Iowa Hills Software Digital and Analog Filters

The Frequency Sampling Method
for FIR Design
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This page shows how to generate an FIR filter with the
frequency **oversampling** method. The samples may be taken from a custom defined
magnitude response, or from a predefined filter polynomial, such as a
Butterworth or Chebyshev.

We also show how to set the phase for a custom response. Defining the gain for a custom filter is usually trivial, but the phase is less trivial because we need to satisfy certain requirements for the Fourier Transform, and we typically want linear phase with a specific slope for optimum FIR performance. We then show how to modify the phase in order to change the filter’s group delay by fractional amounts.

An important aspect of this paper is our use of **oversampling** in the frequency
domain. We always sample the frequency domain at least 1024 times (a large power
of 2 suitable for an FFT) even though we typically want a relatively small
number of FIR taps (e.g. 25 taps).

We don't present any rigorous mathematics here. We simply explain the concepts used in this process. It is assumed the reader understands the properties of FIR filters and the Fourier Transform.

Please take note of our (rather clumsy) notation. First note our use of the terms NumTaps and NUMSAMPLES (from our C code). NUMSAMPLES is defined as a large power of 2 suitable for an FFT, typically 1024. NumTaps is self explanatory.

Also note our use of the term H(s). Sometimes we are working with predefined transfer functions, such as the Butterworth polynomial, and these transfer functions are written in terms of the Laplace variable 's', hence our use of H(s). But we also work in the discrete domain where the frequency domain notation usually takes the form of F(ω). While clumsy, we always use H(s) to refer to the positive frequency domain samples.

In our code snippets, H(s), written as HofS, is a complex array of length NUMSAMPLES / 2. HofS contains the samples from the positive frequency domain, whether they were derived from a continuous polynomial transfer function, or a discrete custom defined transfer function. We then use these positive frequency samples to generate the negative frequency samples, and fill another array, FFTArray, of length NUMSAMPLES for use in an FFT.

For example C code that implements the ideas given here, see our Code Kit.

Note: The Iowa Hills FIR Filter program uses this methd to generate FIR filters with a custom magnitude response, but it does not allow for a custom phase response.

**The Use of Over Sampling**

Many authors discuss frequency sampling as a technique for generating FIR coefficients. In most discussions, however, the author prescribes sampling the frequency response N times to generate N taps.

While this approach is certainly valid, it will only yield a useful result if the frequency response is sufficiently well behaved at the sample locations and between them as well.

A simple example of a poorly behaved frequency response would be one where the magnitude response transitions from 1 to 0 between samples. In this case, the Inverse DFT cannot possibly yield a filter with the correct cutoff frequency.

In general, using N frequency samples to generate an N tap filter is not a good approach to take.

If you are familiar with windowed FIR filters, you know that we start with an ideal low pass frequency response H(ω) and perform the Fourier integral as shown here. This result is documented in many places.

It should be clear from calculus that this integral can be approximated with a summation, and from calculus we know this summation will approach the integral exactly as N, the number of summation points, approaches infinity. This summation would then, in essence, be the Discrete Fourier Transform.

To point, it should be clear that the correct way to approximate this integral would be to use a large number of samples in our DFT, and indeed, one will find that a 1024 point Inverse FFT yields the same results (to several decimal places) as the exact integral.

The obvious question would then be: If we are using 1024 samples to generate a 32 tap filter, which part of the FFT output do we use?

The answer is quite simple. If we are generating a linear phase FIR filter, the output of the FFT will be symmetric about its center bin. Thus if 32 taps are desired, then the center 32 FFT outputs are used. If we need 54 taps, then use the center 54 FFT outputs.

In other words, we simply truncate the beginning and end of the FFT output to obtain the desired length. This implies that the center taps are the same no matter the number of taps used, and this is correct, and also the result the exact integral solution gives.

Please remember that when we use the center of the FFT output, we are assuming that the frequency response used for the Inverse FFT described a linear phase filter. If it doesn’t, then the desired taps will not be centered in the FFT's output. An example of this would be if the frequency response describes an analog filter, such as a Butterworth. Then the desired taps will start at t=0, the beginning of the Inverse FFT output, and as before, we simply use as many taps as desired.

Also remember that the frequency response for an odd tap count filter is not the same as for an even tap count filter. The H(s) for even and odd tap linear phase filters have slightly different phase slopes, which will be discussed below.

**Step 1: Define the Transfer Function H(s)**

H(s) may take on many forms. It may be predefined, such as a Butterworth filter polynomial, or it may be a custom transfer function, where we don’t explicitly define the transfer function in terms of s, but rather simply define the gain and phase of the transfer function in the positive frequency domain.

