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I just read a paper yesterday and there is confusion:

Image

I want to know what is reconstruction fidelity term and the theorem of changing the formula from upper to lower one. Is this theorm from functional analysis?

Could anyone help me ?

EDIT: The link of the paper.

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  • $\begingroup$ a link to the paper would be helpful $\endgroup$ – Fat32 Apr 8 '17 at 0:03
  • $\begingroup$ @Fat32 Ok. I add it $\endgroup$ – stander Qiu Apr 8 '17 at 0:29
  • $\begingroup$ is your question about the relationship of Hermitian symmetry (in the frequency domain) to real signals (in the time domain)? that one is easy. $\endgroup$ – robert bristow-johnson Apr 8 '17 at 0:58
  • $\begingroup$ @robertbristow-johnson I do not understand what highlight part means and how $L_2$ norm becomes integral $\endgroup$ – stander Qiu Apr 8 '17 at 1:07
  • $\begingroup$ actually, given the consistent $L_2$ definition for $$\lVert X(\omega) \rVert_2$$ and its square, that $4$ in Eq. (22) is a mistake. it should be $2$. $$ $$ no, it should be $4$. $2^2=4$. $\endgroup$ – robert bristow-johnson Apr 8 '17 at 4:06
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i have no idea what the "reconstruction fidelity term" is or what it's about.

Hermitian symmetry is a term usually applied to some form the Fourier Transform of a signal that is purely real.

for continuous-time, continuous-frequency Fourier Transform:

$$ X(f) = \int\limits_{-\infty}^{\infty} x(t) \, e^{-j 2 \pi f t} \ dt $$

if $x(t)$ is purely real (that is $\Im\{x(t)\}=0$ for all real $t$), then we know that there is this symmetry about $f=0$:

$$ X(-f) = X(f)^* = \operatorname{conj}\{X(f)\} \qquad \forall f \in \mathbb{R} $$

or

$$\begin{align} \Re\{X(-f)\} &= \Re\{X(f)\} \\ \Im\{X(-f)\} &= -\Im\{X(f)\} \\ |X(-f)| &= |X(f)| \\ \arg\{X(-f)\} &= -\arg\{X(f)\} \\ \end{align}$$

Similarly, for the Discrete Fourier Transform:

$$ X[k] = \sum\limits_{n=0}^{N-1} x[n] \, e^{-j 2 \pi \frac{nk}{N}} $$

if $x[n]$ is purely real (that is $\Im\{x[n]\}=0$ for integer $n \in [0..N-1]$), then we know that there is this symmetry about $k=\tfrac{N}{2}$:

$$ X[N-k] = X[k]^* = \operatorname{conj}\{X[k]\} \qquad 1 \le k \le N-1 $$

or

$$\begin{align} \Re\{X[N-k]\} &= \Re\{X[k]\} \\ \Im\{X[N-k]\} &= -\Im\{X[k]\} \\ |X[N-k]| &= |X[k]| \\ \arg\{X[N-k]\} &= -\arg\{X[k]\} \\ \end{align}$$

Now in my religion regarding the DFT, i insist that

$$\begin{align} x[n+N] &= x[n] \qquad & \forall n \in \mathbb{Z} \\ X[k+N] &= X[k] \qquad & \forall k \in \mathbb{Z} \\ \end{align}$$

this is always the case (the DFT maps a discrete and periodic sequence of period $N$ in one domain to another discrete and periodic sequence of period $N$ in the reciprocal domain). then the Hermitian symmetry takes a simpler form:

if $x[n]$ is purely real (that is $\Im\{x[n]\}=0$ for all integer $n$), then we know that there is this symmetry about $k=0$:

$$ X[-k] = X[k]^* = \operatorname{conj}\{X[k]\} \qquad \forall k \in \mathbb{Z} $$

or

$$\begin{align} \Re\{X[-k]\} &= \Re\{X[k]\} \\ \Im\{X[-k]\} &= -\Im\{X[k]\} \\ |X[-k]| &= |X[k]| \\ \arg\{X[-k]\} &= -\arg\{X[k]\} \\ \end{align}$$

likewise you can identify Hermitian symmetry for the DTFT and relate it to the real-ness of the input to the DTFT. (do i have to do that one, too?)

