Exercise 5.4: Comparison of Rectangular and Hanning Window

Examples for spectral analysis

Let the time course of a periodic signal be given in principle:

$$x(t) = A_1 \cdot \cos (2 \pi \cdot f_1 \cdot t) + A_2 \cdot \cos (2 \pi \cdot f_2 \cdot t) \hspace{0.05cm}.$$

Unknown and thus to be estimated are its parameters $A_1$, $f_1$, $A_2$ and $f_2$.

After weighting the signal with the window function $w(t)$ , the product $y(t) = x(t) \cdot w(t)$ is subjected to a Discrete Fourier Transform (DFT) with the parameters $N = 512$ and $T_{\rm P}$. The time $T_{\rm P}$ of the signal section to be analysed can be set by the user as desired.

Two functions are available for windowing, each of which is zero for $|t| > T_{\rm P}/2$:

The rectangular window:

$${w} (\nu) = \left\{ \begin{array}{c} 1 \\ 0 \\ \end{array} \right.\quad \begin{array}{*{10}c} {\rm{for}} \\ \\ \end{array}\begin{array}{*{20}c} -N/2 \le \nu < N/2 \hspace{0.05cm}, \\ {\rm else} \hspace{0.05cm}, \\ \end{array}$$

$$W(f) ={1}/{f_{\rm A}}\cdot {\rm si}(\pi \cdot {f}/{f_{\rm A}})\hspace{0.05cm},$$

the Hanning window:

$${w} (\nu) = \left\{ \begin{array}{c} 0.5 + 0.5 \cdot \cos (2 \pi \cdot {\nu}/{N}) \\ 0 \\ \end{array} \right.\quad \begin{array}{*{10}c} {\rm{for}} \\ \\ \end{array}\begin{array}{*{20}c} -N/2 \le \nu < N/2 \hspace{0.05cm}, \\ {\rm else} \hspace{0.05cm}, \\ \end{array}$$

$$W(f) ={0.5}/{f_{\rm A}}\cdot {\rm si}(\pi \cdot \frac{f}{f_{\rm A}})+ {0.25}/{f_{\rm A}}\cdot {\rm si}(\pi \cdot \frac{f-f_{\rm A}}{f_{\rm A}})+ {0.5}/{f_{\rm A}}\cdot {\rm si}(\pi \cdot \frac{f+f_{\rm A}}{f_{\rm A}})\hspace{0.05cm}.$$

Here, $W(f)$ is the Fourier transform of the continuous-time window function $w(t)$, while $w(ν)$ indicates the discrete-time weighting function.

In the task, reference is made to various spectral functions $Y(f)$ for example to

$$Y_{\rm A}(f) = 1\, {\rm V}\cdot {\rm \delta} (f \pm 1\,\,{\rm kHz})+ 0.5\,\, {\rm V}\cdot {\rm \delta} (f \pm 1.125\,\,{\rm kHz}) \hspace{0.05cm}.$$

In the graph, two further spectral functions $Y_{\rm B}(f)$ and $Y_{\rm C}(f)$ are shown, which result when a $1 \ \text{kHz}$ signal is analysed by DFT and the DFT parameter $T_{\rm P} = 8.5 \ \text{ms}$ is chosen unfavourably.

For one of the images the rectangular window is used, for the other the Hanning window.
It is not indicated which graph belongs to which window.

Hints:

This task belongs to the chapter Spectrum Analysis.
Note that the frequency resolution $f_{\rm A}$ is equal to the reciprocal of the adjustable parameter $T_{\rm P}$.
Unfortunately, the indices of $f_{\rm A}$ and $Y_{\rm A}(f)$ collide. It is obvious that they are not related. Just to be on the safe side, we point this out.

Questions

	The rectangular window was used for weighting.
	The Hanning window was used for weighting.
	The DFT parameter $T_{\rm P} = 4\ \text{ms}$ was used.
	The DFT spectrum $Y_{\rm A}(f)$ is identical to the actual spectrum $X(f)$.

$G(f_1 = 1.000 \ \text{kHz})\ = \ $

$\text{V}$

$G(f_2 = 1.125 \ \text{kHz})\ = \ $

$\text{V}$

	$Y_{\rm B}(f)$ results with rectangular windowing.
	$Y_{\rm B}(f)$ results with the Hanning window.

Solution

(1) Solutions 1 and 4 are correct:

Using the Hanning window, three Dirac functions should be recognisable even if $x(t)$ contains only one frequency ⇒ the rectangular window was used.
With $T_{\rm P} = 4 \ \text{ms}$ , the frequency resolution is $f_{\rm A}= 1/T_{\rm P} = 0.25 \ \text{kHz}$. Thus the frequency $f_2$ does not lie in the given grid and $Y(f)$ would be composed of very many Dirac lines. This means: the third statement is wrong.

Output signal $y(t)$ with the rectangular window

As can be seen from the graph, $x(t)$ has the period duration $T_{\rm 0} = 8 \ \text{ms}$.
If one chooses the DFT parameter equal to $T_{\rm P} = 4 \ \text{ms}$ (or an integer multiple thereof), the periodic continuation ${\rm P}\{ x(t)\} $ in the interval $|t| \leq T_{\rm P}/2$ coincides with $x(t)$ , so that the weighting function $w(t)$ has no disturbing effect. So:
The DFT spectrum $Y(f)$ thus agrees with the actual spectrum.

(2) Because of $T_{\rm 0} = 8 \ \text{ms}$ , the Hanning spectrum $W(f)$

consists of three Dirac functions at positive frequencies
and three axisymmetrical Diracs at negative frequencies

are composed. For the positive frequencies, the spectral function is:

Output signal $y(t)$ with the Hanning window

$$W(f) =0.5\cdot {\rm \delta}(f) + 0.25\cdot {\rm \delta}(f-f_{\rm A})+ 0.25\cdot {\rm \delta}(f+f_{\rm A})\hspace{0.05cm}.$$

The output spectrum results from the convolution between $X(f)$ and $W(f)$. At positive frequencies, there are now four Diracs with the following weights:

$$\begin{align*} G(f = 0.875\,{\rm kHz}) & = 1\, {\rm V}\cdot 0.25 = 0.250\, {\rm V}, \\ G(f = f_1 = 1.000\,{\rm kHz}) & = 1\, {\rm V}\cdot 0.5 + 0.5\, {\rm V}\cdot 0.25 \hspace{0.15 cm}\underline{ = 0.625\, {\rm V}}, \\ G(f = f_2 = 1.125\,{\rm kHz}) & = 1\, {\rm V}\cdot 0.25 + 0.5\, {\rm V}\cdot 0.5 \hspace{0.15 cm}\underline{= 0.500\, {\rm V}}, \\ G(f = 1.250\,{\rm kHz}) & = 0.5\, {\rm V}\cdot 0.25 = 0.125\, {\rm V} \hspace{0.05cm}.\end{align*}$$

The graph shows the attenuation of the edges by the weighting function $w(t)$ of the Hanning window.

(3) Solution 2 is correct:

The rectangular window delivers a very strongly distorted result if the window width $T_{\rm P}$ (as here) is not adapted to the cosine frequency.
In this case, the Hanning window is more suitable. Then the measured spectrum $Y_{\rm B}(f)$ results.
From the spectrum $Y_{\rm C}(f)$ the searched $1\ \rm kHz$ line is more difficult to detect. The spectrum $Y_{\rm C}(f)$ results after rectangular windowing.