sin(alpha k) is written as

(e^(j alpha k) - e^(-j alpha k))/(2j)

(e^(j alpha k) + e^(-j alpha k))/2

(e^(j alpha k) - e^(-j alpha k))/2

(e^(j alpha k) + e^(-j alpha k))/(2j)

The denominator becomes z^2 - 2z cos alpha + 1 because

(z - e^(j alpha))(z - e^(-j alpha)) = z^2 - 2z cos alpha + 1

e^(j alpha) + e^(-j alpha) = 2 sin alpha

z sin alpha/(z^2 - 2z cos alpha + 1), |z| > 1

z cos alpha/(z^2 - 2z cos alpha + 1), |z| > 1

z sin alpha/(z^2 - 2z sin alpha + 1), |z| > 1

z sin alpha/(z^2 - 2z cos alpha + 1), |z| < 1

2068 Magh · 2.5 marks

z-transform of sin αk

Short answer

Applying Euler's formula and Z[a^k]=z/(z-a) gives Z[sin αk] = z sin α / (z^2 - 2z cos α + 1), for |z| > 1.

Find the z-transform of sin αk.

Idea. Write sin αk using Euler's formula and use Z[aᵏ] = z/(z − a).

sin αk = (e^{jαk} − e^{−jαk}) / (2j)

Z[sin αk] = (1/2j) [ z/(z − e^{jα}) − z/(z − e^{−jα}) ]
          = (1/2j) · z (e^{jα} − e^{−jα}) / [ z² − z(e^{jα} + e^{−jα}) + 1 ]
          = (1/2j) · z (2j sin α) / (z² − 2z cos α + 1)

Answer: Z[sin αk] = z sin α / (z² − 2z cos α + 1), for |z| > 1.

✓ Answer checked by Claude Opus 5.5 (high effort)

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