Steve on Image Processing

Concepts, algorithms & MATLAB

Relationship between continuous-time and discrete-time Fourier transforms 7

Posted by Steve Eddins,

Previously in my Fourier transforms series I've talked about the continuous-time Fourier transform and the discrete-time Fourier transform. Today it's time to start talking about the relationship between these two.

Let's start with the idea of sampling a continuous-time signal, as shown in this graph:

Mathematically, the relationship between the discrete-time signal and the continuous-time signal is given by:

(When I write equations involving both continuous-time and discrete-time quantities, I will sometimes use a subscript "c" to distinguish them.)

The sampling frequency is (in Hz) or (in radians per second).

The discrete-time Fourier transform of is related to the continuous-time Fourier transform of as follows:

But what does that mean? There are two key pieces to this equation. The first is a scaling relationship between and : . This means that the sampling frequency in the continuous-time Fourier transform, , becomes the frequency in the discrete-time Fourier transform. The discrete-time frequency corresponds to half the sampling frequency, or .

The second key piece of the equation is that there are an infinite number of copies of spaced by .

Let's look at a graphical example. Suppose looks like this:

Note that equals zero for all frequencies . This is what we mean when we say a continuous-time signal is band-limited. The frequency is called the bandwidth of the signal.

The discrete-time Fourier transform of looks like this:

where . As I mentioned before, normally only one period of is shown:

For this example, then, between and looks just like a scaled version of .

Next time we'll consider what happens when doesn't look like . In other words, we're about to tackle aliasing.

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7 CommentsOldest to Newest

Mark Andrews replied on : 1 of 7

Steve, I don’t think the terms for Xc have Omega scaled by T otherwise the spectral images would be wider (or narrower) than the base-band spectrum.

Steve replied on : 2 of 7

Mark—The frequency scaling term is correct. That’s what makes the sampling frequency in the Omega domain map to 2*pi in the omega domain.

Mark Andrews replied on : 3 of 7

But doesn’t Omega (big-omega) have units of s^-1 so that Omega/T now has units of s^-2? Or should that be omega (small omega) on the right hand side of the equality?

Steve replied on : 4 of 7

Whoops, there is something wrong with that equation … omega on the left of the equals and Omega on the right, and the periodicity is off. I’ll fix it later on tonight.

Sujith replied on : 6 of 7

Steve, Shouldn’t you categorize this post under “Fourier transforms”? Thanks anyway!