Multiplierless Half-band Filters and Hilbert Transformers

This article provides coefficients of multiplierless Finite Impulse Response 7-tap, 11-tap, and 15-tap half-band filters and Hilbert Transformers.  Since Hilbert transformer coefficients are simply related to half-band coefficients, multiplierless Hilbert transformers are easily derived from multiplierless half-bands.  Image attenuation of the Hilbert transformers presented here is greater than 50 dB over a bandwidth of 0.16*f_s for the 7-tap version, 0.24*f_s for the 11-tap version, and 0.3*f_s for the 15-tap version, where f_s is sample frequency.

The coefficient values and properties of the filters are summarized at the end of the article.  Note that I discussed basics of half-band filters and Hilbert transformers in earlier posts [1,2].

7-Tap Half-band Filter

Here is a set of coefficients for a 7-tap half-band filter:

    b_hb = [-1/28 0 2/7 1/2 2/7 0 -1/28]
         = [-1 0 8 14 8 0 -1]/28              (1)

The sum of b is 1, so this filter has gain of 1 at f = 0 Hz.  For a fixed-point [3,4] multiplierless filter, we’ll need a denominator that is a power of two.  We can obtain this by multiplying the coefficients by 28/32 = 7/8, giving:

    b_hb = [-1 0 8 14 8 0 -1]/32
         = [-1/32 0 1/4 7/16 1/4 0 -1/32]     (2)

The filter now has a gain of 7/8 at f = 0 Hz.The filter block diagram is shown in Figure 1.  All coefficients except for the center tap b₃ are simply signed bit-shifts, while b₃ is the sum of two signed bit-shifts.  We can compute the filter’s magnitude response using Matlab:

    fs= 1;                          % Hz sample frequency
    b_hb= [-1 0 8 14 8 0 -1]/32;    % filter coefficients
    [H,f]= freqz(b_hb,1,256,fs);
    HdB= 20*log10(abs(H));          % dB-magnitude response

The dB-magnitude response is plotted in Figure 2.  As expected for a half-band filter, the response magnitude at f_s/4 with respect to that at f = 0 is -6 dB.  Given that there are only seven taps, performance is modest.  We could obtain a dc gain close to 1.0 by placing a gain of 9/8 in cascade with the filter: overall gain at dc would then be 9/8*7/8 = 63/64.

Figure 1.  7-tap Half-band filter block diagram.  Blocks labeled z^-2 represent a delay of 2 samples.

Figure 2.  Magnitude response of 7-tap half-band filter.  dc gain = 7/8.

7-Tap Hilbert Transformer

We learned about the relationship between the Hilbert Transformer and Half-band filter in an earlier post [5].  There, we found that Hilbert coefficients b_HT(n) can be computed from a half-band design b_HB(n) using the simple relationship:

$$b_{HT}(n)= 2 sin(n\pi /2)*b_{HB}(n),\quad  -N/2 < n < N/2 \qquad (3) $$

where N is even and the number of taps is N + 1.  For our 7-tap case, N = 6, n = -3: 3, and the value of sin(nπ/2) is:

sin(nπ/2) = [1 0 -1 0 1 0 -1]                      (4)

Given half-band coefficients of b_hb = [-1 0 8 14 8 0 -1]/32, Equation 3 results in Hilbert coefficients of:

b_ht = [b_-3 0 b_-1 0 b₁ 0 b₃] = [-1 0 -8 0 8 0 1]/16           (5)

Note that b_HT is odd-symmetric.  All coefficients are realized with signed bit-shifts.  The Hilbert transformer’s block diagram is shown in Figure 3, where we have re-numbered the coefficients to [b₀ 0 b₂ 0 b₄ 0 b₆].  The half-band filter had a gain of 7/8 at f = 0, which causes the Hilbert transformer output x_i to have a gain magnitude of 7/8 at f= f_s/4.  The center tap x_r output must have the same gain as the x_i output, so the center tap must be scaled by G_ct = 7/8, as shown.

