Aufgaben:Exercise 5.3: Mean Square Error

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Gaussian pulse, square pulse, sinc pulse and some parameters

We consider three pulse-like signals, 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:

$$x_3(t) = A \cdot {\rm si}(\pi \cdot t/ T) ,\hspace{0.15cm}{\rm si}(x) = \sin(x)/x\hspace{0.05cm}.$$

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  $\rm si$–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(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&bsp; $N$  uan  $f_{\rm A}$  .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 thenbsp; mean square error  (MSE, here MQF):

$${\rm MQF} = \frac{1}{N}\cdot \sum_{\mu = 0 }^{N-1} \left|X(\mu \cdot f_{\rm A})-\frac{D(\mu)}{f_{\rm A}}\right|^2 \hspace{0.05cm}.$$

The resulting MSE 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 learning video  Possible Errors when Using DFT .

Questions

1

Which range  $|f| \leq f_{\text{max}}$  is covered with  $N = 512$  and  $f_{\rm A} \cdot T = 1/8$ ?

$f_{\text{max}} \cdot T\ = \ $

2

At what time interval  $T_{\rm A}$  are the sampled values of  $x(t)$  available?

$T_{\rm A}/T\ = \ $

3

Due to which effects does the MSE value for the Gaussian pulse increase when using   $f_{\rm A} \cdot T = 1/4$  instead of  $f_{\rm A} \cdot T = 1/8$  verwendet?

	The truncation error is significantly increased.
	The aliasing error is significantly increased.

4

Due to what effects does the MSE value for the Gaussian momentum increase when using  $f_{\rm A} \cdot T = 1/16$  instead of $f_{\rm A} \cdot T = 1/4$  verwendet?

	The termination error is significantly increased.termination
	The aliasing error is significantly increased.

5

Compare the MQF(MSE) values of the rectangular pulse  $x_2(t)$  with those of the Gaussian pulse  $x_1(t)$. Which of the following statements are true?

	$\rm MQF$  becomes larger because the spectral function  $X_2(f)$  decays asymptotically slower than  $X_1(f)$.
	The aliasing error dominates.
	The termination error dominates.

6

Compare the MQF values of the slit pulse  $x_3(t)$  with those of the Gaussian pulse  $x_1(t)$. Which of the following statements are true?

	$\rm MQF$  becomes larger because the spectral function  $X_3(f)$  decays asymptotically slower than  $X_1(f)$.
	The aliasing error dominates.
	The termination error dominates.

Solution

Solution

(1)  With the DFT parameters  $N = 512$  and  $f_{\rm A} \cdot T = 1/8$  the following follows after multiplying the two quantities:

$$f_{\rm P} \cdot T = N \cdot (f_{\rm A} \cdot T) = 64.$$

• 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 termination 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 termination error.
• The mean square error  $(\rm MQF)$  steigt dadurch beim Gaußimpuls  $x_1(t)$  von  $0.15 \cdot 10^{-15}$  auf  $8 \cdot 10^{-15}$, obwohl der Aliasingfehler durch diese Maßnahme sogar etwas kleiner wird.

(4)  Richtig ist der Lösungsvorschlag 2   ⇒   Erhöhung des Aliasingfehlers:

• Durch die Halbierung von  $f_{\rm A}$  wird auch  $f_{\rm P}$  halbiert.
• Dadurch wird der Aliasingfehler etwas größer bei gleichzeitig kleinerem Abbruchfehler.
• Insgesamt steigt beim Gaußimpuls  $x_1(t)$  der mittlere quadratische Fehler  $(\rm MQF)$  von  $1.5 \cdot 10^{-16}$  auf  $3.3 \cdot 10^{-16}$.

(5)  Richtig sind die Lösungsvorschläge 1 und 2:

• Wie aus der Grafik zu ersehen ist, trifft die letzte Aussage nicht zu im Gegensatz zu den ersten beiden.
• Aufgrund des langsamen,  $\rm si$–förmigen Abfalls der Spektralfunktion dominiert der Aliasingfehler.
• Der  $\rm MQF$–Wert ist bei  $f_{\rm A} \cdot T = 1/8$  mit  $1.4 \cdot 10^{-5}$  deshalb deutlich größer als beim Gaußimpuls  $(1.5 \cdot 10^{-16})$.

(6)  Richtig ist der Lösungsvorschlag 3:

• Die Spektralfunktion  $X_3(f)$  hat hier einen rechteckförmigen Vorlauf, so dass die beiden ersten Aussagen nicht zutreffen.
• Dagegen ist bei dieser  $\rm si$–förmigen Zeitfunktion ein Abbruchfehler unvermeidbar. Dieser führt zu den angegebenen großen  $\rm MQF$–Werten.