# Steve on Image Processing

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

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 Responses to “Relationship between continuous-time and discrete-time Fourier transforms”

1. Mark Andrews replied on :

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.

2. Steve replied on :

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.

3. Mark Andrews replied on :

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?

4. Steve replied on :

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.

5. Steve replied on :

Mark—Sorry about my confusion. I corrected the equation and the following two paragraphs.

6. Sujith replied on :

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

7. Steve replied on :

Sujith—Thanks for pointing that out.

Steve Eddins is a software development manager in the MATLAB and image processing areas at MathWorks. Steve coauthored Digital Image Processing Using MATLAB. He writes here about image processing concepts, algorithm implementations, and MATLAB.

These postings are the author's and don't necessarily represent the opinions of MathWorks.