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I am trying to figure out what the Fractional Fourier Transform of the signal $\sqrt{c} x(c(t-\tau))$ would be with respect to that of $x(t)$.

According to the paper "The Fractional Fourier Transform and Time-Frequency Representations" by Luis B. Almeida in IEEE TRANSACTIONS ON SIGNAL PROCESSING, VOL. 42, NO. 11, NOVEMBER 1994, we have:

$$ \mathcal{F}_{\alpha}\left\{ x(t-\tau) \right\} = X_{\alpha}\left( \xi_a - \tau \cdot \text{cos}(\alpha) \right) e^{j \frac{\tau^2}{2} \text{sin}(\alpha) \text{cos}(\alpha) - j \xi_a \tau \text{sin}(\alpha) } $$

where $\alpha$ is the FrFT angle, $a = \frac{2 \alpha}{\pi}$ is the order, $j = \sqrt{-1}$, and $\xi_a$ is the FrFT spectral variable.

Similarly, according to the paper:

$$ \mathcal{F}_{\alpha}\left\{ x(ct) \right\} = \sqrt{\frac{1 - j \text{cot}(\alpha)}{c^2 - j \text{cot}(\alpha)}} X_{\beta} \left( \xi_a \frac{\text{sin}(\beta)}{c \text{sin}(\alpha)} \right) e^{ j \frac{\xi_a}{2} \text{cot}\left( \alpha \left( 1 - \frac{\text{cos}^2(\beta)}{\text{cos}^2(\alpha)} \right) \right)} $$

where $\beta = \text{arctan}(c^2 \text{tan}(\alpha))$.


Question:

So does this mean that:

$$ \mathcal{F}_{\alpha}\left\{ \sqrt{c} x(c(t - \tau)) \right\} = \sqrt{c} \sqrt{\frac{1 - j \text{cot}(\alpha)}{c^2 - j \text{cot}(\alpha)}} X_{\beta} \left( \xi_a \frac{\text{sin}(\beta)}{c \cdot \text{sin}(\alpha)} - c \tau \text{cos}(\alpha) \right) e^{j \frac{c^2 \tau^2}{2} \text{sin}(\alpha) \text{cos}(\alpha) - j c \xi_a \tau \text{sin}(\alpha) } e^{ j \frac{\xi_a}{2} \text{cot}\left( \alpha \left( 1 - \frac{\text{cos}^2(\beta)}{\text{cos}^2(\alpha)} \right) \right)} $$

or do we have something else because $\beta \neq \alpha$?

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