Now we’ll find the frequency response of the Hilbert Transformer.  The transfer function H(z) is defined with respect to the center tap:

$$ H(z)= \frac{H_{FIR}(z)}{z^{-K}} \qquad (6) $$

Where H_FIR(z) is the causal transfer function of the Hilbert transformer and K samples is the delay of the center tap.  We can compute the frequency response of the Hilbert transformer using the following Matlab code:

    b= [-1 0 -8 0 8 0 1]/16;            % HT coefficients
    d= [0 0 0 1 0 0 0];                 % center tap = z^-3
    [H,f]= freqz(b,d,256,'whole',fs);   % freq. response over 0 to fs

Since the coefficients are odd-symmetric with respect to the center tap, the frequency response is pure imaginary, as shown in Figure 4.  As expected, the gain magnitude at f = f_s/4 is 7/8 = 0.875.  The overall gain can be made close to 1.0 by placing a gain of 9/8 in cascade with the Hilbert transformer.  If we choose to place the gain of 9/8 in cascade with the x_r and x_i outputs, a simplification in the x_r path is possible: we can replace G_ct with 9/8*G_ct = 63/64 = 1 – 1/64.

Figure 3.  7-tap Hilbert Transformer Block Diagram.  Blocks labeled z^-2 represent a delay of 2 samples.

Figure 4.  Pure-imaginary frequency response of 7-tap Hilbert transformer, G_ct = 7/8 = 0.875.

7-tap Hilbert Transformer Image Attenuation

We have calculated the Hilbert transformer’s frequency response H(f), but this does not tell us anything about image attenuation.  Let’s take a look at how to compute the image attenuation.  Figure 5a shows a frequency-domain block diagram of a Hilbert Transformer.  Its complex output spectrum is:

$$ X_c= X_r+ jX_i \qquad (7) $$

The frequency response of the complex output is:

$$ G(\omega)= \frac{X_r + jX_i}{X_r} \qquad (8) $$

or,

$$ G(\omega)= 1 + jH(\omega) \qquad (9) $$

where ω = 2πf/f_s is normalized radian frequency.  For an ideal Hilbert transformer, H(ω) = -j for 0 < ω < π and H(ω) = j for π < ω < 2π [2].  Thus G(ω) for the ideal case is given by:

$$G(\omega) = 2, \quad 0 < \omega < \pi  \qquad \quad $$

$$   \qquad \qquad  0, \quad \pi < \omega < 2\pi \qquad (10) $$

Note that G(ω) is pure real.  G(ω) for the ideal Hilbert transformer is plotted in figure 5b. The image band of π < ω < 2π has a response of 0.

Figure 5.  Hilbert transformer.

a) Frequency-domain block diagram.  b) Frequency response of complex output for ideal case.

We can apply Equation 9 for our 7-tap transformer, but first we need to modify it to account for G_ct used in Figure 3 and shown in Figure 6.  G(ω) then becomes:

$$ G(\omega)= G_{ct} + jH(\omega) \qquad (11) $$

The magnitude of G(ω) at ω= π/2 (f = f_s/4) is 2*G_ct.  We define a frequency response relative to this value as:

$$ G_{rel}(\omega) = \frac{G(\omega)}{2G_{ct}} \qquad (12) $$

Figure 6.  Hilbert Transformer frequency-domain block diagram for gain not equal to 1.0.

Now we’ll compute G_rel(f), using G_ct and H(f) that we found in the previous section:

    Gct= 7/8;                % gain at f = fs/4
    G= Gct + j*H;            % freq. response of HT complex output
    Grel= G/(2*Gct);
    Grel_dB= 20*log10(abs(Grel));

Grel_dB is plotted in Figure 7.  The image band is from 0.5*f_s to f_s.  The red line in the image band shows the frequency range for 50 dB image attenuation.  The corresponding passband centered at f_s/4 is also shown: the bandwidth for 50 dB image attenuation is 0.16*f_s.  OK, not very wide bandwidth, but after all, we have only seven taps.

