Difference between revisions of "Aufgaben:Exercise 3.4Z: Trapezoid, Rectangle and Triangle"

From LNTwww
m (Text replacement - "Category:Exercises for Signal Representation" to "Category:Signal Representation: Exercises")
m (Text replacement - "”" to """)
 
(2 intermediate revisions by one other user not shown)
Line 3: Line 3:
 
}}
 
}}
  
[[File:P_ID510__Sig_Z_3_4.png|right|frame|Trapezimpuls und dessen Grenzfälle „Rechteck” und „Dreieck” ]]
+
[[File:P_ID510__Sig_Z_3_4.png|right|frame|Trapezoidal pulse and its limiting cases  "Rectangle"  and  "Triangle" ]]
Three different pulse shapes are considered. The pulse&nbsp; ${x(t)}$&nbsp; is trapezoidal. For&nbsp; $| t | < t_1 = 4 \,\text{ms}$&nbsp;the time course is constant equal to&nbsp; ${A} = 1\, \text{V}$. Afterwards,&nbsp; ${x(t)}$&nbsp; drops linearly to the value zero until the time&nbsp; $t_2 = 6\, \text{ms}$&nbsp;.
+
Three different pulse shapes are considered.&nbsp; The pulse&nbsp; ${x(t)}$&nbsp; is trapezoidal.&nbsp; For&nbsp; $| t | < t_1 = 4 \,\text{ms}$&nbsp;the time course is constant equal to&nbsp; ${A} = 1\, \text{V}$.&nbsp; Afterwards,&nbsp; ${x(t)}$&nbsp; drops linearly to the value zero until the time&nbsp; $t_2 = 6\, \text{ms}$.&nbsp;
With the two derived system quantities, namely
 
  
* the&nbsp; [[Signal_Representation/Fourier_Transform_Laws#Reciprocity_Theorem_of_Time_duration_and_Bandwidth|equivalent bandwidth]]&nbsp;
+
The spectral function of the trapezoidal pulse is
:$$\Delta t = t_1  + t_2$$
 
  
* and the so-called roll-off factor (in the time domain)
+
:$$X( f ) = A \cdot \Delta t \cdot {\mathop{\rm si}\nolimits}( {{\rm \pi} \cdot \Delta t \cdot f} ) \cdot  \hspace{0.1cm}{\mathop{\rm si}\nolimits}( {{\rm \pi}\cdot \Delta t \cdot r_t \cdot  f} ).$$  
:$$r_t = \frac{t_2 - t_1 }{t_2 + t_1 }$$
 
  
is the spectral function of the trapezoidal pulse:
+
with the two derived system quantities, namely
:$$X( f ) = A \cdot \Delta t \cdot {\mathop{\rm si}\nolimits}( {{\rm \pi}  \cdot \Delta t \cdot f} ) \cdot \hspace{0.1cm}{\mathop{\rm si}\nolimits}( {{\rm \pi}\cdot \Delta t \cdot  r_t \cdot f} ).$$
+
 
Furthermore, the rectangular momentum&nbsp; ${r(t)}$&nbsp; and the triangular momentum&nbsp; ${d(t)}$&nbsp; are also shown in the graph, both of which can be interpreted as limiting cases of the trapezoidal momentum&nbsp; ${x(t)}$&nbsp;.
+
* the&nbsp; [[Signal_Representation/Fourier_Transform_Laws#Reciprocity_Theorem_of_Time_duration_and_Bandwidth|equivalent bandwidth]],&nbsp;
 +
:$$\Delta t = t_1  + t_2,$$
 +
 
 +
* and the so-called roll-off factor (in the time domain):
 +
:$$r_t = \frac{t_2 - t_1 }{t_2 + t_1 }.$$
 +
 
 +
Furthermore, the rectangular pulse&nbsp; ${r(t)}$&nbsp; and the triangular pulse&nbsp; ${d(t)}$&nbsp; are also shown in the graph, both of which can be interpreted as limiting cases of the trapezoidal pulse&nbsp; ${x(t)}$.
  
  
Line 24: Line 27:
  
 
''Hints:''  
 
''Hints:''  
*This exercise belongs to the chapter&nbsp; [[Signal_Representation/Fourier_Transform_Laws|Fourier Transform Laws]].
+
*This exercise belongs to the chapter&nbsp; [[Signal_Representation/Fourier_Transform_Theorems|Fourier Transform Theorems]].  
 
