Difference between revisions of "Aufgaben:Exercise 5.4: Comparison of Rectangular and Hanning Window"

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{{quiz-Header|Buchseite=*Buch*/*Kapitel*
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{{quiz-Header|Buchseite=Signal_Representation/Spectrum_Analysis
 
}}
 
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[[File:P_ID1166__Sig_A_5_4_neu.png|250px|right|Beispiel für die Spektralanalyse (Aufgabe A5.4)]]
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[[File:P_ID1166__Sig_A_5_4_neu.png|250px|right|frame|Examples for spectral analysis]]
  
Gegeben sei der prinzipielle Zeitverlauf x(t) eines periodischen Signals. Unbekannt sind die Parameter A1, f1, A2 und f2:
+
Let the time course of a periodic signal be given in principle:
 
   
 
   
$$\begin{align*} x(t) & =  A_1 \cdot \cos (2 \pi \cdot f_1 \cdot t) \\ & +  A_2 \cdot \cos (2 \pi \cdot f_2 \cdot t)
+
:$$x(t)   =  A_1 \cdot \cos (2 \pi \cdot f_1 \cdot t)  +  A_2 \cdot \cos (2 \pi \cdot f_2 \cdot t) \hspace{0.05cm}.$$
\hspace{0.05cm}.\end{align*}$$
+
 
 +
Unknown and thus to be estimated are its parameters  $A_1$,  $f_1$,  $A_2$  and  $f_2$.
 +
 
 +
After weighting the signal with the window function  $w(t)$ , the product  $y(t) = x(t) \cdot w(t)$  is subjected to a  [[Signal_Representation/Discrete_Fourier_Transform_(DFT)|Discrete Fourier Transform]]  (DFT) with the parameters  $N = 512$  and  $T_{\rm P}$.  The time  $T_{\rm P}$  of the signal section to be analyzed can be set by the user as desired.
  
Nach Gewichtung des Signals mit dem Fenster $w(t)$ wird das Produkt $y(t) = x(t) \cdot w(t)$ einer Diskreten Fouriertransformation (DFT) mit den Parametern $N$ = 512 und $T_P$ unterworfen. Die Zeitdauer $T_P$ des analysierten Signalausschnitts kann vom Benutzer beliebig eingestellt werden.
+
Two functions are available for windowing, each of which is zero for  $|t| > T_{\rm P}/2$:
Für die Fensterung stehen folgende Funktionen zur Verfügung, die jeweils für $|t| > T_P/2$ identisch 0 sind:
+
*The  '''rectangular window''':
das Rechteckfenster:
 
 
   
 
   
$${w} (\nu)  = \left\{ \begin{array}{c} 1 \\
+
:$${w} (\nu)  = \left\{ \begin{array}{c} 1 \\
 
  0 \\  \end{array} \right.\quad
 
  0 \\  \end{array} \right.\quad
\begin{array}{*{10}c}    {\rm{f\ddot{u}r}}
+
\begin{array}{*{10}c}    {\rm{for}}
 
\\    \\ \end{array}\begin{array}{*{20}c}
 
\\    \\ \end{array}\begin{array}{*{20}c}
 
-N/2 \le \nu < N/2 \hspace{0.05cm}, \\
 
-N/2 \le \nu < N/2 \hspace{0.05cm}, \\
{\rm sonst} \hspace{0.05cm}, \\
+
{\rm else} \hspace{0.05cm}, \\
 
\end{array}$$
 
\end{array}$$
 
   
 
   
$$W(f) ={1}/{f_{\rm A}}\cdot {\rm si}(\pi \cdot
+
:$$W(f) ={1}/{f_{\rm A}}\cdot {\rm si}(\pi \cdot
 
{f}/{f_{\rm A}})\hspace{0.05cm},$$
 
{f}/{f_{\rm A}})\hspace{0.05cm},$$
  
das Hanning–Fenster:
+
*the&nbsp; '''Hanning window''':
 
   
 
   
$${w} (\nu)  = \left\{ \begin{array}{c} 0.5 + 0.5 \cdot \cos (2 \pi \cdot \frac{\nu}{N}) \\
+
:$${w} (\nu)  = \left\{ \begin{array}{c} 0.5 + 0.5 \cdot \cos (2 \pi \cdot {\nu}/{N}) \\
 
