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ACEEE International Journal on Communication, Vol 1, No. 1, Jan 2010




       A Novel Methodology for Designing Linear
                  Phase IIR Filters
                                         R.Ramanathan1, K.P.Soman2
                                          1
                                        Lecturer, Department of ECE,
                                 Email: r_ramanathan@ettimadai.amrita.edu
       2
         Professor & Head, Center for Excellence in Computational Engineering and Networking (CEN)
                                        Email: kp_soman@amrita.edu
                              Amrita Vishwa Vidyapeetham, Coimbatore, India


Abstract—This paper presents a novel technique for                  programming problem and then we will redefine the same
designing an Infinite Impulse Response (IIR) Filter with            problem with some constraints for solving the linear
Linear Phase Response. The design of IIR filter is always a         programming model. The frequency domain responses of
challenging task due to the reason that a Linear Phase              the filter (both magnitude and phase) are used to model
Response is not realizable in this kind. The conventional           the linear programming problem. Let us consider the
techniques involve large number of samples and higher               standard expression [2] for the filter transfer function,
order filter for better approximation resulting in complex
hardware for implementing the same. In addition, an                            Y ( z)       b0 + b1 z −1 + ... + b N z − N
extensive computational resource for obtaining the inverse                              =                                                   (2)
of huge matrices is required. However, we propose a                            U ( z ) 1 + a1 z −1 + ...a D z − D
technique, which uses the frequency domain sampling along
with the linear programming concept to achieve a filter             Where U (z) is the z-transform of the input signal, Y (z) is
design, which gives a best approximation for the linear
                                                                    the z-transform of the output signal, and ‘a’ and ‘b’
phase response. The proposed method can give the closest
response with less number of samples (only 10) and is               factors are real valued coefficients. From the above
computationally simple. We have presented the filter design         transfer function, we can write the time-domain
along with its formulation and solving methodology.                 difference equation that implements the filter as,
Numerical results are used to substantiate the efficiency of
the proposed method.                                                           y (k ) = − a1 y (k − 1) − ....a D y (k − D )
                                                                                                                                            (3)
                                                                               + b0 u (k ) + ... + bN u (k − N ).
Index  Terms—Infinite         impulse     response,   linear
programming, filter design, linear phase filters                    where the a and b coefficients are exactly the same as in
                                                                    the transfer function above, k is the time index, u (k) and
                     I. INTRODUCTION                                y (k) are the current values of the input and output
   The design of the infinite impulse response filter is            (respectively), u(k− N) was the input value N samples in
often restricted to the specification of the magnitude              the past, and y(k− D) was the output value D samples in
response due to the fact that a linear phase response is not        the past. Generally, if the frequency response of a system
physically realizable. Some methods in the literature try           at a frequency ω1 is given in magnitude/phase form
to achieve a good approximation of the linear phase                 as A1∠φ1 the output amplitude will be A1 times the input
response, but results in extensive and complex hardware.            amplitude and the output phase will be shifted an angle
Thus, the design phase involves huge computational                  φ1 relative to the input phase when a steady-state sine or
resources as well as implementation phase requires                  cosine wave of frequency ω1 is applied to the system.
extensive hardware. We intend to simplify the                       For instance, if the input to the system described above at
computation in design as well as minimize the hardware              time instance ‘k’ is u1(k) = Cos(kω1ts), where ts is the
by just opting for a second order filter. We introduce a            sampling period, then the output will be
concept of using the frequency domain samples of the                y1(k) = A1 Cos(kω1 ts + φ1). The input and output values
desired magnitude and phase response of the filter and              at any sample time can be determined in a similar
using the same to obtain the filter coefficients to                 manner. The final sampled values in the matrix form is,
implement the filter. The digital filter can be modeled as
                                                                                                                                        ⎡a1 ⎤
a linear system represented by                                        ⎡ y1 (0) ⎤ ⎡ − y1 ( −1) ... − y1 ( − D ) u1 (0) ... u1 ( − N ) ⎤ ⎢ ⎥
                                                                      ⎢ y ( 0) ⎥ ⎢                                                    ⎥ ⎢. ⎥
                                                                                ⎥ ⎢ − Y2 ( −1) ... − Y2 ( − D ) u2 (0) ... u2 ( − N ) ⎥ ⎢a ⎥
                             Ax = b                     (1)           ⎢ 2
                                                                      ⎢'.       ⎥=⎢     .       .       .          .    .       .     ⎥*⎢ D⎥
                                                                      ⎢         ⎥ ⎢                                                   ⎥ ⎢b0 ⎥
Where, A is the matrix containing the sampled frequency               ⎢.        ⎥ ⎢     .       .       .          .    .       .     ⎥ ⎢. ⎥
                                                                      ⎢ y N (0) ⎥ ⎢− yM (−1) ... − yM (− D ) uM (0) ... u M (− N )⎥ ⎢ ⎥
responses (magnitude & phase) of the given filter [1].                ⎣         ⎦ ⎣                                                   ⎦
                                                                                                                                        ⎢bN ⎥
                                                                                                                                        ⎣ ⎦
The given magnitude and phase responses of the filter
                                                                                                                                      (4)
can be sampled at a convenient sampling frequency and
the number of samples can be conveniently chosen. We                The above equation is similar to the system A x ≈ b ; as
will now formulate the filter design problem as a linear            given by (1). In the above formulation, we have set the
                                                               19
© 2010 ACEEE
DOI: 01.ijcom.01.01.05
ACEEE International Journal on Communication, Vol 1, No. 1, Jan 2010



