Aufgaben:Exercise 5.5: Fast Fourier Transform - Revision history

Guenter at 16:08, 21 May 2021

2021-05-21T16:08:54Z

Guenter at 16:00, 21 May 2021

2021-05-21T16:00:10Z

Noah at 13:36, 23 March 2021

2021-03-23T13:36:46Z

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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"

Oezdemir: Oezdemir moved page Aufgaben:Aufgabe 5.5: Fast-Fouriertransformation to Aufgaben:Exercise 5.5: Fast Fourier Transform

2020-11-29T21:00:58Z

Oezdemir moved page Aufgabe 5.5: Fast-Fouriertransformation to Exercise 5.5: Fast Fourier Transform

Oezdemir at 21:00, 29 November 2020

2020-11-29T21:00:27Z

Javier: Text replacement - "Signaldarstellung/Fast-Fouriertransformation (FFT)" to "Signal_Representation/Fast_Fourier_Transform_(FFT)"

2020-09-01T13:36:36Z

Text replacement - "Signaldarstellung/Fast-Fouriertransformation (FFT)" to "Signal_Representation/Fast_Fourier_Transform_(FFT)"

Javier: Text replacement - "Signal_Representation/Fast-Fouriertransformation_(FFT)" to "Signal_Representation/Fast_Fourier_Transform_(FFT)"

2020-09-01T13:36:00Z

Text replacement - "Signal_Representation/Fast-Fouriertransformation_(FFT)" to "Signal_Representation/Fast_Fourier_Transform_(FFT)"

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

2020-08-24T08:27:28Z

Text replacement - "Category:Aufgaben zu Signaldarstellung" to "Category:Exercises for Signal Representation"

Guenter at 12:50, 13 July 2020

2020-07-13T12:50:52Z

@@ Line 94: / Line 94: @@
 :$$8 \cdot D(3)  =   (w^0 + w^4) - (w^1 + w^5)+ (w^2 + w^6) - (w^3 + w^7) =   (1 + w + w^2+ w^3) \cdot (w^0 + w^4)\hspace{0.05cm}.$$
-*Thus, because&nbsp; $w_0 = 1$&nbsp; and&nbsp; $w_4 = \text{e}^{-\text{j}\pi } = \hspace{0.08cm} - \hspace{-0.08cm}1$&nbsp;, we obtain&nbsp; $\underline {D(3) = 0}$.
+*Thus, because&nbsp; $w_0 = 1$&nbsp; and&nbsp; $w_4 = \text{e}^{-\text{j}\pi } = \hspace{0.08cm} - \hspace{-0.08cm}1$&nbsp;, we obtain&nbsp; $\underline {D(\mu=3) = 0}$.
@@ Line 101: / Line 101: @@
 :$$ 8 \cdot D(4)  =    w^0 - w^4 + w^8- w^{12}+ w^{16}- w^{20}+
 w^{24}- w^{28}=   4 \cdot (w^0 - w^4)= 8 \hspace{0.3cm}
-\Rightarrow \hspace{0.3cm}\hspace{0.15 cm}\underline{D(4) = 1}\hspace{0.05cm}.$$
+\Rightarrow \hspace{0.3cm}\hspace{0.15 cm}\underline{D(\mu=4) = 1}\hspace{0.05cm}.$$
@@ Line 108: / Line 108: @@
 *The term&nbsp; $w^0 = 1$&nbsp; does not have to be taken into account.
 *All output values with odd indices are zero by subtracting two identical input values.
-*The first statement is not true: &nbsp; Es gilt&nbsp; $X(0) = X(2) = +2$&nbsp; and&nbsp; $X(4) = X(6) = - 2$.
+*The first statement is not true: &nbsp; It holds&nbsp; $X(0) = X(2) = +2$&nbsp; and&nbsp; $X(4) = X(6) = - 2$.
@@ Line 118: / Line 118: @@
-'''(5)'''&nbsp; Because of&nbsp; $Y(5) = Y(6) =Y(7) = 0$&nbsp; , the multiplications with&nbsp; $w$,&nbsp; $w^2$&nbsp; and&nbsp; $w^3$&nbsp; do not matter in the third stage either. All spectral coefficients&nbsp; $D(\mu)$&nbsp; therefore result in zero with the exception of
+'''(5)'''&nbsp; Because of&nbsp; $Y(5) = Y(6) =Y(7) = 0$, the multiplications with&nbsp; $w$,&nbsp; $w^2$&nbsp; and&nbsp; $w^3$&nbsp; do not matter in the third stage either.&nbsp; All spectral coefficients&nbsp; $D(\mu)$&nbsp; therefore result in zero with the exception of
-:$$\hspace{0.15 cm}\underline{D(4)} =  {1}/{N}\cdot \left[Y(0) - Y(4) \right ] \hspace{0.15 cm}\underline{= 1}
+:$$\hspace{0.15 cm}\underline{D(\mu=4)} =  {1}/{N}\cdot \left[Y(0) - Y(4) \right ] \hspace{0.15 cm}\underline{= 1}
 \hspace{0.05cm}.$$
-This result agrees with the results from&nbsp; '''(1)'''&nbsp; und&nbsp; '''(2)'''&nbsp;.
+This result agrees with the results from&nbsp; '''(1)'''&nbsp; und&nbsp; '''(2)'''.
@@ Line 144: / Line 146: @@
 \hspace{0.05cm}.$$
-*In particular, this results in&nbsp; $D(3) \; \underline{=  -1}$&nbsp; und&nbsp; $D(4) \; \underline{= +1}$.
+*In particular, this results in&nbsp; $D(\mu=3) \; \underline{=  -1}$&nbsp; und&nbsp; $D(\mu=4) \; \underline{= +1}$.
 {{ML-Fuß}}
 __NOEDITSECTION__
 [[Category:Signal Representation: Exercises|^5.5 Fast Fourier Transform ^]]