+
*You can check your results using the two interactive applets &nbsp;  
*You can check your results using the two interactive applets &nbsp; [[Applets:Impulse_und_Spektren|Impulses and Spectra]]&nbsp; and&nbsp; [[Applets:Frequenzgang_und_Impulsantwort|Frequenzgang und Impulsantwort]]&nbsp;.
+
:[[Applets:Pulses_and_Spectra|Pulses and Spectra]],  
 +
:[[Applets:Frequency_%26_Impulse_Responses|Frequency & Impulse Responses]].
  
  
Line 33: Line 37:
  
 
<quiz display=simple>
 
<quiz display=simple>
{What is the equivalent impulse duration and the rolloff factor of&nbsp; ${x(t)}$?
+
{What is the equivalent pulse duration and the rolloff factor of&nbsp; ${x(t)}$?
 
|type="{}"}
 
|type="{}"}
 
$\Delta t \ = \ $ { 10 3% } &nbsp;$\text{ms}$
 
$\Delta t \ = \ $ { 10 3% } &nbsp;$\text{ms}$
Line 42: Line 46:
 
|type="[]"}
 
|type="[]"}
 
- The spectral value at frequency&nbsp; $f = 0$&nbsp; is equal to&nbsp; $20 \,\text{mV/Hz}$.
 
- The spectral value at frequency&nbsp; $f = 0$&nbsp; is equal to&nbsp; $20 \,\text{mV/Hz}$.
+ For the phase function the values&nbsp; $0$&nbsp; or&nbsp; $\pi$&nbsp; $(180^{\circ})$&nbsp; are possible.
+
+ For the phase function the values&nbsp; $0$&nbsp; and&nbsp; $\pi$&nbsp; $(180^{\circ})$&nbsp; are possible.
+ ${X(f)}$&nbsp; only has zeros at all multiples of&nbsp; $100 \,\text{Hz}$&nbsp; auf.
+
+ ${X(f)}$&nbsp; only has zeros at all multiples of&nbsp; $100 \,\text{Hz}$.
  
  
{Which statements are true regarding the spectral function&nbsp; ${R(f)}$&nbsp; ?
+
{Which statements are true regarding the spectral function&nbsp; ${R(f)}$&nbsp;?
 
|type="[]"}
 
|type="[]"}
 
+ The spectral value at frequency&nbsp; $f = 0$&nbsp; is equal to&nbsp; ${X(f = 0)}$.
 
+ The spectral value at frequency&nbsp; $f = 0$&nbsp; is equal to&nbsp; ${X(f = 0)}$.
+ The values $0$ or&nbsp; $\pi$&nbsp; $(180^{\circ})$&nbsp; are possible for the phase function.
+
+ The values&nbsp; $0$&nbsp; and&nbsp; $\pi$&nbsp; $(180^{\circ})$&nbsp; are possible for the phase function.
+ ${R(f)}$&nbsp; only has zeros at all multiples of &nbsp; $100 \,\text{Hz}$ auf..
+
+ ${R(f)}$&nbsp; only has zeros at all multiples of&nbsp; $100 \,\text{Hz}$.
  
  
{Which statements are true regarding the spectral function&nbsp; ${D(f)}$&nbsp; ?
+
{Which statements are true regarding the spectral function&nbsp; ${D(f)}$&nbsp;?
 
|type="[]"}
 
|type="[]"}
 
+ The spectral value at frequency&nbsp; $f = 0$&nbsp; is equal to&nbsp; ${X(f = 0)}$.
 
+ The spectral value at frequency&nbsp; $f = 0$&nbsp; is equal to&nbsp; ${X(f = 0)}$.
- The values $0$ or&nbsp; $\pi$&nbsp; $(180^{\circ})$&nbsp; are possible for the phase function.
+
- The values&nbsp; $0$&nbsp; and&nbsp; $\pi$&nbsp; $(180^{\circ})$&nbsp; are possible for the phase function.
+ ${D(f)}$&nbsp; only has zeros at all multiples of &nbsp; $100 \,\text{Hz}$ auf.
+
+ ${D(f)}$&nbsp; only has zeros at all multiples of&nbsp; $100 \,\text{Hz}$.
  
