Exercise 5.3: Mean Square Error

Gaussian pulse, square pulse,
sinc pulse and some parameters

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

(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 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.

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

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