current sample index as zero. Our aim now is to solve the          Here ‘.*’ is a Matlab operation, which is an element by
above problem, and get the filter coefficients given by the        element multiplication. Rewriting the above constraint
vector ‘x’. Standard methods use the inverse of the                using w1 and w2,
matrix ‘A’ which involves a heavy computation for huge
matrices [3],[4],[5],[6],[7],[8]. If the inverse of the matrix                         0 ≤ w 1 ≤ a. * y                            219
does not exist, then Singular Value Decomposition can be
used for Pseudo inverses [9]. To avoid this and make the                               0 ≤ w 2 ≤ a. * y                   (10)
procedure computationally simple, we introduce the
concept of linear programming into this context. We now            Now the problem, which we need to solve, is given by
formulate the above as a simple optimization problem
                                                                                       1                 1
                                                                                            ∑w                ∑ w +∑ a
with few constraints.
                                                                            Min                  1   +            2         i
               II. PROBLEM FORMULATION                                  x, w1 , w2 , a C1    i           C2   i       i

   In the above model discussed, the matrix ‘A’ contains                         S.t   Ax + w 1 − w 2 = y;
the sampled coefficients at different frequencies and the
column vector ‘x’ contains the filter coefficients. The                  0 ≤ w 1 ≤ a. * y   , 0 ≤ w 2 ≤ a. * y and a ≥ 0;
first ‘D’ values in ‘x’ represent the coefficients a1, a2, etc                                                              (11)
in denominator and the next ‘N’ values give the b0, b1, b2,
etc coefficients in numerator. The column vector ‘y’ is            Here C1 and C2 are control parameters that can be
the desired output value (both amplitude and phase) at             adjusted to vary the depth of optimization of w1 and w2
specific frequencies. Now we can move on to formulate              respectively. By the above formulation, we try to reduce
the optimization problem to find the filter coefficients.          the value of error ‘ε’ along with the maximum
We first aim to find the best possible solution by                 permissible limit defined by ‘a’. When we attempt to
introducing a factor ‘ε’, which is vector containing               reduce the value of ‘a’, indirectly the maximum value
allowable error. Then we try to reduce the error to get the        that w1 and w2 can assume is reduced which in turn
best solution by minimizing this error parameter.                  results in the value of ‘ε’. The problem is a chain reaction
                                                                   one. That mutually reduces the above said parameters.
                 Ax + ε = y;                                       This formulation of the optimization problem is easily
                                                          (5)      solvable by any simple tool, even with a spreadsheet
                 ⇒ ε = y − Ax                                      package like Microsoft Excel. This formulation results in
                                                                   reduced computational complexity and less number of
Now our goal is to frame a minimization problem to
                                                                   frequency domain samples (eg: 8 or 16).
reduce the error. Here our variables are ‘x’ and ‘ε ’.

                Min       ∑ε    i   +   ∑a   i
                                                                                   III. NUMERICAL RESULTS
                x,ε , a     i           i                              We have taken few examples [10], [11], [12] of low
                                                                   pass; band pass and high pass filter and attempted to
S.t   Ax + ε = y; 0 ≤ ε ≤ a. * y            and a ≥ 0;    (6)      solve for the filter coefficients. The knowledge of desired
                                                                   frequency response (both magnitude as well as phase) is
Instead of directly manipulating the error parameter, we           the only requirement. We have considered a second order
can define two variables w1 and w2 (essentially vectors of         filter in all the cases discussed with 10 sample values.
length same as ‘ε’) such that                                      The number of samples is left to the choice of the
                                                                   designer with only condition that minimum number of
                                                                   samples be 10 for a good approximation. The proper
                ε = w1 − w 2 ;                            (7)
                                                                   choices of sample points play an important role in the
                                                                   response approximation. The sampling is done at the rate
                                                                   of 1000Hz or 10-3 seconds. Since we have chosen a
So our equation (5) becomes,
                                                                   second order filter, we have N=D=2. We have solved
                Ax + w 1 − w 2 = y;                       (8)
                                                                   the optimization problem using Microsoft Excel and the
                                                                   results are verified by using Matlab. The obtained
We also have a possibility to restrict the maximum value           coefficients are used to plot the magnitude and phase
error to a limit say a fraction of the actual response value,      response and the results are verified.
so that relatively there is no adverse change in the overall
response. Letting that positive factor to be ‘a’ that              A. Illustration 1:
contains the fraction (eg: 0.01 or 1/100 th), where ‘a’ is
                                                                      A Low pass IIR Filter was designed for the given
also a vector.
                                                                   Specifications as shown in figure 1. The signal is
                                                                   sampled and the sample values corresponding to
                                                                   amplitude and phase are tabulated in Table 1. The filter
                  0 ≤ ε ≤ a. * y                         (9)       problem is formulated as a linear problem using the
                                                                   above discussed formulation and is solved using Excel
                                                                   and the coefficients obtained are listed in Table 2, also
                                                                   the response obtained using Matlab is shown in figure 1.
                                                               - 20 -
© 2010 ACEEE
DOI: 01.ijcom.01.01.05
ACEEE International Journal on Communication, Vol 1, No. 1, Jan 2010




