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Linear Phase Filters

Module by: Douglas L. Jones. E-mail the author

In general, for πωπ ω Hω=|Hω|e(iθω) H ω H ω θ ω Strictly speaking, we say Hω H ω is linear phase if Hω=|Hω|e(iωK)e(i θ 0 ) H ω H ω ω K θ 0 Why is this important? A linear phase response gives the same time delay for ALL frequencies! (Remember the shift theorem.) This is very desirable in many applications, particularly when the appearance of the time-domain waveform is of interest, such as in an oscilloscope. (see Figure 1)

Figure 1
Figure 1 (imag001.png)

Restrictions on h(n) to get linear phase

Hω=h=0M1hne(iωn)=h0+h1e(iω)+h2e(i2ω)++hM1e(iω(M1))=e(iωM12)(h0eiωM12++hM1e(iωM12))=e(iωM12)((h0+hM1)cosM12ω+(h1+hM2)cosM32ω++i(h0sinM12ω+)) H ω h 0 M 1 h n ω n h 0 h 1 ω h 2 2 ω h M 1 ω M 1 ω M 1 2 h 0 ω M 1 2 h M 1 ω M 1 2 ω M 1 2 h 0 h M 1 M 1 2 ω h 1 h M 2 M 3 2 ω h 0 h M 1 M 1 2 ω
(1)
For linear phase, we require the right side of Equation 1 to be e(i θ 0 )(real,positive function of ω) θ 0 (real,positive function of ω) . For θ 0 =0 θ 0 0 , we thus require h0+hM1=real number h 0 h M 1 real number h0hM1=pure imaginary number h 0 h M 1 pure imaginary number h1+hM2=pure real number h 1 h M 2 pure real number h1hM2=pure imaginary number h 1 h M 2 pure imaginary number Thus hk= h * M1k h k h * M 1 k is a necessary condition for the right side of Equation 1 to be real valued, for θ 0 =0 θ 0 0 .

For θ 0 =π2 θ 0 2 , or e(i θ 0 )=i θ 0 , we require h0+hM1=pure imaginary h 0 h M 1 pure imaginary h0hM1=pure real number h 0 h M 1 pure real number hk=( h * M1k) h k h * M 1 k

Usually, one is interested in filters with real-valued coefficients, or see Figure 2 and Figure 3.

Figure 2: θ 0 =0 θ 0 0 (Symmetric Filters). hk=hM1k hk h M1 k.
Figure 2 (imag002.png)
Figure 3: θ 0 =π2 θ 0 2 (Anti-Symmetric Filters). hk=hM1k hk h M1 k.
Figure 3 (imag003.png)
Filter design techniques are usually slightly different for each of these four different filter types. We will study the most common case, symmetric-odd length, in detail, and often leave the others for homework or tests or for when one encounters them in practice. Even-symmetric filters are often used; the anti-symmetric filters are rarely used in practice, except for special classes of filters, like differentiators or Hilbert transformers, in which the desired response is anti-symmetric.

So far, we have satisfied the condition that Hω=Aωe(i θ 0 )e(iωM12) H ω A ω θ 0 ω M 1 2 where Aω A ω is real-valued. However, we have not assured that Aω A ω is non-negative. In general, this makes the design techniques much more difficult, so most FIR filter design methods actually design filters with Generalized Linear Phase: Hω=Aωe(iωM12) H ω A ω ω M 1 2 , where Aω A ω is real-valued, but possible negative. Aω A ω is called the amplitude of the frequency response.

excuse:

Aω A ω usually goes negative only in the stopband, and the stopband phase response is generally unimportant.

note:

|Hω|=±Aω=Aωe(iπ12(1signAω)) H ω ± A ω A ω 1 2 1 sign A ω where signx={1  if  x>0-1  if  x<0 sign x 1 x 0 -1 x 0

Example 1

Lowpass Filter

Figure 4
Desired |H(ω)|
Desired |H(ω)| (imag004.png)
Figure 5: The slope of each line is M12 M1 2.
Desired ∠H(ω)
Desired ∠H(ω) (imag005.png)
Figure 6: AωAω goes negative.
Actual |H(ω)|
Actual |H(ω)| (imag006.png)
Figure 7: 2π 2 phase jumps due to periodicity of phase. π phase jumps due to sign change in Aω A ω .
Actual ∠H(ω)
Actual ∠H(ω) (imag007.png)
Time-delay introduces generalized linear phase.
note:
For odd-length FIR filters, a linear-phase design procedure is equivalent to a zero-phase design procedure followed by an M12 M 1 2 -sample delay of the impulse response. For even-length filters, the delay is non-integer, and the linear phase must be incorporated directly in the desired response!

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