  
Line 65: Line 69:
 
===Solution===
 
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp;  The equivalent pulse duration is&nbsp; $\Delta t = t_1 + t_2 \;\underline{= 10 \,\text{ms}}$&nbsp; and the rolloff factor&nbsp; $r_t = 2/10 \;\underline{= 0.2}$.
+
'''(1)'''&nbsp;  The equivalent pulse duration is&nbsp; $\Delta t = t_1 + t_2 \;\underline{= 10 \,\text{ms}}$&nbsp; and the rolloff factor is&nbsp; $r_t = 2/10 \;\underline{= 0.2}$.
  
  
Line 72: Line 76:
 
*The spectral value at&nbsp; $f = 0$&nbsp; is&nbsp; $A \cdot \Delta t = 10 \,\text{mV/Hz}$.  
 
*The spectral value at&nbsp; $f = 0$&nbsp; is&nbsp; $A \cdot \Delta t = 10 \,\text{mV/Hz}$.  
 
*Since&nbsp; ${X(f)}$&nbsp; is real and can assume both positive and negative values, only the two phase values&nbsp; $0$&nbsp; und&nbsp; $\pi$&nbsp; are possible.
 
*Since&nbsp; ${X(f)}$&nbsp; is real and can assume both positive and negative values, only the two phase values&nbsp; $0$&nbsp; und&nbsp; $\pi$&nbsp; are possible.
*Zeros exist due to the first si-function at all multiples of&nbsp; $1/\Delta t = 100\, \text{Hz}$.  
+
*Zeros exist due to the first si&ndash;function at all multiples of&nbsp; $1/\Delta t = 100\, \text{Hz}$.  
*The second si function leads to zero crossings at intervals of&nbsp; $1/(r_t \cdot \Delta t) = 500 \,\text{Hz}$. These coincide exactly with the zeros of the first si-function.  
+
*The second si&ndash;function leads to zero crossings at intervals of&nbsp; $1/(r_t \cdot \Delta t) = 500 \,\text{Hz}$.&nbsp; These coincide exactly with the zeros of the first si&ndash;function.  
  
  
Line 85: Line 89:
 
'''(4)'''&nbsp; Proposed <u>solutions 2 and 3</u> are correct:
 
'''(4)'''&nbsp; Proposed <u>solutions 2 and 3</u> are correct:
 
*For the triangular pulse, the rolloff factor is&nbsp; $r_t = 1$.  
 
*For the triangular pulse, the rolloff factor is&nbsp; $r_t = 1$.  
*The equivalent pulse duration is&nbsp; $\Delta t = 10 \,\text{ms}$. It follows that &nbsp; $D( f ) = A \cdot \Delta t \cdot {\mathop{\rm si}\nolimits} ^2 ( {{\rm{\pi }} \cdot \Delta t \cdot f} )$&nbsp; and&nbsp; $D( f = 0) = A \cdot \Delta t  = X( f = 0)$.  
+
*The equivalent pulse duration is&nbsp; $\Delta t = 10 \,\text{ms}$.&nbsp; It follows that &nbsp; $D( f ) = A \cdot \Delta t \cdot {\mathop{\rm si}\nolimits} ^2 ( {{\rm{\pi }} \cdot \Delta t \cdot f} )$&nbsp; and&nbsp; $D( f = 0) = A \cdot \Delta t  = X( f = 0)$.  
*Since&nbsp; ${D(f)}$&nbsp; cannot become negative, the phase&nbsp; $[{\rm arc} \; {D(f)}]$&nbsp; is always zero. The phase value&nbsp; $\pi$&nbsp; $(180°)$&nbsp; is therefore not possible with the triangular form.  
+
*Since&nbsp; ${D(f)}$&nbsp; cannot become negative, the phase&nbsp; $[{\rm arc} \; {D(f)}]$&nbsp; is always zero.&nbsp; The phase value&nbsp; $\pi$&nbsp; $(180°)$&nbsp; is therefore not possible with the triangular pulse.  
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  

Latest revision as of 15:39, 28 May 2021

Trapezoidal pulse and its limiting cases  "Rectangle"  and  "Triangle"

Three different pulse shapes are considered.  The pulse  ${x(t)}$  is trapezoidal.  For  $| t | < t_1 = 4 \,\text{ms}$ the time course is constant equal to  ${A} = 1\, \text{V}$.  Afterwards,  ${x(t)}$  drops linearly to the value zero until the time  $t_2 = 6\, \text{ms}$. 

