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?

Answer 1

\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}

Answer 2

\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.