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Altera_Forum
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17 years ago

Low Pass IIR filter using cyclone II FPGA

Hi everyone,

I hava a task to implement low pass iir filter on cyclone II FPGA.

The parameters of filter are the following:

fp=5kHz; pass band frequency

fs=8kHz; stop band frequency

Fs=44kHz; sampling frequency

amax=1db; maximal attenuation in a pass band

anim=50db; minimal attenuation in a stop band

as a refferent filter I use Chebyshev filter.

After a few calculation I got N=7 for a order of my filter. After that, I got the transffer function

H(z) = 8*10^(-6)*(1+z)^7/(z^7-5.515z^6+13.761z^5-20.023z^4+18.29*z^3....

As you see I have to implement multiplication of usigned numbers.

I need advice: which is the best way to implement this recursive formula?

I also need to know how many bits-wide operands must be and how to implement multiplication.

Thank you very much for your efforts to help me !

Bojan

25 Replies

  • Altera_Forum's avatar
    Altera_Forum
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    17 years ago

    Than you for advice Micheel but I must to use Cyclone II FPGA this time and VHDL programming language. So... if you have ideas please share with me.

    Thank you very much,

    Bojan
  • Altera_Forum's avatar
    Altera_Forum
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    17 years ago

    U.Meyer-Baeses digital signal processing with fpga is a profound DSP textbook with a detailed IIR filter chapter. It also has VHDL examples.

    I can't see, that the filter is defined recursively in your example (and it probably shouldn't be). But it's strongly recommended to scale the coefficients to a fixed point representation to allow a FPGA suited filter implementation with signed signals. Some filter tools, e. g. Nuhertz Filter Solutions, perform this operation in filter design automaticly. Also MatLab/Octave can do, of course.
  • Altera_Forum's avatar
    Altera_Forum
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    17 years ago

    Yes, filter is recursive and the 7th order.

    b0y(n)=a0x(n)+a1x(n-1)+...+a7x(n-7)+b1y(n-1)+b2y(n-2)+...+b7y(n-7)

    I want to know which is the best way to represent coefficients ai, bi, i=0...7?

    Maybe like a signed number with a fixed point? And with how many bits? Is it 16bits enough?

    Thank you for you effort,

    Bojan
  • Altera_Forum's avatar
    Altera_Forum
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    17 years ago

    The resoution of the coefficients (16 bits or less or more) depends on the wanted accuracy of the filter output, e.g. if you have a mask which the filter must fit. So I suggest you to use Matlab/Octave or any other math tool to perform a fixed-point finite-resolution simulation. You will be able to insert your coefficients with 16bits (or different) and see the result on the output. The minimum needed resolution will be the one which fits your output mask. After that you can go to the the FPGA resolution.

  • Altera_Forum's avatar
    Altera_Forum
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    17 years ago

    --- Quote Start ---

    Yes, filter is recursive and the 7th order.

    b0y(n)=a0x(n)+a1x(n-1)+...+a7x(n-7)+b1y(n-1)+b2y(n-2)+...+b7y(n-7)

    --- Quote End ---

    The filter is recursive has any IIR filter, but the formula is a standard form of z transfer function, I wouldn't regard it as recursive.

    Coefficient resolution utilisation is ineffective in the standard form, cause a large range has to be covered. I found 18 bit resolution (13.5) is required for an acceptable frequency response in the present case, 16 Bit has an considerable magnitude error. In contrast, a cascaded form using first and second order building blocks achieves similar results with only 10 Bit (9.1) coefficient resolution.
  • Altera_Forum's avatar
    Altera_Forum
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    17 years ago

    Use the direct ii-transpose structure. Not the direct form.

    It's ideal on fixed-point IIR filtering, it minimizes quantization errors better than others.

    In order to improve the performance use a rounding escheme.

    Download the Mentor Graphics C header files ac_datatypes to test the arithmetic.
  • Altera_Forum's avatar
    Altera_Forum
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    17 years ago

    Ok guys,

    I did just like you told me: maximum of my ai, bi, i=1,...,7 coeff is -20.023...

    I did the simulations and 18 [6.12] coefficients bit resolution fits my output mask. So I overcome the problem with coeff representation and I represent the coeff in a two's complement representation with 18bits (6.12).

    I'm trying to understand how to perform multiplication of sampled digitalized signal x(n-i) and coefficient ai, bi. How to perform the multiplication when I have coefficient ai, bi in a two's complement representation with a fixed point and digitalized semple (output of A/D converter) x(n-i) which is not in a two's complement and don't have fractional part.

    If you understand my problem and have any idea please advice me.

    Thank you very very much!

    Bojan
  • Altera_Forum's avatar
    Altera_Forum
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    17 years ago

    The transition from 2s complement to unsigned is performed by inverting the MSB (this is the same as adding half dinamic range), and everything works if the original 2s complement signal is not overflowing or underflowing.

    So before your multiplication you have to perform the conversion and your multiplication will work in unsigned mode.
  • Altera_Forum's avatar
    Altera_Forum
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    17 years ago

    OK, you gave me idea. I'will perform multiplication with all positive coefficients, and after multiplication I'will perform substraction instead of addition.

    a1*x(n-1)-b1*y(n-1) =(a1*x(n-1)) - (b1y(n-1)) !

    So my coeff will be +20.023... And now... how to perform multiplication of 18bit sampled digitalized value of input signal x and coefficient which have 18bits (6bits for number and 12bits for fractional part - 6.12)?

    Sorry if I ask too much and if I ask nonsenses !

    Bojan
  • Altera_Forum's avatar
    Altera_Forum
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    17 years ago

    It is a classical problem of fixed point mathematics. The ideal Altera tool for your job is the altmult_accum megafunction which multiplies and accumulates in a single module.

    The fractional number should be transformed to integer numbers (you may use Matlab/Octave for a simulation before) and the result should be scaled back considering the multiplication factor that you have applied to one of the operands before the multiplication.

    If you need help about the Altera megafunction you can have a look at:

    http://www.altera.com/literature/ug/ug_altmult_accum.pdf