Difference between revisions of "Aufgaben:Exercise 2.1Z: Sum Signal"

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===Solution===
 
===Solution===
 
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'''(1)'''   The following applies to the square wave signal:  $T_x = 1 \,\text{ms}$    ⇒   $f_x \hspace{0.15cm}\underline{= 1 \, \text{kHz}}$.
+
'''(1)'''   The following applies to the rectangular signal:  $T_x = 1 \,\text{ms}$    ⇒  
 +
:$$f_x \hspace{0.15cm}\underline{= 1 \, \text{kHz}}.$$
  
 +
'''(2)'''   The following applies to the triangular signal:  $T_y = 2.5 \,\text{ms}$  und 
 +
:$$f_y \hspace{0.15cm}\underline{= 0.4\,  \text{kHz}}.$$
  
 
+
'''(3)'''   The basic frequency  $f_s$  of the sum signal  $s(t)$  is the greatest common divisor  $f_x = 1 \,\text{kHz}$  and  $f_y = 0.4 \,\text{kHz}$.  
'''(2)'''   The following applies to the triangular signal:  $T_y = 2.5 \,\text{ms}$  und  $f_y \hspace{0.15cm}\underline{= 0.4\,  \text{kHz}}$.
 
 
 
 
 
 
 
'''(3)'''   The base frequency  $f_s$  of the sum signal  $s(t)$  is the greatest common divisor  $f_x = 1 \,\text{kHz}$  and  $f_y = 0.4 \,\text{kHz}$.  
 
 
*Daraus folgt  $f_s = 200 \,\text{Hz}$  und die Periodendauer  $T_s\hspace{0.15cm}\underline{ = 5 \,\text{ms}}$, wie auch aus der grafischen Darstellung des Signals  ${s(t)}$  auf der Angabenseite hervorgeht.
 
*Daraus folgt  $f_s = 200 \,\text{Hz}$  und die Periodendauer  $T_s\hspace{0.15cm}\underline{ = 5 \,\text{ms}}$, wie auch aus der grafischen Darstellung des Signals  ${s(t)}$  auf der Angabenseite hervorgeht.
  

Revision as of 13:17, 12 April 2021

Rectangular signal, triangular signal, sum signal

The adjacent diagram shows the periodic signals  ${x(t)}$  and  ${y(t)}$,  from which the sum   ${s(t)}$  – sketched in the lower diagram – and the difference  ${d(t)}$  are formed.

Furthermore, in this task we consider the signal  ${w(t)}$, which results from the sum of two periodic signals  ${u(t)}$  and  $v(t)$ .  The base frequencies of these signals are

  • $f_u = 998 \,\text{Hz},$
  • $f_v = 1002 \,\text{Hz}.$

That is all we know about the signals  ${u(t)}$  and  $v(t)$.




Hints:


Questions

1

What is the period duration  $T_x$  and the basic frequency  $f_x$  of the signal  ${x(t)}$?

$f_x\ = \ $

  $\text{kHz}$

2

What is the period duration  $T_y$  and the basic frequency  $f_y$  of the signal  ${y(t)}$?

$f_y\ = \ $

  $\text{kHz}$

3

Determine the basic frequency  $f_s$  and the period duration  $T_s$  of the sum signal  ${s(t)}$. 
Verify your results with the help of the sketched signal.

$T_s\ = \ $

  $\text{ms}$

4

What is the period duration  $T_d$  of the difference signal  ${d(t)}$ ?

$T_d\ = \ $

  $\text{ms}$

5

What is the period duration  $T_w$  of the signal  ${w(t)} = {u(t)} + v(t)$?

$T_w\ = \ $

  $\text{ms}$


Solution

(1)  The following applies to the rectangular signal:  $T_x = 1 \,\text{ms}$   ⇒  

$$f_x \hspace{0.15cm}\underline{= 1 \, \text{kHz}}.$$

(2)  The following applies to the triangular signal:  $T_y = 2.5 \,\text{ms}$  und 

$$f_y \hspace{0.15cm}\underline{= 0.4\, \text{kHz}}.$$

(3)  The basic frequency  $f_s$  of the sum signal  $s(t)$  is the greatest common divisor  $f_x = 1 \,\text{kHz}$  and  $f_y = 0.4 \,\text{kHz}$.

  • Daraus folgt  $f_s = 200 \,\text{Hz}$  und die Periodendauer  $T_s\hspace{0.15cm}\underline{ = 5 \,\text{ms}}$, wie auch aus der grafischen Darstellung des Signals  ${s(t)}$  auf der Angabenseite hervorgeht.


Difference $d(t) = x(t) - y(t)$

(4)  The period duration  $T_d$  does not change compared to the period duration  $T_s$ , if the signal  ${y(t)}$  is not added but subtracted:     $T_d = T_s \hspace{0.15cm}\underline{= 5\, \text{ms}}$.


(5)  The greatest common divisor of  $f_u = 998 \,\text{Hz}$  and  $f_{v} = 1002 \,\text{Hz}$  is  $f_w = 2 \,\text{Hz}$.

  • The inverse of this gives the period duration $T_w \hspace{0.15cm}\underline{= 500 \,\text{ms}}$.