Outline. FIR Filter Characteristics Linear Phase Windowing Method Frequency Sampling Method Equiripple Optimal Method Design Examples

Transcription

1 FIR Filter Design

2 Outline FIR Filter Characteristics Linear Phase Windowing Method Frequency Sampling Method Equiripple Optimal Method Design Examples

3 FIR Filter Characteristics FIR difference equation The output only depends on the previous input samples FIR transfer function All poles are at the origin and also called all zeros filter = = = = ) ( 1 0 ) ( ) ( ) ( N n n N N N n n z n h z z n h z H = = = = M k M k k k n x k h k n x b n y 0 0 ) ( ) ( ) ( ) (

4 Advantages of FIR Filter Stability is never an issue All poles at origin => must within unit circle Linear Phase FIR Filter is very easy to design A lot of applications require linear phase: Digital Communication System Audio Signal Processing Image Processing

5 Properties of Linear Phase Phase delay of a system is defined as T p = θ ( ω) ω A linear phase system has phase response of the following relationships: θ(ω) = -αω or θ(ω) = β - αω where α is a constant phase delay and β =±π

6 Linear Phase Filters: Type 1 and 2 It can be shown that for θ(ω) = -αω the filter impulse response must have positive symmetry: h( n) = h( N n 1), n = 0,1,...,( N 1) / 2 n = 0,1,...,( N / 2) 1 ( N odd) ( N even) Type 1 linear phase FIR filter: N is odd number and α = (N-1)/2 is an integer. Type 2 linear-phase filter N is even number and α = (N-1)/2 is not an integer

7 Linear Phase Filters: Type 3 and 4 It can be shown that for θ(ω) = β - αω the filter impulse response must have negative symmetry: h(n) = -h(n-n-1) α = (N-1)/2 β = ±π/2 Type 3 linear phase FIR filter: N is odd number Type 4 linear-phase filter N is even number

8 Example 1 An FIR digital filter has impulse response, h(n), defined over the interval 0 n N - 1. Show that if N = 7 and h(n) satisfies the positive symmetry condition h(n) = h(n - n - 1) the filter has a linear phase characteristic.

9 Types of linear phase FIR filters

10 Impulse response comparison of the 4 linear phase filters

11 FIR Filter Coefficient Design Method Given filter specification, Determine the impulse response h(n) Popular FIR Design Methods: - Window Method - Frequency Sampling Method - Optimal Filter Design Matlab functions for FIR filter design fir1( ), fir2( ), remez( )

12 Window Method : ideal (desired) frequency response where w(n) is a window sequence H d (ω) = π π ω ω ω π d e H n h n j d d ) ( 2 1 ) ( ) ( )* ( ) ( ) ( ) ( ) ( ω ω ω W H H n w n h n h d d = =

13 Ideal Lowpass Filter rule) Hopital's 0(using L' 2 0,, ) sin( ) ( = = = = = = n f n n n n f d e d e n h c c c c n j n j D ω ω ω π ω π π π ω π π ω

14 Ideal Impulse Responses for Standard Frequency Selective filters

15 Gibb s Effect

16 Comment Note that ripple in the neighbourhood of 10% overshoot in the passband represents only 0.7 db, but that 10% ripple in the stop represents 20 db.

17 We would probably find this filter in adequate to meet filtering requirements in the stop band. How do we improve performance? We have commented that the poor ripple characteristics are due to the bumpy function used in the convolution To reduce the ripple, we need a truncation function whose Fourier Transform is not as bumpy (i.e. smoother) The bumpy behaviour of the transform is related to the discontinuity of the truncation function.

18 Window Design Example

19 Important Features of Common Window Functions

20 Rectangular Window

21 Triangular Window

22 Hamming Window

23 Blackman Window

24 Kaiser Window

25 Window functions comparison (a) Rectangular (b) Hamming (c) Blackman

26 Comments Windows with smoother behavious in the timedomain are smoother in the frequency domain Windows that are smoother in the time-domain tend to have narrower time duration Windows that are smoother in the domain tend to have wider bandwidths When convolving with a wider function, the resultant function will have wider transitions

27 Summary of the Window method Step 1: Specify the ideal or desired frequency response of filter, H D (ω). Step 2: Obtain the impulse response h D (n) by evaluation the inverse Fourier transform. Step 3: Select a window function that satisfies the specifications Determine the filter order based on f Step4. Calculate the filter coefficients, h(n), by h(n) = h D (n)w(n)

28 Example Obtain the coefficients of an FIR lowpass to meet the specifications given below using the window method. Passband edge frequency Transition width Stopband attenuation Sampling frequency 1.5 khz 0.5 khz >50 db 8 khz

29 Solution The unit sample response of an ideal lowpass filter is given by h d ( n) = ωc / π, sin( ωcn) πn, n n = 0 0

30 Solution Hamming window is used for simplicity. To find the normalized transition width as f = transition width sampling frequency The minimum filter length of the FIR filter using Hamming window is determined by = = 3.3 N = f = =

31 Solution The filter coefficients can be obtained by where w( n) h d ( n) w( n) 26 n 2πn = cos N The cutoff frequency f c transition band: 26 is on the center of the f c = f F p s + f 2 = =

32 Matlab Solution fc=2*( )/8; % calculate the cutoff frequency h=fir1(53,fc); % generate the filter coefficients [H, F]=freqz(h,1,100,8000); magh=abs(h); db=20*log10(magh); phase=unwrap(angle(h))*(180/pi); subplot(2,1,1); plot(f, db, 'b'); grid; axis([0,4000,-90,10]); xlabel('frequency in Hz'); ylabel ('Magnitude in db'); subplot(2,1,2); plot(f,phase, 'b'); grid; xlabel('frequency in Hz'); ylabel ('Phase in degrees');

33 Matlab Results

34 Advantages and Disadvantages of the window method Advantage: Simplicity: it is simple to apply and simple to understand. Disadvantages: Lack of flexibility. Both the peak passband and stopband ripples are approximately equal. The passband and stopband edge frequencies cannot be precissely specified.