         TABLE 1: SAMPLED DATA :( 10 POINTS) - LPF


   Amplitude            Phase (radians)        Frequency (Hz)

        1.0                  -0.1                  2.7400

        1.0                  -0.5                  50.0400

        1.0                  -0.97                 97.9600

        1.0                  -1.01                101.8700

        0.9                  -1.1                 110.0000               Fig 1b. Comparison of desired and obtained Phase Responses of the low
                                                                                                       pass filter
        0.6                  -1.25                125.0000
                                                                        The desired and obtained magnitude and phase responses
        0.5                  -1.25                150.0000              are comparatively studied and the figure 1 depicts the
                                                                        same. It is quite clear that the obtained response better
        0.3                  -1.25                178.2800
                                                                        approximates the desired one with just 10 samples and a
        0.1                  -1.25                300.4200
                                                                        second order IIR filter. With this method, a linear phase
                                                                        response can be achieved even with an IIR filter.
       0.001                 -1.25                450.5600

                                                                        B. Illustration 2:
                                                                           We now discuss the second example of designing a
              TABLE 2: OBTAINED COEFFICIENTS - LPF                      band pass filter to operate in the band from 100 Hz – 200
                                                                        Hz with a linear phase response as depicted in fig 2b. The
 a0             a1          a2            b1       b2          b3       signal is sampled and the sample values corresponding to
                                                                        amplitude and phase are tabulated in Table 3. The filter
1.00          -0.8663     0.3141     0.2141     0.1122       0.1209     problem is formulated as a linear problem using the
                                                                        above discussed formulation and is solved using Excel
                                                                        and the coefficients obtained are listed in Table 4 and the
It can be inferred that the phase response exhibits a linear            response obtained using Matlab is shown in figure 2.
characteristics in our pass band which is 0 – 100 Hz. The
phase is exactly ‘-1 radians’ at 100 Hz and it exactly fits a                      TABLE 3: SAMPLED DATA (10 POINTS) - BPF
straight line with slope angle 45 degrees. We are not
concerned of the phase relation in the stop band, but the                                                                 Frequency
                                                                                Amplitude          Phase (radians)           (Hz)
phase seems to fit a linear curve in stop band also. The
small deviation in the response in the transition from the                          0.0                  0.5                 0.0000
pass band to stop band can be rectified by increasing the
number of sample points in this region. We have                                     0.25                 0.5                50.0000
experimented with 10 samples and tried to prove that
with just 10 samples, we are able to get a good                                     0.5                  0.5                95.0000
approximation. Therefore, by suitably increasing the
number form 10 to 16 or 32, the obtained response will                              0.9                  0.5               100.0000
exactly follow the desired one.
                                                                                    1.0                 -0.11              150.0000

                                                                                    1.0                 -0.63              200.0000

                                                                                    0.75                 -1.0              250.0000

                                                                                    0.5                 -1.46              300.0000

                                                                                    0.3                  -1.5              350.0000

                                                                                    0.2                  -1.5              450.5600




Fig1a. Comparison of desired and obtained Magnitude Responses of the
                           low pass filter


                                                                    - 21 -
© 2010 ACEEE
DOI: 01.ijcom.01.01.05
ACEEE International Journal on Communication, Vol 1, No. 1, Jan 2010


          TABLE 4: OBTAINED COEFFICIENTS - BPF                                        TABLE 5: SAMPLED DATA (10 POINTS) - HPF


220                                                                                Amplitude        Phase (radians)        Frequency (Hz)

  a0          a1          a2         b1          b2          b3                       0.0                   1.7                 0.0000

 1.00      -0.7502      0.2644     0.2617      0.2471      -0.5010                    0.1                   1.7                50.0000

                                                                                      0.2                   1.7                95.0000
The comparison plots (figure 2) between the desired and
                                                                                      0.3                   1.7               100.0000
obtained responses are presented. The phase response in
the pass band, that is 100 Hz – 200 Hz, is quite linear                               0.5                   1.5               150.0000
with respect to frequency. This region is of interest to us
and is the key feature to be noticed.                                                 1.0                   1.2               200.0000

