One of things that was mentioned when we went through the DFT and explaining what it was, was that the sampled input signal becomes periodic. So what does this mean?

First, it means you have to be careful how
you interpret the results, but just as importantly
it can distort the frequency response. This happens because the DFT assumes
that the input signal is periodic, so if the input sequence finishes on
a whole number of periods everything is fine. But if your input sequence
is a sine wave oscillating at 8000 times a second, unless you calculate
the exact number of samples to give you an exact period, then the odds
on you getting the right amount are very slim. What happens then is that
a discontinuity occurs, the signal value suddenly jumps. The input sequence
that the DFT "*sees*" for a sequence consisting of one and a half
periods of a sine wave is shown in the image below.

This is just for a regular sine wave, imagine a signal that is not predictable there is no way that you will ever get an exact match in your input sequence.

So what happens in the frequency domain when you get these discontinuities? In the case shown above, with a single sine wave, you would normally expect to find a single sample of value 1 at the frequency of the sine wave, if and only if this value lies directly on a frequency bin.

is called a

In order to sample signals with very high frequencies you need a very high sampling rate. But to get useful resolution, you need to increase the number of your samples. This is an example of a classic engineering compromise, do you have good resolution or have masses and masses of data that take up your disk space? Its up to you and what you want it for.

An ideal frequency response.

A frequency response that has suffered leakage

Note that the magnitude of each of the samples is less than the magnitude of the ideal sample.

Since the energy has leaked out of the samples to the other bins, so
the amplitude of the peak will be less than the amplitude of the original
frequency sample (i.e. < 1). But if you add the squares of all the amplitudes
(to find how much energy they have) they *should* all add up to the
original amplitude (energy) (i.e. = 1)

To complicate matters further, if you increase the number of samples to give you greater resolution, the largest peak will get larger, indicating that it is nearer the true frequency, but the side samples where the energy has leaked out to, will get larger as well, making the response not as accurate as it should be.

These images show the frequency response of a sine wave, sin(*pi*/4),
both real and imaginary parts (remember
that DFT response is often complex).

- The first one is of a sine wave where the input sequence consists of FOUR complete periods.
- The second example is a sine wave where the input sequence consists of 4.4 periods (not a whole number of periods).

Imaginary Freq. Response

In the first example the frequency response is purely real, and
has two peaks, one at 4 (*pi*/4) and at 28 (the
DFT gives symmetric results)

Real Freq. Response

Imaginary Freq. Response

The next point is that the Fourier transform has developed a non-zero imaginary frequency response. The best way to explain describe it is that in the first example the imaginary freq response was there all along, it's just that when the samples occur, it always has the value of ZERO. BUT when the samples occur in the second example (remember that we went through the bit about freq resolution on this page ?) the frequency response isn't ZERO and so we now see the imaginary response in the discrete world.

If that last bit has left you a little bit confused then don't worry about it, just as long as you grasp the idea of the energy leaking between bins.

On to *Windowing* or back to *DFT
Properties*

or back to *DFT Contents* or back
to *Main Contents*