If we are working with a predefined polynomial, such as a Butterworth, then this step is essentially complete. We just need to show how to sample H(s) appropriately in order to do the required frequency warping. Remember that H(s) may define any type of response, whether it be low pass or band pass or a differentiator.

We start with the most common application for this technique, where we define H(s) for a custom FIR filter. We will start with a simple windowed filter where the magnitude response is trivial.

For the sake of this paper, we sample the positive frequency domain
NUMSAMPLES / 2 times. Ω_{c} is defined as a fraction of Pi and with a
range of 0.0 < Ω_{c} < 1.0. Then the magnitude of H(s) for this example is 1 for the
samples 0<= n <= Ω_{c}·NUMSAMPLES/2 and 0 outside this range.

The phase of H(s) is only slightly more complicated in that it depends on the number of taps we intend to use. For odd tap counts, the phase changes at the rate of -π per sample. For even tap counts, the phase changes at -(N-1)/N·π per sample where N is NUMSAMPLES.

This code snippet shows how we set the phase for a linear phase low pass filter. The WinMag array was filled in a previous function call and contains the desired magnitude for H(s). The HofS array is complex valued.

// Set the Radians per Sample

RadPerSample = -M_PI * (NUMSAMPLES - 1.0)/NUMSAMPLES; //
Even tap count.

if(NumTaps % 2 == 1) RadPerSample = -M_PI;
// Odd tap count.

// Fill the H(s) array. This is for a low pass response.

for(j=0; j<NUMSAMPLES/2; j++)

{

Arg = RadPerSample * (double)j;

HofS[j] = WinMag[j] * ComplexD( cos(Arg), sin(Arg)
);

}

This low pass filter is now fully defined in the positive frequency domain. The magnitude response for this example is trivial, but it can take on any shape. The phase however, will be as shown here for a low pass filter, no matter the shape of magnitude response. The details on setting the phase for other filter types, such as a high pass filter, are given in our example code.

**H(s) defined by a polynomial**

It is sometimes desirable to obtain an FIR response for an analog polynomial,
such as a Butterworth filter. Then both the magnitude and phase are
predefined.

The only problem then is to properly define the frequencies used to sample the polynomial. We have two problems to solve. The first is that the polynomial is defined for frequencies out to infinity and we need to compress this range down to Pi for the sake of Discrete Fourier Transform (warp the frequencies). The second is that filter polynomials, such as the Butterworth, are defined to have a 3 dB cutoff at 1 Hz, and we need to move this to our desired cutoff frequency.

This code snippet shows how this is done, where
OmegaC is the desired 3 dB frequency 0.0 <Ω_{c} < 1.0

// Fill SimOmega with the frequencies used to
evaluate the polynomial.

Omega0 = 1.0 / tan(OmegaC * M_PI_2);

for(j=0; j<NUMSAMPLES/2; j++)

{

SimOmega[j] = Omega0 * tan(M_PI * (double)j / NUMSAMPLES);

}

// Evaluate the 2nd order sections in a manner similar to
this.

// NSections is the number of 2nd order sections.

// N2, N1, N0, D2, D1, D0 are the 2nd order polynomial
coefficients.

for(j=0; j<NUMSAMPLES/2; j++)

{

s = ComplexD(0.0, SimOmega[j]);

s_sq = s * s;

H = ComplexD(1.0,0.0);

for(n=0; n

{

Numerator = N2[n] * s_sq +
N1[n] * s + N0[n];

Denominator = D2[n] * s_sq + D1[n] * s
+ D0[n];

H = H * Numerator / Denominator;

}

HofS[j] = H;

}

**Step 2: Create a NUMSAMPLES Complex Array for use in an FFT**

Let us start by reminding the reader that the value of H(s) at the negative frequencies is the complex conjugate of the positive frequency values. With this in mind, we want to point out some implications of this for filter design. Here we show the bin locations for an 8 point FFT

The DC term, 0, and the Nyquist Frequency term, N/2, do not have a conjugate match. As such, these two terms must be real valued. Let us discuss some valid values for these two terms in the context of FIR filter design.

For low pass filters, the 0 bin represents the DC gain of the filter. Thus for unity gain filters, the only valid values for bin zero are 1, or -1 for inverting filters. The Nyquist bin represents the gain of the analog filter at ±∞ so it must be 0. These two bins may not contain a complex value.

For a high pass filter, the DC gain is zero, so bin 0 must be zero. The Nyquist bin represents the gain of the analog filter at ±∞ so the only valid values for this bin are 1, or -1 for inverting filters (assuming unity gain).

For band pass filters, both the DC and Nyquist values are 0.

For Notch filters with unity gain, the DC and Nyquist bins must be either 1 or -1, but need not be the same.

