Page 389 - Analog and Digital Filter Design
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386 Analog and Digital Filter Design




                       power since the first and last filter taps are having no effect. In other words, the
                       filter length has effectively been reduced by a factor of 2. For example, a 21-tap
                       filter will only have 19 nonzero coefficients.

                       A better algorithm assumes that the window is 2 taps longer than the number
                       of  taps actually available. The zero-valued coefficients are then placed outside
                       the array of tap multipliers; that is, they are not used. Thus all taps have non-
                       zero value multipliers and contribute to the filter.

                              h(n) = 1.0 - llZl/(N + 1)/2,
                              IZ = -(N - 1)/2, - (N - 3)/2, . . . - 1,0,1, . . . (N - 3)/2,  (N - 1)/2.

                       Now, when n = -(N-1)/2  and (N-1)/2, the window coefficient is equal to:

                              h(n) = 1 .O - (N - 1)/(N + l),  when n = -(N - 1)/2 and (N - 1)/2

                       In a 21-tap filter the ftnal tap’s coefficient is h(n) = 1 .O - 20/22 = 1/11. Thus, using
                       the revised equation, the coefficients reduce by 1/11 per tap on either side of the
                       zero time coefficient.


                       3 Von  Hann (Raised Cosine) Window
                       The Von  Hann  window  is  sometimes known  as the  Raised  Cosine  window
                       because its values are calculated from a cosine raised to the power two. It is
                       derived from a simple expression:
                                       [ : I
                                           ,
                              h(n) = cos2 - alternatively this is given as:
                                             [
                                              2;n],
                              h(n)=O.5+0.5~0~ -
                              where n = -(N - 1)/2, . . . - 1,0,1, . . . (N - 3)/2, (N - 1)/2,

                       When time-shifted, so that the window edges occur at n = 0 and iz  = N - 1, the
                       second half  of the equation changes sign:


                              h(n) =0.5-0.5~0~        >  where n = 0,1, . . . (N - 3)/2, (N - 1)/2.

                       Notice that it is necessary to add  1 to the value of n in the numerator, so that
                       the end values of the window are not zero. The denominator also has to become
                       N + 1, so that h(n) = 1 when n = (N - 1)/2 (the zero time value).

                       With this window, the first side-lobe stopband attenuation is 35 dB, increasing
                       by  18 dB per octave at higher frequencies.
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