Aufgaben:Exercise 5.4Z: On the Hanning Window

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Characterisation of the Hanning window

In this task, important properties of the frequently used Hanning window are to be derived. The continuous-time representation in the interval from  $-T_{\rm P}/2$  to  $+T_{\rm P}/2$  is here as follows:

$$w(t)= {\rm cos}^2(\pi \cdot

{t}/{T_{\rm P}})= 0.5\cdot \big(1 + {\rm cos}(2\pi \cdot {t}/{T_{\rm P}}) \big ) \hspace{0.05cm}.$$ Outside the symmetric time domain of duration  $T_{\rm P}$  ist $w (t) \equiv 0$.

The upper graph shows the discrete-time representation  $w(\nu) = w({\nu} \cdot T_{\rm A})$, where  $T_{\rm A}$  is smaller than   $T_{\rm P}$ by a factor  $N = 32$ . The definition range of the discrete time variable  $ν$  extends from  $-16$  to  $+15$.

In the lower graph, the Fourier transform  $W(f)$  of the continuous-time window function  $w(t)$  is shown logarithmically. The abscissa is normalised to  $f_{\rm A} = 1/T_{\rm P}$ . One can see:

• The equidistant values  $W({\mu} \cdot f_{\rm A})$  are zero except for  $μ = 0$  and  $μ = ±1$.
• The main lobe thus extends to the frequency range  $|f| ≤ 2 · f_{\rm A}$.
• $W(f)$  is largest in magnitude outside the main lobe for  $f = ±2.5 · f_{\rm A}$ ..
• Thus, the minimum distance between the main and side lobes applies here:

$$A_{\rm H/S} = 20 \cdot {\rm lg}\hspace{0.15cm}

\frac{|W(0)|}{|W(2.5 \cdot f_{\rm A})|} \hspace{0.15cm}{\rm (in}\hspace{0.1cm}{\rm dB)}\hspace{0.05cm}.$$ 

Hints:

• This task belongs to the chapter  Spectral Analysis.
• Note that the frequency resolution  $f_{\rm A}$  is equal to the reciprocal of the adjustable parameter  $T_{\rm P}$ .

Questions

1 Give the discrete-time coefficients  $w(ν)$  of the Hanning window analytically.
What are the numerical values for  $ν = 0$,  $ν = 1$  and  $ν = -\hspace{0.05cm}8$?

$w(ν = 0) \hspace{0.37cm} = \ $

$w(ν = 1) \hspace{0.37cm} = \ $

$w(ν = -8) \hspace{0.03cm} = \ $

2 Calculate the spectral function  $W(f)$  in general. Which of the following statements are correct?

	$W(f)$  yields complex results for special frequency values.
	$W(f)$  is even with respect to  $f$ , i.e.  $W(-f) = W(+f)$.
	The spectral value  $W(f = 0)$  is equal to  $0.5/f_{\rm A}$  and thus real.

3 What are  $W(f = ±f_{\rm A})$  and the $\text{6 dB}$bandwidth normalised to   $f_{\rm A}$ ?

$W(±f_{\rm A}) \hspace{0.15cm} = \ $ $\ \cdot \ 1/f_{\rm A}$

$B_{\rm 6\hspace{0.05cm}dB}\hspace{-0.05cm}/\hspace{-0.05cm}f_{\rm A} \hspace{0.2cm} = \ $

4 What is the minimum distance between the main lobe and the side lobe.

$A_{\rm H/S} \ = \ $ $\ \rm dB$

Solution

Solution

(1)  After trigonometric transformation, the result for the continuous-time window function is:

$$w(t) = {\rm cos}^2(\pi \cdot

{t}/{T_{\rm P}}) = 0.5+ 0.5\cdot {\rm cos}(2\pi \cdot {t}/{T_{\rm P}})\hspace{0.05cm}.$$

