1 / 55

Chapter 5

Chapter 5. FIR Filter Design. Objectives. Describe the general approach to filter design and the design equation for FIR filters. Define phase distortion and show how it affects signals. Demonstrate how linear phase response eliminates phase distortion.

ambrose
Download Presentation

Chapter 5

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 5 FIR Filter Design

  2. Objectives • Describe the general approach to filter design and the design equation for FIR filters. • Define phase distortion and show how it affects signals. • Demonstrate how linear phase response eliminates phase distortion. • Demonstrate the symmetry condition on FIR impulse response that results in linear phase. • Derive the impulse responses of ideal low-pass and high-pass linear phase filters. • Describe and demonstrate the window design method for linear phase FIR filters. • Demonstrate the design of low-pass, high-pass, band-pass, and band-reject FIR filters. • Describe and demonstrate the sampling method of linear phase FIR filter design. • Demonstrate the tools in MATLAB for optimized FIR filter design using the Parks-McClellan algorithm.

  3. Design of Frequency Selective FIR Filters • For frequency selective filters the design problem is to find the impulse response, h[n], for a desired frequency response, H(Ω) • In principle, this can be done using the inverse discrete-time Fourier transform (DTFT):

  4. Phase Distortion • Phase distortion results from a variable time delay (phase delay) for different frequency components of a signal. • If the phase response of a filter is a linear function of frequency, then the phase delay is constant for all frequencies and no phase distortion occurs.

  5. Linear Phase An input signal (blue sinusoid) processed in the pass-band of a filter (output red sinusoid) will, in general, experience a phase change (θ1 to θ2) and a time delay (t1 to t2).

  6. Linear Phase (Analog Filter) The output amplitude equals the input amplitude in the pass-band This is the phase change condition for constant phase delay independent of frequency: the phase change is a linear function of frequency

  7. Linear Phase (Digital Filter) The output amplitude equals the input amplitude in the pass-band Δn is a constant if Δt is constant The condition for Δn being constant is that Δθ be a linear function of Ω A constant group delay is a signature of linear phase

  8. Constant Phase vs. Linear Phase >> t=0:.04/1000:.04; % The small increment of t approximates an analog signal >> f0=50; >> term1=(2/pi)*sin(2*pi*(f0)*t); >> term2=(2/(3*pi))*sin(2*pi*(3*f0)*t); >> term3=(2/(5*pi))*sin(2*pi*(5*f0)*t); >> term4=(2/(7*pi))*sin(2*pi*(7*f0)*t); >> s=term1+term2+term3+term4; >> plot(t*1000,s);title('4 Term Square Wave');xlabel('milliseconds'); Zero phase change for each sinusoid

  9. Constant Phase vs. Linear Phase >> term1=(2/pi)*sin(2*pi*(f0)*t+pi/6); >> term2=(2/(3*pi))*sin(2*pi*(3*f0)*t+pi/6); >> term3=(2/(5*pi))*sin(2*pi*(5*f0)*t+pi/6); >> term4=(2/(7*pi))*sin(2*pi*(7*f0)*t+pi/6); >> s=term1+term2+term3+term4; >> plot(t*1000,s); >> title('4 Term Square Wave with Constant 30 deg Phase');xlabel('milliseconds'); A constant phase change of +30 degrees for each sinusoid results in phase distortion.

  10. Constant Phase vs. Linear Phase >> term1=(2/pi)*sin(2*pi*(f0)*t-.01*pi*(f0)); >> term2=(2/(3*pi))*sin(2*pi*(3*f0)*t-.01*pi*(3*f0)); >> term3=(2/(5*pi))*sin(2*pi*(5*f0)*t-.01*pi*(5*f0)); >> term4=(2/(7*pi))*sin(2*pi*(7*f0)*t-.01*pi*(7*f0)); >> s=term1+term2+term3+term4; >> plot(t*1000,s);title('4 Term Square Wave with 5ms Phase Delay');xlabel('milliseconds') Computed with a constant phase (time) delay of 5 ms and linear phase. Linear phase results in no phase distortion.

