# Bytefield: How to add multiple \underbrace below bits diagram

I am new to LaTeX and I am trying to put multiple underbrace under a package using bytefield package. I want to have 4 underbrace consecutive to each other, but I am only able to put 2 and the other 2 appear with a space in between.

First I define the bitbox of the packet, then I put the underbrace below them. I have review the bytefield documentation and there is no underbrace example. I found this question Bytefield: Word Group below blocks but it only puts one underbrace. With that answer I was able to put 2 together and 2 together but not consecutive to each other. The code I have is the following:

\documentclass[border=10pt,png]{standalone}
\usepackage{amsmath}
\usepackage[endianness=big]{bytefield}
\bytefieldsetup{boxformatting={\centering\footnotesize}}
\begin{document}

\begin{bytefield}[bitwidth=1.1em]{21}
\bitbox{1}{1} & \bitbox{1}{2} & \bitbox{1}{3} & \bitbox{1}{4} & \bitbox{1}{5} &
\bitbox{1}{6} & \bitbox{1}{7} & \bitbox{1}{8} & \bitbox{1}{9} & \bitbox{1}{10} &
\bitbox{1}{11} & \bitbox{1}{12} & \bitbox{1}{13} & \bitbox{1}{14} & \bitbox{1}{15} &
\bitbox{1}{16} & \bitbox{1}{17} & \bitbox{1}{18} & \bitbox{1}{19} & \bitbox{1}{20} &
\bitbox{1}{...}\\

\bitbox[]{0}{$\underbrace{\hspace{6.7em}}_{\text{\normalsize$A_0$}}$} &
\bitbox[]{15}{$\underbrace{\hspace{6.7em}}_{\text{\normalsize$A_1$}}$} &
\bitbox[]{0}{$\underbrace{\hspace{6.7em}}_{\text{\normalsize$A_2$}}$}     &
\bitbox[]{15}{$\underbrace{\hspace{6.7em}}_{\text{\normalsize$A_3$}}$} \\
\end{bytefield}
\end{document}


The result I obtain is: And what I wish to obtain is this:

I am not sure why this space appears between the first 2 underbrace and the third one. I have change the \hspace, the value inside the \bitbox{}, put the bitbox inside the \underbrace, but nothing seems to do the trick.

Don't alternate between 0 and 15:

\documentclass{article}
\usepackage{amsmath}
\usepackage[endianness=big]{bytefield}

\bytefieldsetup{boxformatting={\centering\footnotesize}}

\begin{document}

\begin{bytefield}[bitwidth=1.1em]{21}
\bitbox{1}{1} & \bitbox{1}{2} & \bitbox{1}{3} & \bitbox{1}{4} & \bitbox{1}{5} &
\bitbox{1}{6} & \bitbox{1}{7} & \bitbox{1}{8} & \bitbox{1}{9} & \bitbox{1}{10} &
\bitbox{1}{11} & \bitbox{1}{12} & \bitbox{1}{13} & \bitbox{1}{14} & \bitbox{1}{15} &
\bitbox{1}{16} & \bitbox{1}{17} & \bitbox{1}{18} & \bitbox{1}{19} & \bitbox{1}{20} &
\bitbox{1}{...\vphantom{1}}\\

\bitbox[]{5}{$\underbrace{\hspace{6.2em}}_{\text{\normalsize$A_0$}}$} &
\bitbox[]{5}{$\underbrace{\hspace{6.2em}}_{\text{\normalsize$A_1$}}$} &
\bitbox[]{5}{$\underbrace{\hspace{6.2em}}_{\text{\normalsize$A_2$}}$} &
\bitbox[]{5}{$\underbrace{\hspace{6.2em}}_{\text{\normalsize$A_3$}}$} \\
\end{bytefield}

\end{document}


The 6.2 em has been computed by trial and error. Note the phantom in the last cell in order to set the dots at the right height.