  0 \\  \end{array} \right.\quad
 
  0 \\  \end{array} \right.\quad
\begin{array}{*{10}c}    {\rm{f\ddot{u}r}}
+
\begin{array}{*{10}c}    {\rm{for}}
 
\\    \\ \end{array}\begin{array}{*{20}c}
 
\\    \\ \end{array}\begin{array}{*{20}c}
 
-N/2 \le \nu < N/2 \hspace{0.05cm}, \\
 
-N/2 \le \nu < N/2 \hspace{0.05cm}, \\
{\rm sonst} \hspace{0.05cm}, \\
+
{\rm else} \hspace{0.05cm}, \\
 
\end{array}$$  
 
\end{array}$$  
  
$$W(f) ={0.5}/{f_{\rm A}}\cdot {\rm si}(\pi \cdot
+
:$$W(f) ={0.5}/{f_{\rm A}}\cdot {\rm si}(\pi \cdot
 
\frac{f}{f_{\rm A}})+ {0.25}/{f_{\rm A}}\cdot {\rm si}(\pi
 
\frac{f}{f_{\rm A}})+ {0.25}/{f_{\rm A}}\cdot {\rm si}(\pi
 
\cdot \frac{f-f_{\rm A}}{f_{\rm A}})+ {0.5}/{f_{\rm A}}\cdot
 
\cdot \frac{f-f_{\rm A}}{f_{\rm A}})+ {0.5}/{f_{\rm A}}\cdot
 
{\rm si}(\pi \cdot \frac{f+f_{\rm A}}{f_{\rm A}})\hspace{0.05cm}.$$
 
{\rm si}(\pi \cdot \frac{f+f_{\rm A}}{f_{\rm A}})\hspace{0.05cm}.$$
  
Beachten Sie, dass die Frequenzauflösung $f_A$ gleich dem Kehrwert des einstellbaren Parameters $T_P$ ist. $W(f)$ ist die Fouriertransformierte der zeitkontinuierlichen Fensterfunktion $w(t)$, während die oben angegebene Funktion $w(ν)$ die zeitdiskrete Gewichtungsfunktion angibt.
+
Here, $W(f)$&nbsp; is the Fourier transform of the continuous-time window function&nbsp; $w(t)$, while&nbsp; $w(ν)$&nbsp; indicates the discrete-time weighting function.
Im Laufe der Aufgabe wird auf verschiedene Spektralfunktionen $Y(f)$ Bezug genommen, zum Beispiel auf
+
 
 +
In the task, reference is made to various spectral functions&nbsp; $Y(f)$&nbsp; for example to
 
   
 
   
$$Y_{\rm A}(f) = 1\, {\rm V}\cdot {\rm \delta} (f \pm  1\,\,{\rm kHz})+
+
:$$Y_{\rm A}(f) = 1\, {\rm V}\cdot {\rm \delta} (f \pm  1\,\,{\rm kHz})+
 
  0.5\,\, {\rm V}\cdot {\rm \delta} (f \pm  1.125\,\,{\rm kHz})
 
  0.5\,\, {\rm V}\cdot {\rm \delta} (f \pm  1.125\,\,{\rm kHz})
 
  \hspace{0.05cm}.$$
 
  \hspace{0.05cm}.$$
  
In der obigen Grafik sind zwei weitere Spektralfunktionen $Y_B(f)$ und $Y_C(f)$ abgebildet, die sich ergeben, wenn ein 1 kHz–Signal mittels DFT analysiert wird und der DFT–Parameter $T_P$ = 8.5 ms ungünstig gewählt ist.
+
In the graph, two further spectral functions&nbsp; $Y_{\rm B}(f)$&nbsp; and&nbsp; $Y_{\rm C}(f)$&nbsp; are shown, which result when a&nbsp; $1 \ \text{kHz}$&nbsp; signal is analyzed by DFT and the DFT parameter&nbsp; $T_{\rm P} = 8.5 \ \text{ms}$&nbsp; is chosen unfavourably.
Für eines der Bilder ist das Rechteckfenster zugrundegelegt, für das andere das Hanning–Fenster. Nicht angegeben wird, welche Spektralfunktion zu welchem Fenster gehört.
+
 
Hinweis: Diese Aufgabe bezieht sich auf den Theorieteil von Kapitel 5.4.
+
*For one of the images the rectangular window is used, for the other the Hanning window.
 +
*It is not indicated which graph belongs to which window.
 +
 
