A quick question I was asked by a friend for homework. We have an LTI system and we needed an example for reconstruction and such...
Anyway, he asked me if the constant signal is bandwidth limited.
Since the Fourier Transform of a constant is $\delta(\omega)$ with a constant. And for $\omega=0$ we will receive $\infty$ and for others, $0$
I think it is not because the signal is $\infty$, although bandwidth limited talks about $\omega$ axis only and there it is limited.
Will be happy to know, thanks!.


2 Answers 2


Is the constant signal $x(t)=1$ band-width limited?


For mathematical proof, the trick is that the Dirac delta function is defined only under integration, and you need to start from its Fourier transform, which is well-defined,

$$ f_\delta(t)=\mathscr{F}(\delta(\omega))=\int_{-\infty}^{+\infty} \delta(\omega) e^{-j \omega t} d\omega = e^{-j \omega t} |_{w=0} = 1 $$

Then, by the Fourier inversion theorem $F_1(\omega)=\mathscr{F}(f_\delta(t))=c \delta(\omega)$, with $c$ being a constant.

The question is how $X(\omega) = F_1(\omega)$ is used.

From the engineering perspective, you can only observe that constant signal $x(t)$ through a bandlimited system with response $H(\omega)$. Therefore, in most cases including LTI systems, the observed signal $y(t)$ has spectrum $Y(\omega=0)= c H(0)$ and $Y(\omega \neq 0)=0$, which can be proved by writing down the Fourier transform of the convolution and can also be loosely thought as $H(\omega) \times c\delta(\omega)$, that satisfies the bandlimited signal definition. There is no problem of infinite power at $0$. You can do sampling, discrete fourier transform, system analysis and many more signal processing stuffs with such $y(t)$ that represents perfectly $x(t)$.


A signal is considered bandwidth-limited if its Fourier Transform (FT) is non-zero only within a finite range of frequencies. For a signal to be perfectly bandwidth-limited, its FT should exist and be zero beyond a certain frequency. This concept is crucial in understanding signal reconstruction, particularly when discussing the Nyquist-Shannon sampling theorem, which provides a criterion for perfectly reconstructing a bandwidth-limited signal from its samples.

When considering a constant signal, say $x(t) = C$ where $C$ is a constant, the Fourier Transform of this signal is indeed a Dirac delta function $\delta(\omega)$ at $\omega = 0$. This result suggests that the signal contains energy only at zero frequency. In other words, its spectral content is "limited" to a single point at $\omega = 0$.

However, there is a particular issue with your under standing of $\delta(\omega)$:

The Dirac delta function, $\delta(\omega)$ is not a function in the traditional sense of the term. The statement:

And for $\omega=0$ we will receive $\infty$ and for others, $0$

is wholly incorrect. The most rigorous definition of the Dirac delta is as follows:

$$\int_{-\infty}^{\infty}f(x)g(x)dx = f(0) \tag{1}$$

where any function, $g(x)$, satisfying the above Inner Product definition is an impulse. Hence, it is not unique. It can take tons of values at points other than $x=0$ and still satisfy eqn. $(1)$ as the area underneath a single point in a Riemann integral is $0$. So a better wording would be to say that its area is concentrated at $x=0$ rather than saying "And for $\omega=0$ we will receive $\infty$ and for others, $0$".

  • $\begingroup$ Hi, thanks for the explanation. Regarding the bandwidth, so you say the signal is indeed bandwidth-limited, but not perfected since its Fourier transform is not bounded from $Y$ axis. Did I understand correctly?......... Regarding the energy, yea...... regarding the incorrect, I understand what you say, but basically, according to definition, it is $\infty$, so I really am troubled now :\............... $\endgroup$ Feb 7 at 9:20
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    $\begingroup$ @BenShaines no, you din't get that right. Ahsan nowhere says that boundedness in the Y axis is relevant to whether it's bandlimited. Ahsan says that the concetration around $\omega=0$ is what counts for bandlimitation. $\endgroup$ Feb 7 at 14:18
  • $\begingroup$ Oh I see, thanks! sorry for the confusion $\endgroup$ Feb 7 at 17:00

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