Exercise 2.2: DC Component of Signals

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Square wave signal with/ without DC component

The graph shows six time signals defined for all times $($from  $-\infty$  to  $+\infty)$.  For all sample signals  $x_i(t)$  the associated spectral function can be written as:

$$X_i(f)=A_0\cdot{\rm \delta}(f)+\Delta X_i(f).$$


  • $A_0$  is the DC component of the signal.
  • $\Delta X_i(f)$  is the spectrum of the residual signal reduced by the DC component: 
$$\Delta x_i(t) = x_i(t) - A_0.$$




Which of the signals contains a DC component, i.e. for which signals is   $A_0 \neq 0$?

Signal  $x_1(t),$
signal  $x_2(t),$
signal  $x_3(t),$
signal  $x_4(t),$
signal  $x_5(t),$
signal  $x_6(t).$


For which of the signals is the „residual spectrum”  $\Delta X_i(f) =0$?

Signal  $x_1(t),$
signal  $x_2(t),$
signal  $x_3(t),$
signal  $x_4(t),$
signal  $x_5(t),$
signal  $x_6(t).$


What is the DC component of the signal  $x_3(t)$?

$x_3(t)\hspace{-0.1cm}:\,\,A_0 \ = \ $

  ${\rm V}$


What is the DC component of the signal  $x_4(t)$?

$x_4(t)\hspace{-0.1cm}:\,\,A_0\ = \ $

  ${\rm V}$


What is the DC component of the signal  $x_6(t)$?

$x_6(t)\hspace{-0.1cm}:\,\,A_0\ = \ $

  ${\rm V}$


(1)  The correct answers are 1, 3, 4, 5 and 6.

  • All signals except  $x_2(t)$  contain a DC signal component.

(2)  Only solution 5 is correct:

  • If the DC component   $1\text{V}$ is subtracted from the signal   $x_5(t)$,  the residual signal  $\Delta x_5(t) = x5(t) - 1\text{V}$  is zero.
  • Accordignly, the spectral function is  $\Delta X_5(f) = 0$.
  • For all other time courses  $\Delta x_i(t)ßne 0$  and thus the associated spectral function   $\Delta X_i(f)\ne 0$,  too.

(3)  Given a periodic signal, averaging over a period duration is sufficient to calculate the DC signal component  $A_0$ .

  • For signal  $x_3(t)$  the period duration is  $T_0 = 3\,\text{ms}$.  This results in the required DC component:$$A_0=\rm \frac{1}{3\,ms}\cdot \big[1\,V\cdot 1\,ms+(-1\,V)\cdot 2\,ms \big] \hspace{0.15cm}\underline{=-0.333\,V}.$$

(4)  The signal  $x_4(t)$  can be written as:  $x_4(t) = 0.5 \,{\rm V} + Δx_4(t)$.

  • Here  $Δx_4(t)$  denotes a rectangular pulse with amplitude  $0.5 \,{\rm V} $  and duration  $4 \,{\rm ms} $,
  • which due to its finite duration does not contribute to the DC signal component.
  • Therefore  $A_0 \hspace{0.15cm}\underline{=0.5 \,{\rm V}}$ applies here.

(5)  The general equation for calculating the DC signal component is:

$$A_0=\lim_{T_{\rm M}\to \infty}\frac{1}{T_{\rm M}}\int_{-T_{\rm M}/2}^{+T_{\rm M}/2}x(t)\, {\rm d }t.$$
  • If one splits this integral into two partial integrals, one obtains:
$$A_0=\lim_{T_{\rm M}\to \infty}\frac{1}{T_{\rm M}}\int _{-T_{\rm M}/2}^{0}0 {\rm V} \cdot\, {\rm d } {\it t }+\lim_{T_{\rm M}\to \infty}\frac{1}{T_{\rm M}}\int _{0}^{+T_{\rm M}/2}1 \rm V \ {\rm d }{\it t }.$$
  • Only the second term makes a contribution.  From this follows again :  $A_0 \hspace{0.15cm}\underline{=0.5 \,{\rm V}}$.