 +
 
 +
 
 +
 
 +
 
 +
 
 +
 
 +
 
 +
''Hints:''
 +
*This task belongs to the chapter&nbsp; [[Signal_Representation/Spectrum_Analysis|Spectrum Analysis]].
 +
*Note that the frequency resolution&nbsp; $f_{\rm A}$&nbsp; is equal to the reciprocal of the adjustable parameter&nbsp; $T_{\rm P}$.
 +
*Unfortunately, the indices of&nbsp; $f_{\rm A}$&nbsp; and&nbsp; $Y_{\rm A}(f)$ collide.&nbsp; It is obvious that they are not related.&nbsp; Just to be on the safe side, we point this out. 
 +
 +
 
  
===Fragebogen===
+
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
  
{Welche der folgenden Aussagen treffen mit Sicherheit zu, wenn die DFT das Ausgangsspektrum $Y_A(f)$ anzeigt?
+
{Which of the following statements are true with certainty when the DFT displays the output spectrum&nbsp; $Y_{\rm A}(f)$&nbsp;?
 
|type="[]"}
 
|type="[]"}
+ Zur Gewichtung wurde das Rechteckfenster verwendet.
+
+ The rectangular window was used for weighting.
- Zur Gewichtung wurde das Hanning–Fenster verwendet.
+
- The Hanning window was used for weighting.
- Es wurde der DFT–Parameter $T_P$ = 4 ms verwendet.
+
- The DFT parameter&nbsp; $T_{\rm P} = 4\ \text{ms}$&nbsp; was used.
+ Das DFT–Spektrum $Y_A(f)$ ist identisch mit dem Spektrum $X(f)$.
+
+ The DFT spectrum&nbsp; $Y_{\rm A}(f)$&nbsp; is identical to the actual spectrum&nbsp; $X(f)$.
  
{Wie lautet $Y(f)$ bei Verwendung des Hanning–Fensters und $T_P$ = 8 ms, wenn das Eingangsspektrum $X(f) = Y_A(f)$ anliegt? Geben Sie die Gewichte der Diraclinien bei $f_1$ = 1 kHz und $f_2$ = 1.125 kHz an.
+
{Using the Hanning window and&nbsp;  $T_{\rm P} = 8 \ \text{ms} $, what is&nbsp; $Y(f)$&nbsp; when the input spectrum&nbsp; $X(f) = Y_{\rm A}(f)$&nbsp; is applied? <br>Give the weights of the Dirac lines at&nbsp; $f_1= 1\ \text{kHz}$ &nbsp; &rArr; &nbsp; $G(f_1)$&nbsp; and at &nbsp; $f_2 = 1.125\ \text{kHz}$&nbsp; &rArr; &nbsp; $G(f_2)$.
 
|type="{}"}
 
|type="{}"}
$G(f_1 = 1 \text{kHz}) =$ { 0.625 3% } V
+
$G(f_1 = 1.000 \ \text{kHz})\ = \ $ { 0.625 3% } &nbsp;$\text{V}$
$G(f_1 = 1.125 \text{kHz}) =$ { 0.5 3% } V
+
$G(f_2 = 1.125 \ \text{kHz})\ = \ $ { 0.5 3% } &nbsp;$\text{V}$
  
{Wir betrachten das 1 kHz–Cosinussignal $x(t)$. Welches Spektrum $– Y_B(f)$ oder $Y_C(f)$ – ergibt sich mit dem Rechteck– bzw. dem Hanning–Fenster, wenn der DFT-Parameter $T_P$ = 8.5 ms ungünstig gewählt ist?
+
{We consider the&nbsp; $1\ \text{kHz}$ cosine signal&nbsp; $x(t)$.&nbsp; Which spectrum -&nbsp; $Y_{\rm B}(f)$&nbsp; or&nbsp; $Y_{\rm C}(f)$&nbsp; results with the rectangular or the Hanning window, respectively, if the DFT parameter&nbsp;  $T_{\rm P} = 8.5 \ \text{ms}$&nbsp; is chosen unfavourably?
|type="[]"}
+
|type="()"}
- YB(f) ergibt sich bei Rechteckfensterung.
+
- $Y_{\rm B}(f)$&nbsp; results with rectangular windowing.
+ YB(f) ergibt sich mit dem Hanning-Fenster.
+
+ $Y_{\rm B}(f)$&nbsp; results with the Hanning window.
  