                                                                                      1.0                   0.9               250.0000

                                                                                      1.0                   0.6               300.0000
                                                                                      1.0                   0.4               350.0000

                                                                                      1.0                   0.1               450.5600

                                                                                       TABLE 6: OBTAINED COEFFICIENTS - HPF


                                                                              a0            a1       a2             b1        b2           b3

                                                                          1.00        -0.2194      0.1662         0.4975   -0.7182       0.2208


 Fig 2a. Comparison of desired and obtained Magnitude Responses of
                        the Band pass filter




                                                                              Fig 3a. Comparison of desired and obtained Magnitude Responses of
                                                                                                      the high pass filter

 Fig 2b. Comparison of desired and obtained Phase Responses of the
                          Band pass filter

C. Illustration 3:
   The Third example we have considered is that of a
high pass filter. The procedure similar to the previous
examples is followed here also. The Desired responses
(figure 3), sampled data (table 5), the obtained
coefficients (table 6) and obtained responses (figure 3)
are presented below. The comparison plot (figure 3)
clearly describes the linear phase characteristics of the
designed filter. By increasing the number of samples, we
can make the obtained response exactly follow the
designed one. Here we have focused on the linear phase                        Fig 3b. Comparison of desired and obtained Phase Responses of the
response, hence chosen just 10 samples. By choosing 32                                                 high pass filter
samples, the obtained response is found to be the exact
copy of the desired response.                                                  IV. PARAMETERS INFLUENCING THE FILTER DESIGN
                                                                           Firstly, the filters designed by the above method solely
                                                                         depend upon the number of samples and the sample

                                                                     - 22 -
© 2010 ACEEE
DOI: 01.ijcom.01.01.05
ACEEE International Journal on Communication, Vol 1, No. 1, Jan 2010



values. The key point is the choice of the sample values.       [4]Ricardo A. Vargas and C. Sidney Burrus. On the design of
The frequency points where there is a significant change        IIR filters with arbitrary frequency response. In Proceedings of
in the response either magnitude or the phase should be         IEEE International Conference on Acoustics Speech and Signal
captured by the sample values we choose. For a normal           Processing , ICASSP-01, Salt Lake City, May 7-11 2001.
                                                                [5] C. Sidney Burrus and Ricardo A. Vargas. The direct design
filter, we have considered (low pass or high pass or band       of recursive or IIR digital filters. In Proceedings of International
pass); four points in the transition band, three points in      Symposium on Communications, Control, and Signal
the pass and stop band will do the trick. However, if the       Processing, ISCCSP-08, Malta, March 2008.
filter should provide specific responses, then the number       [6] L. B. Jackson. Frequency-domain steiglitz-mcbride method
of samples should be chosen accordingly depending upon          for least-squares IIR filters design, arma modeling, and
the level of approximation we require. However, as a            periodogram smoothing. IEEE Signal Processing Letters,
general case a filter designed by this proposed                 15:498211;52, 2008.
methodology with 10 samples should yield a good                 [7] Takao Kobayashi and Satoshi Imai. Design of IIR digital
approximation of the magnitude and phase response               filters with arbitrary log magnitude function by WLS
                                                                techniques. IEEE Trans. on ASSP, 38(2):2478211;252,
when compared to the design by existing methods. In             February 1990.
addition, we can achieve a linear phase Infinite Impulse        [8] Atmadji W. Soewito. Least Square Digital Filter Design in
Response (IIR) filter which is normally not realizable.         the Frequency Domain. PhD. thesis, Rice University, ECE
Hence, this method can be effectively used to achieve the       Department, Houston, TX 77251, December 1990.
desired response using an IIR filter.                           [9] G.Strang, “Linear Algebra and its Applications”,3rd
    Secondly, we are interested only in the phase response      ed.,Thomson Brooks/Cole.
in the pass band and the phase in the stop band is not of       [10] S.K.Mitra, J. Kaiser, Handbook for Digital Signal
our concern. Yet, the literatures reveal that the phase         Processing, 1993 John Wiley and Sons, Inc.
                                                                [11] T. W. Parks and C. S. Burrus. Digital Filter Design. John
change in stop band will influence the overall response of
                                                                Wiley & Sons, New York, 1987
the filter response. These gives a cushion of changing the      [12] L. R. Rabiner and B. Gold. Theory and Application of
phase in the stop band and further achieve better results.      Digital Signal Processing. Prentice-Hall, Engle-wood Cliffs, NJ,
Therefore, we can introduce few more variables that are         1975.
phase values in stop band in our formulation, even
provide some lower and upper bounds, and further refine
the formulation.

