Let x(n) is a discrete time signal. We have done the Discrete Fourier Transform of it. Then the spectrum we get in frequency domain is continuous. How it happens?
Approach Number one: Think about it the other way around: A line spectrum represents in the time domain continuous sinusoidals with the periods defined by the spectral lines. Now you can use one of the characteristics of the Fourier Transform- its symmetry: What was frequency domain is now time domain and vice versa..... That way you see, that the "lines in the time domain" are continuous waves in the frequency. (see my Screen Shot 1.
Apporach number 2, closer to your desired "proof" (ScreenShot 2):
A time discrete signal consists of a series of amplitude modulated, time shifted dirac pulses: See the "Sum over s(nT)*delta(t-nT)"
If you now Fourier-Transform the whole term, you must not forget, that the FT of a shifted dirac delta in time itself is already a continuous wave in the frequency. Like the dirac at t=0 produces the "White Noise Spectrum". Now all those continuous waves from that many shifted diracs will sum up to a continuous, periodically repeated spectrum- et voila! (You may try to read the second line from the right....)
This is not a stringent proof, but I hope it helps!
uli
PS: There may be typos in the figs ;-)
The Discrete-time Fourier Transform (DTFT) of a discrete time series produces a frequency spectrum that is continuous and periodic. It is the dual of the Fourier series that from a continuous and periodic signal in time produces a discrete spectrum in frequency. What is the problem with that? In fact there are 4 Fourier transformatons. See attached.
The Discrete-Time Fourier Transform (DTFT) of a discrete time signal is continuous.
But de DFT (Discret Fourier Trasform) is Discrete. The DFT of a discret time signal of 100 samples produce a DFT composed of 100 complex coefficients.
Only a periodic Signal gives a discrete spectrum. This must not be the case when the signal is discrete. If a DFT shall give a correct spectrum it is assumed, that the signal is discrete AND periodic. Only if both conditions are fulfilled then it gives also a correct diskrete spectrum
Its due to the dual nature of Fourier transform. DTFT and DFT are different. As DTFT is a continues function of frequency it can't be stored in finite computer memory. To process a signal which is continues function of time with finite memory we do sampling. Similarly DFT is the sampled version of DTFT.
For a periodic continues time domain(continues function of time) signal the spectrum is discrete and it is dual in nature. So in digital, when signal becomes discrete in time domain its spectrum(DTFT) becomes continues and periodic with period 2pi. That is the reason why our range of vision in for spectrum is from 0 to 2pi. We can not store a continues function with finite memory. So we sample the DTFT and depending upon the sample points(N) we have N point DFT. DFT is also periodic but discrete and finite.
The reason for this circumstance is the duality between discrete and periodic functions. A periodic function Fourier transformed yields a discrete function and a discrete function Fourier transformed yields a periodic function, see e.g http://www.mdpi.com/2227-7390/3/2/299.
Please help me. why the magnitude spectrum of DTFS is discrete while the spectrum of DTFT is continuous?
Dear Sujan Mahmud, this is because "DTFS" and "DTFT" are misleading terms.
There are merely the (1) Fourier transform (for non-discrete non-periodic functions), (2) Fourier series analysis (Fourier transform for periodic functions) and Fourier series synthesis (Fourier transform for discrete functions, known as the "DTFT") and the Discrete Fourier Transform (for discrete peridic functions).
The meaning of "DTFS", in contrast, may differ from one author to another.
Data Fourier Transforms and Fourier Series and Their Interrelations
Article Four Particular Cases of the Fourier Transform
Jens Fisher
There are 4 cases of the same transformation. Think of them as binary possibilities 00, 01, 10 and 11, where 0 and 1 relate to continuous (0) or discrete (1). And of course, if something is discrete in one domsin it is periodic in the other. It took me a while, when I was a student to see that clearly. See the equations on the document I posted a while ago here (FOURALL) attached here again. By the way, when you like a post hit 'Recommend'. As I did with yours.
Dear Fernando Soares Schlindwein, I fully agree to what you are writing.
The reason why four Fourier transform definitions are required is the fact that "discrete functions" and "periodic functions" do only coexist in "generalized functions theory". Here, it can be shown that "discrete functions" are the Fourier transform of "periodic functions" and, vice versa, the Fourier transform of "periodic functions" are "discrete functions". Moreover, the Fourier transform of "discrete periodic functions" are "discrete periodic functions".
More generally, the Fourier transform of "periodization" is "discretization" and, vice versa, the Fourier transform of "discretization" is "periodization".
I have recently shown this in
Article Four Particular Cases of the Fourier Transform