Aufgaben:Exercise 5.4Z: On the Hanning Window - Revision history

Hwang at 03:18, 18 September 2022

2022-09-18T03:18:18Z

Guenter at 13:23, 20 May 2021

2021-05-20T13:23:57Z

Guenter at 13:10, 20 May 2021

2021-05-20T13:10:47Z

Guenter at 13:09, 20 May 2021

2021-05-20T13:09:09Z

Guenter: Guenter moved page Aufgaben:Exercise 5.4Z: On the Hanning window to Aufgaben:Exercise 5.4Z: On the Hanning Window

2021-05-19T12:38:01Z

Guenter moved page Exercise 5.4Z: On the Hanning window to Exercise 5.4Z: On the Hanning Window

Guenter: Guenter moved page Aufgaben:Exercise 5.4Z: On The Hanning Window to Aufgaben:Exercise 5.4Z: On the Hanning window

2021-05-19T12:36:36Z

Guenter moved page Exercise 5.4Z: On The Hanning Window to Exercise 5.4Z: On the Hanning window

Javier: Text replacement - "Category:Exercises for Signal Representation" to "Category:Signal Representation: Exercises"

2021-03-23T12:38:46Z

Text replacement - "Category:Exercises for Signal Representation" to "Category:Signal Representation: Exercises"

Noah at 19:42, 22 March 2021

2021-03-22T19:42:28Z

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Oezdemir: Oezdemir moved page Aufgaben:Aufgabe 5.4Z: Zum Hanning-Fenster to Aufgaben:Exercise 5.4Z: On The Hanning Window

2020-11-29T20:58:59Z

Oezdemir moved page Aufgabe 5.4Z: Zum Hanning-Fenster to Exercise 5.4Z: On The Hanning Window

Oezdemir at 20:57, 29 November 2020

2020-11-29T20:57:02Z

@@ Line 115: / Line 115: @@
-'''(4)'''&nbsp; The largest spectral amount outside the main lobe occurs at&nbsp; $f = ±2.5 f_{\rm A}$.&nbsp; With the result of subtask&nbsp; '''(2)'''&nbsp; holds:
+'''(4)'''&nbsp; The largest spectral magnitude outside the main lobe occurs at&nbsp; $f = ±2.5 f_{\rm A}$.&nbsp; With the result of subtask&nbsp; '''(2)'''&nbsp; holds:
 :$$W(f = 2.5 \cdot f_{\rm A}) = \frac{0.5}{f_{\rm A}}\cdot {\rm si}(2.5 \pi )
   +\frac{0.25}{f_{\rm A}}\cdot {\rm si}(1.5 \pi )+\frac{0.25}{f_{\rm A}}\cdot {\rm si}(3.5 \pi )=  \frac{0.25}{\pi \cdot f_{\rm A}}\left[ \frac{2}{2.5}-\frac{1}{1.5}-\frac{1}{3.5}\right] \approx -\frac{0.0121}{ f_{\rm A}}\hspace{0.05cm}.$$

@@ Line 32: / Line 32: @@
 ''Hints:''
 *This task belongs to the chapter&nbsp; [[Signal_Representation/Spectrum_Analysis|Spectral Analysis]].
-*Note that the frequency resolution $f_{\rm A}$&nbsp; is equal to the reciprocal of the adjustable parameter&nbsp; $T_{\rm P}$&nbsp;.
+*Note that the frequency resolution $f_{\rm A}$&nbsp; is equal to the reciprocal of the adjustable parameter&nbsp; $T_{\rm P}$.
@@ Line 72: / Line 73: @@
 {t}/{T_{\rm P}}) = 0.5+ 0.5\cdot {\rm
 cos}(2\pi \cdot {t}/{T_{\rm P}})\hspace{0.05cm}.$$
-*After time discretisation with&nbsp; $ν = t/T_{\rm A}$&nbsp; and&nbsp; $T_{\rm P}/T_{\rm A} = N = 32$&nbsp; , one obtains for the discrete-time window:
+*After time discretisation with&nbsp; $ν = t/T_{\rm A}$&nbsp; and&nbsp; $T_{\rm P}/T_{\rm A} = N = 32$, one obtains for the discrete-time window:
 :$$w(\nu)  =  w(\nu \cdot T_{\rm A}) = 0.5+ 0.5\cdot {\rm
 cos}(2\pi \cdot {\nu}/{N})\hspace{0.8cm}
@@ Line 114: / Line 115: @@
-'''(4)'''&nbsp; The largest spectral amount outside the main lobe occurs at&nbsp; $f = ±2.5 f_{\rm A}$&nbsp;. With the result of subtask&nbsp; '''(2)'''&nbsp; holds:
+'''(4)'''&nbsp; The largest spectral amount outside the main lobe occurs at&nbsp; $f = ±2.5 f_{\rm A}$.&nbsp; With the result of subtask&nbsp; '''(2)'''&nbsp; holds:
 :$$W(f = 2.5 \cdot f_{\rm A}) = \frac{0.5}{f_{\rm A}}\cdot {\rm si}(2.5 \pi )
   +\frac{0.25}{f_{\rm A}}\cdot {\rm si}(1.5 \pi )+\frac{0.25}{f_{\rm A}}\cdot {\rm si}(3.5 \pi )=  \frac{0.25}{\pi \cdot f_{\rm A}}\left[ \frac{2}{2.5}-\frac{1}{1.5}-\frac{1}{3.5}\right] \approx -\frac{0.0121}{ f_{\rm A}}\hspace{0.05cm}.$$
 *This gives the minimum distance between the main lobe and the side lobes:
-:$$A_{\rm H/S} = 20 \cdot {\rm lg}\hspace{0.15cm}
+:$$D_{\text{min (main lobe / side lobes) } } = 20 \cdot {\rm lg}\hspace{0.15cm}
   \frac{|W(0)|}{|W(2.5 \cdot f_{\rm A})|} = 20 \cdot {\rm lg}\hspace{0.15cm}
   \frac{0.5}{0.0121}\hspace{0.15 cm}\underline{\approx 32.3\,\,{\rm dB}}\hspace{0.05cm}.$$

