Estou tentando compilar o texto abaixo (é uma versão muito reduzida do meu documento original). Está me dando 2 erros:
Erro 1: Erro LaTeX: Algo está errado - talvez um item faltando. (linha 317)
Erro 2: Extra } ou \endgroup esquecido. (linha 372)
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\section{System model}~\label{sec:sysmodl}
\subsection{asfgjkfd}
models. In adaptive GT, the previous test results can be used to design the future tests. In non-adaptive setting, all group tests are designed independent of each other.
$\textbf{X}^{i} \in \{0,\sqrt{P}\}^{n}, \forall i \in \mathcal{D}$. The BS measures received energy during each channel-use and produces a binary 0-1 output $Z_t$ as in (\ref{threshold}). These 1-bit energy measurements at the receiver side corresponds to the group test results.
framework corresponds to non-adaptive GT since the testing/grouping pattern is based on predefined binary preambles and is not updated based on the GT results of previous channel-uses.
We will observe in Section \ref{sec3} that the GT model can be leveraged to provide an achievability scheme to the minimum user identification cost for the non-coherent $(\ell,k)-$MnAC when $\alpha = 0 $, ie., $k= O(1)$ and a corresponding lower bound when $\alpha \neq 0 $.
\section{dfhjlu}
\label{sec3}
em in a point-to-point vector input - scalar output channel whose inputs correspond to the active users as shown in Fig. \ref{fig:redalpha}. Thereafter, we derive the maximum rate of the equivalent channel by exploiting its cascade structure.
\subsection{Efghk}
\vspace{0.05cm}
Considering the channel in Fig. 1, since the set of active users is $\mathcal{A}=\{a_1,\ldots a_k\}$, the input to the non-coherent $(\ell,k)-$MnAC is $\Tilde{\textbf{X}}=(X^{a_1},\ldots X^{a_k})$. Thus, the signal at the input of envelope detector is
$ S=\sqrt{P}\sum_{m=1}^{k}h^{a_m} +W.$
Let $V=\frac{\sum_{i=1}^{k}X^{a_i}}{\sqrt{P}}$ denote the Hamming weight of $\Tilde{\textbf{X}}$ which is the number of active users transmitting `On' signal during the particular channel-use we have at hand.
Thus, conditioned on $V=v$, $U:=|S|^{2}$ follows an exponential distribution given by
\begin{equation}
f_{U|V}(u|v)=\frac{1}{v \sigma^{2}P+\sigma_{w}^{2}} e^{-\frac{u}{v \sigma^{2}P+\sigma_{w}^{2}}}, u \geq 0.
\label{expdis}
\end{equation} As evident from (\ref{threshold}) and (\ref{expdis}), we have $\Tilde{\textbf{X}} \rightarrow V\rightarrow Z$, i.e., the transition probability $p\left(z\mid \Tilde{\textbf{x}},v\right)$ is only dependent on the channel input $\Tilde{\textbf{X}}=(X^{a_1},X^{a_2},\ldots X^{a_k})$ through its Hamming weight $V$. Also, since $ f_{U|V}$ is an exponential distribution, $p_v:=p\left(Z=0\mid V=v\right)$ can be expressed as
\begin{equation}
p_v=1-e^{-\frac{\gamma}{v \sigma^{2}P+\sigma_{w}^{2}}}. \label{channeleq}
\end{equation} Similarly, $\operatorname{Pr}\left(Z=1\mid V=v\right)=1-p_v.$ Note that $p_v$ is a strictly decreasing function of $v$ where $v \in\{0,1,..,k\}$ assuming w.l.o.g. positive values for $\sigma^2, \sigma_w^2$ and $\gamma$.