@@ Line 5: / Line 5: @@
 [[File:EN_Sig_A_5_5.png|right|frame|FFT algorithm for&nbsp; $N=8$]]
-The graph shows the signal flow diagram of the FFT for&nbsp; $N = 8$. The associated spectral coefficients&nbsp; $D(0), \hspace{0.03cm}\text{...} \hspace{0.1cm} , D(7)$&nbsp; are determined from the time coefficients&nbsp; $d(0), \hspace{0.03cm}\text{...} \hspace{0.1cm}, d(7)$&nbsp;. The following applies to these with&nbsp; $0 ≤ μ ≤ 7$:
+The graph shows the signal flow diagram of the Fast Fourier Transform&nbsp; $\rm (FFT)$&nbsp; for&nbsp; $N = 8$.
 :$$D(\mu) =  \frac{1}{N}\cdot \sum_{\nu = 0 }^{N-1}
@@ Line 19: / Line 21: @@
-It holds that&nbsp; $b(κ) = d(ν)$, if&nbsp; $ν$&nbsp; is represented as a dual number and the resulting three bits are written as&nbsp; $κ$&nbsp; in reverse order. For example
+It holds that&nbsp; $b(κ) = d(ν)$, if&nbsp; $ν$&nbsp; is represented as a dual number and the resulting three bits are written as&nbsp; $κ$&nbsp; in reverse order.&nbsp; For example
 * &nbsp; $ν = 1$&nbsp; $($binary&nbsp; $001)$&nbsp; is followed by&nbsp; $κ = 4$&nbsp; $($binary&nbsp; $100)$,
 * &nbsp; $d(2)$&nbsp; remains at the same position&nbsp; $2$&nbsp; $($binary&nbsp; $010)$.
-The actual FFT algorithm happens for the example&nbsp; $N = 8$&nbsp; in&nbsp; $\log_2 N = 3$&nbsp; stages, denoted&nbsp; $L = 1$,&nbsp; $L =2$&nbsp; and&nbsp; $L = 3$&nbsp;. Further:
+The actual FFT algorithm happens for the example&nbsp; $N = 8$&nbsp; in&nbsp; $\log_2 N = 3$&nbsp; stages, denoted&nbsp; $L = 1$,&nbsp; $L =2$&nbsp; and&nbsp; $L = 3$.&nbsp; Further:
-* In each stage, four basic operations - so-called ''butterflies'' - are to be performed.
+* In each stage, four basic operations - so-called&nbsp; '''butterflies'''&nbsp; - are to be performed.
-* The values at the output of the first stage are designated in this task as&nbsp; $X(0),\hspace{0.03cm}\text{...} \hspace{0.1cm} , X(7)$&nbsp; , those of the second as&nbsp; $Y(0), \hspace{0.03cm}\text{...} \hspace{0.1cm}  , Y(7)$.
+* The values at the output of the first stage are designated in this task as&nbsp; $X(0),\hspace{0.03cm}\text{...} \hspace{0.1cm} , X(7)$, <br>those of the second as&nbsp; $Y(0), \hspace{0.03cm}\text{...} \hspace{0.1cm}  , Y(7)$.
-* After the third and last stage, all values must be divided by&nbsp; $N$&nbsp;. The final result&nbsp; $D(0), \hspace{0.03cm}\text{...} \hspace{0.1cm}  , D(7)$&nbsp; is then available here.
+* After the third and last stage, all values must be divided by&nbsp; $N$.&nbsp; <br>The final result&nbsp; $D(0), \hspace{0.03cm}\text{...} \hspace{0.1cm}  , D(7)$&nbsp; is available here.
@@ Line 45: / Line 47: @@
 <quiz display=simple>
-{Calculate the DFT coefficient&nbsp; $D(3)$.
+{Calculate the DFT coefficient&nbsp; $D(\mu=3)$.
 |type="{}"}
-$D(3) \ = \ $ { 0. }
+$D(\mu=3) \ = \ $ { 0. }
-{Calculate the DFT coefficient&nbsp; $D(4)$.
+{Calculate the DFT coefficient&nbsp; $D(\mu=4)$.
 |type="{}"}
-$D(4) \ = \ $ { 1 3% }
+$D(\mu=4) \ = \ $ { 1 3% }
-{Determine the initial values&nbsp; $X(0)$, ... , $X(7)$&nbsp; of the first stage. Which of the following statements are true?
+{Determine the initial values&nbsp; $X(0)$, ... , $X(7)$&nbsp; of the first stage.&nbsp; Which of the following statements are true?
 |type="[]"}
-- All&nbsp; $X$&ndash;values with even indices are equal to&nbsp; $2$.
+- All&nbsp; $X$ values with even indices are equal to&nbsp; $2$.
-+ All&nbsp; $X$&ndash;values with odd indices are equal to&nbsp; $0$.
++ All&nbsp; $X$ values with odd indices are equal to&nbsp; $0$.
-{Determine the initial values&nbsp; $Y(0)$, ... , $Y(7)$&nbsp; of the second stage. Enter the values&nbsp; $Y(0)$&nbsp; and&nbsp; $Y(4)$&nbsp; as a check.
+{Determine the initial values&nbsp; $Y(0)$, ... , $Y(7)$&nbsp; of the second stage.&nbsp; Enter the values&nbsp; $Y(0)$&nbsp; and&nbsp; $Y(4)$&nbsp; as a check.
 |type="{}"}
@@ Line 66: / Line 68: @@
 {Calculate all&nbsp; $N$&nbsp; spectral values&nbsp; $D(\mu)$, in particular
 |type="{}"}
 $D(\mu = 4) \ = \ $ { 1 3% }
-{What would be the spectral coefficients for&nbsp; $d(ν = 4) = 1$&nbsp; and&nbsp; $d(ν \neq 4) = 0$&nbsp;? <br>Enter the values&nbsp; $D(3)$&nbsp; and&nbsp; $D(4)$&nbsp; as a check.
+{What would be the spectral coefficients for&nbsp; $d(ν = 4) = 1$&nbsp; and&nbsp; $d(ν \neq 4) = 0$&nbsp;? <br>Enter the values&nbsp; $D(\mu =3)$&nbsp; and&nbsp; $D(\mu =4)$&nbsp; as a check.
 |type="{}"}
 $D(\mu = 3) \ = \ $ { -1.03--0.97 }