@@ Line 18: / Line 18: @@
 *The main lobe thus extends to the frequency range&nbsp; $|f| ≤ 2 · f_{\rm A}$.
 *$W(f)$&nbsp; is largest in magnitude outside the main lobe for&nbsp; $f = ±2.5 · f_{\rm A}$.
-*Thus, the minimum distance between the main and side lobes applies here:
+*Thus, the minimum distance between the main and the side lobes applies here:
 :$$D_{\text{min (main lobe / side lobes) } } = 20 \cdot {\rm lg}\hspace{0.15cm}

@@ Line 5: / Line 5: @@
 [[File:P_ID1168__Sig_Z_5_4.png|250px|right|frame|Characterisation of the Hanning window]]
-In this task, important properties of the frequently used Hanning window are to be derived. The continuous-time representation in the interval from&nbsp; $-T_{\rm P}/2$&nbsp; to&nbsp; $+T_{\rm P}/2$&nbsp; is here as follows:
+In this task, important properties of the frequently used Hanning window are to be derived.&nbsp; The continuous-time representation in the interval from&nbsp; $-T_{\rm P}/2$&nbsp; to&nbsp; $+T_{\rm P}/2$&nbsp; is here as follows:
 :$$w(t)= {\rm cos}^2(\pi \cdot
 {t}/{T_{\rm P}})=  0.5\cdot \big(1 + {\rm cos}(2\pi \cdot
 {t}/{T_{\rm P}}) \big )
 \hspace{0.05cm}.$$
-Outside the symmetric time domain of duration&nbsp; $T_{\rm P}$&nbsp; ist $w&nbsp;(t) \equiv 0$.
+Outside the symmetric time domain of duration&nbsp; $T_{\rm P}$:&nbsp; &nbsp; $w(t) \equiv 0$.
-The upper graph shows the discrete-time representation&nbsp; $w(\nu) = w({\nu}  \cdot T_{\rm A})$, where&nbsp; $T_{\rm A}$&nbsp; is smaller than &nbsp; $T_{\rm P}$ by a factor&nbsp; $N = 32$&nbsp;. The definition range of the discrete time variable&nbsp; $ν$&nbsp; extends from&nbsp; $-16$&nbsp; to&nbsp; $+15$.
+The upper graph shows the discrete-time representation&nbsp; $w(\nu) = w({\nu}  \cdot T_{\rm A})$, where&nbsp; $T_{\rm A}$&nbsp; is smaller than &nbsp; $T_{\rm P}$ by a factor&nbsp; $N = 32$.&nbsp; The definition range of the discrete time variable&nbsp; $ν$&nbsp; extends from&nbsp; $-16$&nbsp; to&nbsp; $+15$.
-In the lower graph, the Fourier transform&nbsp; $W(f)$&nbsp; of the continuous-time window function&nbsp; $w(t)$&nbsp; is shown logarithmically. The abscissa is normalised to&nbsp; $f_{\rm A} = 1/T_{\rm P}$&nbsp;. One can see:
+In the lower graph, the Fourier transform&nbsp; $W(f)$&nbsp; of the continuous-time window function&nbsp; $w(t)$&nbsp; is shown logarithmically.&nbsp; The abscissa is normalised to&nbsp; $f_{\rm A} = 1/T_{\rm P}$.&nbsp; One can see:
 *The equidistant values&nbsp; $W({\mu}  \cdot f_{\rm A})$&nbsp; are zero except for&nbsp; $μ = 0$&nbsp; and&nbsp; $μ = ±1$.
 *The main lobe thus extends to the frequency range&nbsp; $|f| ≤ 2 · f_{\rm A}$.
-*$W(f)$&nbsp; is largest in magnitude outside the main lobe for&nbsp; $f = ±2.5 · f_{\rm A}$&nbsp;..
+*$W(f)$&nbsp; is largest in magnitude outside the main lobe for&nbsp; $f = ±2.5 · f_{\rm A}$.
 *Thus, the minimum distance between the main and side lobes applies here:
-:$$A_{\rm H/S} = 20 \cdot {\rm lg}\hspace{0.15cm}