Thus, the non-coherent $(\ell,k)-$MnAC can be equivalently viewed as a traditional point-to-point communication channel whose input is the active user set as in Fig. \ref{fig:redalpha}. Moreover, this equivalent channel can be modeled as a cascade of two channels; the first channel computes the Hamming weight $V$ of the input $\Tilde{\textbf{X}}$ whereas the second channel translates the Hamming weight $V$ into a binary output $Z$ depending on the fading statistics $(\sigma^2)$, noise variance $(\sigma_w^2)$ of the wireless channel and the non-coherent detector threshold $\gamma$. We exploit this cascade channel structure to establish the minimum user identification cost $n(\ell)$ for the non-coherent $(\ell,k)-$MnAC.
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} \setlength{\belowcaptionskip}{-38pt}
\caption{Equivalent channel with only the active users of the $(\ell,k)-$MnAC as inputs.}
\label{fig:redalpha}
\end{figure}
\subsection{ Maximum rate of the equivalent channel}
\begin{lemma}
For a given fading statistics $\sigma^2$, noise variance $\sigma_w^2$, and non-coherent detector threshold $\gamma$ for the $(\ell,k)$-MnAC, the maximum rate of the equivalent point-to-point channel in Fig. \ref{fig:redalpha} is
\begin{equation}
C=\max_{(\gamma,q_{sp})}h\Big(E\Big[e^{-\frac{\gamma}{V \sigma^{2}P+\sigma_{w}^{2}}}\Big]\Big)-E\Big[h\Big(e^{-\frac{\gamma}{V \sigma^{2}P+\sigma_{w}^{2}}}\Big)\Big] \label{jengap2}
\end{equation} where $E(\cdot)$ denotes expectation w.r.t $ V$, $h(x)=-x \log x-(1-x) \log (1-x)$ is the binary entropy function and $q_{sp}$ denotes the optimal sampling probability used for i.i.d preamble generation across all users.
\label{lem1}
\end{lemma}
\begin{proof} The equivalent point-to-point communication channel in Fig. \ref{fig:redalpha} has two tunable parameters, viz, the sampling probability vector $\textbf{q} =\{q_{a_1} \ldots q_{a_k}\}$ at the user side and the non-coherent detector threshold $\gamma$ at the BS. Thus, the maximum rate of this equivalent channel between the binary vector input $\Tilde{\textbf{X}}$ and binary scalar output $Z$ is
completing the proof.\end{proof}
\vspace{-0.3cm}\subsection{st}
where $n(\ell)$ is as given in Theorem 1. This is because $C$ in Lemma 1 is smaller than $\frac{1}{2}\log (1+ kP_{av})$ where $P_{av}:= q_{sp}^*P$, with $ q_{sp}^*$ being the optimal $ q_{sp}$ in Theorem 1. Clearly, the user identification in a Gaussian MnAC requires lesser number of channel uses due to the fact that the model does not incorporate fading and is not constrained to OOK signaling and non-coherent detection as in our non-coherent $(\ell,k)-$MnAC.
\section{Pghhjkkkff}
\begin{definition}
\textbf{ $\zeta \% -$ partial recovery}: For a true active set $\mathcal{A}$ and an estimated active set $\hat{\mathcal{A}}$, consider an error event $E_1$ defined as
\begin{equation}
E_1 := \left\{\left|\hat{\mathcal{A}}^{\mathrm{c}} \cap \mathcal{A}\right| > k\left(1-\frac{\zeta}{100}\right) \right\}. \label{errev}
\end{equation} We have $\zeta \%$-partial recovery, if \begin{equation}
\mathbb{P}_{e,\zeta}^{(\ell)}:= P( E_1) \rightarrow 0 \text { as } \ell \rightarrow \infty, \nonumber
\end{equation}
where $\mathbb{P}_{e,\zeta}^{(\ell)}$ denotes the probability of error for $\zeta \% -$ partial recovery. Probability of successful identification for $\zeta \% -$ recovery is defined as $\mathbb{P}_{succ,\zeta}^{(\ell)}:= 1 -\mathbb{P}_{e,\zeta}^{(\ell)}.$
\end{definition}
The error event $E_1$ considers if the fraction of true active devices in $\mathcal{A}$ that are misdetected exceeds $\left(1-\frac{\zeta}{100}\right)$. Note that since our proposed strategies output a set of size $k$, both false positives and misdetections occur in equal numbers. Thus, the error event $E_1$ in (\ref{errev}) is essentially equivalent to $\left\{\left|\hat{\mathcal{A}}^{\mathrm{c}} \cap \mathcal{A}\right| > k\left(1-\frac{\zeta}{100}\right) \bigcup \left|\hat{\mathcal{A}} \cap \mathcal{A}^c\right| > k\left(1-\frac{\zeta}{100}\right) \right\}$. Moreover, when $\zeta = 100$, the partial recovery setting boils down to our original setting of exact recovery given in Def. 2. The case when the number of active users $k$ is unknown at the BS will be discussed in Section IV.C.
