Suppose $\mathcal { F } \{ f ( x ) \} = F ( \alpha ) , \mathcal { F } \{ g ( x ) \} = G ( \alpha )$

question 9

Multiple Choice

Suppose $\mathcal { F } \{ f ( x ) \} = F ( \alpha ) , \mathcal { F } \{ g ( x ) \} = G ( \alpha )$ . In the convolution theorem, the formula for the Fourier transform is Select all that apply.

• A) \(\int _ { - \infty } ^ { \infty } f ( \tau ) g ( t - \tau ) d \tau = \mathcal { F } ^ { - 1 } \{ F ( \alpha ) G ( \alpha ) \}\)
• B) \(\int _ { - \infty } ^ { \infty } f ( t - \tau ) g ( \tau ) d \tau = \mathcal { F } ^ { - 1 } \{ F ( \alpha ) G ( \alpha ) \}\)
• C) \(\mathcal { F } \left\{ \int _ { - \infty } ^ { \infty } f ( \tau ) g ( t - \tau ) d \tau \right\} = F ( \alpha ) G ( \alpha )\)
• D) \(\mathcal { F } \left\{ \int _ { - \infty } ^ { \infty } f ( t - \tau ) g ( \tau ) d \tau \right\} = F ( \alpha ) G ( \alpha )\)
• E) none of the above

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