Difference between revisions of "Aufgaben:Exercise 5.3: Mean Square Error"
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− | [[File:P_ID1145__Sig_A_5_3.png|250px|right|frame| | + | [[File:P_ID1145__Sig_A_5_3.png|250px|right|frame|Gaussian pulse, square pulse, <br>sinc pulse and some parameters]] |
− | + | We consider three pulses, namely | |
− | * | + | *a [[Signal_Representation/Special_Cases_of_Pulses#Gaussian_pulse|Gaussian pulse]] with amplitude A and equivalent duration T: |
:x1(t)=A⋅e−π(t/T)2, | :x1(t)=A⋅e−π(t/T)2, | ||
− | * | + | *a [[Signal_Representation/Special_Cases_of_Pulses#Rectangular_pulse|rectangular pulse]] x2(t) with amplitude A and (equivalent) duration T: |
:$$x_2(t) = \left\{ \begin{array}{c} A \\ | :$$x_2(t) = \left\{ \begin{array}{c} A \\ | ||
Line 20: | Line 20: | ||
\end{array}$$ | \end{array}$$ | ||
− | * | + | *a so called "sinc pulse" according to the following definition: |
− | :$$x_3(t) = A \cdot {\rm | + | :$$x_3(t) = A \cdot {\rm sinc}(t/ T) ,\hspace{0.15cm}{\rm sinc}(x) = |
− | \sin(x)/x\hspace{0.05cm}.$$ | + | \sin(\pi x)/(\pi x)\hspace{0.05cm}.$$ |
− | + | Let the signal parameters be A=1 V and T=1 ms in each case. | |
− | + | The conventional [[Signal_Representation/Fourier_Transform_and_Its_Inverse|Fourier transform]] leads to the following spectral functions: | |
− | * X1(f) | + | * X1(f) is also Gaussian, |
− | * X2(f) | + | * X2(f) runs according to the $\rm sinc$ function, |
− | * X3(f) | + | * X3(f) is constant for |f|<1/(2T) and outside zero. |
− | + | For all spectral functions, X(f=0)=A⋅T. | |
− | + | If the discrete-frequency spectrum is determined by the [[Signal_Representation/Discrete_Fourier_Transform_(DFT)|Discrete Fourier Transform]] $\rm (DFT)$ with the DFT parameters | |
− | * N=512 ⇒ | + | * N=512 ⇒ number of samples considered in the time and frequency domain, |
− | *fA ⇒ | + | *fA ⇒ interpolation distance in the frequency domain, |
− | + | this will lead to distortions due to truncation and/or aliasing errors. | |
− | + | ||
− | + | The other DFT parameters are clearly fixed withn N and fA. The following applies to these: | |
:$$f_{\rm P} = N \cdot f_{\rm A},\hspace{0.3cm}T_{\rm P} = 1/f_{\rm A},\hspace{0.3cm}T_{\rm A} = T_{\rm | :$$f_{\rm P} = N \cdot f_{\rm A},\hspace{0.3cm}T_{\rm P} = 1/f_{\rm A},\hspace{0.3cm}T_{\rm A} = T_{\rm | ||
P}/N | P}/N | ||
\hspace{0.05cm}.$$ | \hspace{0.05cm}.$$ | ||
− | + | The accuracy of the respective DFT approximation is captured by the "mean square error" $\rm (MSE). <br>Here, we use the designation \rm MQF$ ⇒ (German: "Mittlerer Quadratischer Fehler"): | |
− | |||
:$${\rm MQF} = \frac{1}{N}\cdot \sum_{\mu = 0 }^{N-1} | :$${\rm MQF} = \frac{1}{N}\cdot \sum_{\mu = 0 }^{N-1} | ||
\left|X(\mu \cdot f_{\rm A})-\frac{D(\mu)}{f_{\rm A}}\right|^2 \hspace{0.05cm}.$$ | \left|X(\mu \cdot f_{\rm A})-\frac{D(\mu)}{f_{\rm A}}\right|^2 \hspace{0.05cm}.$$ | ||
− | + | The resulting MQF values are given in the graph above, valid for N=512 as well as for | |
*fA⋅T=1/4, | *fA⋅T=1/4, | ||
*fA⋅T=1/8, | *fA⋅T=1/8, | ||
Line 65: | Line 64: | ||
− | '' | + | ''Hints:'' |
− | * | + | *This task belongs to the chapter [[Signal_Representation/Possible_Errors_When_Using_DFT|Possible errors when using DFT]]. |
− | * | + | *The theory for this chapter is summarised in the (German language) learning video <br> [[Fehlermöglichkeiten_bei_Anwendung_der_DFT_(Lernvideo)|Fehlermöglichkeiten bei Anwendung der DFT]] ⇒ "Possible errors when using DFT". |
− | === | + | ===Questions=== |
<quiz display=simple> | <quiz display=simple> | ||
− | { | + | {Which range |f|≤fmax is covered with N=512 and fA⋅T=1/8 ? |
|type="{}"} | |type="{}"} | ||
fmax⋅T = { 32 3% } | fmax⋅T = { 32 3% } | ||
− | { | + | {At what time interval TA are the sampled values of x(t) available? |
|type="{}"} | |type="{}"} | ||
TA/T = { 0.01562 3% } | TA/T = { 0.01562 3% } | ||
− | { | + | {Due to which effect does the MQF value for the Gaussian pulse increase when using fA⋅T=1/4 instead of fA⋅T=1/8? |
|type="()"} | |type="()"} | ||
− | + | + | + The truncation error is significantly increased. |
− | - | + | - The aliasing error is significantly increased. |
− | { | + | {Due to what effect does the MQF value for the Gaussian pulse increase when using fA⋅T=1/16 instead of fA⋅T=1/4? |
|type="()"} | |type="()"} | ||
− | - | + | - The truncation error is significantly increased. |
− | + | + | + The aliasing error is significantly increased. |
− | { | + | {Compare the MQF values of the rectangular pulse x2(t) with those of the Gaussian pulse x1(t). Which of the following statements are true? |
|type="[]"} | |type="[]"} | ||
− | + MQF | + | + MQF becomes larger because the spectral function X2(f) decays asymptotically slower than X1(f). |
− | + | + | + The aliasing error dominates. |
− | - | + | - The truncation error dominates. |
− | { | + | {Compare the MQF values of the "sinc pulse" x3(t) with those of the Gaussian pulse x1(t). Which of the following statements are true? |
|type="[]"} | |type="[]"} | ||
− | - MQF | + | - MQF becomes larger because the spectral function X3(f) decays asymptotically slower than X1(f). |
− | - | + | - The aliasing error dominates. |
− | + | + | + The truncation error dominates. |
</quiz> | </quiz> | ||
− | === | + | ===Solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
− | '''(1)''' | + | '''(1)''' With the DFT parameters N=512 and fA⋅T=1/8 the following follows after multiplying the two quantities: |
:fP⋅T=N⋅(fA⋅T)=64. | :fP⋅T=N⋅(fA⋅T)=64. | ||
− | * | + | *This covers the frequency range $-f_{\rm P}/2 \leq f < +f_{\rm P}/2$: |
:fmax⋅T=32_. | :fmax⋅T=32_. | ||
− | '''(2)''' | + | '''(2)''' The periodisation of the time function is based on the parameter TP=1/fA=8T. |
− | * | + | *The distance between two samples is therefore |
:TA/T=TP/TN=8512=0.015625_. | :TA/T=TP/TN=8512=0.015625_. | ||
− | '''(3)''' | + | '''(3)''' Correct is the <u>proposed solution 1 ⇒ increase of the truncation error</u>: |
− | * | + | *This measure simultaneously halves TP from 8T to 4T . |
− | * | + | *Thus, only samples in the range –2T \leq t < 2T are taken into account, which increases the truncation error. |
− | * | + | *The mean square error (\rm MQF) increases from 0.15 \cdot 10^{-15} to $8 \cdot 10^{-15} for the Gaussian pulse x_1(t)$, |
+ | *although the aliasing error actually decreases slightly by this measure. | ||
− | '''(4)''' | + | '''(4)''' Correct is the <u>proposed solution 2 ⇒ increase of the aliasing error:</u>: |
− | * | + | *By halving f_{\rm A} ⇒ f_{\rm P} is also halved. |
− | * | + | *As a result, the aliasing error becomes somewhat larger with a smaller truncation error at the same time. |
− | * | + | *Overall, for the Gaussian pulse x_1(t), the mean square error (\rm MQF) increases from 1.5 \cdot 10^{-16} to 3.3 \cdot 10^{-16}. |
− | '''(5)''' | + | '''(5)''' <u>Proposed solutions 1 and 2</u> are correct: |
− | * | + | *As can be seen from the graph, the last statement is not true in contrast to the first two. |
− | * | + | *Due to the slow ($\rm sinc$–shaped) decay of the spectral function, the aliasing error dominates. |
− | * | + | *The \rm MQF value at f_{\rm A} \cdot T = 1/8 with 1.4 \cdot 10^{-5} is therefore significantly larger than for the Gaussian pulse (1.5 \cdot 10^{-16}). |
− | '''(6)''' | + | '''(6)''' <u>Proposed solution 3</u> is correct: |
− | * | + | *The spectral function X_3(f) here has a rectangular lead, so that the first two statements do not apply. |
− | * | + | *On the other hand, a truncation error is unavoidable with this $\rm sinc$–shaped time function. This leads to the large \rm MQF values given. |
{{ML-Fuß}} | {{ML-Fuß}} | ||
__NOEDITSECTION__ | __NOEDITSECTION__ | ||
− | [[Category: | + | [[Category:Signal Representation: Exercises|^5.3 Possible DFT Errors^]] |
Latest revision as of 17:14, 17 May 2021
We consider three pulses, namely
- a Gaussian pulse with amplitude A and equivalent duration T:
- x_1(t) = A \cdot {\rm e}^{- \pi (t/T)^2} \hspace{0.05cm},
- a rectangular pulse x_2(t) with amplitude A and (equivalent) duration T:
- x_2(t) = \left\{ \begin{array}{c} A \\ 0 \\ \end{array} \right.\quad \begin{array}{*{10}c} {\rm{f\ddot{u}r}} \\ {\rm{f\ddot{u}r}} \\ \end{array}\begin{array}{*{20}c} |t| < T/2 \hspace{0.05cm}, \\ |t| > T/2 \hspace{0.05cm}, \\ \end{array}
- a so called "sinc pulse" according to the following definition:
- x_3(t) = A \cdot {\rm sinc}(t/ T) ,\hspace{0.15cm}{\rm sinc}(x) = \sin(\pi x)/(\pi x)\hspace{0.05cm}.