\subsection{N-COMP based user identification}
primarily analyzed in the context of symmetric noise
models wherein errors in test results are equally likely \cite{6120373,6763117}. r is \cite{4797638} where the focus is on the problem of neighbor discovery in a wireless sensor network rather than the massive random access setting considered in this paper.
Let $\mathcal{G}_i:=\frac{\sum_{t=1}^{n}X^{i}_t}{\sqrt{P}$ denote the Hamming weight of $\textbf{X}^{i}$, the ymbols if it is in active state. Let $\mathcal{R}_i:=\frac{\sum_{t=1}^{n}X^{i}_tZ_t}{\sqrt{P}}$ denote the number of these channel uses in which the received energy at the detector exceeds a predetermined threshold. In N-COMP based user identification, our strategy is to classify the $k$
\begin{figure}
\centering
\begin{minipage}{.5\textwidth}
\centering
\begin{tikzpicture}
\sbox0{\includegraphics[width=.9\linewidth,height=65mm,trim={1.3cm 0.55cm 0 0},clip]{ICASSP_NCOMP_EXACT.png}}}% get width and height
\node[above right,inner sep=0pt] at (0,0) {\usebox{0}};
\node[black] at (0.5\wd0,-0.06\ht0) {\normalsize{Number of channel-uses, $n$}};
\node[black,rotate=90] at (-0.04\wd0,0.5\ht0) {\normalsize{Probability of successful identification}};
\end{tikzpicture}
\setlength{\belowcaptionskip}{-15pt}
\caption{NCOMP: Exact recovery for $(1000,25)$-MnAC. }
\label{fig:2usercap1}
\end{minipage}%
\begin{minipage}{.5\textwidth}
\centering
\begin{tikzpicture}
\sbox0{\includegraphics[width=.9\linewidth,height=65mm,trim={1.3cm 0.55cm 0 0},clip]{ICASSP_NCOMP_90perc.png}}% get width and height
\node[above right,inner sep=0pt] at (0,0) {\usebox{0}};
\node[black] at (0.5\wd0,-0.06\ht0) {\normalsize{Number of channel-uses, $n$}};
\node[black,rotate=90] at (-0.04\wd0,0.5\ht0) {\normalsize{Probability of successful identification}};
\end{tikzpicture}
\setlength{\belowcaptionskip}{-15pt}
\caption{NCOMP: 90$\%$ recovery for $(1000,25)$-MnAC. }
\label{fig:2usercap2}
\end{minipage}
\end{figure}
\subsection{BP based user identification}
\begin{figure}
\centering
\begin{minipage}{.5\textwidth}
\begin{tikzpicture}
\sbox0{\includegraphics[width=.9\linewidth,height=65mm,trim={1.4cm 0.8cm 0 0},clip]{BP_STvsNCOMP_exact.png}}% get width and height
\node[above right,inner sep=0pt] at (0,0) {\usebox{0}};
\node[black] at (0.5\wd0,-0.06\ht0) {\normalsize{Number of channel-uses, $n$}};
\node[black,rotate=90] at (-0.04\wd0,0.5\ht0) {\small{Probability of successful identification}};
\end{tikzpicture}
\setlength{\belowcaptionskip}{-15pt}
\caption{\footnotesize{Exact recovery: $(1000,25)$-MnAC at SNR = 10 dB.}}
\label{fig:z5}
\end{minipage}%
\begin{minipage}{.5\textwidth}
\centering
\begin{tikzpicture}
