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We know that Fourier Transform of a signal exists if it is absolutely integrable and it exists for periodic signals if impulse functions are allowed.

If we consider the fourier transform of $\text{rect}(t)$ , we get $\text{sinc}(f)$ in frequency domain. Which has got zero frequency component as $1$. But we all know, DC value of $\text{rect}(t)$ is zero.

My question is:

  1. If a signal has got a zero frequency component in frequency domain ,There must be DC value in time domain. But why there is no DC value in case of $\text{rect}(t)$ in time domain?
  2. What is the difference between DC component and zero frequency component?
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  • $\begingroup$ You argue about properties of periodic signals, yet none of your examples is periodic. $\endgroup$
    – Jazzmaniac
    Feb 17, 2015 at 15:22
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    $\begingroup$ The DC value of rect(t) is not zero. $\endgroup$
    – Jim Clay
    Feb 17, 2015 at 15:49
  • $\begingroup$ yeah,What i am saying is , if a signal is bandlimited,then its DC value should be zero, which is not true in the above case@Jazzmaniac $\endgroup$
    – spectre
    Feb 17, 2015 at 15:51
  • $\begingroup$ can you please elaborate on that ?@JimClay $\endgroup$
    – spectre
    Feb 17, 2015 at 15:52
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    $\begingroup$ @JimClay: Its Fourier transform at DC is non-zero, but its DC-value is zero, in the sense that its average over time, as the interval approaches infinity, is zero. Please see my answer below for more detail. $\endgroup$
    – Matt L.
    Feb 17, 2015 at 16:03

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Let's first have a look at the rectangular signal given as an example in your question. If you have a rectangle $s(t)$ in the time domain which is $1$ in the interval $[-T/2,T/2]$ and zero elsewhere, its Fourier transform is $S(f)=T\text{sinc}(Tf)$, where I use $\text{sinc}(x)=\sin(\pi x)/(\pi x)$. The value of its Fourier transform at $f=0$ equals $S(0)=T$, which corresponds to

$$\int_{-\infty}^{\infty}s(t)dt=T\tag{1}$$

Its time average (or mean, or DC value) is given by

$$\bar{s}=\lim_{T_0\rightarrow\infty}\frac{1}{T_0}\int_{-T_0/2}^{T_0/2}s(t)dt=0\tag{2}$$

It is clear that any function for which the integral in (1) is finite, must have a DC-value of zero. The integral in (1) is the value of the Fourier transform of the signal at DC, and this is probably what confuses you. The DC value of a signal, and the value of its Fourier transform at DC are not the same. Any signal with a finite Fourier transform at DC has a DC value of zero, i.e. $\bar{s}=0$. Any signal with a non-zero DC value $\bar{s}\neq 0$ has a Dirac delta impulse component in its Fourier transform at DC.

If you write a signal as

$$s(t)=\bar{s}+\tilde{s}(t)$$

where $\bar{s}$ is the DC component as computed from (2), and, consequently, $\tilde{s}(t)$ has a DC component of zero, then its Fourier transform is

$$S(f)=\bar{s}\delta(f)+\tilde{S}(f)$$

where $\tilde{S}(0)$ is finite.

EDIT: Also note that when the Fourier transform of a signal $s(t)$ has a certain non-zero value at a frequency $f_0$, then this does not entail that the signal has a pure sinusoidal component at that frequency. The same is true for DC. If the Fourier transform has a finite value at DC, the time-domain signal has no DC component, otherwise there would be a Dirac impulse at $f=0$, just as there would be a Dirac impulse at $f_0$ if the signal contained a sinusoid at the frequency.

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  • $\begingroup$ thanks a lot for such a clear explanation,But my queastion is not solved,1) Are you saying zero freq component is not same as DC value? $\endgroup$
    – spectre
    Feb 17, 2015 at 16:15
  • $\begingroup$ @spectre: These are just names used incoherently by people. What I'm saying is that the DC value of a signal is not the same as the value of its Fourier transform at DC. You always need to check what people actually mean by the terms they use. As far as I know there is no clear definition for what is meant by "zero frequency component". I would guess that they mean the value of the Fourier transform at DC, and then the answer you're looking for would be: NO, those two are not the same. $\endgroup$
    – Matt L.
    Feb 17, 2015 at 16:18
  • $\begingroup$ But intuitively it doesnot make any sence right? $\endgroup$
    – spectre
    Feb 17, 2015 at 16:48
  • $\begingroup$ @spectre: As long as you use it with care, you are ok. DC value of a signal mathematically refers to average value of x(t) over infinite duration. However in practice intuitively you may call it just as if its the DC voltage of a 3.6 V li-ion rechargeable cell phone battery. Even though it wont exist forever long, it will provide that DC value in practice... then you are ok. $\endgroup$
    – Fat32
    Feb 17, 2015 at 17:01
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    $\begingroup$ @spectre I think that there are two sources of confusion. Most of us are engineers, so we are used to dealing with the practical, not abstract theory. Integrating from $-\infty$ to $+\infty$ is theory, not reality. In any real scenario, rect(t) will have non-zero mean. The other difference is that most of us deal with discrete Fourier transforms, not continuous transforms. In the realm of the discrete Fourier transform the zero frequency bin would diminish as the mean diminished- i.e. the limit of the zero frequency component of the transform would also be zero. $\endgroup$
    – Jim Clay
    Feb 17, 2015 at 20:49
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There is no difference between DC component and zero frequency component. They are two different names for the same thing.

Your mistake is in thinking that sinc(t) does not have a non-zero mean. sinc(t) does have a non-zero mean.

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  • $\begingroup$ if a signal has got DC value ,then it must be power signal and its FT should be impulse,As you said , if signal has got DC value ,why there is no impulse in freq domain? $\endgroup$
    – spectre
    Feb 17, 2015 at 15:05
  • $\begingroup$ please check the edit $\endgroup$
    – spectre
    Feb 17, 2015 at 15:18
  • $\begingroup$ @spectre It is not a power signal, it is an energy signal. See Bulent's answer. $\endgroup$
    – Jim Clay
    Feb 17, 2015 at 15:44
  • $\begingroup$ yes,what i am saying is why an energy signal has got DC component in freq domain , but it has got no DC component in time domain?@Jim clay $\endgroup$
    – spectre
    Feb 17, 2015 at 15:55
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    $\begingroup$ I think the important distinction to make is between a signal's DC value given by $\lim_{T\rightarrow\infty}\frac{1}{T}\int_{-T/2}^{T/2}s(t)dt$ and the value of its Fourier transform at DC. The signal's DC value can be zero, even though its FT at DC has a finite value. $\endgroup$
    – Matt L.
    Feb 17, 2015 at 20:22
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I am going to offer a very simple intuitive explanation to add to the excellent and detailed mathematical answers already given. I believe the question being asked comes down to the confusion of observing that the transform of the rect function (a Sinc function) has a value of 1 when the frequency component is 0, but intuitively we know that a rect function has no DC component (it's average goes to zero as time goes to infinity).

To resolve this quite simply, realize that the transform for any non-repeating waveform is a continuous function in frequency. The transform of the rect function represents an energy density in frequency, and a non-zero frequency range is always required to quantify non-zero density in frequency. "DC" is a point on the frequency domain which has zero width, and therefore would have zero energy in this case. To really observe DC with "zero width" in frequency, implies that we would have to observe it for an infinite amount of time. This is consistent with our first explanation that the mean of the rect function in time approaches zero as time goes to infinity. Further, if we observe the mean of the rect function for any shorter duration of time than infinity, then we are observing over an actual width in frequency (approximately 1/T where T is the observation time), and we will also see that the mean over a finite time interval is also non-zero.

With that DC can only be represented in a Fourier Transform as an impulse at $f=0$.

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For periodic signals (or integrating a finite window, with the outsides unspecified), the DC value of a 50% duty cycle rect function depends on the sum of the top level and the base level. It's only zero if the base is at a level inverse to the top.

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  • $\begingroup$ yeah, you are right, But rect(t) in the queastion is aperiodic and exists for -1/2<t<1/2 $\endgroup$
    – spectre
    Feb 17, 2015 at 15:58
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A brief answer.

Rect does NOT contain any DC component = No component at frequency = 0. But the fourier transform does tell there is a DC component at frequency = 0, and the rect function associated with that inverse fourier transform does have a DC component.

How the hell! How can the same rect function both have a DC component and not? It does not make sense. Well, the catch lies in the details.

In the first case, you are viewing a zero mean rect function - DC component 0.

To calculate the fourier transform of the same function, you need to have only positive values, so the entire rect function is shifted up (DC) to create the fourier transform graph that you see. Thus the DC component.

The shifting is usually done. I do not why.

Nice question!

P.S : To understand fourier transform as an engineer, there is one video which is better than 3blue1brown's. Check this out : https://www.youtube.com/watch?v=Oi0DrIf7Ipc&list=PL8mQv9_ssjDbntpOMevDqi1arvDjxfQsK&index=12 from 50:00 and the next couple videos in that playlist to understand how it works. You can design the DFT algorithm by yourself in python following this approach. Will hardly take you ten minutes.

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  • $\begingroup$ I think you should consider the difference between finite energy signals and finite power signals. In the sense of a finite-energy signal, $\text{rect}(t)$ most certainly has a DC component. It is the value of the Fourier transform evaluated at 0. $\endgroup$
    – robert bristow-johnson
    Nov 7, 2023 at 21:26
  • $\begingroup$ How does the definition of whether it is a finite energy signal affect whether rect(t) has a DC component? And does that mean all finite signals (which are bound for any particular time instant f(t) < infinity for all t) have a DC component? $\endgroup$
    – Siddhanth Ramani
    Nov 7, 2023 at 22:15
  • $\begingroup$ Well, mathematically, what is a (non-zero) DC signal? $\endgroup$
    – robert bristow-johnson
    Nov 7, 2023 at 22:44
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    $\begingroup$ I understand what you're saying now, thanks! $\endgroup$
    – Siddhanth Ramani
    Dec 9, 2023 at 6:17

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