          V. CONCLUSION AND FUTURE SCOPE
    The formulation of the IIR filter design problem, with
both the magnitude and phase specifications, is easily
done using this proposed technique of linear
programming. This application of linear programming for
design of linear phase IIR filters is a novel methodology.
This requires only very low computational resources in
spite of a very good approximation of the linear phase
response. The results obtained by the proposed method
are quite encouraging. The results can be further
improved by proper choice of the number of samples and
position of sample points. These results presented above
guarantee the efficiency of the proposed method. The
literature shows that by slightly tuning the phase in the
stop band (which is of no significance) the phase
response in the pass band can be changed accordingly.

These phenomena can be incorporated in the proposed
technique to achieve best results. The combination is
implemented and the performance is being investigated.
It is very much conducive that the proposed method may
be effectively adopted by the design engineers. The
proposed method proves to be a simple method when
compared to the counterparts for providing better results.

                      REFERENCES
[1] G.Berchin,”Precise Filter Design” IEEE Signal Processing
Magazine, January 2007.
[2] R. Lyons, Understanding Digital Signal Processing, 2nd
ed., Upper Saddle River, NJ:Prentice-Hall, pp. 232–240, 2004.
[3] C. Sidney Burrus,”Direct Frequency Domain IIR Filter
Design Methods”Version 1.1: Jun 24, 2008


                                                           - 23 -
© 2010 ACEEE
DOI: 01.ijcom.01.01.05

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A Novel Methodology for Designing Linear Phase IIR Filters

