Phaseless Signal Recovery in Infinite Dimensional Spaces Using Structured Modulations

Abstract

This paper considers the recovery of continuous signals in infinite dimensional spaces from the magnitude of their frequency samples. It proposes a sampling scheme which involves a combination of oversampling and modulations with complex exponentials. Sufficient conditions are given such that almost every signal with compact support can be reconstructed up to a unimodular constant using only its magnitude samples in the frequency domain. Finally it is shown that an average sampling rate of four times the Nyquist rate is enough to reconstruct almost every time-limited signal.

Appendix: An Auxiliary Lemma

Lemma 7.1

Let $S$ be a sine-type function of type $\sigma$ and let $\Lambda = \{\lambda _n\}_{n\in \mathbb {Z} }$ be its zero set. If $\lambda _{n_{0}} \in \Lambda$ is an arbitrary zero of $S$ then

$$\begin{aligned} \sum _{n\ne n_{0}} \frac{1}{|\lambda _n - \lambda _{n_{0}}|^{2}} \le C < \infty \end{aligned}$$

with a constant $C$ which depends only on $S$ but not on $n_{0}$.

Proof

Since $S$ is a sine-type function there are constants $A,B,H$ such that inequalities (6) hold. Furthermore there exists a constant $\delta > 0$ such that $|\lambda _{m} - \lambda _{n}| \ge \delta$ for all $m\ne n$ and a constant $\alpha >0$ such that $|S'(\lambda _{n})| \ge \alpha$ for all $\lambda _{n} \in \Lambda$ (see, e.g., [20, 37]). In particular, $S$ is an entire function of exponential type $\sigma$. Then the Phragmén–Lindelöf principle and (6) imply that $S$ is bounded on every line parallel to $\mathbb {R}$. Therefore there exists a constant $M$ such that

$$\begin{aligned} |S(\xi + \mathrm {i} \eta )| \le M\, \mathrm {e} ^{\sigma |\eta |}\, \quad \text {for all}\ \xi , \eta \in \mathbb {R} . \end{aligned}$$

(32)

Set $\widetilde{S}(z) := S(z + \lambda _{n_{0}})$ and write $\lambda _{n_{0}} = \xi _{0} + \mathrm {i} \eta _{0}$. It follows from (6) that $|\eta _{0}| \le H$ and (32) gives

$$\begin{aligned} \big |\widetilde{S}(\xi + \mathrm {i} \eta )\big | \le M\, \mathrm {e} ^{\sigma |\eta + \eta _{0}|} \le M\, \mathrm {e} ^{\sigma H}\, \mathrm {e} ^{\sigma |\eta |} \quad \text {for all}\ \xi , \eta \in \mathbb {R} . \end{aligned}$$

Consequently $|\widetilde{S}(z)| \le \widetilde{M}\, \mathrm {e} ^{\sigma |z|}$ for all $z\in \mathbb {C}$ and with the constant $\widetilde{M} = M \mathrm {e} ^{\sigma H}$ which depends only on $S$ but not on $n_{0}$. The zeros of $\widetilde{S}$ are $\widetilde{\lambda }_{n} = \lambda _{n} - \lambda _{n_0}$, and we assume that they are ordered increasingly by their absolute values, i.e. such that $0 = |\widetilde{\lambda }_{0}| < \delta \le |\widetilde{\lambda }_1| \le |\widetilde{\lambda }_{2}| \le \cdots$. Then we define $Q(z) := \widetilde{S}(z)/z$. This is again an entire function of exponential type $\sigma$ which satisfies

$$\begin{aligned}&|Q(z)| \le P\, \mathrm {e} ^{\sigma |z|}\quad \text {for all}\ z\in \mathbb {C} \quad \text {and with}\ P = \widetilde{M}\, \mathrm {e} ^{\sigma } = M\, \mathrm {e} ^{\sigma (H+1)},\nonumber \\&\big | Q(0) \big | = \big | \widetilde{S}'(0) \big | = \big | \widetilde{S}'(\widetilde{\lambda }_{0}) \big | \ge \alpha , \end{aligned}$$

(33)

and the zero set of $Q$ is obviously $\{\widetilde{\lambda }_n\}^{\infty }_{n=1}$. If $n(r)$ denotes the number of zeros of $Q$ for which $|\widetilde{\lambda }_n| \le r$, then Jensen’s formula [20] and (33) imply

$$\begin{aligned} N(r) := \int ^{r}_{0} \frac{n(\tau )}{\tau }\, \mathrm {d} \tau = \frac{1}{2\pi } \int ^{\pi }_{-\pi } \ln |Q(r \mathrm {e} ^{ \mathrm {i} \theta })|\, \mathrm {d} \theta - \ln |Q(0)| \le \ln (P/\alpha ) + \sigma \, r. \end{aligned}$$

Since $n(r)$ is non-decreasing, we have $N(e r) \ge \int ^{e r}_{r} n(\tau )\, \tau ^{-1}\, \mathrm {d} \tau \ge n(r)$, where $e = \mathrm {e} ^{1}$. This yields the upper bound $n(r) \le \ln (P/\alpha ) + e\, \sigma \, r$. If we take $r = |\widetilde{\lambda }_{n}|$ then $n(r) = n$ and one gets $n = n(r) \le \ln (P/\alpha ) + e\, \sigma \, |\widetilde{\lambda }_n|$. Now we have

$$\begin{aligned} \sum _{n\ne n_{0}} \big |\lambda _n - \lambda _{n_{0}}\big |^{-2} = \sum ^{\infty }_{n=1} \big |\widetilde{\lambda }_{n}\big |^{-2} = \sum ^{N-1}_{n=1} \big |\widetilde{\lambda }_{n}\big |^{-2} + \sum ^{\infty }_{n=N} \big |\widetilde{\lambda }_{n}\big |^{-2} \end{aligned}$$

where $N \in \mathbb {N}$ was chosen as the smallest integer such that $N \ge \ln (P/\alpha ) + 1$. Next we use for the first sum on the right hand side that $|\widetilde{\lambda }_{n}| \ge \delta$ for all $n \ge 1$. In the second sum we apply the bound $|\widetilde{\lambda }_{n}| \ge [ n-\ln (P/\alpha ) ]/[e\, \sigma ]$ from above. This gives

$$\begin{aligned} \sum _{n\ne n_{0}} \frac{1}{\big |\lambda _n - \lambda _{n_{0}}\big |^{2}}&\le \frac{N-1}{\delta ^{2}} + \sum ^{\infty }_{n=N} \frac{ \mathrm {e} ^{2} \sigma ^{2}}{[n - \ln (P/\alpha )]^{2}}\\&\le \frac{\ln (P/\alpha ) + 1}{\delta ^{2}} + \mathrm {e} ^{2} \sigma ^{2} \sum ^{\infty }_{m=1} \frac{1}{m^2} = \frac{\ln (e P / \alpha )}{\delta ^{2}} + \frac{ \mathrm {e} ^{2} \pi ^{2}}{6}\, \sigma ^{2} =: C \end{aligned}$$

where the constant $C<\infty$ is independent of $n_0$ and depends only on $S$. $\square$