35 Frequency Sampling Method H d (ω) = H d (ω) with linear phase H(k) = H d (ω) ω = (2π/N)k IDFT h( n) = 1 N N 1 k = 0 H ( k) e j (2π / N ) nk

36 Example A requirement exists for a lowpass FIR satisfying the following specifications: Passband Sampling frequency Filter length khz 18kHz Obtain the filter coefficients using the frequency sampling method.

37 Solution

38 Parks-McClellan Method Often called the Remez exchange method. This method designs an optimal linear phase filter directly from the design specifications. This is the standard method for FIR filter design. In matlab, this method is available as remez()

39 Approximation Errors From the theory of the Fourier series, the rectangular window design method (truncation of the impulse response) gives the best mean square approximation to a desired frequency response for a given filter length M.

40 Minimax Design However simple truncation leads to adverse behaviour near discontinuities and in the stopband. Better filters generally result from minimization of the maximum error (L ) or a frequency weighed error criterion.

41 Design Procedures To specify the frequency response To estimate the filter order: Equations are available to estimate the filter order Matlab function: remezord() An optimization program is run to compute the filter coefficients. Matlab functon: remez() to check the result meets the requirements or not Matlab functon: freqz()

42 Example: Lowpass Filter Design a digital FIR lowpass filter to meet the specifications given below using the equiripple method. Sampling frequency Passband Stopband Passband ripple Stopband attenuation 16 khz Hz Hz less than 0.1 db > 40 db

43 Solution The passband ripple to be less than 0.1 db, we have R p = 0.1 = 20log 10 (1+ δ p ) δ p = /20 1 = The minimum stopband attenuation at least 40 db, A p = 40 = -20log 10 (δ s ) δ s = 10-40/20 = 0.01

44 Matlab for Filter Order Estimation F=[ ]; M=[1 0]; dev=[ ]; Fs=16000; [N Fo Mo W]=remezord(F, M, dev, Fs) Results: N = 154 F o = {0, , , 1.000}, M o = {1, 1, 0, 0} W = {1.0, 1.0}.

45 Matlab codes for filter design h=remez(n, Fo, Mo, W); [H, F]=freqz(h, 1, [0:10:8000], Fs); magh=abs(h); db=20*log10(magh); phase=unwrap(angle(h))*(180/pi); subplot(2,1,1); plot(f, db, 'b'); grid; xlabel('frequency in Hz'); ylabel ('Magnitude in db'); axis([0,8000,-110,10]); subplot(2,1,2); plot(f,phase, 'b'); grid; xlabel('frequency in Hz'); ylabel ('Phase in degrees');

46 Frequency Response of the Designed FIR Filter

47 Frequency Response between 0 to 400 Hz The plots show that the designed filter could meet the passband ripple of

48 Example: Band-Reject Filter This example is to design a 1 KHz band-reject filter with a 40 Hz bandwidth (980 to 1020 Hz) and 8 KHz sampling rate. The over all specifications are: Sampling frequency Passband Stopband Passband ripple Stopband attenuation 8 khz Hz and Hz Hz less than 1 db > 60 db

49 Solution The passband ripple to be less than 1 db, we have R p = 1 = 20log 10 (1+ δ p ) δ p = 10 1/20 1 = 0.12 The minimum stopband attenuation at least 60 db, A p = 60 = -20log 10 (δ s ) δ s = 10-60/20 = 0.001

50 Matlab for Filter Order Estimation F = [ ]; M = [1 0 1]; dev = [ ]; Fs = 8000; [N Fo Mo W] = remezord(f, M, dev, Fs); Results: N = 67 F o = {0, , , , , 1.000}, M o = {1, 1, 0, 0, 1, 1} W = {1, 120, 1.0}.

51 Matlab codes for filter design h = remez(n, Fo, Mo, W); [H, F]=freqz(h, 1, [500:10:1500], Fs); magh=abs(h); db=20*log10(magh); phase=unwrap(angle(h))*(180/pi); subplot(2,1,1); plot(f, db, 'b'); grid; xlabel('frequency in Hz'); ylabel ('Magnitude in db'); subplot(2,1,2); plot(f,phase, 'b'); grid xlabel('frequency in Hz'); ylabel ('Phase in degrees');

52 Frequency Response of the Designed FIR Filter The plots shows the initial design with the estimated order N=69 failed to converge to an equiripple solution.

53 Frequency Response with N = 80

54 Finite Wordlength Effects of FIR Filter ADC noise Coefficient quantization errors Roundoff errors from quantizing results of arithmetic operations Arithmetic overflow

55 Example Determine the effects of quantizing, by rounding, the coefficients of the following filter to 8 bits: stopband attenuation > 90 db passband ripple < db passband edge frequency khz stopband edge frequency khz sampling frequency 20 khz number of coefficients 45

56 Solution Use the design program optimal.c with the following input: number of filter coefficients 45 band edge frequencies 0, , , 0.5 weight 1, 7.28 The coefficients of the filter, before and after rounding to 8 bits, are listed in Table The corresponding frequency responses are given in Figure It is seen that, after quantization, the minimum stopband attenuation is 36 db, a degradation of more than 58 db. Clearly, more than 8 bits of resolution is required for the coefficients in this particular example.

57 Filter coefficients before and after quantization to 8 bits.

58 (a) Effects of coefficient quantization. (b) Passbad., unquantized;, quantized.