I know that $X(f)$ gives the amplitude associated with the frequency component $f$ of a signal $x(t)$.

Now, a sinusoidal signal in time $x(t) = A \cos (2 \pi f_0 t)$, has a Fourier transform $X(f) = \frac{A}{2}[\delta(f-f_0) + \delta(f+f_0) ]$.

My question is that the Dirac Delta funcion tends to $\infty$ at $0$. Then, multiplying it by $A/2$ should also result $\infty$. If this was the case, then what is the significance of $X(f)$ for $x(t) = A \cos(2 \pi f_0 t)$.

More specifically what is the significance of the impulses (having infinite amplitudes) in the frequency domain?

Thank you. :)


1 Answer 1


The Dirac delta is not strictly a function but a distribution. The Dirac delta is such that $\delta(x)=0 \ \forall x\neq0$ and it has to meet the following restriction:

$$\int_{-\infty}^{\infty} \delta(x) \ \mathrm{d}x = 1$$

This means that the unit impulse must integrate $1$ over all the real numbers.

Let's define this other "function" $$\tilde{\delta}(x)=2\delta(x)$$

As we can easily see, this new "function" must integrate $2$:

$$\int_{-\infty}^{\infty} \tilde{\delta}(x) \ \mathrm{d}x = 2\int_{-\infty}^{\infty} \delta(x) \ \mathrm{d}x = 2$$

And this is what we get when multiplying an impulse by a constat number. The area under the "function" changes and, due to the Dirac delta being non-zero only at the origin, this means that the impulse must change its height (if we can call it so) in order to change the area it integrates.

In the frequency domain, these constants act as weights that determine something like the relevance of a frequency in a given time-domain signal. If we have a small (i.e. it integrates a small area) impulse in a determined frequency, then there is a pure sinusoid of that frequency present in the signal whose amplitude is rather small. On the other hand, a large impulse would correspond to a pure sinusoid with a large amplitude.


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