The spectral function of the trapezoidal pulse is

$$X( f ) = A \cdot \Delta t \cdot {\mathop{\rm si}\nolimits}( {{\rm \pi} \cdot \Delta t \cdot f} ) \cdot \hspace{0.1cm}{\mathop{\rm si}\nolimits}( {{\rm \pi}\cdot \Delta t \cdot r_t \cdot f} ).$$

with the two derived system quantities, namely

$$\Delta t = t_1 + t_2,$$
  • and the so-called roll-off factor (in the time domain):
$$r_t = \frac{t_2 - t_1 }{t_2 + t_1 }.$$

Furthermore, the rectangular pulse  ${r(t)}$  and the triangular pulse  ${d(t)}$  are also shown in the graph, both of which can be interpreted as limiting cases of the trapezoidal pulse  ${x(t)}$.




Hints:

  • This exercise belongs to the chapter  Fourier Transform Theorems.
  • You can check your results using the two interactive applets  
Pulses and Spectra,
Frequency & Impulse Responses.


Questions

1

What is the equivalent pulse duration and the rolloff factor of  ${x(t)}$?

$\Delta t \ = \ $

 $\text{ms}$
$r_t\hspace{0.3cm} = \ $

2

Which statements are true regarding the spectral function  ${X(f)}$ ?

The spectral value at frequency  $f = 0$  is equal to  $20 \,\text{mV/Hz}$.
For the phase function the values  $0$  and  $\pi$  $(180^{\circ})$  are possible.
${X(f)}$  only has zeros at all multiples of  $100 \,\text{Hz}$.

3

Which statements are true regarding the spectral function  ${R(f)}$ ?

The spectral value at frequency  $f = 0$  is equal to  ${X(f = 0)}$.
The values  $0$  and  $\pi$  $(180^{\circ})$  are possible for the phase function.
${R(f)}$  only has zeros at all multiples of  $100 \,\text{Hz}$.

4

Which statements are true regarding the spectral function  ${D(f)}$ ?

The spectral value at frequency  $f = 0$  is equal to  ${X(f = 0)}$.
The values  $0$  and  $\pi$  $(180^{\circ})$  are possible for the phase function.
${D(f)}$  only has zeros at all multiples of  $100 \,\text{Hz}$.


Solution

(1)  The equivalent pulse duration is  $\Delta t = t_1 + t_2 \;\underline{= 10 \,\text{ms}}$  and the rolloff factor is  $r_t = 2/10 \;\underline{= 0.2}$.


(2)  Proposed solutions 2 and 3 are correct:

  • The spectral value at  $f = 0$  is  $A \cdot \Delta t = 10 \,\text{mV/Hz}$.
  • Since  ${X(f)}$  is real and can assume both positive and negative values, only the two phase values  $0$  und  $\pi$  are possible.
  • Zeros exist due to the first si–function at all multiples of  $1/\Delta t = 100\, \text{Hz}$.
  • The second si–function leads to zero crossings at intervals of  $1/(r_t \cdot \Delta t) = 500 \,\text{Hz}$.  These coincide exactly with the zeros of the first si–function.


(3)  All proposed solutions are correct:

  • With the equivalent pulse duration  $\Delta t = 10 \,\text{ms}$  and the rolloff factor  $r_t = 0$  one obtains:   $R( f ) = A \cdot \Delta t \cdot {\mathop{\rm si}\nolimits} ( {{\rm{\pi }} \cdot \Delta t \cdot f} ).$
  • It follows that  $R( f = 0) = A \cdot \Delta t = X( f = 0).$


(4)  Proposed solutions 2 and 3 are correct:

  • For the triangular pulse, the rolloff factor is  $r_t = 1$.
  • The equivalent pulse duration is  $\Delta t = 10 \,\text{ms}$.  It follows that   $D( f ) = A \cdot \Delta t \cdot {\mathop{\rm si}\nolimits} ^2 ( {{\rm{\pi }} \cdot \Delta t \cdot f} )$  and  $D( f = 0) = A \cdot \Delta t = X( f = 0)$.
  • Since  ${D(f)}$  cannot become negative, the phase  $[{\rm arc} \; {D(f)}]$  is always zero.  The phase value  $\pi$  $(180°)$  is therefore not possible with the triangular pulse.