  11. Sufficient Condition for Linear Phase • If a FIR filter consists of an odd number of coefficients, M + 1, where M is even, and is symmetrical about the M/2 term, the filter has a linear phase response of Δθ = −(M/2)Ω. The filter will have a group delay of M/2 • This is termed a “type I” filter (odd length and positive symmetry). Other types have different permutations of length and symmetry. This type is the most common and easiest to design.

  12. Sufficient Condition for Linear Phase • Odd number of impulse response values • Symmetry point: n=2 • Equal values of the impulse response about the symmetry point • Group delay = 2

  13. Sufficient Condition for Linear Phase >> h=[-1,1,2,1,-1]; >> fvtool(h,1)

  14. Sufficient Condition for Linear PhaseThe Running Average Filter >>n=0:6; >> raf7=(1/7)*ones(1,7); >> stem(n,raf7) >> xlabel('n');ylabel('h[n]'); >> title('Impulse Response of a 7-Point Running Average Filter') >> axis([-1 7 0 .16]) >> fvtool(raf7,1)

  15. The Ideal Low-Pass Filter

  16. Impulse Response of the Ideal LP Filter

  17. Impulse Response of the Ideal LP Filter • The impulse response of the ideal low pass filter is: • It is easily seen that this impulse response has linear phase because it has the symmetry property of h[-n] = h[n] • As given, however, this impulse response is not computable: it is infinite and anti-causal • A finite set of M+1 h[n] values can be delayed by M/2 samples to create a finite and causal h[n] • Creating the finite and causal impulse response will affect some filter properties, but not its linear phase response

  18. Example:21 Coefficient LP Filter with Ω0 = π/4 >> n=0:20; >> omega=pi/4; % This is the cut-off frequency >> h=(omega/pi)*sinc(omega*(n-10)/pi); % Note the 10 step shift >> stem(n,h) >> title('Sample-Shifted LP Impulse Response') >> xlabel('n') >> ylabel('h[n]') >> fvtool(h,1) Note the linear phase properties of the impulse response

  19. Example:201 Coefficient LP Filter with Ω0 = π/4 >> n=0:200; % This sets the order of the filter where length(n)=201 >> omega=pi/4; >> h=(omega/pi)*sinc((n-100)*omega/pi); %Note the sample shift of 100 samples >> fvtool(h,1) Higher order = sharper transition Side-lobe “ripple” (Gibbs phenomenon) due to abrupt truncation of the impulse response

  20. Tapering Windows • Used to taper the abrupt truncation of the impulse response towards zero • Example: Hamming window

  21. >> n=0:20; >> omega=pi/4; >> h=(omega/pi)*sinc((n-10)*omega/pi); >> w=0.54+0.46*cos(2*pi*(n-10)/20); % This is the Hamming window. Note that % N-1 = 20 in this case >> hw=h.*w; % Note the use of “ .* “ to multiply h and w % sample-by-sample >> stem(n,w) >> title('Hamming Window') >> figure,stem(n,h,'ko') >> hold >> stem(n,hw,'bd') >> title('Comparison of Rectangular and Hamming Window Impulse Response') >> legend('Rectangular','Hamming') >> fvtool(hw,1) Effect of Tapering Windows

  22. Effect of Tapering Windows

  23. Effect of Tapering Windows

  24. The Ideal High Pass Filter

  25. The Ideal High Pass Filter

  26. Design Example:201 Coefficient HP with Ω0 = 3π/4 >> n=0:200; >> omega=3*pi/4; >> h=sinc(n-100)-(omega/pi)*sinc(omega*(n-100)/pi); % Both terms shifted by 100 % samples >> hw=h.*blackman(201)'; >> fvtool(hw,1) The Blackman tapering window provides greater suppression of stop-band side-lobes