  • 1. ACEEE International Journal on Communication, Vol 1, No. 1, Jan 2010 A Novel Methodology for Designing Linear Phase IIR Filters R.Ramanathan1, K.P.Soman2 1 Lecturer, Department of ECE, Email: r_ramanathan@ettimadai.amrita.edu 2 Professor & Head, Center for Excellence in Computational Engineering and Networking (CEN) Email: kp_soman@amrita.edu Amrita Vishwa Vidyapeetham, Coimbatore, India Abstract—This paper presents a novel technique for programming problem and then we will redefine the same designing an Infinite Impulse Response (IIR) Filter with problem with some constraints for solving the linear Linear Phase Response. The design of IIR filter is always a programming model. The frequency domain responses of challenging task due to the reason that a Linear Phase the filter (both magnitude and phase) are used to model Response is not realizable in this kind. The conventional the linear programming problem. Let us consider the techniques involve large number of samples and higher standard expression [2] for the filter transfer function, order filter for better approximation resulting in complex hardware for implementing the same. In addition, an Y ( z) b0 + b1 z −1 + ... + b N z − N extensive computational resource for obtaining the inverse = (2) of huge matrices is required. However, we propose a U ( z ) 1 + a1 z −1 + ...a D z − D technique, which uses the frequency domain sampling along with the linear programming concept to achieve a filter Where U (z) is the z-transform of the input signal, Y (z) is design, which gives a best approximation for the linear the z-transform of the output signal, and ‘a’ and ‘b’ phase response. The proposed method can give the closest response with less number of samples (only 10) and is factors are real valued coefficients. From the above computationally simple. We have presented the filter design transfer function, we can write the time-domain along with its formulation and solving methodology. difference equation that implements the filter as, Numerical results are used to substantiate the efficiency of the proposed method. y (k ) = − a1 y (k − 1) − ....a D y (k − D ) (3) + b0 u (k ) + ... + bN u (k − N ). Index Terms—Infinite impulse response, linear programming, filter design, linear phase filters where the a and b coefficients are exactly the same as in the transfer function above, k is the time index, u (k) and I. INTRODUCTION y (k) are the current values of the input and output The design of the infinite impulse response filter is (respectively), u(k− N) was the input value N samples in often restricted to the specification of the magnitude the past, and y(k− D) was the output value D samples in response due to the fact that a linear phase response is not the past. Generally, if the frequency response of a system physically realizable. Some methods in the literature try at a frequency ω1 is given in magnitude/phase form to achieve a good approximation of the linear phase as A1∠φ1 the output amplitude will be A1 times the input response, but results in extensive and complex hardware. amplitude and the output phase will be shifted an angle Thus, the design phase involves huge computational φ1 relative to the input phase when a steady-state sine or resources as well as implementation phase requires cosine wave of frequency ω1 is applied to the system. extensive hardware. We intend to simplify the For instance, if the input to the system described above at computation in design as well as minimize the hardware time instance ‘k’ is u1(k) = Cos(kω1ts), where ts is the by just opting for a second order filter. We introduce a sampling period, then the output will be concept of using the frequency domain samples of the y1(k) = A1 Cos(kω1 ts + φ1). The input and output values desired magnitude and phase response of the filter and at any sample time can be determined in a similar using the same to obtain the filter coefficients to manner. The final sampled values in the matrix form is, implement the filter. The digital filter can be modeled as ⎡a1 ⎤ a linear system represented by ⎡ y1 (0) ⎤ ⎡ − y1 ( −1) ... − y1 ( − D ) u1 (0) ... u1 ( − N ) ⎤ ⎢ ⎥ ⎢ y ( 0) ⎥ ⎢ ⎥ ⎢. ⎥ ⎥ ⎢ − Y2 ( −1) ... − Y2 ( − D ) u2 (0) ... u2 ( − N ) ⎥ ⎢a ⎥ Ax = b (1) ⎢ 2 ⎢'. ⎥=⎢ . . . . . . ⎥*⎢ D⎥ ⎢ ⎥ ⎢ ⎥ ⎢b0 ⎥ Where, A is the matrix containing the sampled frequency ⎢. ⎥ ⎢ . . . . . . ⎥ ⎢. ⎥ ⎢ y N (0) ⎥ ⎢− yM (−1) ... − yM (− D ) uM (0) ... u M (− N )⎥ ⎢ ⎥ responses (magnitude & phase) of the given filter [1]. ⎣ ⎦ ⎣ ⎦ ⎢bN ⎥ ⎣ ⎦ The given magnitude and phase responses of the filter (4) can be sampled at a convenient sampling frequency and the number of samples can be conveniently chosen. We The above equation is similar to the system A x ≈ b ; as will now formulate the filter design problem as a linear given by (1). In the above formulation, we have set the 19 © 2010 ACEEE DOI: 01.ijcom.01.01.05
  • 2. ACEEE International Journal on Communication, Vol 1, No. 1, Jan 2010 current sample index as zero. Our aim now is to solve the Here ‘.*’ is a Matlab operation, which is an element by above problem, and get the filter coefficients given by the element multiplication. Rewriting the above constraint vector ‘x’. Standard methods use the inverse of the using w1 and w2, matrix ‘A’ which involves a heavy computation for huge matrices [3],[4],[5],[6],[7],[8]. If the inverse of the matrix 0 ≤ w 1 ≤ a. * y 219 does not exist, then Singular Value Decomposition can be used for Pseudo inverses [9]. To avoid this and make the 0 ≤ w 2 ≤ a. * y (10) procedure computationally simple, we introduce the concept of linear programming into this context. We now Now the problem, which we need to solve, is given by formulate the above as a simple optimization problem 1 1 ∑w ∑ w +∑ a with few constraints. Min 1 + 2 i II. PROBLEM FORMULATION x, w1 , w2 , a C1 i C2 i i In the above model discussed, the matrix ‘A’ contains S.t Ax + w 1 − w 2 = y; the sampled coefficients at different frequencies and the column vector ‘x’ contains the filter coefficients. The 0 ≤ w 1 ≤ a. * y , 0 ≤ w 2 ≤ a. * y and a ≥ 0; first ‘D’ values in ‘x’ represent the coefficients a1, a2, etc (11) in denominator and the next ‘N’ values give the