Consider H_{a}(s)=\;{1}{(s+1)²}\;{and}\;T=0.1\;{s} Using bilinear transformation, H(z) is

Correct Answer is : 0.0476(1+z−1)2(1−0.9048z−1)2\;{0.0476(1+z^{-1})²}{(1-0.9048z^{-1})²}(1−0.9048z−1)20.0476(1+z−1)2

\;{0.0476(1+z)²}{(1-0.9048z^{-1})²}

\;{0.0476(1+z^{-1})²}{(1-0.9048z^{-1})²}

Correct Answer is : 0.0476(1+z−1)2(1−0.9048z−1)2\;{0.0476(1+z^{-1})²}{(1-0.9048z^{-1})²}(1−0.9048z−1)20.0476(1+z−1)2

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UPSC ESE 2025 Electronics and Telecommunications Paper

Question : 49 of 150

Marks: +1, -0

Consider

H_{a}(s)=\;\frac{1}{(s+1)^{2}}\;\text{and}\;T=0.1\;\text{s}

H(z)

Solution:

(d)

Step 1: Bilinear Transformation Formula

s=\;\frac{2}{T}\cdot\;\frac{1-z^{-1}}{1+z^{-1}}

Step 2: Substitute s into

H_{a}(s)

Given:

H_{a}(s)=\;\frac{1}{(s+1)^{2}}

Substitute s from bilinear transform:

H(z)=\left[\;\frac{1}{\left(\frac{2}{T}\cdot\frac{1-z^{-1}}{1+z^{-1}}+1\right)^{2}}\right]

Plug in

T=0.1

, Simplify and after algebraic manipulation and normalization we get:

H(z)=\;\frac{0.0476(1+z^{-1})^{2}}{(1-0.9048z^{-1})^{2}}

Hence, the correct option is (d).

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