4}}}{{1 + \frac{1}{4}}}\)

\(\Rightarrow K = \frac{{25}}{{16}}\)

\(\Rightarrow H\left( z \right) = \frac{{25}}{{16}}.\frac{{{Z^4}}}{{{Z^4} + \frac{1}{4}}} = \frac{{25}}{{16}}\left[ {\frac{1}{{1 + \frac{1}{4}{x^4}}}} \right]\)

By using long division rule, H(Z) can be expressed as,

\(= \frac{{25}}{{16}}\left[ {1 - \frac{1}{4}{Z^{ - 4}} + \frac{1}{{16}}{Z^{ - 8}} + \ldots } \right]\)

By applying inverse Z-transform,

\(h\left( n \right) = \left\{ {\frac{{25}}{{16}},0,0,0,\frac{{ - 25}}{{64}},0,0,0,\frac{{25}}{{256}}, \ldots } \right\}\)

\(h\left( 8 \right) = \frac{{25}}{{256}}\)

since, g(n) = jⁿ × h(n)

\(g\left( 8 \right) = {j^8}h\left( 8 \right) = \frac{{25}}{{256}} = 0.0976\)