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Does the energy signal or finite energy signal have 0 dc component always?How to confirm this from the frequency spectrum?

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  • $\begingroup$ Think about it this way. Set a signal generator to generate a sinusoid with 1 V peak-to-peak voltage and 2 V DC bias. This signal has a DC component, and it is obviously an energy signal (otherwise it would be impossible to generate). $\endgroup$
    – MBaz
    Jun 23 '15 at 14:49
  • $\begingroup$ @MBaz Oh, please! How about a remark that is a little less confusing for a beginner, and possibly even correct? $\endgroup$ Jun 23 '15 at 15:04
  • $\begingroup$ Well, I'm sorry. It helps me to relate the equations to physical signals, and to think in terms of finite intervals before moving to infinite ones, but not everyone is the same. I may also be spending too much time in the lab lately. $\endgroup$
    – MBaz
    Jun 23 '15 at 15:43
  • $\begingroup$ @MBaz: But you do agree that the signal $x(t)=A\sin(\omega_0t+\phi)+B$ is NOT an energy signal? (even though you won't see it very often in your lab) $\endgroup$
    – Matt L.
    Jun 23 '15 at 16:13
  • $\begingroup$ @MattL. Of course, I agree. $\endgroup$
    – MBaz
    Jun 23 '15 at 20:20
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An energy signal $f(t)$ satisfies

$$\int_{-\infty}^{\infty}|f(t)|^2dt<\infty\tag{1}$$

i.e. it has finite energy. If $f(t)$ had a non-zero DC component $f_0$ defined by

$$f_0=\lim_{T\rightarrow\infty}\frac{1}{T}\int_{-T/2}^{T/2}f(t)dt\tag{2}$$

the integral in (1) wouldn't converge. So an energy signal must have a zero DC component.

Note that there is some confusion about the term "DC component". Some people would call $F(0)$, i.e. the value of the Fourier transform of $f(t)$ at frequency $\omega=0$, the "DC component" of $f(t)$. However, note that $F(0)$ given by

$$F(0)=\int_{-\infty}^{\infty}f(t)dt\tag{3}$$

can be non-zero (but finite), even if (1) is satisfied. One example of such a signal satisfying (1) but with $F(0)\neq 0$ is a sinc pulse.

Also have a look a this answer to a related question.

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  • $\begingroup$ So,finally what do i call as a dc component?Eq (2) or Eq (3)?Do u say rect(t) has a dc component? $\endgroup$
    – skt9
    Jun 23 '15 at 16:05
  • $\begingroup$ @skt9: I would say the correct definition is Eq. (2). According to that definition rect(t) has a zero DC component (because it is an energy signal). $\endgroup$
    – Matt L.
    Jun 23 '15 at 16:10

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