Figure 7.  7-tap Hilbert transformer.  Frequency response G_rel(f) of complex output, showing bandwidth for 50 dB image attenuation.

11-Tap Half-band Filter

Here is a set of coefficients for an 11-tap half-band filter:

    b_hb = [8 0 -40 0 192 319 192 0 -40 0 8]/1024
         = [1/128 0 -5/128 0 3/16 319/1024 3/16 0 -5/128 0 1/128]        (13)

These coefficients were synthesized as a Canonic Signed Digit (CSD) design.  I discussed CSD filters in an earlier post [6].  The filter has gain of 639/1024 at f = 0.  The coefficients can be realized as signed bit-shifts or sums of signed bit-shifts as indicated:

b0 = b10 = 1/128

b2 = b8 = -5/128 = -1/32 - 1/128

b4 = b6= 3/16 = 1/4 - 1/16 (or 1/8 + 1/16)

b5 = 319/1024 = 1/4 + 1/16 – 1/1024               (14)

The filter block diagram is shown in Figure 8, and the dB-magnitude response is plotted in Figure 9.  The dc gain of 639/1024 causes the 4.1 dB insertion loss shown in the magnitude response.  We could obtain a dc gain close to 1.0 by placing a gain of 3/2 in cascade with the filter: overall gain at dc would then be 1.5*639/1024 = 0.936.

Figure 8.  11-tap half-band filter block diagram.

Figure 9.  Magnitude response of 11-tap half-band filter.  dc gain = 639/1024.

11-Tap Hilbert Transformer

Using Equation 3, we can derive a Hilbert transformer from the 11-tap half-band filter.  The Hilbert transformer coefficients are:

    b_ht = [-1 0 -5 0 -24 0 24 0 5 0 1]/64
         = [-1/64 0 -5/64 0 -3/8 0 3/8 0 5/64 0 1/64]        (15)

As was the case for the half-band filter, the coefficients are sums of signed bit-shifts, as shown in Figure 10.  The half-band filter had a gain of 639/1024 at f = 0, which causes the Hilbert transformer output x_i to have a gain magnitude of 639/1024 at f= f_s/4.  The center tap x_r output must have the same gain as the x_i output, so the center tap must be scaled by G_ct = 639/1024, as shown.  The overall gain can be made close to 1.0 by placing a gain of 3/2 in cascade with the Hilbert transformer.

Figure 10.  11-tap Hilbert transformer block diagram.

The pure-imaginary frequency response of the Hilbert transformer is shown in Figure 11, where the gain magnitude at f = f_s/4 is 639/1024 = 0.624.  Grel_dB is plotted in Figure 7.  The red line in the image band shows the frequency range for 50 dB image attenuation.  The corresponding passband centered at f_s/4 is also shown: the bandwidth for 50 dB image attenuation is 0.24*f_s.

Figure 11.  Pure-imaginary frequency response of 11-tap Hilbert transformer, G_ct = 639/1024 = 0.624.

Figure 12.  11-tap Hilbert transformer.  Frequency response G_rel(f) of complex output, showing bandwidth for 50 dB image attenuation.

15-Tap Half-band Filter

Here is a set of coefficients for a 15-tap half-band filter:

    b_hb = [-3 0 15 0 -48 0 194 316 194 0 -48 0 15 0 -3]/1024          (16)

As for the 11-tap case, these coefficients were synthesized as a CSD design.  The filter has gain of 632/1024 at f = 0.   The coefficients can be realized as sums of signed bit-shifts as indicated:

b0 = b14 = -3/1024 = -1/512 – 1/1024

b2 = b12 = 15/1024 = 1/64 – 1/1024

b4 = b10 = -48/1024 = -1/32 – 1/64

b6 = b8 = 194/1024 = 1/4 – 1/16 + 1/512

b7 = 316/1024 = 1/4 + 1/16 – 1/256                   (17)

I have not included a block diagram of this filter.  The filter’s dB-magnitude response is plotted in Figure 13, where we see that the dc gain of 632/1024 has caused an insertion loss of 4.2 dB.  We could obtain a dc gain close to 1.0 by placing a gain of 3/2 in cascade with the filter: overall gain at dc would then be 1.5*632/1024 = 0.9258.