← Older revision		Revision as of 12:38, 23 March 2021
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−	[[Category:~~Exercises for~~ Signal Representation\|^5.5 Fast Fourier Transform ^]]	+	[[Category:Signal Representation: Exercises\|^5.5 Fast Fourier Transform ^]]

← Older revision		Revision as of 21:00, 29 November 2020
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−	[[Category:Exercises for Signal Representation\|^5. ~~Zeit- und frequenzdiskrete Signaldarstellung~~^]]	+	[[Category:Exercises for Signal Representation\|^5.5 Fast Fourier Transform ^]]

← Older revision		Revision as of 13:36, 1 September 2020
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−	{{quiz-Header\|Buchseite=~~Signaldarstellung~~/~~Fast-Fouriertransformation~~ (FFT)	+	{{quiz-Header\|Buchseite=Signal_Representation/Fast_Fourier_Transform_(FFT)
	}}		}}

← Older revision		Revision as of 13:36, 1 September 2020
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	''Hinweis:''		''Hinweis:''
−	*Die Aufgabe gehört zum Kapitel  [[Signal_Representation/~~Fast-Fouriertransformation_~~(FFT)\|Fast-Fouriertransformation (FFT)]].	+	*Die Aufgabe gehört zum Kapitel  [[Signal_Representation/Fast_Fourier_Transform_(FFT)\|Fast-Fouriertransformation (FFT)]].

← Older revision		Revision as of 08:27, 24 August 2020
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−	[[Category:~~Aufgaben zu Signaldarstellung~~\|^5. Zeit- und frequenzdiskrete Signaldarstellung^]]	+	[[Category:Exercises for Signal Representation\|^5. Zeit- und frequenzdiskrete Signaldarstellung^]]

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID1177__Sig_A_5_5_neu.png|right|frame|FFT-Algorithmus für&nbsp; $N=8$]]
+[[File:EN_Sig_A_5_5.png|right|frame|FFT-Algorithmus für&nbsp; $N=8$]]
 Die Grafik zeigt den Signalflussplan der FFT für&nbsp; $N = 8$. Aus den Zeitkoeffizienten&nbsp; $d(0), \hspace{0.03cm}\text{...} \hspace{0.1cm}, d(7)$&nbsp; werden die dazugehörigen Spektralkoeffizienten&nbsp; $D(0), \hspace{0.03cm}\text{...} \hspace{0.1cm} , D(7)$&nbsp; ermittelt. Für diese gilt mit&nbsp; $0 ≤ μ ≤ 7$: