How to set the space between rows in a table

Question

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}
\begin{tabular}{l@\quad|@\quad l@\quad|@\quad l@\quad|@\quad 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. & $f(t)\cos(\omega_0 t)$ & $\cfrac{\hat{f}(\omega-\omega_0)+\hat{f}(\omega+\omega_0)}{2}$ \\
    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?

score 38 · Accepted Answer · edited Jun 09 '17 at 16:26

38

\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}

enter image description here

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}

enter image description here

edited Jun 09 '17 at 16:26

Moriambar

11,466

answered Mar 20 '13 at 06:39

1

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 09:22
@Paul: see my edit and the 2nd example – Mar 20 '13 at 09:29
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 – 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
An aesthetic comment. The 'i' symbol and 'e' symbol are operators. Yours expression are incorrect and we would be \textrm{i} and \textrm{e}. – jpayansomet Mar 20 '13 at 19:29
Great! (+1). For those searching a solution in xtable like me, this works fine if you add an extra slash and paste something like "\\rule{0pt}{5ex}" to your first column. – Frans Rodenburg Jul 29 '19 at 12:05

score 15 · Answer 2 · edited Jun 16 '17 at 15:54

15

\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}
\begin{tabular}{l@\quad|@\quad l@\quad|@\quad l@\quad|@\quad 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. & $f(t)\cos(\omega_0 t)$ &
        $\left[\hat{f}(\omega-\omega_0)+\hat{f}(\omega+\omega_0)\right]/\,2$ \\
    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. enter image description here

edited Jun 16 '17 at 15:54

Moriambar

11,466

answered Mar 18 '13 at 03:35

jub0bs

58,916

5

I would much rather go with (...)/2. – Werner Mar 18 '13 at 04:38
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 04:41
1

I agree with @Werner. You could also consider $[...]/2$ (brackets as delimeters without scaling). – Mar 20 '13 at 09:55
2

$\left[...\right]$ makes more sense – Mar 20 '13 at 10:12
@Herbert Alright, but I suggest adding a thin space in front of the denominator, then. – jub0bs Mar 20 '13 at 10:16
6

Seriously, people, this is not an answer to my question. Removing the \frac obviously causes the spacing problem to disappear, you don't have to tell me that. My code is just an example of a lot of tables I'm going to make, and in some of those tables I really want a \frac. – Paul Mar 20 '13 at 19:00

How to set the space between rows in a table

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