Figure 13.  Magnitude response of 15-tap half-band filter.  dc gain = 632/1024.

15-Tap Hilbert Transformer

Again using Equation 3, the 15-tap Hilbert transformer coefficients are derived from the 15-tap half-band coefficients as:

    b_ht = [-3 0 -15 0 -48 0 -194 0 194 0 48 0 15 0 3]/512           (18)

These coefficients can be realized as sums of signed bit-shifts as indicated:

b0 = -b14 = -3/512 = -1/256 – 1/512

b2 = -b12 = -15/512 = -1/32 + 1/512

b4 = -b10 = -48/512 = -1/16 – 1/32

b6 = -b8 = -194/512 = -1/2 + 1/8 - 1/256                      (19)

The half-band filter had a gain of 632/1024 at f = 0, which causes the Hilbert transformer output x_i to have a gain magnitude of 632/1024 at f= f_s/4.  The center tap x_r output must have the same gain as the x_i output, so the center tap must be scaled by:

G_ct = 632/1024 = 1/2 + 1/8 – 1/128                 (20)

The overall gain can be made close to 1.0 by placing a gain of 3/2 in cascade with the Hilbert transformer.

The pure-imaginary frequency response of the Hilbert transformer is shown in Figure 14, where the gain magnitude at f = f_s/4 is, as expected, 632/1024 = 0.6172.  Grel_dB is plotted in Figure 15.  The red line in the image band shows the frequency range for 50 dB image attenuation.   The corresponding passband centered at f_s/4 is also shown: the bandwidth for 50 dB image attenuation is 0.3*f_s.

Figure 14.  Pure-imaginary frequency response of 15-tap Hilbert transformer, G_ct = 632/1024 = 0.6172

Figure 15.  15-tap Hilbert transformer.  Frequency response G_rel(f) of complex output, showing bandwidth for 50 dB image attenuation.

Summary

Tables 1 and 2 summarize the coefficients and properties of the filters we have discussed.  All coefficients can be implemented as sums of (three or fewer) signed powers of 2, called “signed digits”.  The tables include optional external gain for each filter to achieve an overall passband gain close to 1.0.

I have also included 3-tap versions of the half-band filter and Hilbert Transformer.  The 3-tap half-band filter is identical to a linear interpolate-by-2 filter.  The 3-tap Hilbert transformer’s coefficients are the negative of those of a central-difference differentiator.

References

1.  Robertson, Neil, “Simplest Calculation of Half-band Filter Coefficients”, DSPrelated.com, Nov., 2017, https://www.dsprelated.com/showarticle/1113.php

2.  Robertson, Neil, “Add the Hilbert Transformer to Your DSP Toolkit, Part 1”, DSPrelated.com, Nov, 2022, https://www.dsprelated.com/showarticle/1486.php

3.  Felton, Christopher, “A Fixed-Point Introduction by Example”, DSPrelated.com, April, 2011, https://www.dsprelated.com/showarticle/139.php

4.  Lyons, Richard, Understanding Digital Signal Processing, 3^rd Ed., Pearson, 2011, section 12.1.

5.  Robertson, Neil, “Add the Hilbert Transformer to Your DSP Toolkit, Part 2”, DSPrelated.com, Dec, 2022, https://www.dsprelated.com/showarticle/1487.php

6.  Robertson, Neil, “Matlab Code to Synthesize Multiplierless FIR Filters”, DSPrelated.com, Oct, 2016, https://www.dsprelated.com/showarticle/1011.php

Neil Robertson          October, 2023

Sign in