# Difference between DC component and zero frequency component of signal

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?
• You argue about properties of periodic signals, yet none of your examples is periodic. – Jazzmaniac Feb 17 '15 at 15:22
• The DC value of rect(t) is not zero. – Jim Clay Feb 17 '15 at 15:49
• 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 – spectre Feb 17 '15 at 15:51
• can you please elaborate on that ?@JimClay – spectre Feb 17 '15 at 15:52
• @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. – Matt L. Feb 17 '15 at 16:03

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.

• 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? – spectre Feb 17 '15 at 16:15
• @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. – Matt L. Feb 17 '15 at 16:18
• But intuitively it doesnot make any sence right? – spectre Feb 17 '15 at 16:48
• @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. – Fat32 Feb 17 '15 at 17:01
• @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. – Jim Clay Feb 17 '15 at 20:49

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.

• 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? – spectre Feb 17 '15 at 15:05
• please check the edit – spectre Feb 17 '15 at 15:18
• @spectre It is not a power signal, it is an energy signal. See Bulent's answer. – Jim Clay Feb 17 '15 at 15:44
• 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 – spectre Feb 17 '15 at 15:55
• 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. – Matt L. Feb 17 '15 at 20:22

for CTFT $X(j0)=\int x(t)dt$ , if this integral is bounded (converges) then X(j0) will also be finite nonzero value. However when this integral is unbounded, does not converge to a finite value, then $X(j\omega)$ is said to have a weighted impulse $K\delta(\omega)$ at the origin, which means there is a constant term (which is periodic) in the signal x(t) such as $x_{dc}(t)=K/2\pi$ for all t.

Any signal x(t) can be decomposed as a DC component plus an AC component as $$x(t) = X_{dc} + x_{ac}(t)$$ where the DC component $X_{dc}$ does not depend on time and the AC component $x_{ac}(t)$ has zero DC value. For example for the signal $x(t) = 2\cos^2(t) = 1 + cos(2t)$ , $X_{dc} = 1$ and $x_{ac}(t)=\cos(2t)$

The CTFT of such a signal will be $$X(j\omega) = (2\pi X_{dc}) \delta(\omega) + X_{ac}(j\omega)$$ Where the impulse term represents the DC value of the signal x(t) and the second term repsents the CTFT of the AC signal. Then we can identify three cases:
1- $X(j\omega)$ has an impulse at the origin, then $X_{dc}$ is nonzero, signal has a DC

2- $X(j0)$ is nonzero but finite, then $X_{dc}$ is zero but $\int {x_{ac}(t) dt}$ is nonzero and finite

3- $X(j0)$ is zero, then both $X_{dc}$ and $\int {x_{ac}(t) dt}$ are zero.

In many practical applications such pulsed binary signal transmission through bandlimited AWGN channels we shall practically refer to DC transmission capability of the channel in order to mean that the channel can pass a pulse p(t) with $\int p(t)dt$ being nonzero, which is said to be a dc-signalling case. Even though formally the pulse does not have a DC value, the frequency response of the channel $H(jw)$ must be non-zero at $\omega = 0$ so that the pulse will transmit without being distorted. If the transmission channel is bandpass rather than strict lowpass with $H(j0) = 0$, then you should consider using ac-pulses for which $\int p(t)dt =0$

• thanks for the info, and yeah, i am aware that X(j0)='area of time domain signal'.But it is true mathematically,Intuitively zero freq component should be DC value right? – spectre Feb 17 '15 at 16:03
• @spectre: First of all, value of the CTFT X(jw) at a specific single frequency $\omega_0$ will not make that intuition if it not an impulse there. I mean consider $x(t)=cos(\omega_0 t)$ whose CTFT is $X(j\omega)=\pi \delta(\omega -\omega_0) + \pi \delta(\omega +\omega_0)$ When you see these impulses, you can argue that there is a sine wave of infinite duration inside the signal x(t), but if $X(j\omega_0)$ is a finite, nonimpulse, value then there is no such sine wave present inside x(t), the same applies to DC consideration as a limit of frequency approaching 0. – Fat32 Feb 17 '15 at 16:17
• sorry ,I couldnot follow you,and in case of cos(Wot), Fourier transform is infinite ,which is impulse – spectre Feb 17 '15 at 16:23

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. I hope this helped.

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.

• yeah, you are right, But rect(t) in the queastion is aperiodic and exists for -1/2<t<1/2 – spectre Feb 17 '15 at 15:58