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Let $\ell^1(\mathbb{Z})$ be the space of biinfinite sequences $f = (f(n))_{n \in \mathbb{Z}} \subset \mathbb{C}$ such that it is absolutely summable. The discrete Fourier transform or Fourier series is defined by, $$ \mathcal{F}(f)(e^{i \theta}) = \sum_{n \in \mathbb{Z}} f(n) e^{-in\theta}, \quad \theta \in (-\pi,\pi]. $$ The inverse discrete Fourier transform is defined as, $$ \mathcal{F}^{-1}(F)(n) = \frac{1}{2\pi} \int_{-\pi}^\pi F(e^{i \theta}) e^{in\theta} \, d\theta, \quad n \in \mathbb{Z}. $$

On the other hand, in asymptotics theory, we have the stationary phase approximation, where integrals of the form, $$ \int_\mathbb{R} g(x) e^{ikf(x)} \, dx, $$ can be estimated when $k \to \infty$, provided all critical points of $f$ are nondegenerate, i.e., $f''(x) \neq 0$. Hence, this theory doesn't directly apply to the inverse discrete Fourier transform above.

Question

Is there a theory that applies to the inverse discrete Fourier transform when $n \to \infty$ for asymptotic results? Any book or article recommendations on this topic?

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There is no asymptotic expression for $f[\infty]$ under DTFT (Discrete Time Fourier Transform). However, there is an expression for $f[\infty]$ under Z Transform, which is the equivalent of Laplace Transform for discrete sequences. DTFT is Z Transform when evaluated on the unit circle in the complex plane.

Final Value Theorem with Z Transform can be expressed using the following under suitable constraints noted in the Wikipedia page:

$$f[\infty] = \lim\limits_{z \to 1} (z-1) F(z)$$

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  • $\begingroup$ Thank you for your response! If I understand correctly, you're suggesting that I convert my DTFT to the Z-transform and then apply the asymptotic behavior you're describing. Is that correct? $\endgroup$ Commented Oct 16, 2024 at 13:31

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