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Z-transform of a conjugated sequence ("a straightforward" exercise)
The conjugation property is expressed as [tex]x^*[n] \stackrel{Z}{\leftrightarrow} X^*(z^*)[/tex]
This property follows in a straightforward maner from the definition of the [itex]z[/itex]-transform, the details of which are left as an exercise.
Z-transform definition: [tex]X(z)=\sum_{n=-\infty}^\infty x[n]z^{-n}[/tex]
Given a complex sequence, its z-transform is [tex]Z\{x[n]\} = \sum_{n=-\infty}^\infty (x_R[n] + jx_I[n]) z^{-n} = X_R(z) + jX_I(z) = X(z)[/tex]
Hence, the z-transform of a conjugated sequence [tex]Z\{x^*[n]\} = \sum_{n=-\infty}^\infty (x_R[n] - jx_I[n]) z^{-n} = X_R(z) - jX_I(z) = X^*(z)[/tex]
Now, how come I didn't get the [itex]z^*[/itex], as in [itex]X^*(z^*)[/itex]?
Homework Statement
The conjugation property is expressed as [tex]x^*[n] \stackrel{Z}{\leftrightarrow} X^*(z^*)[/tex]
This property follows in a straightforward maner from the definition of the [itex]z[/itex]-transform, the details of which are left as an exercise.
Homework Equations
Z-transform definition: [tex]X(z)=\sum_{n=-\infty}^\infty x[n]z^{-n}[/tex]
The Attempt at a Solution
Given a complex sequence, its z-transform is [tex]Z\{x[n]\} = \sum_{n=-\infty}^\infty (x_R[n] + jx_I[n]) z^{-n} = X_R(z) + jX_I(z) = X(z)[/tex]
Hence, the z-transform of a conjugated sequence [tex]Z\{x^*[n]\} = \sum_{n=-\infty}^\infty (x_R[n] - jx_I[n]) z^{-n} = X_R(z) - jX_I(z) = X^*(z)[/tex]
Now, how come I didn't get the [itex]z^*[/itex], as in [itex]X^*(z^*)[/itex]?
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