 
</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''1.''' Bei Verwendung des Hanning–Fensters müssten zunächst 3 Diracfunktionen zu erkennen sein, auch wenn $x(t)$ nur eine Frequenz beinhaltet es wurde das Rechteckfenster verwendet.
+
'''(1)'''&nbsp;  <u>Solutions 1 and 4</u> are correct:
Mit $T_P$ = 4 ms ergibt sich für die Frequenzauflösung $f_A = 1/T_P = 0.25$ kHz. Damit liegt die Frequenz $f_2$ nicht im vorgegebenen Raster und $Y(f)$ würde sich aus sehr vielen Diraclinien zusammensetzen. Das heißt: die dritte Aussage ist falsch.
+
*Using the Hanning window, three Dirac functions should be recognisable even if&nbsp; $x(t)$&nbsp; contains only one frequency &nbsp; &nbsp; the rectangular window was used.
Wie aus der nachfolgenden Grafik hervorgeht, hat $x(t)$ die Periodendauer $T_0$ = 8 ms. Wählt man den DFT–Parameter gleich $T_P$ = 8 ms (oder ein ganzzahliges Vielfaches davon), so stimmt die periodische Fortsetzung $P\{x(t)\}$ im Intervall $|t| \leq T$P/2$ mit $x(t)$ überein, so dass sich die Gewichtungsfunktion $w(t)$ nicht störend auswirkt: Das DFT–Spektrum $Y(f)$ stimmt somit mit dem tatsächlichen Spektrum überein. Richtig sind somit die Lösungsvorschläge 1 und 4.
+
*With&nbsp; $T_{\rm P} = 4 \ \text{ms}$&nbsp;, the frequency resolution is&nbsp; $f_{\rm A}= 1/T_{\rm P} = 0.25 \ \text{kHz}$.&nbsp; Thus the frequency&nbsp; $f_2$&nbsp; does not lie in the given grid and&nbsp; $Y(f)$&nbsp; would be composed of very many Dirac lines.&nbsp; This means: &nbsp; the third statement is wrong.
 +
[[File:P_ID1167__Sig_A_5_4a.png|right|frame|Output signal&nbsp; $y(t)$&nbsp; with the rectangular window]]
  
[[File:P_ID1167__Sig_A_5_4a.png|250px|right|Beispielsignal 1 zur Spektralanalyse (ML zu Aufgabe A5.4)]]
+
*As can be seen from the graph,&nbsp; $x(t)$&nbsp; has the period duration&nbsp; $T_{\rm 0} = 8 \ \text{ms}$.  
 +
*If one chooses the DFT parameter equal to&nbsp; $T_{\rm P} = 4 \ \text{ms}$&nbsp; (or an integer multiple thereof), the periodic continuation&nbsp; ${\rm P}\{ x(t)\} $&nbsp;  in the interval&nbsp; $|t| \leq T_{\rm P}/2$&nbsp; coincides with&nbsp; $x(t)$&nbsp;, so that the weighting function&nbsp; $w(t)$&nbsp; has no disturbing effect.&nbsp; So:
 +
*The DFT spectrum&nbsp; $Y(f)$&nbsp; thus agrees with the actual spectrum.
  
'''2.''' Wegen $T_P$ = 8 ms setzt sich das Hanning–Spektrum $W(f)$ aus drei Diracfunktionen bei positiven Frequenzen und drei dazu achsensymmetrischen Diracs bei negativen Frequenzen zusammen. Für die positiven Frequenzen lautet die Spektralfunktion:
+
 
 +
 
 +
'''(2)'''&nbsp; Because of&nbsp; $T_{\rm 0} = 8 \ \text{ms}$&nbsp;, the Hanning spectrum&nbsp; $W(f)$&nbsp;
 +
*consists of three Dirac functions at positive frequencies
 +
*and three axisymmetrical Diracs at negative frequencies
 +
 
 +
 
 +
are composed. For the positive frequencies, the spectral function is:
 +
[[File:P_ID1169__Sig_A_5_4b.png|right|frame|Output signal&nbsp; $y(t)$&nbsp; with the Hanning window]]
 
