Question

The Z-transform of f(k)=sin5k, k20 is​

Answer 1

Step-by-step explanation:

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Answer 2

Correct question :

A sequence x(k), where k is an integer, is given as:

$x(k) = \left \{ {{0, k <0} \atop {sin5k, k\geq 0}} \right.$

Suppose,

y(k) = Σx(m)

Without using Z- transform table, determine Z-transform of y(k) based on the definition and properties  of the transform.

Solution :

x(k) = sin5k u(k)

sin5k = e^(j5k) - e^(-j5k)

                    2j

z transform of eᵃⁿ u(n)

= ᵃⁿ eᵃⁿ z⁻¹ u(n)

= Σ eᵃⁿ z⁻ⁿ

= Σ(eᵃz⁻¹)ⁿ = 1 / 1 - eᵃz⁻¹ = 1 / 1 - eᵃ / z

z{eᵃⁿ u(n)} = z / z - eᵃ

z{e^(j5k)} = z / z - e^(j5)

z{e⁻ᵃⁿ} = Σe⁻ᵃⁿz⁻ⁿ = Σ(e⁻ᵃz⁻¹)ⁿ = 1/(1-e⁻ᵃz⁻¹) = 1/(1-1/eᵃz) = eᵃz/(eᵃz-1) = z/z-e⁻ᵃ

z{e⁻ᵃⁿ} = z / z - e⁻ᵃ

z{e^(-j5k)} = z / z - e^(-j5)

$z(sin wku(k)) = \frac{1}{2j}[\frac{z}{z-e^{j5} } -\frac{z}{z-e^{-j5} } ]$

$=\frac{1}{2j} [\frac{z(z-e^{j5})-z(z-e^{-j5}) }{(z-e^{j5})(z-e^{-j5})} ]$

$=\frac{1}{2j}\frac{[z(z-e^{-j5})-(z-e^{j5} ) ] }{z^{2}-ze^{-j5} -ze^{j5}+1 }$

$=\frac{1}{2j}\frac{z[z-e^{-j5}-z+e^{j5} }{z^{2}-z[e^{j5} +e^{-j5} +1] }$

$=\frac{z\frac{e^{j5} -e^{-j5} }{2j} }{z^{2} -z\frac{2}{2}(e^{j5+e^{-j5} } )+1 }$

$=\frac{zsin5}{z^{2}-2zcos5+1 }$