b0, b1, b2, etc coefficients in numerator. The column vector ‘y’ is Here C1 and C2 are control parameters that can be the desired output value (both amplitude and phase) at adjusted to vary the depth of optimization of w1 and w2 specific frequencies. Now we can move on to formulate respectively. By the above formulation, we try to reduce the optimization problem to find the filter coefficients. the value of error ‘ε’ along with the maximum We first aim to find the best possible solution by permissible limit defined by ‘a’. When we attempt to introducing a factor ‘ε’, which is vector containing reduce the value of ‘a’, indirectly the maximum value allowable error. Then we try to reduce the error to get the that w1 and w2 can assume is reduced which in turn best solution by minimizing this error parameter. results in the value of ‘ε’. The problem is a chain reaction one. That mutually reduces the above said parameters. Ax + ε = y; This formulation of the optimization problem is easily (5) solvable by any simple tool, even with a spreadsheet ⇒ ε = y − Ax package like Microsoft Excel. This formulation results in reduced computational complexity and less number of Now our goal is to frame a minimization problem to frequency domain samples (eg: 8 or 16). reduce the error. Here our variables are ‘x’ and ‘ε ’. Min ∑ε i + ∑a i III. NUMERICAL RESULTS x,ε , a i i We have taken few examples [10], [11], [12] of low pass; band pass and high pass filter and attempted to S.t Ax + ε = y; 0 ≤ ε ≤ a. * y and a ≥ 0; (6) solve for the filter coefficients. The knowledge of desired frequency response (both magnitude as well as phase) is Instead of directly manipulating the error parameter, we the only requirement. We have considered a second order can define two variables w1 and w2 (essentially vectors of filter in all the cases discussed with 10 sample values. length same as ‘ε’) such that The number of samples is left to the choice of the designer with only condition that minimum number of samples be 10 for a good approximation. The proper ε = w1 − w 2 ; (7) choices of sample points play an important role in the response approximation. The sampling is done at the rate of 1000Hz or 10-3 seconds. Since we have chosen a So our equation (5) becomes, second order filter, we have N=D=2. We have solved Ax + w 1 − w 2 = y; (8) the optimization problem using Microsoft Excel and the results are verified by using Matlab. The obtained We also have a possibility to restrict the maximum value coefficients are used to plot the magnitude and phase error to a limit say a fraction of the actual response value, response and the results are verified. so that relatively there is no adverse change in the overall response. Letting that positive factor to be ‘a’ that A. Illustration 1: contains the fraction (eg: 0.01 or 1/100 th), where ‘a’ is A Low pass IIR Filter was designed for the given also a vector. Specifications as shown in figure 1. The signal is sampled and the sample values corresponding to amplitude and phase are tabulated in Table 1. The filter 0 ≤ ε ≤ a. * y (9) problem is formulated as a linear problem using the above discussed formulation and is solved using Excel and the coefficients obtained are listed in Table 2, also the response obtained using Matlab is shown in figure 1. - 20 - © 2010 ACEEE DOI: 01.ijcom.01.01.05
  • 3. ACEEE International Journal on Communication, Vol 1, No. 1, Jan 2010 TABLE 1: SAMPLED DATA :( 10 POINTS) - LPF Amplitude Phase (radians) Frequency (Hz) 1.0 -0.1 2.7400 1.0 -0.5 50.0400 1.0 -0.97 97.9600 1.0 -1.01 101.8700 0.9 -1.1 110.0000 Fig 1b. Comparison of desired and obtained Phase Responses of the low pass filter 0.6 -1.25 125.0000 The desired and obtained magnitude and phase responses 0.5 -1.25 150.0000 are comparatively studied and the figure 1 depicts the same. It is quite clear that the obtained response better 0.3 -1.25 178.2800 approximates the desired one with just 10 samples and a 0.1 -1.25 300.4200 second order IIR filter. With this method, a linear phase response can be achieved even with an IIR filter. 0.001 -1.25 450.5600 B. Illustration 2: We now discuss the second example of designing a TABLE 2: OBTAINED COEFFICIENTS - LPF band pass filter to operate in the band from 100 Hz – 200 Hz with a linear phase response as depicted in fig 2b. The a0 a1 a2 b1 b2 b3 signal is sampled and the sample values corresponding to amplitude and phase are tabulated in Table 3. The filter 1.00 -0.8663 0.3141 0.2141 0.1122 0.1209 problem is formulated as a linear problem using the above discussed formulation and is solved using Excel and the coefficients obtained are listed in Table 4 and the It can be inferred that the phase response exhibits a linear response obtained using Matlab is shown in figure 2. characteristics in our pass band which is 0 – 100 Hz. The phase is exactly ‘-1 radians’ at 100 Hz and it exactly fits a TABLE 3: SAMPLED DATA (10 POINTS) - BPF straight line with slope angle 45 degrees. We are not concerned of the phase relation in the stop band, but the Frequency Amplitude Phase (radians) (Hz) phase seems to fit a linear curve in stop band also. The small deviation in the response in the transition from the 0.0 0.5 0.0000 pass band to stop band can be rectified by increasing the number of sample points in this region. We have 0.25 0.5 50.0000 experimented with 10 samples and tried to prove that with just 10 samples, we are able to get a good 0.5 0.5 95.0000 approximation. Therefore, by suitably increasing the number form 10 to 16 or 32, the obtained response will 0.9 0.5 100.0000 exactly follow the desired one. 1.0 -0.11 150.0000 1.0 -0.63 200.0000 0.75 -1.0 250.0000 0.5 -1.46 300.0000 0.3 -1.5 350.0000 0.2 -1.5 450.5600 Fig1a. Comparison of desired and obtained Magnitude Responses of the low pass filter - 21 - © 2010 ACEEE DOI: 01.ijcom.01.01.05