   
 
   
$$W(f) =0.5\cdot {\rm \delta}(f) + 0.25\cdot {\rm \delta}(f-f_{\rm A})+ 0.25\cdot {\rm \delta}(f+f_{\rm A})\hspace{0.05cm}.$$
+
:$$W(f) =0.5\cdot {\rm \delta}(f) + 0.25\cdot {\rm \delta}(f-f_{\rm A})+ 0.25\cdot {\rm \delta}(f+f_{\rm A})\hspace{0.05cm}.$$
  
Das Ausgangsspektrum ergibt sich aus der Faltung zwischen $X(f)$ und $W(f)$. Bei positiven Frequenzen ergeben sich nun vier Diracs mit folgenden Gewichten:
+
The output spectrum results from the convolution between&nbsp; $X(f)$&nbsp; and&nbsp; $W(f)$.&nbsp; At positive frequencies, there are now four Diracs with the following weights:
 
   
 
   
$$G(f = 0.875\,{\rm kHz}) & = & 1\, {\rm V}\cdot 0.25 = 0.250\, {\rm
+
:$$\begin{align*} G(f = 0.875\,{\rm kHz}) & = 1\, {\rm V}\cdot 0.25 = 0.250\, {\rm
 
  V}, \\
 
  V}, \\
  G(f = f_1 = 1.000\,{\rm kHz}) & = & 1\, {\rm V}\cdot 0.5 + 0.5\, {\rm V}\cdot 0.25 \hspace{0.15 cm}\underline{ = 0.625\, {\rm
+
  G(f = f_1 = 1.000\,{\rm kHz}) & = 1\, {\rm V}\cdot 0.5 + 0.5\, {\rm V}\cdot 0.25 \hspace{0.15 cm}\underline{ = 0.625\, {\rm
 
  V}}, \\
 
  V}}, \\
  G(f = f_2 = 1.125\,{\rm kHz}) & = & 1\, {\rm V}\cdot 0.25 + 0.5\, {\rm V}\cdot 0.5  \hspace{0.15 cm}\underline{= 0.500\, {\rm
+
  G(f = f_2 = 1.125\,{\rm kHz}) & = 1\, {\rm V}\cdot 0.25 + 0.5\, {\rm V}\cdot 0.5  \hspace{0.15 cm}\underline{= 0.500\, {\rm
 
  V}}, \\
 
  V}}, \\
  G(f = 1.250\,{\rm kHz}) & = & 0.5\, {\rm V}\cdot 0.25 = 0.125\, {\rm
+
  G(f = 1.250\,{\rm kHz}) & = 0.5\, {\rm V}\cdot 0.25 = 0.125\, {\rm
 
  V}
 
  V}
  \hspace{0.05cm}.$$
+
  \hspace{0.05cm}.\end{align*}$$
  
Die folgende Grafik zeigt die Abschwächung der Ränder durch die Gewichtungsfunktion $w(t)$ des Hanning–Fensters.
+
The graph shows the attenuation of the edges by the weighting function&nbsp; $w(t)$&nbsp; of the Hanning window.
  
[[File:P_ID1169__Sig_A_5_4b.png|250px|right|Beispielsignal 2 zur Spektralanalyse (ML zu Aufgabe A5.4)]]
 
  
'''3.''' Das Rechteck–Fenster liefert dann ein sehr stark verfälschtes Ergebnis, wenn die Fensterbreite $T_P$ (wie hier) nicht an die Frequenz des Cosinussignals angepasst ist. In diesem Fall ist das Hanning–Fenster besser geeignet. Daraus folgt: Richtig ist der zweite Lösungsvorschlag.
+
'''(3)'''&nbsp; <u> Solution 2</u> is correct:
 +
*The rectangular window delivers a very strongly distorted result if the window width&nbsp; $T_{\rm P}$&nbsp; (as here) is not adapted to the cosine frequency.
 +
*In this case, the Hanning window is more suitable.&nbsp; Then the measured spectrum&nbsp; $Y_{\rm B}(f)$ results.
 +
* From the spectrum&nbsp; $Y_{\rm C}(f)$&nbsp; the searched $1\ \rm kHz$ line is more difficult to detect.&nbsp; The spectrum&nbsp; $Y_{\rm C}(f)$&nbsp; results after rectangular windowing.  
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  
 
__NOEDITSECTION__
 
__NOEDITSECTION__
[[Category:Aufgaben zu Signaldarstellung|^5. Zeit- und frequenzdisktrete Signaldarstellung^]]
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[[Category:Signal Representation: Exercises|^5.4 Spectrum Analysis^]]

Revision as of 12:47, 22 September 2021

Examples for spectral analysis

Let the time course of a periodic signal be given in principle:

$$x(t) = A_1 \cdot \cos (2 \pi \cdot f_1 \cdot t) + A_2 \cdot \cos (2 \pi \cdot f_2 \cdot t) \hspace{0.05cm}.$$

Unknown and thus to be estimated are its parameters  $A_1$,  $f_1$,  $A_2$  and  $f_2$.

