I heard somewhere that Fourier transform only works on a signal that
contains parts that are harmonic to each other (i.e. f, 2f, 3f etc.).
No. What you say would imply that the signal must be periodic, and that is not true. The Fourier transform can be computed for any signal, periodic or not, with finite energy or even with finite power. In the latter case (of finite power), the transform will, in most cases, be unbounded.
What you say is correct for Fourier series, not for the Fourier transform. The Fourier series decomposition is defined only for periodic signals. The spectrum of a periodic signal will be able to have contents only at \$f_0\$, \$2f_0\$, \$3f_0\$, etc. That does not mean that it will always have contents at all those frequencies. It means that it cannot have contents outside that set of frequencies.
Can anyone explain me on proof of this?
That was not right, so it cannot be proved.
What is the sufficient condition for Fourier transform?
There is no "the sufficient condition". There are several sufficient conditions.
For instance, one sufficient condition is that it satisfies both Dirichlet conditions:
1) Over any time interval of finite length, the function w(t) is single valued with a finite number of maxima and minima, and the number of discontinuities (if any) is finite.
2) w(t) is absolutely integrable.
That condition is sufficient, but not necessary.
A weaker sufficient condition for the existence of the Fourier transform is that the signal has finite energy. All physically realizable waveforms are finite-energy, so that means that all physical waveforms encountered in engineering practice are Fourier transformable. However, that is to have a Fourier transform with bounded values. If you allow yourself to work with unbounded values, then you can also compute the Fourier transform of a signal that has finite power (which is a less strict requirement). For instance, a sin(\$2\pi f_0·t\$) that exists for all t has finite power, but not finite energy. You can compute the Fourier transform of such a signal, but it will have unbounded values (in fact, "deltas") at \$f_0\$ and \$-f_0\$.
The discrete Fourier transform works on the assumption that your signal is periodic.
So, say we start with this arbitrary time-domain signal:
If it's periodic, we should be able to repeat it:
Well I guess that works, but we've introduced a discontinuity. This is like adding a square wave to your signal: you are going to see a new frequency component emerge equal to the DFT period, plus all of its odd harmonics.
In other words, the DFT sees any discontinuities even if they are at the ends of the signal. In fact, since the signal is periodic, it doesn't matter if we rotate all the inputs. If we do that with our original, we end up with:
This is exactly the same input as the first, as far as the DFT is concerned.
A window function works by tapering the ends to some similar value (usually 0) gradually, thereby making them equal. But it does so gradually, so that a minimum of extra frequency components are introduced. If we apply a window function to our original signal, you get something like this:
Which when duplicated, gives you:
or rotated:
No discontinuities! Now our non-periodic signal looks like a periodic signal, and we made it so while introducing a minimum of frequency-domain distortion. Of course, different window functions define "minimum distortion" in different ways, according to what you are trying to accomplish with the transform.
Best Answer
Your system is a modulator.
Since \$\cos(2\pi f_0 t)= \dfrac{e^{\,j2\pi f_0 t} + e^{\,-j2\pi f_0 t}}{2} \$ it follows (\$f_0=1\$):
\begin{align*} y(t) = \cos(2\pi t)\, x(t) = \dfrac{e^{\,j2\pi f_0 t} + e^{\,-j2\pi f_0 t}}{2} x(t) = \dfrac 1 2 x(t) e^{\,j2\pi f_0 t} + \dfrac 1 2 x(t) e^{\,-j2\pi f_0 t} \end{align*}
Given the following property of the Fourier transform:
\begin{align*} F\{x(t)e^{\,j2\pi f_0 t}\} = X(f-f_0) \end{align*}
Transforming with Fourier y(t) gives:
\begin{align*} Y(f) = \dfrac 1 2 X(f-f_0) + \dfrac 1 2 X(f + f_0) = \dfrac 1 2 X(f-1) + \dfrac 1 2 X(f + 1) \end{align*}