+:$$D_{\text{min (main lobe / side lobes) } } = 20 \cdot {\rm lg}\hspace{0.15cm}
   \frac{|W(0)|}{|W(2.5 \cdot f_{\rm A})|} \hspace{0.15cm}{\rm (in}\hspace{0.1cm}{\rm dB)}\hspace{0.05cm}.$$
@@ Line 32: / Line 32: @@
 ''Hints:''
 *This task belongs to the chapter&nbsp; [[Signal_Representation/Spectrum_Analysis|Spectral Analysis]].
-*Note that the frequency resolution&nbsp; $f_{\rm A}$&nbsp; is equal to the reciprocal of the adjustable parameter&nbsp; $T_{\rm P}$&nbsp;.
+*Note that the frequency resolution $f_{\rm A}$&nbsp; is equal to the reciprocal of the adjustable parameter&nbsp; $T_{\rm P}$&nbsp;.
@@ Line 42: / Line 42: @@
 <quiz display=simple>
-{Give the discrete-time coefficients&nbsp; $w(ν)$&nbsp; of the Hanning window analytically. <br>What are the numerical values for&nbsp; $ν = 0$,&nbsp; $ν = 1$&nbsp; and&nbsp; $ν =  -\hspace{0.05cm}8$?
+{Give the discrete-time coefficients&nbsp; $w(ν)$&nbsp; of the Hanning window analytically.&nbsp; What are the numerical values for&nbsp; $ν = 0$,&nbsp; $ν = 1$&nbsp; and&nbsp; $ν =  -\hspace{0.05cm}8$?
 |type="{}"}
 $w(ν = 0) \hspace{0.37cm} = \ $ { 1 1% }
@@ Line 48: / Line 48: @@
 $w(ν =  -8) \hspace{0.03cm} = \ $ { 0.5 1% }
-{Calculate the spectral function&nbsp; $W(f)$&nbsp; in general. Which of the following statements are correct?
+{Calculate the spectral function&nbsp; $W(f)$&nbsp; in general.&nbsp; Which of the following statements are correct?
 |type="[]"}
 - $W(f)$&nbsp; yields complex results for special frequency values.
 + $W(f)$&nbsp; is even with respect to&nbsp; $f$&nbsp;, i.e.&nbsp; $W(-f) = W(+f)$.
-+ The spectral value&nbsp; $W(f = 0)$&nbsp; is equal to&nbsp; $0.5/f_{\rm A}$&nbsp; and thus real.
++ The spectral value&nbsp; $W(f = 0)=0.5/f_{\rm A}$&nbsp; and thus real.
-{What are&nbsp; $W(f = ±f_{\rm A})$&nbsp; and the $\text{6 dB}$bandwidth normalised to &nbsp; $f_{\rm A}$&nbsp;?
+{What are&nbsp; $W(f = ±f_{\rm A})$&nbsp; and the&nbsp; $\text{6dB}$&nbsp; bandwidth normalised to &nbsp; $f_{\rm A}$?
 |type="{}"}
 $W(±f_{\rm A})  \hspace{0.15cm} = \ $ { 0.25 1% } $\ \cdot \ 1/f_{\rm A}$
 $B_{\rm 6\hspace{0.05cm}dB}\hspace{-0.05cm}/\hspace{-0.05cm}f_{\rm A}   \hspace{0.2cm} = \ $ { 2 1% }
-{What is the minimum distance between the main lobe and the side lobe.
+{What is the minimum distance between the main lobe and the side lobes.
 |type="{}"}
-$A_{\rm H/S} \ = \ $ { 32.3 1% } $\ \rm dB$
+$D_{\text{min (main lobe / side lobes) } } \ = \ $ { 32.3 1% } $\ \rm dB$

← Older revision		Revision as of 12:38, 23 March 2021
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−	[[Category:~~Exercises for~~ Signal Representation\|^5.4 Spectrum Analysis^]]	+	[[Category:Signal Representation: Exercises\|^5.4 Spectrum Analysis^]]

← Older revision		Revision as of 20:57, 29 November 2020
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−	[[Category:Exercises for Signal Representation\|^5. ~~Zeit- und frequenzdiskrete Signaldarstellung~~^]]	+	[[Category:Exercises for Signal Representation\|^5.4 Spectrum Analysis^]]