After weighting the signal with the window function  $w(t)$ , the product  $y(t) = x(t) \cdot w(t)$  is subjected to a  Discrete Fourier Transform  (DFT) with the parameters  $N = 512$  and  $T_{\rm P}$.  The time  $T_{\rm P}$  of the signal section to be analyzed can be set by the user as desired.

Two functions are available for windowing, each of which is zero for  $|t| > T_{\rm P}/2$:

  • The  rectangular window:
$${w} (\nu) = \left\{ \begin{array}{c} 1 \\ 0 \\ \end{array} \right.\quad \begin{array}{*{10}c} {\rm{for}} \\ \\ \end{array}\begin{array}{*{20}c} -N/2 \le \nu < N/2 \hspace{0.05cm}, \\ {\rm else} \hspace{0.05cm}, \\ \end{array}$$
$$W(f) ={1}/{f_{\rm A}}\cdot {\rm si}(\pi \cdot {f}/{f_{\rm A}})\hspace{0.05cm},$$
  • the  Hanning window:
$${w} (\nu) = \left\{ \begin{array}{c} 0.5 + 0.5 \cdot \cos (2 \pi \cdot {\nu}/{N}) \\ 0 \\ \end{array} \right.\quad \begin{array}{*{10}c} {\rm{for}} \\ \\ \end{array}\begin{array}{*{20}c} -N/2 \le \nu < N/2 \hspace{0.05cm}, \\ {\rm else} \hspace{0.05cm}, \\ \end{array}$$
$$W(f) ={0.5}/{f_{\rm A}}\cdot {\rm si}(\pi \cdot \frac{f}{f_{\rm A}})+ {0.25}/{f_{\rm A}}\cdot {\rm si}(\pi \cdot \frac{f-f_{\rm A}}{f_{\rm A}})+ {0.5}/{f_{\rm A}}\cdot {\rm si}(\pi \cdot \frac{f+f_{\rm A}}{f_{\rm A}})\hspace{0.05cm}.$$

Here, $W(f)$  is the Fourier transform of the continuous-time window function  $w(t)$, while  $w(ν)$  indicates the discrete-time weighting function.

In the task, reference is made to various spectral functions  $Y(f)$  for example to

$$Y_{\rm A}(f) = 1\, {\rm V}\cdot {\rm \delta} (f \pm 1\,\,{\rm kHz})+ 0.5\,\, {\rm V}\cdot {\rm \delta} (f \pm 1.125\,\,{\rm kHz}) \hspace{0.05cm}.$$

In the graph, two further spectral functions  $Y_{\rm B}(f)$  and  $Y_{\rm C}(f)$  are shown, which result when a  $1 \ \text{kHz}$  signal is analyzed by DFT and the DFT parameter  $T_{\rm P} = 8.5 \ \text{ms}$  is chosen unfavourably.

  • For one of the images the rectangular window is used, for the other the Hanning window.
  • It is not indicated which graph belongs to which window.





Hints:

  • This task belongs to the chapter  Spectrum Analysis.
  • Note that the frequency resolution  $f_{\rm A}$  is equal to the reciprocal of the adjustable parameter  $T_{\rm P}$.
  • Unfortunately, the indices of  $f_{\rm A}$  and  $Y_{\rm A}(f)$ collide.  It is obvious that they are not related.  Just to be on the safe side, we point this out.


Questions

1

Which of the following statements are true with certainty when the DFT displays the output spectrum  $Y_{\rm A}(f)$ ?

The rectangular window was used for weighting.
The Hanning window was used for weighting.
The DFT parameter  $T_{\rm P} = 4\ \text{ms}$  was used.
The DFT spectrum  $Y_{\rm A}(f)$  is identical to the actual spectrum  $X(f)$.

