Aufgaben:Exercise 5.4: Comparison of Rectangular and Hanning Window - Revision history

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	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:		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:

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

	The graph shows the attenuation of the edges by the weighting function  $w(t)$  of the Hanning window.		The graph shows the attenuation of the edges by the weighting function  $w(t)$  of the Hanning window.
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	[[Category:Signal Representation: Exercises\|^5.4 Spectrum Analysis^]]		[[Category:Signal Representation: Exercises\|^5.4 Spectrum Analysis^]]
	[[de:Aufgaben:~~Aufgabe_5~~.4:~~_Spektralanalyse~~]]		[[de:Aufgaben:Aufgabe 5.4: Vergleich von Rechteck- und Hanningfenster]]

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 *The&nbsp; '''rectangular window''':
-:$${w} (\nu)   = \left\{ \begin{array}{c} 1 \\
+:$${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},$$*the&nbsp; '''Hanning window'''::$${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, $W(f)$&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.

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← Older revision		Revision as of 16:17, 16 March 2026
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	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:		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:

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

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

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	-N/2 \le \nu < N/2 \hspace{0.05cm}, \\		-N/2 \le \nu < N/2 \hspace{0.05cm}, \\
	{\rm else} \hspace{0.05cm}, \\		{\rm else} \hspace{0.05cm}, \\
	\end{array}$$		\end{array}$$:$$W(f) ={1}/{f_{\rm A}}\cdot {\rm si}(\pi \cdot
			{f}/{f_{\rm A}})\hspace{0.05cm},$$*the  '''Hanning window'''::$${w} (\nu) = \left\{ \begin{array}{c} 0.5 + 0.5 \cdot \cos (2 \pi \cdot {\nu}/{N}) \\
	:$$W(f) ={1}/{f_{\rm A}}\cdot {\rm si}(\pi \cdot
	{f}/{f_{\rm A}})\hspace{0.05cm},$$

	*the  '''Hanning window''':

	:$${w} (\nu) = \left\{ \begin{array}{c} 0.5 + 0.5 \cdot \cos (2 \pi \cdot {\nu}/{N}) \\
	0 \\ \end{array} \right.\quad		0 \\ \end{array} \right.\quad
	\begin{array}{*{10}c} {\rm{for}}		\begin{array}{*{10}c} {\rm{for}}
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	-N/2 \le \nu < N/2 \hspace{0.05cm}, \\		-N/2 \le \nu < N/2 \hspace{0.05cm}, \\
	{\rm else} \hspace{0.05cm}, \\		{\rm else} \hspace{0.05cm}, \\
	\end{array}$$		\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
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	In the task, reference is made to various spectral functions  $Y(f)$  for example to		In the task, reference is made to various spectral functions  $Y(f)$  for example to

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

	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.		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.

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	[[Category:Signal Representation: Exercises\|^5.4 Spectrum Analysis^]]		[[Category:Signal Representation: Exercises\|^5.4 Spectrum Analysis^]]
			[[de:Aufgaben:Aufgabe_5.4:_Spektralanalyse]]

Hwang at 13:24, 18 January 2023

2023-01-18T13:24:10Z

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	+ The DFT spectrum  $Y_{\rm A}(f)$  is identical to the actual spectrum  $X(f)$.		+ The DFT spectrum  $Y_{\rm A}(f)$  is identical to the actual spectrum  $X(f)$.

	{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? <br>Give the weights of the Dirac lines at  $f_1= 1\ \text{kHz}$   ⇒   $G(f_1)$  and at   $f_2 = 1.125\ \text{kHz}$  ⇒   $G(f_2)$.		{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? <br>Give the weights of the Dirac delta lines at  $f_1= 1\ \text{kHz}$   ⇒   $G(f_1)$  and at   $f_2 = 1.125\ \text{kHz}$  ⇒   $G(f_2)$.
	\|type="{}"}		\|type="{}"}
	$G(f_1 = 1.000 \ \text{kHz})\ = \ $ { 0.625 3% }  $\text{V}$		$G(f_1 = 1.000 \ \text{kHz})\ = \ $ { 0.625 3% }  $\text{V}$
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	'''(1)'''  <u>Solutions 1 and 4</u> are correct:		'''(1)'''  <u>Solutions 1 and 4</u> 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.		*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.		*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 delta lines.  This means:   the third statement is wrong.
	[[File:P_ID1167__Sig_A_5_4a.png\|right\|frame\|Output signal  $y(t)$  with the rectangular window]]		[[File:P_ID1167__Sig_A_5_4a.png\|right\|frame\|Output signal  $y(t)$  with the rectangular window]]

Javier: Text replacement - "analyse" to "analyze"

2021-09-22T11:47:37Z

Text replacement - "analyse" to "analyze"

← Older revision		Revision as of 13:47, 22 September 2021
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	Unknown and thus to be estimated are its parameters  $A_1$,  $f_1$,  $A_2$  and  $f_2$.		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 ~~analysed~~ can be set by the user as desired.		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.

	Two functions are available for windowing, each of which is zero for  $\|t\| > T_{\rm P}/2$:		Two functions are available for windowing, each of which is zero for  $\|t\| > T_{\rm P}/2$:
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	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	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 ~~analysed~~ by DFT and the DFT parameter  $T_{\rm P} = 8.5 \ \text{ms}$  is chosen unfavourably.		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.		*For one of the images the rectangular window is used, for the other the Hanning window.

Guenter at 15:16, 19 May 2021

2021-05-19T15:16:33Z

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Guenter: Guenter moved page Aufgaben:Exercise 5.4: Comparison of rectangular and Hanning window to Aufgaben:Exercise 5.4: Comparison of Rectangular and Hanning Window

2021-05-19T12:37:34Z

Guenter moved page Aufgaben:Exercise 5.4: Comparison of rectangular and Hanning window to Aufgaben:Exercise 5.4: Comparison of Rectangular and Hanning Window

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Guenter: Guenter moved page Aufgaben:Exercise 5.4: Comparison of Rectangular And Hanning Window to Aufgaben:Exercise 5.4: Comparison of rectangular and Hanning window

2021-05-19T12:36:00Z

Guenter moved page Aufgaben:Exercise 5.4: Comparison of Rectangular And Hanning Window to Aufgaben:Exercise 5.4: Comparison of rectangular and Hanning window

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