\sbox0{\includegraphics[width=.9\linewidth,height=65mm,trim={1.4cm 0.8cm 0 0},clip]{BP_STvsNCOMP_90.png}}% get width and height
\node[above right,inner sep=0pt] at (0,0) {\usebox{0}};
\node[black] at (0.5\wd0,-0.06\ht0) {\normalsize{Number of channel-uses, $n$}};
\node[black,rotate=90] at (-0.04\wd0,0.5\ht0) {\small{Probability of successful identification}};
\end{tikzpicture}
\setlength{\belowcaptionskip}{-15pt}
\caption{\footnotesize{$90\%$ recovery: $(1000,25)$-MnAC at SNR = 10 dB.} }
\label{fig:z6}
\end{minipage}
\end{figure}
\begin{figure}
\centering
\begin{tikzpicture}
\sbox0{\includegraphics[width=.55\linewidth,height=70mm,trim={1.4cm 0.8cm 0 0},clip]{BP_ST_SHT_AHT_exact_10dB.png}}% get width and height
\node[above right,inner sep=0pt] at (0,0) {\usebox{0}};
\node[black] at (0.5\wd0,-0.06\ht0) {\small{\small{Number of channel-uses $n$}}};
\node[black,rotate=90] at (-0.04\wd0,0.5\ht0) {\small{Probability of successful identification}};
\end{tikzpicture}
\setlength{\belowcaptionskip}{-25pt}
\caption{ Comparison of various BP algorithms for exact recovery in $(1000,25)$-MnAC at SNR $=$ 10 dB. }
\label{fig:q4}
\end{figure}
\begin{figure}
\centering
\begin{tikzpicture}
\sbox0{\includegraphics[width=.55\linewidth,height=70mm,trim={1.4cm 0.8cm 0 0},clip]{BP_ST_SHT_AHT_90_10dB.png}}% get width and height
\node[above right,inner sep=0pt] at (0,0) {\usebox{0}};
\node[black] at (0.5\wd0,-0.06\ht0) {\small{\small{Number of channel-uses $n$}}};
\node[black,rotate=90] at (-0.04\wd0,0.5\ht0) {\small{Probability of successful identification}};
\end{tikzpicture}
\setlength{\belowcaptionskip}{-25pt}
\caption{ Comparison of various BP algorithms for $90\%$ recovery in $(1000,25)$-MnAC at SNR $=$ 10 dB. }
\label{fig:q5}
\end{figure}
\section{a}
t.
\bibliographystyle{IEEEtranTCOM}
\bibliography{IEEE_TCOM}
\end{document}
Descobri que a linha 372 \sbox0{\includegraphics[width=.9\linewidth,height=65mm,trim={1.3cm 0.55cm 0 0},clip]{ICASSP_NCOMP_EXACT.png}}}% get width and heighttem um extra
'}' que tentei remover. Mas isso não resolve o problema. Alguém pode me informar o que há de errado no meu código?
IEEETranTCOM.cls está disponível aqui:https://www.comsoc.org/media/1381/download
Responder1
Seus erros estão aqui:
Linha 284: Você não pode ter um nó com quebra de linha que também não possua a
alignchave ment. Com base no seu código, usei[align=left]. Referência:Quebras de linha manuais/automáticas e alinhamento de texto em nós TikZLinha 364: Você precisa de um conjunto balanceado de chaves para
\frac. Você está perdendo a chave de fechamento do denominador em\frac{\sum_{t = 1}^n X^i_t}{\sqrt{P}}.Linha 371: Você tem uma chave de fechamento extra
}com\sbox0{...}}.