5.4 Linear-phase fir filters: amplitude formulas

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Summary: amplitude formulas

Type $()$ $A()$
I $-(M)$ $h(M)+2\sum_{n=0}^{M-1} h(n)\cos ((M-n))$
II $-(M)$ $2\sum_{n=0}^{\frac{N}{2}-1} h(n)\cos ((M-n))$
III $-(M)+\frac{\pi }{2}$ $2\sum_{n=0}^{M-1} h(n)\sin ((M-n))$
IV $-(M)+\frac{\pi }{2}$ $2\sum_{n=0}^{\frac{N}{2}-1} h(n)\sin ((M-n))$

where $M=\frac{N-1}{2}$

Amplitude response characteristics

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

Type Properties
I $A()$ is even about $=0$ $A()=A(-)$
$A()$ is even about $=\pi$ $A(\pi +)=A(\pi -)$
$A()$ is periodic with $2\pi$ $A(+2\pi )=A()$
II $A()$ is even about $=0$ $A()=A(-)$
$A()$ is odd about $=\pi$ $A(\pi +)=-A(\pi -)$
$A()$ is periodic with $4\pi$ $A(+4\pi )=A()$
III $A()$ is odd about $=0$ $A()=-A(-)$
$A()$ is odd about $=\pi$ $A(\pi +)=-A(\pi -)$
$A()$ is periodic with $2\pi$ $A(+2\pi )=A()$
IV $A()$ is odd about $=0$ $A()=-A(-)$
$A()$ is even about $=\pi$ $A(\pi +)=A(\pi -)$
$A()$ is periodic with $4\pi$ $A(+4\pi )=A()$

Evaluating the amplitude response

The frequency response ${H}^{f}()$ of an FIR filter can be evaluated at $L$ equally spaced frequencies between 0 and $\pi$ using the DFT. Consider a causal FIR filter with an impulse response $h(n)$ of length- $N$ , with $N\le L$ . Samples of the frequency response of the filter can be written as $H(\frac{2\pi }{L}k)=\sum_{n=0}^{N-1} h(n)e^{-i\frac{2\pi }{L}nk}$ Define the $L$ -point signal $\{g(n)\colon 0\le n\le L-1\}$ as $g(n)=\begin{cases}h(n) & \text{if 0\le n\le N-1}\\ 0 & \text{if N\le n\le L-1}\end{cases}$ Then $H(\frac{2\pi }{L}k)=G(k)={\mathrm{DFT}}_{L}(g(n))$ where $G(k)$ is the $L$ -point DFT of $g(n)$ .

Types i and ii

Suppose the FIR filter $h(n)$ is either a Type I or a Type II FIR filter. Then we have from above ${H}^{f}()=A()e^{-iM}$ or $A()={H}^{f}()e^{iM}$ Samples of the real-valued amplitude $A()$ can be obtained from samples of the function ${H}^{f}()$ as: $A(\frac{2\pi }{L}k)=H(\frac{2\pi }{L}k)e^{iM\frac{2\pi }{L}k}=G(k){W}_{L}^{Mk}$ Therefore, the samples of the real-valued amplitude function can be obtained by zero-padding $h(n)$ , taking the DFT, and multiplying by the complex exponential. This can be written as:

$A(\frac{2\pi }{L}k)={\mathrm{DFT}}_{L}()$
h n 0 L - N
W L M k

Types iii and iv

For Type III and Type IV FIR filters, we have ${H}^{f}()=ie^{-iM}A()$ or $A()=-i{H}^{f}()e^{iM}$ Therefore, samples of the real-valued amplitude $A()$ can be obtained from samples of the function ${H}^{f}()$ as: $A(\frac{2\pi }{L}k)=-iH(\frac{2\pi }{L}k)e^{iM\frac{2\pi }{L}k}=-iG(k){W}_{L}^{Mk}$ Therefore, the samples of the real-valued amplitude function can be obtained by zero-padding $h(n)$ , taking the DFT, and multiplying by the complex exponential.

$A(\frac{2\pi }{L}k)=-i{\mathrm{DFT}}_{L}()$
h n 0 L - N
W L M k

Evaluating the amp resp (type i)

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

The following Matlab code fragment for the plot of $A()$ 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 computerprecision. Without this command, Matlab takes A to be a complex vector and the following plot command will not be right.

Observe the symmetry of $A()$ due to $h(n)$ being real-valued. Because of this symmetry, $A()$ is usually plotted for $0\le \le \pi$ only.

Evaluating the amp resp (type ii)

The following Matlab code fragment produces a plot of $A()$ 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 $A(\pi )=0$ . 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 $A()$ for Type III and Type IV FIR filters. Modify the Matlab code above for these types. Do you notice that $A()=0$ always for special values of  ?

Modules for further study

• Zero Locations of Linear-Phase Filters
• Design of Linear-Phase FIR Filters by Interpolation
• Linear-Phase FIR Filter Design by Least Squares

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