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  • $\begingroup$ I see. No need to do the DTFT, they are similar. So the $L_2$ norm in the $\mathbb{R}$ changes to the nonnegative part because of the Hermitian symmetry. $\endgroup$ – stander Qiu Apr 8 '17 at 2:52
  • $\begingroup$ evidently, the $L_2$ norm becomes this integral because of its definition. the reason why the integral does not include the negative frequencies is because $$ 1+\operatorname{sgn}(\omega) = \begin{cases} 2 \qquad & \text{for } \omega>0 \\ 1 \qquad & \text{for } \omega=0 \\ 0 \qquad & \text{for } \omega<0 \\ \end{cases} $$ there is an additional factor of 2 that i do not understand where it comes from. otherwise i do not know how Eq. (21) becomes Eq. (22). $\endgroup$ – robert bristow-johnson Apr 8 '17 at 3:45
  • $\begingroup$ ostensibly, just from the context, $$ L_2\{X\} \triangleq \lVert X(\omega) \rVert_2 \triangleq \sqrt{\int\limits_{-\infty}^{\infty} |X(\omega)|^2 \, d\omega }$$ and $|X(-\omega)| = |X(\omega)|$ so $$ \lVert X(\omega) \rVert_2 = \sqrt{\int\limits_{0}^{\infty} 2|X(\omega)|^2 \, d\omega }$$ or $$ \lVert X(\omega) \rVert_2^2 = \int\limits_{0}^{\infty} 2|X(\omega)|^2 \, d\omega $$ so now i understand where the extra factor of 2 comes from. $\endgroup$ – robert bristow-johnson Apr 8 '17 at 4:00
  • $\begingroup$ It is from the $1 + sgn(w)$ $\endgroup$ – stander Qiu Apr 8 '17 at 13:46
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Let us write $$g(\omega) = \hat{f}(\omega) - \sum_i \hat{u}_i(\omega)+\frac{\hat{\lambda}(\omega)}{2}\,.$$ This term is a typical "reconstruction error" term: it denotes the error made (pointwise, i.e. for each $\omega$) when trying to reconstruct $\hat{f}$ (or find an estimator of $\hat{f}$) as a sum of modes $\hat{u}_i$ plus a term related to the Lagrangian (here in the Fourier domain).

The fidelity term, often encountered in optimization, is a quantitative measure of this error, computed here as a squared $L_2$ norm:

$$ \int_{-\infty}^{\infty} |g(\omega)|^2d\omega = \int_{-\infty}^{0} |g(\omega)|^2d\omega + \int_{0}^{\infty} |g(\omega)|^2d\omega = \int_{0}^{\infty} |g(-\omega)|^2d\omega + \int_{0}^{\infty} |g(\omega)|^2d\omega \,.$$

using the $\omega \to -\omega$ change of variable in the first integral.

All the $\lambda(t)$, $f(t)$, and $u_i(t)$ are real, hence their Fourier transforms possess Hermitian (or conjugate) symmetry. For instance, $\overline{\hat{f}}(\omega) = {\hat{f}}(-\omega)$, thus it applies to $g$ as well:

$$ \overline{g}(\omega) = g(-\omega)\,.$$

Remembering that $$|g(\omega)|^2 = g(\omega)\overline{g}(\omega)\,,$$ you observe that

$$\int_{0}^{\infty} |g(-\omega)|^2d\omega = \int_{0}^{\infty} |\overline{g}(\omega)|^2d\omega = \int_{0}^{\infty} \overline{g}(\omega) \overline{\overline{g}}(\omega)d\omega = \int_{0}^{\infty} \overline{g}(\omega) g(\omega)d\omega\,,$$ hence the second term in Equation (22) is simply:

$$2\int_{0}^{\infty}| {g}(\omega) |^2d\omega\,.$$

Being Hermitian is more a property that a theorem. For the first term in the RHS of Equation (22), the $ \int_{-\infty}^{\infty} $ is only computed as $ \int_{0}^{\infty} $, since $1+\mathrm{sgn}(\omega)$ vanishes on the interval $]-\infty,0[$. On the interval $]0,\infty,[$, $1+\mathrm{sgn}(\omega)=2$ and it is squared out as the 4 factor.

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