Difference between revisions of "Aufgaben:Exercise 2.2: DC Component of Signals"

From LNTwww
 
(7 intermediate revisions by 3 users not shown)
Line 3: Line 3:
 
}}
 
}}
  
[[File:P_ID273__Sig_A_2_2.png|right|frame|Rechtecksignale mit und ohne Gleichanteil]]
+
[[File:P_ID273__Sig_A_2_2.png|right|frame|Square wave signal with/ without DC component]]
  
The graph shows some time signals defined for all times $($from  $-\infty$  to  $+\infty)$ . For all six sample signals  $x_i(t)$  the associated spectral function can be written as:
+
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:
 
   
 
   
 
:$$X_i(f)=A_0\cdot{\rm \delta}(f)+\Delta X_i(f).$$
 
:$$X_i(f)=A_0\cdot{\rm \delta}(f)+\Delta X_i(f).$$
  
Here
+
Here:
*$A_0$  is the DC component, and
+
*$A_0$  is the DC component of the signal.
*$\Delta X_i(f)$  is the spectrum of the residual signal reduced by the DC component  $\Delta x_i(t) = x_i(t) - A_0$.
+
*$\Delta X_i(f)$  is the spectrum of the residual signal reduced by the DC component:   
 
+
:$$\Delta x_i(t) = x_i(t) - A_0.$$
 
 
 
 
 
 
 
 
  
  
  
 
''Hint:''  
 
''Hint:''  
*This exercise belongs to the chapter  [[Signal_Representation/Direct_Current_Signal_-_Limit_Case_of_a_Periodic_Signal|Direct Current Signal - Limit Case of a Periodic Signal]].
+
*This exercise belongs to the chapter&nbsp; <br>[[Signal_Representation/Direct_Current_Signal_-_Limit_Case_of_a_Periodic_Signal|Direct Current Signal - Limit Case of a Periodic Signal]].
  
  
 
 
  
  
Line 72: Line 66:
  
 
'''(2)'''&nbsp; Only <u>solution 5 is correct</u>:
 
'''(2)'''&nbsp; Only <u>solution 5 is correct</u>:
*If the DC component &nbsp; $1\text{V}$ is subtracted from the signal &nbsp; $x_5(t)$&nbsp;, the residual signal&nbsp; $\Delta x_5(t) = x5(t) - 1\text{V}$&nbsp; is equal to zero.  
+
*If the DC component &nbsp; $1\text{V}$ is subtracted from the signal &nbsp; $x_5(t)$,&nbsp; the residual signal&nbsp; $\Delta x_5(t) = x5(t) - 1\text{V}$&nbsp; is zero.  
 
*Accordignly, the spectral function is&nbsp; $\Delta X_5(f) = 0$.  
 
*Accordignly, the spectral function is&nbsp; $\Delta X_5(f) = 0$.  
*For all other time courses&nbsp; $\Delta x_i(t)$&nbsp; is not equal to zero and thus the associated spectral function &nbsp; $\Delta X_i(f)$ is also not equal to zero.  
+
*For all other time courses&nbsp; $\Delta x_i(t)ßne 0$&nbsp; and thus the associated spectral function &nbsp; $\Delta X_i(f)\ne 0$,&nbsp; too.  
  
  
  
 
'''(3)'''&nbsp; Given a periodic signal, averaging over a period duration is sufficient to calculate the DC signal component&nbsp; $A_0$&nbsp;.  
 
'''(3)'''&nbsp; Given a periodic signal, averaging over a period duration is sufficient to calculate the DC signal component&nbsp; $A_0$&nbsp;.  
*For the sample signal&nbsp;  $x_3(t)$&nbsp; the period duration is&nbsp; $T_0 = 3\,\text{ms}$. This results in the required DC component:$$A_0=\rm \frac{1}{3\,ms}\cdot \big[1\,V\cdot 1\,ms+(-1\,V)\cdot 2\,ms \big]
+
*For signal&nbsp;  $x_3(t)$&nbsp; the period duration is&nbsp; $T_0 = 3\,\text{ms}$.&nbsp; This results in the required DC component:$$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}.$$
 
\hspace{0.15cm}\underline{=-0.333\,V}.$$
  
Line 85: Line 79:
  
 
'''(4)'''&nbsp; The signal&nbsp; $x_4(t)$&nbsp; can be written as:&nbsp; $x_4(t) = 0.5 \,{\rm V} + Δx_4(t)$.  
 
'''(4)'''&nbsp; The signal&nbsp; $x_4(t)$&nbsp; can be written as:&nbsp; $x_4(t) = 0.5 \,{\rm V} + Δx_4(t)$.  
*Here&nbsp; $Δx_4(t)$&nbsp; denotes a square wave pulse with amplitude&nbsp; $0.5 \,{\rm V} $&nbsp; and duration&nbsp; $4 \,{\rm ms} $, which due to its finite duration does not contribute to the DC signal component..  
+
*Here&nbsp; $Δx_4(t)$&nbsp; denotes a rectangular pulse with amplitude&nbsp; $0.5 \,{\rm V} $&nbsp; and duration&nbsp; $4 \,{\rm ms} $,  
 +
*which due to its finite duration does not contribute to the DC signal component.  
 
*Therefore&nbsp; $A_0 \hspace{0.15cm}\underline{=0.5 \,{\rm V}}$ applies here.
 
*Therefore&nbsp; $A_0 \hspace{0.15cm}\underline{=0.5 \,{\rm V}}$ applies here.
  
Line 98: Line 93:
 
:$$A_0=\lim_{T_{\rm M}\to \infty}\frac{1}{T_{\rm M}}\int _{-T_{\rm M}/2}^{0}0 {\rm V} \cdot\, {\rm d } {\it t }+\lim_{T_{\rm M}\to \infty}\frac{1}{T_{\rm M}}\int _{0}^{+T_{\rm M}/2}1 \rm V \ {\rm d }{\it t }.$$
 
:$$A_0=\lim_{T_{\rm M}\to \infty}\frac{1}{T_{\rm M}}\int _{-T_{\rm M}/2}^{0}0 {\rm V} \cdot\, {\rm d } {\it t }+\lim_{T_{\rm M}\to \infty}\frac{1}{T_{\rm M}}\int _{0}^{+T_{\rm M}/2}1 \rm V \ {\rm d }{\it t }.$$
  
*Only the second term makes a contribution. From this follows again :&nbsp; $A_0 \hspace{0.15cm}\underline{=0.5 \,{\rm V}}$.
+
*Only the second term makes a contribution.&nbsp; From this follows again :&nbsp; $A_0 \hspace{0.15cm}\underline{=0.5 \,{\rm V}}$.
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  
 
__NOEDITSECTION__
 
__NOEDITSECTION__
[[Category:Exercises for Signal Representation|^2.2 Direct Current Signal - Limit Case of a Periodic Signal^]]
+
[[Category:Signal Representation: Exercises|^2.2 Direct Current Signal^]]

Latest revision as of 17:30, 17 May 2021

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:

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

Here:

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


Hint:



Questions

1

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  $x_6(t).$

2

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  $x_6(t).$

3

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

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

  ${\rm V}$

4

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

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

  ${\rm V}$

5

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

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

  ${\rm V}$


Solution

(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 :  $A_0 \hspace{0.15cm}\underline{=0.5 \,{\rm V}}$.