periodicity of discrete versus continuous

[syedahmar]

Why is it that continuous signals have infinite frequencies possible while discrete time signals have frequencies only in a 2pi range??
whats the logic behind this?
???

 

[shakeebh]

take a close look at the analysis equation of DFT and then tell us if u r unable to make it out?

 

[echo47]

A discrete time signal can represent pretty much any signal you want, if you choose enough points (or infinite points) and an appropriate sample rate.

However, if you acquire a fixed number of points at a fixed sample rate, as is commonly done in DSP designs, then you limit your possible signals by those constraints. For example, how many different frequency waves can you identify in a set of only 8 data points. Not very many!

 

[syedahmar]

i still dont get it...looking at DTFT...where will i find this thing????which book??

 

[emrek]

I think such questions always arise due to a lack of intuitive understanding. Unfortunately many books do not deal with this side of the problem. Giving technical and mathematical details is not teaching in my opinion. Also due to this fact there are a lot of requests for a good book in this forum. All books in a subject tell similar things. But it is the effort of making people really understand the concepts and relations that differs the book from the others. Of course you must also have such an effort.

About the question, I can not say that I really understood the subject. But i think the answer is related to the question: into how many pieces you can divide a real thing (matter, signal ...). 2 pi (or actually pi) is showing the limit that you can sample a signal and deal with this sampling rate. Also we dont know that continuous signals have infinite frequencies. Have you count it. There may be an atomic pieces of frequency. My answer may be absurd. But I think this may be a good way of understanding a concept. Thinking very general, even making random relations with other known concepts. After some time you will find the right relation.

 

[abhishek_elec]

syedahmar, If you look at the digital signal, there is no concept of time in it. The X-axis is not the continuous time axis, instead it's the sample axis. X axis is discretized.

When you are representing digital signal in a equation,
for ex.
continuous signal : y(t) = Sin wt
becomes : y = Sin w' n

Here w' can not be in the units of 1/time , its unitless. If you compare values of y with y(t) at sampled points, you have to put w' = 2*pi*f' = 2*pi* f / Fs ; Fs is sampling freq.

If you apply sampling theorem here, f' <= Fs
So we say that w' is in the range of 0 to 2*pi.

If you want to understand intuitively, as you increase the Fs , the rate of variation in consecutive digital samples becomes less. As the Fs tends to f (signal's max frequency) , the rate of variation in the consecutive sample values increases and digital frequency approaches 1 (or w' approaches 2*pi).

 

[nitthilan]

Hi,
It must be seen from fourier transform that when ever you sample a signal (multiplication by a periodic pulse train of period 'T' ) it results in periodic repetition in the frequency domain (convolution of the signal by a periodic pulse train of period '1/T').
Since it is periodic in frequency domain, the represent it between 0-2*pi and assume it to be repeating after it. If you observe properly the periodicity in frequency domain would be (2*pi/T). Since T is constant it is left out.

Hope this clears the doubt.

Regards,
KJN

 

[abhishek_elec]

KJN, after sampling the spectrum becomes periodic but this is not the reason for range of 2*pi.
If you represent the sampled signal on continuous frequency axis, the period is Fs , not 2*pi.

 

[checkmate]

Digital spectrums are always represented in terms of normalized frequencies (wrt to sampling frequency), not absolute ones. Two common represntations are w/fs and f/fs.

In the first form, w/fs=2pi.f/fs. Since nyquist theorem requires f to be less than 0.5fs, the only valid range is -pi to pi.

Similarly, when f/fs is used instead, the valid range is -0.5 to 0.5.

If nyquist theorem is violated in the first place, then the digital spectrum is not a true representation of the original signal, and hence it's useless.

 

[abhishek_elec]

You are right, checkmate. I forgot to write 2*Fs in my explanation.
Nyquist theorem has to be satisfied for correct reconstruction.

 

[nitthilan]

hi abishek_elec,
I had mentioned it in my previous post itself that the periodicity is 0 - 2*pi/T were T the time interval between sampling and since it is constant it is left out.

If you observe properly the periodicity in frequency domain would be (2*pi/T). Since T is constant it is left out.

Regards,
KJN