  • 4. ACEEE International Journal on Communication, Vol 1, No. 1, Jan 2010 TABLE 4: OBTAINED COEFFICIENTS - BPF TABLE 5: SAMPLED DATA (10 POINTS) - HPF 220 Amplitude Phase (radians) Frequency (Hz) a0 a1 a2 b1 b2 b3 0.0 1.7 0.0000 1.00 -0.7502 0.2644 0.2617 0.2471 -0.5010 0.1 1.7 50.0000 0.2 1.7 95.0000 The comparison plots (figure 2) between the desired and 0.3 1.7 100.0000 obtained responses are presented. The phase response in the pass band, that is 100 Hz – 200 Hz, is quite linear 0.5 1.5 150.0000 with respect to frequency. This region is of interest to us and is the key feature to be noticed. 1.0 1.2 200.0000 1.0 0.9 250.0000 1.0 0.6 300.0000 1.0 0.4 350.0000 1.0 0.1 450.5600 TABLE 6: OBTAINED COEFFICIENTS - HPF a0 a1 a2 b1 b2 b3 1.00 -0.2194 0.1662 0.4975 -0.7182 0.2208 Fig 2a. Comparison of desired and obtained Magnitude Responses of the Band pass filter Fig 3a. Comparison of desired and obtained Magnitude Responses of the high pass filter Fig 2b. Comparison of desired and obtained Phase Responses of the Band pass filter C. Illustration 3: The Third example we have considered is that of a high pass filter. The procedure similar to the previous examples is followed here also. The Desired responses (figure 3), sampled data (table 5), the obtained coefficients (table 6) and obtained responses (figure 3) are presented below. The comparison plot (figure 3) clearly describes the linear phase characteristics of the designed filter. By increasing the number of samples, we can make the obtained response exactly follow the designed one. Here we have focused on the linear phase Fig 3b. Comparison of desired and obtained Phase Responses of the response, hence chosen just 10 samples. By choosing 32 high pass filter samples, the obtained response is found to be the exact copy of the desired response. IV. PARAMETERS INFLUENCING THE FILTER DESIGN Firstly, the filters designed by the above method solely depend upon the number of samples and the sample - 22 - © 2010 ACEEE DOI: 01.ijcom.01.01.05
  • 5. ACEEE International Journal on Communication, Vol 1, No. 1, Jan 2010 values. The key point is the choice of the sample values. [4]Ricardo A. Vargas and C. Sidney Burrus. On the design of The frequency points where there is a significant change IIR filters with arbitrary frequency response. In Proceedings of in the response either magnitude or the phase should be IEEE International Conference on Acoustics Speech and Signal captured by the sample values we choose. For a normal Processing , ICASSP-01, Salt Lake City, May 7-11 2001. [5] C. Sidney Burrus and Ricardo A. Vargas. The direct design filter, we have considered (low pass or high pass or band of recursive or IIR digital filters. In Proceedings of International pass); four points in the transition band, three points in Symposium on Communications, Control, and Signal the pass and stop band will do the trick. However, if the Processing, ISCCSP-08, Malta, March 2008. filter should provide specific responses, then the number [6] L. B. Jackson. Frequency-domain steiglitz-mcbride method of samples should be chosen accordingly depending upon for least-squares IIR filters design, arma modeling, and the level of approximation we require. However, as a periodogram smoothing. IEEE Signal Processing Letters, general case a filter designed by this proposed 15:498211;52, 2008. methodology with 10 samples should yield a good [7] Takao Kobayashi and Satoshi Imai. Design of IIR digital approximation of the magnitude and phase response filters with arbitrary log magnitude function by WLS techniques. IEEE Trans. on ASSP, 38(2):2478211;252, when compared to the design by existing methods. In February 1990. addition, we can achieve a linear phase Infinite Impulse [8] Atmadji W. Soewito. Least Square Digital Filter Design in Response (IIR) filter which is normally not realizable. the Frequency Domain. PhD. thesis, Rice University, ECE Hence, this method can be effectively used to achieve the Department, Houston, TX 77251, December 1990. desired response using an IIR filter. [9] G.Strang, “Linear Algebra and its Applications”,3rd Secondly, we are interested only in the phase response ed.,Thomson Brooks/Cole. in the pass band and the phase in the stop band is not of [10] S.K.Mitra, J. Kaiser, Handbook for Digital Signal our concern. Yet, the literatures reveal that the phase Processing, 1993 John Wiley and Sons, Inc. [11] T. W. Parks and C. S. Burrus. Digital Filter Design. John change in stop band will influence the overall response of Wiley & Sons, New York, 1987 the filter response. These gives a cushion of changing the [12] L. R. Rabiner and B. Gold. Theory and Application of phase in the stop band and further achieve better results. Digital Signal Processing. Prentice-Hall, Engle-wood Cliffs, NJ, Therefore, we can introduce few more variables that are 1975. phase values in stop band in our formulation, even provide some lower and upper bounds, and further refine the formulation. V. CONCLUSION AND FUTURE SCOPE The formulation of the IIR filter design problem, with both the magnitude and phase specifications, is easily done using this proposed technique of linear programming. This application of linear programming for design of linear phase IIR filters is a novel methodology. This requires only very low computational resources in spite of a very good approximation of the linear phase response. The results obtained by the proposed method are quite encouraging. The results can be further improved by proper choice of the number of samples and position of sample points. These results presented above guarantee the efficiency of the proposed method. The literature shows that by slightly tuning the phase in the stop band (which is of no significance) the phase response in the pass band can be changed accordingly. These phenomena can be incorporated in the proposed technique to achieve best results. The combination is implemented and the performance is being investigated. It is very much conducive that the proposed method may be effectively adopted by the design engineers. The proposed method proves to be a simple method when compared to the counterparts for providing better results. REFERENCES [1] G.Berchin,”Precise Filter Design” IEEE Signal Processing Magazine, January 2007. [2] R. Lyons, Understanding Digital Signal Processing, 2nd ed., Upper Saddle River, NJ:Prentice-Hall, pp. 232–240, 2004. [3] C. Sidney Burrus,”Direct Frequency Domain IIR Filter Design Methods”Version 1.1: Jun 24, 2008 - 23 - © 2010 ACEEE DOI: 01.ijcom.01.01.05