# Get equal D and T slopes with tikz-timing

I'm using the tikz-timing package to create a bus timing diagram:

\begin{tikztimingtable}[timing/slope=0.5]
DT & 2D{}12D{}1D\\
CK & [H] 1H2{6T}2T\\
\end{tikztimingtable}


Now it occurs that the slope of D{}D is twice as long as the slope of T. That is in a way logical, because T is only one character (is that the reason?), but I now want a neat way to get the slopes even long.

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The reason is not because D{}D are two characters and T is just one. As Qrrbrbirlbel's answer shows different slopes are used for different types of transitions. This was done by me (the package author) by design. The D transitions are double as long by default because the resulting cross would be too narrow IMHO otherwise. The Z transitions are only half, because the Y distance is only half either. –  Martin Scharrer Jan 9 '13 at 16:23
Yes, I made a mind-note about that after reading Qrrbrbirlbel's answer. Thanks for the explanation! –  Camil Staps Jan 9 '13 at 16:37

You need to set the individual slopes. The package manual states in Table 4, page 7:

timing/slope=<0.0 – 1.0>    Sets slope for logic transitions.
This also sets dslope = 2 * slope,
zslope = slope / 2.
timing/lslope=<0.0 – 1.0>   Sets slope for logic transitions only. Default: 0.1
timing/dslope=<0.0 – 1.0>   Sets slope for data transitions.       Default: 0.2
timing/zslope=<0.0 – 1.0>   Sets slope for Z transitions.          Default: 0.05


## Code

\documentclass[tikz]{standalone}
\usepackage{tikz-timing}
\begin{document}
\begin{tikztimingtable}[
timing/lslope=.5,
timing/dslope=.5,
]
DT & 2D{}12D{}1D   \\
CK & [H] 1H2{6T}2T \\
\end{tikztimingtable}
\end{document}


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