2

Using the Hanning window and  $T_{\rm P} = 8 \ \text{ms} $, what is  $Y(f)$  when the input spectrum  $X(f) = Y_{\rm A}(f)$  is applied?
Give the weights of the Dirac lines at  $f_1= 1\ \text{kHz}$   ⇒   $G(f_1)$  and at   $f_2 = 1.125\ \text{kHz}$  ⇒   $G(f_2)$.

$G(f_1 = 1.000 \ \text{kHz})\ = \ $

 $\text{V}$
$G(f_2 = 1.125 \ \text{kHz})\ = \ $

 $\text{V}$

3

We consider the  $1\ \text{kHz}$ cosine signal  $x(t)$.  Which spectrum -  $Y_{\rm B}(f)$  or  $Y_{\rm C}(f)$  – results with the rectangular or the Hanning window, respectively, if the DFT parameter  $T_{\rm P} = 8.5 \ \text{ms}$  is chosen unfavourably?

$Y_{\rm B}(f)$  results with rectangular windowing.
$Y_{\rm B}(f)$  results with the Hanning window.


Solution

(1)  Solutions 1 and 4 are correct:

  • Using the Hanning window, three Dirac functions should be recognisable even if  $x(t)$  contains only one frequency   ⇒   the rectangular window was used.
  • With  $T_{\rm P} = 4 \ \text{ms}$ , the frequency resolution is  $f_{\rm A}= 1/T_{\rm P} = 0.25 \ \text{kHz}$.  Thus the frequency  $f_2$  does not lie in the given grid and  $Y(f)$  would be composed of very many Dirac lines.  This means:   the third statement is wrong.
Output signal  $y(t)$  with the rectangular window
  • As can be seen from the graph,  $x(t)$  has the period duration  $T_{\rm 0} = 8 \ \text{ms}$.
  • If one chooses the DFT parameter equal to  $T_{\rm P} = 4 \ \text{ms}$  (or an integer multiple thereof), the periodic continuation  ${\rm P}\{ x(t)\} $  in the interval  $|t| \leq T_{\rm P}/2$  coincides with  $x(t)$ , so that the weighting function  $w(t)$  has no disturbing effect.  So:
  • The DFT spectrum  $Y(f)$  thus agrees with the actual spectrum.


(2)  Because of  $T_{\rm 0} = 8 \ \text{ms}$ , the Hanning spectrum  $W(f)$ 

  • consists of three Dirac functions at positive frequencies
  • and three axisymmetrical Diracs at negative frequencies


are composed. For the positive frequencies, the spectral function is:

Output signal  $y(t)$  with the Hanning window
$$W(f) =0.5\cdot {\rm \delta}(f) + 0.25\cdot {\rm \delta}(f-f_{\rm A})+ 0.25\cdot {\rm \delta}(f+f_{\rm A})\hspace{0.05cm}.$$

The output spectrum results from the convolution between  $X(f)$  and  $W(f)$.  At positive frequencies, there are now four Diracs with the following weights:

$$\begin{align*} G(f = 0.875\,{\rm kHz}) & = 1\, {\rm V}\cdot 0.25 = 0.250\, {\rm V}, \\ G(f = f_1 = 1.000\,{\rm kHz}) & = 1\, {\rm V}\cdot 0.5 + 0.5\, {\rm V}\cdot 0.25 \hspace{0.15 cm}\underline{ = 0.625\, {\rm V}}, \\ G(f = f_2 = 1.125\,{\rm kHz}) & = 1\, {\rm V}\cdot 0.25 + 0.5\, {\rm V}\cdot 0.5 \hspace{0.15 cm}\underline{= 0.500\, {\rm V}}, \\ G(f = 1.250\,{\rm kHz}) & = 0.5\, {\rm V}\cdot 0.25 = 0.125\, {\rm V} \hspace{0.05cm}.\end{align*}$$

The graph shows the attenuation of the edges by the weighting function  $w(t)$  of the Hanning window.


(3)  Solution 2 is correct:

  • The rectangular window delivers a very strongly distorted result if the window width  $T_{\rm P}$  (as here) is not adapted to the cosine frequency.
  • In this case, the Hanning window is more suitable.  Then the measured spectrum  $Y_{\rm B}(f)$ results.
  • From the spectrum  $Y_{\rm C}(f)$  the searched $1\ \rm kHz$ line is more difficult to detect.  The spectrum  $Y_{\rm C}(f)$  results after rectangular windowing.