Let the signal parameters be A = 1\ {\rm V} and T = 1\ {\rm ms} in each case.
The conventional Fourier transform leads to the following spectral functions:
- X_1(f) is also Gaussian,
- X_2(f) runs according to the \rm sinc function,
- X_3(f) is constant for |f| < 1/(2 T) and outside zero.
For all spectral functions, X(f = 0) = A \cdot T.
If the discrete-frequency spectrum is determined by the Discrete Fourier Transform \rm (DFT) with the DFT parameters
- N = 512 ⇒ number of samples considered in the time and frequency domain,
- f_{\rm A} ⇒ interpolation distance in the frequency domain,
this will lead to distortions due to truncation and/or aliasing errors.
The other DFT parameters are clearly fixed withn N and f_{\rm A}. The following applies to these:
- f_{\rm P} = N \cdot f_{\rm A},\hspace{0.3cm}T_{\rm P} = 1/f_{\rm A},\hspace{0.3cm}T_{\rm A} = T_{\rm P}/N \hspace{0.05cm}.
The accuracy of the respective DFT approximation is captured by the "mean square error" \rm (MSE).
Here, we use the designation \rm MQF ⇒ (German: "Mittlerer Quadratischer Fehler"):
- {\rm MQF} = \frac{1}{N}\cdot \sum_{\mu = 0 }^{N-1} \left|X(\mu \cdot f_{\rm A})-\frac{D(\mu)}{f_{\rm A}}\right|^2 \hspace{0.05cm}.
The resulting MQF values are given in the graph above, valid for N = 512 as well as for
- f_{\rm A} \cdot T = 1/4,
- f_{\rm A} \cdot T = 1/8,
- f_{\rm A} \cdot T = 1/16.
Hints:
- This task belongs to the chapter Possible errors when using DFT.
- The theory for this chapter is summarised in the (German language) learning video
Fehlermöglichkeiten bei Anwendung der DFT ⇒ "Possible errors when using DFT".
Questions
Solution
- f_{\rm P} \cdot T = N \cdot (f_{\rm A} \cdot T) = 64.
- This covers the frequency range -f_{\rm P}/2 \leq f < +f_{\rm P}/2:
- f_{\rm max }\cdot T \hspace{0.15 cm}\underline{= 32}\hspace{0.05cm}.
(2) The periodisation of the time function is based on the parameter T_{\rm P} = 1/f_{\rm A} = 8T.
- The distance between two samples is therefore
- T_{\rm A}/T = \frac{T_{\rm P}/T}{N} = \frac{8}{512}\hspace{0.15 cm}\underline{ = 0.015625}\hspace{0.05cm}.
(3) Correct is the proposed solution 1 ⇒ increase of the truncation error:
- This measure simultaneously halves T_{\rm P} from 8T to 4T .
- Thus, only samples in the range –2T \leq t < 2T are taken into account, which increases the truncation error.
- The mean square error (\rm MQF) increases from 0.15 \cdot 10^{-15} to 8 \cdot 10^{-15} for the Gaussian pulse x_1(t),
- although the aliasing error actually decreases slightly by this measure.
(4) Correct is the proposed solution 2 ⇒ increase of the aliasing error::
- By halving f_{\rm A} ⇒ f_{\rm P} is also halved.
- As a result, the aliasing error becomes somewhat larger with a smaller truncation error at the same time.
- Overall, for the Gaussian pulse x_1(t), the mean square error (\rm MQF) increases from 1.5 \cdot 10^{-16} to 3.3 \cdot 10^{-16}.
(5) Proposed solutions 1 and 2 are correct:
- As can be seen from the graph, the last statement is not true in contrast to the first two.
- Due to the slow (\rm sinc–shaped) decay of the spectral function, the aliasing error dominates.
- The \rm MQF value at f_{\rm A} \cdot T = 1/8 with 1.4 \cdot 10^{-5} is therefore significantly larger than for the Gaussian pulse (1.5 \cdot 10^{-16}).
(6) Proposed solution 3 is correct:
- The spectral function X_3(f) here has a rectangular lead, so that the first two statements do not apply.
- On the other hand, a truncation error is unavoidable with this \rm sinc–shaped time function. This leads to the large \rm MQF values given.