These requirements can also be thought of as restrictions on the phase at DC and Nyquist. For low pass and notch filters, the phase must be either 0 or 180° at DC, while for high pass and notch filters, the phase must be either 0 or 180° at Nyquist.

Band pass filters are unique in this respect. Since the magnitude of a band pass filter equals 0 at both DC and Nyquist, the phase at these two frequencies is ignored, but this allows us to set the phase in the bins adjacent to DC and Nyquist to any value without presenting a phase discontinuity to the DFT. This means we can give the phase curve in a band pass filter any offset we desire, which isn’t true for low pass, high pass, and notch filters for the reasons given in the preceding paragraph. While a band pass filter's phase may have any absolute value, the slope of the phase curve is not arbitrary, and must be set as for other FIR filters in order to maintain optimal performance.

**Step 3: Generate the FIR coefficients by doing an Inverse FFT**

In the preceding step, we generated an input array for the FFT of length NUMSAMPLES (a large power of 2) which will generate NUMSAMPLES of FIR coefficients. If the phase of H(s) was set to change at the rate of -π per sample for odd tap count filters, and at -(N-1)/N·π per sample for even tap count filters, then the FIR taps are located in the center FFT bins.

This code snippet shows the details. The integer variable StartT is the location of the first tap. Note this integer division NumTaps/2 handles the case for even and odd tap counts appropriately.

Also note that the variable FFTArray[] is a complex array that was used as both the input and output of the FFT. We use the real part of the return value because the imaginary part is essentially zero (the result of setting the negative frequency values to the conjugate of the positive frequency values). Since our FFT doesn’t scale the output of an inverse FFT, which is typical of most FFTs, we must scale the coefficients by NUMSAMPLES, the length of the FFT.

// This is where the FIR taps are located in the
FFT’s return.

StartT = NUMSAMPLES/2 - NumTaps/2;

for(t=0; t<NumTaps;
t++)

{

FirCoeff[t] = FFTArray[t+StartT].real() / (double)NUMSAMPLES;

}

If the input to the FFT represented an analog filter, such as a Butterworth, then the FIR taps are located at the beginning of the FFTArray ( at t=0) and we would set StartT=0 in this code.

**Step 4: Window the FIR Coefficients**

It is not the purpose of this paper to explain the windowing techniques typically used for FIR filters, which are typically symmetric about their center tap. We do however want to show a windowing technique for a filter derived from an analog transfer function.

This plot shows a typical impulse response and window for a linear phase FIR filter. Both the impulse response and the window are symmetric about the center.

This plot shows a typical impulse response generated by an analog filter polynomial, such as the Butterworth. These impulse responses are not symmetric and are best windowed with something similar to a half cosine window W(n) = cos(n/NumTaps · π/2). It isn’t complicated, but this window is quite effective and usually sufficient for these analog style impulse responses.

**Step 5: Adjust the Group Delay as Needed **

Occasionally we need to adjust the filter’s group delay as a way to align signals. A filter’s group delay can typically be changed by ± TapCount/20 without seriously degrading the filter’s performance.

There are many ways to adjust the filter’s delay time. Here are three ways,
which **are not recommended**.

- After H(s) is formed in step 1, whether it be a custom transfer function or the
samples from a filter polynomial, simply multiply H(s) by the Laplacian time
delay operator e
^{-a·s}. The value of ‘a‘ is in terms of fraction of samples. - In step 3, when we are extracting the coefficients from the output of the FFT, simply move the starting bin for the coefficients by ±1 to affect a delay of ±1. This will not allow us to do a fractional delay change, and can only be used with a linear phase filters where the impulse response is centered in the FFT’s output, so this isn’t a particularly useful method.
- If we are generating the FIR filter coefficients with a formula similar
to:

h(n) = 2·Ω_{c}·sinc(2·Ω_{c}·π·n)

then a fractional delay can be introduced by adding a fraction to n within the sinc function.

The problem with these three methods is that they generate an impulse
response which is not symmetric about its center before we have had a chance to
apply a window, such as a Kaiser. This is a significant problem without a good
solution because there isn’t a good way to generate an effective non-symmetric
window. To avoid this problem, we use the following procedure as a final step to
adjust a filter’s group delay.

- After the coefficients have been generated and windowed appropriately, take an FFT of the taps.
- Multiply the output of the FFT by the Laplacian time delay operator e
^{-a·s}where ‘s’ is the FFT’s bin frequency and the value of ‘a’ represents the fraction of a sample. In other words, ‘a’ is not in terms of the sampling frequency.

- Do an Inverse FFT.

This simple process allows you to make fractional group delay changes to any type of FIR impulse response.

This plot shows an example of of an FIR filter generated with this method. 8 Pole Inv Cheby Coefficients.

For example C code that implements the ideas given here, see our Code Kit.