What is the physical meaning of Fourier Transform having only $\delta(\omega)$ functions?

While calculating the Fourier Transform of the signal $$x(t) = 1 + \cos(6 \pi t + \pi/8)$$ I found that its Fourier Transform is purely in terms of Dirac Delta $\delta(\omega)$, that is its transform is only defined for certain values of frequency. What is the physical meaning of such a Fourier Transform?

• it's defined for almost all points. what you mean is that it's nonzero only for countably many points. – Marcus Müller Oct 1 '17 at 10:40
• Just to be complete: Mathematical statements by themselves do not have any physical meaning or significance. The meaning is associated by the mind of an engineer or a scientist who interprets the results according to the known physical laws... – Fat32 Oct 1 '17 at 14:27
• the mathematical meaning to having only $\delta(\omega-\omega_0)$ in the Fourier transform is that the time-domain signal is $e^{j \omega_0 t}$. – robert bristow-johnson Oct 2 '17 at 7:09

Explaining the physical meaning of something without the physical context in which it is encountered is going to be not particularly satisfactory.

$1+\cos(\omega t)$ is no more physical than $\delta( \omega)$ is, it’s just that the former has some mathematical properties that some consider having a more rigorous foundation. Without stating what time interval it is defined on, the most parsimonious frequency domain representation will default to time stretching from $-\infty$ to $\infty$ and results in a sum of delta functions.

When time is constrained to something closer to reality, the frequency domain representation becomes more physically reasonable. You can start thinking about things like the uncertainty principle and energy.

The interesting thing about the delta function is that it enters into more realistic calculations in straight forward ways so it isn’t fake but more along the lines of an ideal that can’t be reached.

Instead of Dirac delta functions, I would rather use the term distributions or generalized functions. Dirac deltas are not functions. In the most standard sense of Fourier transforms and integration, a non-zero constant function or a cosine is neither integrable, nor square integrable. It is thus difficult to give then a true physical meaning. However, cosines and sines play a central role in Fourier theory.

Hence, basically, the Fourier transform of a complex exponential $e^{ i \omega t}$ in the sense of distributions is a shifted Dirac pulse.

Going back to your question, if your FT is only composed of Dirac pulses, then your signal is composed of a linear combination of complex exponentials.

• So there is no physical interpretation of this phenomenon? Hence, from such a Fourier transform, I can only say that my original signal is composed of linear combination of complex exponentials. Nothing physical can be stated about the original signal? – Himanshu Sharma Oct 1 '17 at 9:57
• I believe that stating it is a linear combination of complex exponentials is a strong characterization already. However, such signals are not physically realizable. Could you tell some "physical interpretation" you would have expected? – Laurent Duval Oct 1 '17 at 10:00
• What I thought was that you can only view the signal at certain frequencies given within the argument of Dirac pulse. Is it correct? – Himanshu Sharma Oct 1 '17 at 10:01
• I feel I don't really understand your comment: "only view the signal at certain frequencies given within the argument of Dirac pulse." This could be understood differently, depending on the type of frequency analysis you are talking about – Laurent Duval Oct 1 '17 at 12:43

Intuitively speaking, $x(t)$ is periodic signal with infinite extend and consists of a DC term plus a cosine term. Those terms, alone, has infinite energy but finite average powers just as the total signal has.

There are a number of approaches in defining the Fourier transform of such signals. First one is to consider the continuous-time Fourier series (CTFS) of the periodic signal. As one can imagine, such a series decomposition of the periodic signal consists of a discrete set of complex coefficients each multiplying the corresponding exponential terms. For real signals the coefficients will be conjugate symmetric.

On the other hand, if you insist on using the conventional Fourier transform approach that was suited to finite energy signals, then the formal integral does not converge, but by the allowence of generalized functions such as the dirac delta $\delta(\omega-\omega_0)$ which represents the presence of a single periodic component with a frequency of $\omega=\omega_0$, then the periodic , infinite energy, finite average power, signals will have a valid Fourier transforms composed of discrete set of frequencies signified with the impulse positions.

It can be shown that the Fourier transform expression will have a very close relation to the Fourier series relation and the Fourier transform weights will be a direct function of the Fourier series coefficients at the corresponding dicerete frequencies.

On the other hand, when you consider a practical periodic signal which has a finite duration (therefore which mathematically is not periodic at all) then the Fourier transform will always converge and will have continuous set of frequencies. However you may still observe a very sharp transition at the periodicity frequencies which are not dirac pulses anymore but very narrow sinc pulses.