1
$\begingroup$

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.

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

1 Answer 1

2
$\begingroup$

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 $a<b$):

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

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:

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

$\endgroup$
2
  • 1
    $\begingroup$ 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 $x=0$. $\endgroup$ Mar 31, 2023 at 18:19
  • $\begingroup$ @DilipSarwate: Yes, added. $\endgroup$
    – Matt L.
    Mar 31, 2023 at 20:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.