Let x(n) be an arbitrary signal, not necessarily real valued, with DTFT X(ω). What is the DTFT of x*(–n)?
Question
Let x(n) be an arbitrary signal, not necessarily real valued, with DTFT X(ω). What is the DTFT of x*(–n)?
Solution
The Discrete Time Fourier Transform (DTFT) of a signal x(n) is given by X(ω).
The DTFT of the complex conjugate of the time-reversed signal, x*(-n), is given by the complex conjugate of the frequency-reversed spectrum, X*(-ω).
Here's the step-by-step process:
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The DTFT of x(n) is X(ω) = Σ x(n) e^(-jωn) where the sum is over all n.
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The time-reversed signal is x(-n).
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The complex conjugate of a signal is denoted by , so the complex conjugate of the time-reversed signal is x(-n).
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The DTFT of x*(-n) is then X*(-ω) = Σ x*(-n) e^(jωn) where the sum is over all n.
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This is the complex conjugate of the frequency-reversed spectrum, X*(-ω).
So, the DTFT of x*(-n) is X*(-ω).
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