# Sampling Theorem

So I have a function $$f \in L^{2}(\mathbb{R} )$$ which can be reconstructed from its sample values if a sample rate is: $$\frac{1}{T}=2\Omega$$. The continuous function $$f \in L^{2}(\mathbb{R} );f:\mathbb{R} \rightarrow \mathbb{C},\enspace \mbox{be a 3\Omega bandlimited}$$ $\hat{g}$ is the Fourier transform of the function $g \in L^{2}(\mathbb{R} )$ which is represented by the T-sampled version of $f$. We consider the T-sampled version of $f$ for $T \doteqdot \frac{1}{2\Omega }$

How should I sketch $|\hat{g}|$ if $|\hat{f}|$ has the following appearance:

• Wait, $\hat g$ and $\hat f$are the Fourier transforms of what? – Marcus Müller Dec 16 '16 at 9:21
• @MarcusMüller $$\hat{g}\enspace is \enspace the\enspace Fourier\enspace transform\enspace of\enspace the\enspace function\enspace g \in L^{2}(\mathbb{R} )$$ and $$\hat{f}\enspace is \enspace the\enspace Fourier\enspace transform\enspace of\enspace the\enspace function\enspace f \in L^{2}(\mathbb{R} )$$ – Jishan Dec 16 '16 at 9:34
• You can write inline math simply by using one $instead of two. Anyway, where does g come from? How does it relate to f? – Marcus Müller Dec 16 '16 at 9:36 • You don't mention g anywhere in your question before you declare it to be the transform of f, which you know contradict in your comment... – Marcus Müller Dec 16 '16 at 9:37 • @MarcusMüller updated the question – Jishan Dec 16 '16 at 9:56 ## 1 Answer I'm answering the following problem: Given a function$ f $which has a Bandwidth (Half side) of$ 3 \Omega $and the following Fourier Transform: How would look like the Fourier Transform of$ g $which is a samples version of$ f $at rate$ T = \frac{1}{2 \Omega} $. Sampling in the Time Domain creates Replications in the Fourier Domain. The replications are$ \frac{1}{T} = 2 \Omega $apart. Now, just take the function you have in your drawing and add to it a replication of it centered at$ 2 \Omega \$.

It means that the information marked as A and B will be added to C and D.

• could you show me the drawing please. I am still a bit confused – Jishan Dec 16 '16 at 13:18