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Linear-Phase FIR Filters: Amplitude Formulas

Module by: Ivan Selesnick. E-mail the author

Summary: (Blank Abstract)

SUMMARY: AMPLITUDE FORMULAS

Table 1
Type θω θ ω Aω A ω
I (Mω) M ω hM+2n=0M1hncos(Mn)ω h M 2 n 0 M 1 h n M n ω
II (Mω) M ω 2n=0N21hncos(Mn)ω 2 n 0 N 2 1 h n M n ω
III (Mω)+π2 M ω 2 2n=0M1hnsin(Mn)ω 2 n 0 M 1 h n M n ω
IV (Mω)+π2 M ω 2 2n=0N21hnsin(Mn)ω 2 n 0 N 2 1 h n M n ω

where M=N12 M N 1 2

AMPLITUDE RESPONSE CHARACTERISTICS

To analyze or design linear-phase FIR filters, we need to know the characteristics of the amplitude response Aω A ω .

Table 2
Type Properties
I Aω A ω is even about ω=0 ω 0 Aω=Aω A ω A ω
Aω A ω is even about ω=π ω Aπ+ω=Aπω A ω A ω
Aω A ω is periodic with 2π 2 Aω+2π=Aω A ω 2 A ω
II Aω A ω is even about ω=0 ω 0 Aω=Aω A ω A ω
Aω A ω is odd about ω=π ω Aπ+ω=Aπω A ω A ω
Aω A ω is periodic with 4π 4 Aω+4π=Aω A ω 4 A ω
III Aω A ω is odd about ω=0 ω 0 Aω=Aω A ω A ω
Aω A ω is odd about ω=π ω Aπ+ω=Aπω A ω A ω
Aω A ω is periodic with 2π 2 Aω+2π=Aω A ω 2 A ω
IV Aω A ω is odd about ω=0 ω 0 Aω=Aω A ω A ω
Aω A ω is even about ω=π ω Aπ+ω=Aπω A ω A ω
Aω A ω is periodic with 4π 4 Aω+4π=Aω A ω 4 A ω
Figure 1
Figure 1 (fourAmps.png)

EVALUATING THE AMPLITUDE RESPONSE

The frequency response H f ω H f ω of an FIR filter can be evaluated at LL equally spaced frequencies between 0 and π using the DFT. Consider a causal FIR filter with an impulse response hn h n of length-NN, with NL N L . Samples of the frequency response of the filter can be written as H2πLk=n=0N1hne(i)2πLnk H 2 L k n 0 N 1 h n 2 L n k Define the LL-point signal gn 0nL1 g n 0 n L 1 as gn={hn  if  0nN10  if  NnL1 g n h n 0 n N 1 0 N n L 1 Then H2πLk=Gk= DFT L gn H 2 L k G k DFT L g n where Gk G k is the LL-point DFT of gn g n .

Types I and II

Suppose the FIR filter hn h n is either a Type I or a Type II FIR filter. Then we have from above H f ω=Aωe(i)Mω H f ω A ω M ω or Aω= H f ωeiMω A ω H f ω M ω Samples of the real-valued amplitude Aω A ω can be obtained from samples of the function H f ω H f ω as: A2πLk=H2πLkeiM2πLk=Gk W L M k A 2 L k H 2 L k M 2 L k G k W L M k Therefore, the samples of the real-valued amplitude function can be obtained by zero-padding hn h n , taking the DFT, and multiplying by the complex exponential. This can be written as:

A2πLk= DFT L hn 0 L - N W L M k A 2 L k DFT L h n 0 L - N W L M k
(1)

Types III and IV

For Type III and Type IV FIR filters, we have H f ω=ie(i)MωAω H f ω M ω A ω or Aω=(i) H f ωeiMω A ω H f ω M ω Therefore, samples of the real-valued amplitude Aω A ω can be obtained from samples of the function H f ω H f ω as: A2πLk=(i)H2πLkeiM2πLk=(i)Gk W L M k A 2 L k H 2 L k M 2 L k G k W L M k Therefore, the samples of the real-valued amplitude function can be obtained by zero-padding hn h n , taking the DFT, and multiplying by the complex exponential.

A2πLk=(i) DFT L hn 0 L - N W L M k A 2 L k DFT L h n 0 L - N W L M k
(2)

Example 1: EVALUATING THE AMP RESP (TYPE I)

In this example, the filter is a Type I FIR filter of length 7. An accurate plot of Aω A ω can be obtained with zero padding.

Figure 2
Figure 2 (type1.png)

The following Matlab code fragment for the plot of Aω A ω for a Type I FIR filter.


	  h = [3 4 5 6 5 4 3]/30;
	  N = 7;
	  M = (N-1)/2;
	  L = 512;
	  H = fft([h zeros(1,L-N)]);
	  k = 0:L-1;
	  W = exp(j*2*pi/L);
	  A = H .* W.^(M*k);
	  A = real(A);

	  figure(1)
	  w = [0:L-1]*2*pi/(L-1);
	  subplot(2,1,1)
	  plot(w/pi,abs(H))
	  ylabel('|H(\omega)| = |A(\omega)|')
	  xlabel('\omega/\pi')
	  subplot(2,1,2)
	  plot(w/pi,A)
	  ylabel('A(\omega)')
	  xlabel('\omega/\pi')
	  print -deps type1
	

The command A = real(A) removes the imaginary part which is equal to zero to within computer precision. Without this command, Matlab takes A to be a complex vector and the following plot command will not be right.

Observe the symmetry of Aω A ω due to hn h n being real-valued. Because of this symmetry, Aω A ω is usually plotted for 0ωπ 0 ω only.

Example 2: EVALUATING THE AMP RESP (TYPE II)

Figure 3
Figure 3 (type2.png)

The following Matlab code fragment produces a plot of Aω A ω for a Type II FIR filter.


	  h = [3 5 6 7 7 6 5 3]/42;
	  N = 8;
	  M = (N-1)/2;
	  L = 512;
	  H = fft([h zeros(1,L-N)]);
	  k = 0:L-1;
	  W = exp(j*2*pi/L);
	  A = H .* W.^(M*k);
	  A = real(A);

	  figure(1)
	  w = [0:L-1]*2*pi/(L-1);
	  subplot(2,1,1)
	  plot(w/pi,abs(H))
	  ylabel('|H(\omega)| = |A(\omega)|')
	  xlabel('\omega/\pi')
	  subplot(2,1,2)
	  plot(w/pi,A)
	  ylabel('A(\omega)')
	  xlabel('\omega/\pi')
	  print -deps type2
	

The imaginary part of the amplitude is zero. Notice that Aπ=0 A 0 . In fact this will always be the case for a Type II FIR filter.

An exercise for the student: Describe how to obtain samples of Aω A ω for Type III and Type IV FIR filters. Modify the Matlab code above for these types. Do you notice that Aω=0 A ω 0 always for special values of ωω?

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