Intuitive or physical explanation of DTFT$\{1\}=2\pi\delta(\omega)$

I am trying to understand the fact that

"The DTFT of 1 (an infinite discrete sequence of unit impulses from from $$-\infty$$ to $$+\infty$$) is $$2\pi\delta(\omega)$$"

in an intuitive or physical manner so that the significance of this result and its application in signal processing is understood.

just to show how the above result is obtained ( in a rather indirect manner )

the IDFT of $$2\cdot\pi\cdot\delta(\omega) is = \frac{1}{2\pi} \int_{-\pi}^{\pi} 2\cdot\pi\cdot\delta(\omega)\cdot e^{{j\omega n}}d\omega$$

now , $$\delta(\omega)$$ is defined for $$\omega=0$$ and is $$=1$$ , and $$0$$ everywhere else. The $$R.H.S$$ works out to $$1$$.

so ,the DTFT of 1 is $$2\cdot\pi\cdot\delta(\omega)$$.

What does the above statement mean physically/intuitively ? , that the only frequency component of the signal ( an infinite discrete sequence of unit impulses from from $$-\infty to +\infty$$) is an impulse of magnitude of $$2\cdot\pi$$ at $$\omega = 0$$. So is it being interpreted as a zero frequency (DC component) signal with an average magnitude of $$2\cdot\pi$$.

Where does the magnitude $$2\cdot\pi$$ come from ?.

Am I right or missing something here ? - Thanks.

• If it were me, I would say that $$\mathrm{DTFT}\big\{ x[n]=1 \big\} = 2 \pi \sum_{k=-\infty}^{\infty} \delta(\omega - 2\pi k)$$ Mar 31, 2023 at 20:41

You're right that an infinitely long constant sequence has just one frequency component at DC. That's why its spectrum must be concentrated at DC and must be zero for all $$\omega\neq 0$$.

Note that the Dirac delta impulse $$\delta(\omega)$$ is zero for all $$\omega\neq 0$$, but it doesn't make sense to talk about its value at $$\omega=0$$, because $$\delta(\omega)$$ is not a conventional function but a generalized function, also called a distribution. For your intuition, you might want to think about $$\delta(\omega)$$ being infinite at $$\omega=0$$, but definitely not equal to $$1$$ as you've mentioned in your question. It is the Kronecker delta that equals $$1$$ for zero argument, but that's totally different from the Dirac delta impulse. The latter only makes sense under an integral, and the property that you need to know is (assuming $$a):

$$\int_a^bf(x)\delta(x)dx=\begin{cases}f(0),& a<0

provided that $$f(x)$$ is continuous at $$x=0$$.

The factor $$2\pi$$ in the transform of a constant value is a consequence of dealing with angular frequency $$\omega=2\pi f$$. If you used $$f$$ instead you would get rid of that scaling factor:

$$\textrm{DTFT}\{1\}=\delta(f)$$

• You might want to add, for the sake of completeness, that the integral that you have written has value $f(0)$ provided that $f(x)$ is continuous at $x=0$. It doesn't matter for the application here since $f(x)=1 \forall x$ is continuous everywhere including at $x=0$. Mar 31, 2023 at 18:19
• @DilipSarwate: Yes, added. Mar 31, 2023 at 20:30