• After time discretisation with  $ν = t/T_{\rm A}$  and  $T_{\rm P}/T_{\rm A} = N = 32$  , one obtains for the discrete-time window:

$$w(\nu) = w(\nu \cdot T_{\rm A}) = 0.5+ 0.5\cdot {\rm

cos}(2\pi \cdot {\nu}/{N})\hspace{0.8cm} \Rightarrow \hspace{0.3cm}w(\nu = 0) \hspace{0.15 cm}\underline{= 1},$$

$$w(\nu = 1) = 0.5+ 0.5\cdot {\rm cos}(

\frac{\pi}{16})\hspace{0.15 cm}\underline{ = 0.99}, $$

$$w(\nu = -8)=0.5+

0.5\cdot {\rm cos}( \frac{-\pi}{2}) \hspace{0.15 cm}\underline{= 0.5}\hspace{0.05cm}.$$

(2)  The correct solutions are 2 and 3::

• The periodic continuation of  $w(t)$  corresponding to the period  $T_{\rm P}$  yields a (periodic) signal with a DC and a cosine component.
• From this follows with   $f_{\rm A} = 1/T_{\rm P}$:

$${\rm P}\{w(t)\} = 0.5+0.5\cdot {\rm

cos}(2\pi \cdot f_{\rm A} \cdot t) \hspace{0.2cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \hspace{0.2cm}0.5\cdot {\rm \delta}(f) + 0.25\cdot {\rm \delta}(f \pm f_{\rm A}))\hspace{0.05cm}.$$

• The time-limited signal  $w(t)$  is obtained from  ${\rm P}\{w(t)\}$  by multiplication with a rectangle of amplitude  $1$  and duration  $T_{\rm P}$.
• Its spectrum  $W(f)$  is thus obtained from the convolution of the above spectral function with the function  $T_{\rm P} · {\rm si}(π \cdot f \cdot T_{\rm P}) = 1/f_{\rm A} · {\rm si}(π \cdot f/f_{\rm A})$:

$$w(t)

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

• This spectral function is even and also real for all frequencies  $f$ . The spectral value at frequency  $f = 0$  gives the window area:

$$W(f=0) =

\frac{0.5}{f_{\rm A}}= \int_{-\infty}^{+\infty}w(t)\hspace{0.05cm}{\rm d}t\hspace{0.05cm}.$$

(3)  From the result of sub-task  (2)  it also follows:

$$W(f = ±f_{\rm A}) = W(0)/2\hspace{0.15cm}\underline{ = 0.25} \cdot 1/{f_{\rm A}}.$$

• Due to the monotonic course in the range  $|f| < f_{\rm A}$ , the magnitude function  $|W(f)|$  has dropped to half of the maximum for the first time exactly at  $± f_{\rm A}$ .
• Thus,  $B_{\rm 6\hspace{0.05cm}dB}\hspace{-0.05cm}/\hspace{-0.05cm}f_{\rm A} \;\underline{=2}$.

(4)  The largest spectral amount outside the main lobe occurs at  $f = ±2.5 f_{\rm A}$ . With the result of subtask  (2)  holds:

$$W(f = 2.5 \cdot f_{\rm A}) = \frac{0.5}{f_{\rm A}}\cdot {\rm si}(2.5 \pi )

+\frac{0.25}{f_{\rm A}}\cdot {\rm si}(1.5 \pi )+\frac{0.25}{f_{\rm A}}\cdot {\rm si}(3.5 \pi )=  \frac{0.25}{\pi \cdot f_{\rm A}}\left[ \frac{2}{2.5}-\frac{1}{1.5}-\frac{1}{3.5}\right] \approx -\frac{0.0121}{ f_{\rm A}}\hspace{0.05cm}.$$

• This gives the minimum distance between the main lobe and the side lobes:

$$A_{\rm H/S} = 20 \cdot {\rm lg}\hspace{0.15cm}

\frac{|W(0)|}{|W(2.5 \cdot f_{\rm A})|} = 20 \cdot {\rm lg}\hspace{0.15cm}
\frac{0.5}{0.0121}\hspace{0.15 cm}\underline{\approx 32.3\,\,{\rm dB}}\hspace{0.05cm}.$$