# How to set the space between rows in a table

This is the Latex code I have to create a table:

\documentclass{article}

\usepackage{amsmath}
\usepackage{tabularx}
\usepackage[top=0.5cm, bottom=0.5cm, left=0.5cm, right=0.5cm, columnsep=0.75cm]{geometry}

\begin{document}

\subsubsection*{Fourier Transform properties}
Property & Time domain & Frequency domain & Condition \\
\hline
Time-shift & $f(t - \tau)$ & $\hat{f}(\omega)e^{-i \omega \tau}$ \\
Frequency-shift & $f(t) e^{i \omega_0 t}$ & $\hat{f}(\omega - \omega_0)$ \\
Modulation Thm. & $f(t)\cos(\omega_0 t)$ & $\cfrac{\hat{f}(\omega-\omega_0)+\hat{f}(\omega+\omega_0)}{2}$ \\
Differentiation (time) & $f^{(n)}(t)$ & $(i\omega)^n \hat{f}(\omega)$ & $\lim_{t \to \pm \infty} f(t) = 0$
\end{tabular}

\end{document}


With this code there is no space between the rows in the table. How can I add some space (let's say 0.3 cm) between the rows? I've tried a few things, but they don't work as expected.

When using \setlength{\extrarowheight}{0.3cm}, the table looks like this:

As you can see, no space is added between the third and the fourth row. Also, the text in the header row is aligned at the bottom, which does not look nice.

When using \renewcommand{\arraystretch}{1.8}, the table looks like this:

This looks better, but the space between the second and third row is twice as big as the space between the third and fourth row.

Which command can I use that always puts the same amount of spacing between two rows?

-

\documentclass{article}

\usepackage{mathtools}

\begin{document}

\subsubsection*{Fourier Transform properties}
{\def\arraystretch{2}\tabcolsep=10pt
\begin{tabular}{@{}l | l | l | l @{}}
Property & Time domain & Frequency domain & Condition \\
\hline
Time-shift & $f(t - \tau)$ & $\hat{f}(\omega)e^{-i \omega \tau}$ \\
Frequency-shift & $f(t) e^{i \omega_0 t}$ & $\hat{f}(\omega - \omega_0)$ \\
\rule{0pt}{5ex}%  EXTRA vertical height
Modulation Thm. & $f(t)\cos(\omega_0 t)$ & $\dfrac{ \hat{f}(\omega-\omega_0)+\hat{f}(\omega+\omega_0) }{2}$ \\
Differentiation (time) & $f^{(n)}(t)$ & $(i\omega)^n \hat{f}(\omega)$ & $\displaystyle\lim_{\mathclap{t \to \pm \infty}} f(t) = 0$
\end{tabular}%
}
\end{document}


another possibility is to use package tabls. But this may cause problems when using other tabular packages. Try it and maybe the possible optional arguments are of interest, as minimal distance between tabulkar lines.

\documentclass{article}

\usepackage{mathtools}
\usepackage{tabls}

\begin{document}

\subsubsection*{Fourier Transform properties}
{\tablinesep=2ex\tabcolsep=10pt
\begin{tabular}{@{}l | l | l | l @{}}
Property & Time domain & Frequency domain & Condition \\
\hline
Time-shift & $f(t - \tau)$ & $\hat{f}(\omega)e^{-i \omega \tau}$ \\
Frequency-shift & $f(t) e^{i \omega_0 t}$ & $\hat{f}(\omega - \omega_0)$ \\
Modulation Thm. & $f(t)\cos(\omega_0 t)$ & $\dfrac{ \hat{f}(\omega-\omega_0)+\hat{f}(\omega+\omega_0) }{2}$ \\
Differentiation (time) & $f^{(n)}(t)$ & $(i\omega)^n \hat{f}(\omega)$ & $\displaystyle\lim_{\mathclap{t \to \pm \infty}} f(t) = 0$
\end{tabular}%
}
\end{document}


-
This works, but I still hope someone has a solution that doesn't involve doing something special for one row (like the \rule{0pt}{5ex}). And thanks for correcting my ugly way of doing horizontal spacing. –  Paul Mar 20 '13 at 9:22
@Paul: see my edit and the 2nd example –  Herbert Mar 20 '13 at 9:29
All right, one last request: It looks like that in the horizontal direction sort-of the same problem occurs now. The "t → ±∞" in the formula in the fourth column is ignored in determining the position of the formula. Would it be possible to move that formula a bit to the right. Again, with a command for the table, not for the specific column or cell. –  Paul Mar 20 '13 at 12:17
for the last column specifier: .. @{\kern2\tabcolsep}l @{}} or any other length. You can also use as an alternative \mathrlap{t ...} instead of \mathclap –  Herbert Mar 20 '13 at 12:37
Thanks Herbert! I've I discovered that simply removing \mathclap solves the second problem. –  Paul Mar 20 '13 at 18:28

\extrarowheight does add the same amount of space to each row, but the fraction \frac{\hat{f}(\omega-\omega_0)+\hat{f}(\omega+\omega_0)}{2} makes for a rather unappealing table. For better appearance, you might prefer something like that:

\documentclass{article}

\usepackage{amsmath}
\usepackage{tabularx}
\usepackage{array}
\usepackage[top=0.5cm, bottom=0.5cm, left=0.5cm, right=0.5cm, columnsep=0.75cm]{geometry}

\begin{document}

\subsubsection*{Fourier Transform properties}
{
\setlength{\extrarowheight}{.5em}
Property & Time domain & Frequency domain & Condition \\
\hline
Time-shift & $f(t - \tau)$ & $\hat{f}(\omega)e^{-i \omega \tau}$ \\
Frequency-shift & $f(t) e^{i \omega_0 t}$ & $\hat{f}(\omega - \omega_0)$ \\
Modulation Thm. & $f(t)\cos(\omega_0 t)$ &
$\left[\hat{f}(\omega-\omega_0)+\hat{f}(\omega+\omega_0)\right]/\,2$ \\
Differentiation (time) & $f^{(n)}(t)$ &
$(i\omega)^n \hat{f}(\omega)$ & $\lim_{t \to \pm \infty} f(t) = 0$
\end{tabular}

\end{document}


EDIT: thanks to those users who posted constructive suggestions below.

-
I would much rather go with (...)/2. –  Werner Mar 18 '13 at 4:38
I added an image to show the problem with extrarowheight. Maybe writing the fraction differently works for this case, but I want a solution that works in every case. I don't want a workaround. –  Paul Mar 18 '13 at 4:41
I agree with @Werner. You could also consider $[...]/2$ (brackets as delimeters without scaling). –  Marc van Dongen Mar 20 '13 at 9:55
$\left[...\right]$ makes more sense –  Herbert Mar 20 '13 at 10:12
@Herbert Alright, but I suggest adding a thin space in front of the denominator, then. –  Jubobs Mar 20 '13 at 10:16