at the moment I am reading the following paper

Benno, Steven A., and José MF Moura. "On translation invariant subspaces and critically sampled wavelet transforms." Multidimensional Systems and Signal Processing 8.1-2 (1997): 89-110.

The step between the equations (10) and (11) I can't comprehend. The step is as follows

$$\int_R G(\omega) \overline{\tilde{G}(\omega)} e^{-j2\pi\omega\tau} \Big(\sum_k e^{-j2\pi(f-\omega)k}\Big)d\omega = $$ $$\int_R G(\omega) \overline{\tilde{G}(\omega)} e^{-j2\pi\omega\tau} \Big(\sum_k \delta(f-\omega+k)\Big) d\omega. $$

I know that $e^{-i2\pi k}$ for $k\in\mathbb{Z}$ is an orthnormal basis, but not over $\mathbb{R}$ and I have no idea, why $k$ comes into the dirac function with an "+". At most I woud expect something like $\delta(f-\omega)$ since it is in a product with k.

Has someone an idea about this?

Thanks Matthias

  • 1
    $\begingroup$ good catch. interesting too. $\endgroup$ – Nick Jan 5 '19 at 20:56

Clearly there is a typo in equation (10), the correct expression is $$\int_R G(\omega) \overline{\tilde{G}(\omega)} e^{-j2\pi\omega\tau} \Big(\sum_k e^{-j2\pi(f-\omega+k)}\Big)d\omega, $$ which can be seen by substituting the definition of $a_k$ exactly as the authors describe.

  • $\begingroup$ Hi, thank you for your answer. I checked equation (10) if $a_k$ is inserted. The sum is reordered by k and (10) should be correct. $\endgroup$ – Matthias Lauber Jan 6 '19 at 11:19

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