  27. Band-Stop and Band-Pass Filters hbr = hLP + hHP hbp = hLPH - hLPL

  28. Band-Stop and Band-Pass Examples >> n=0:200; >> fs=2000; >> omegaL=2*pi*400/fs; >> omegaH=2*pi*600/fs; >> hLPL=(omegaL/pi)*sinc(omegaL*(n-100)/pi).*hamming(201)'; >> hLPH=(omegaH/pi)*sinc(omegaH*(n-100)/pi).*hamming(201)'; >> hHP=(sinc(n-100)-(omegaH/pi)*sinc(omegaH*(n-100)/pi)).*hamming(201)'; >> h_bandpass=hLPH-hLPL; >> h_bandreject=hLPL+hHP; >> fvtool(h_bandpass,1) >> fvtool(h_bandreject,1)

  29. Effect of Group Delay(Order 200 Filter) >> m=1:250; % The sinusoid will have a length of 250 samples >> f=2*pi*100/fs; >> x=sin(f*m); >> y=filter(hLPL,1,x); % The filter command executes the filter on the signal x >> subplot(2,1,1),plot(x),title('Input 100 Hz Sinusoid'),axis([0,250,-2,2]); >> subplot(2,1,2),plot(y),title('Low-Pass Filter Output'),axis([0,250,-2,2]);

  30. Sampling Method of FIR DesignBasic Theory • The sampling method is based on the principle that the DFT is a sample of the DTFT • To see this, take the DFT of the frequency response of a filter: >> n=0:10; % Design a low pass filter by the window method >> omega=pi/4; >> h1=(omega/pi)*sinc(omega*(n-5)/pi); >> dtft_demo(h1,0,2*pi,512); % Display the DTFT of the filter >> hold >> [H1,f]=dft_demo(h1); % Take the DFT of the filter >> stem(f/pi,abs(H1),'k'); >> legend('DTFT of h1','DFT of h1') >> title('Fourier Transforms of the Impulse Response h1') >> hold off

  31. DFT and DTFT of a Filter

  32. Sampling Method of FIR DesignBasic Theory • Evenly spaced samples of the frequency response in the frequency range Ω = 0 to 2π represents the discrete Fourier transform (DFT) of a finite impulse response of the same length • Therefore, given the frequency response, the impulse response can be computed from the inverse DFT of the frequency response

  33. Steps in a Sampling DesignUsing the Custom M-files • Step 1: Determine the critical frequency Ω0 • Step 2: Determine the order M, where M is even • Step 3: Construct a vector of M+1 real-valued frequency response values evenly spaced from Ω = 0 to 2π. (Custom M-files help do this) [H,omega]=selectH_lp(Ω0,M+1) or [H,omega]=selectH_hp(Ω0,M+1)

  34. Steps in a Sampling Design • Step 4: Create a causal frequency response by delaying the response by M/2 steps H_delay=exp(-j*omega*M/2).*H • Step 5: Compute the impulse response of the delayed frequency response using the inverse discrete Fourier transform h=inv_dft_demo(H_delay) • Step 6: Window the impulse response with a tapering window function; for example: hw=h.*hamming(length(h))'

  35. Sampling Design Example 1 Design an order 24 HP filter with a cut-off frequency of π/2 • Generate and plot the sampled frequency response: >> [x,f]=selectH_hp(pi/2,25); >> stem(f/pi,x); >> title('Samples of the Frequency Response of a N=25 HP Filter with pi/2 Cutoff') >> xlabel('Digital Frequency in Units of Pi') >> axis([0 2 0 1.5]);

  36. Design Example 1Sampled Frequency Response

  37. Design Example 1Impulse Response • Compute the causal frequency response: >> M=length(x)-1; >> H=exp(-j*f*M/2).*x; • Compute the impulse response with the inverse DFT and plot the resulting frequency response: >> hhp1=inv_dft_demo(H); >> fvtool(hhp1,1)

  38. Design Example 1Frequency Response

  39. Design Example 1Compare the sampling design with an ideal window design >> n=0:24; >> omega=pi/2; >> hhp2=sinc(n-12)-(omega/pi)*sinc(omega*(n-12)/pi); >> [Hhp1,freq]=dtft_demo(hhp1,0,pi,512); >> [Hhp2,freq]=dtft_demo(hhp2,0,pi,512); >> plot(freq/pi,abs(Hhp1)); %This plots the magnitude of Hhp1 versus frequency >> hold >> plot(freq/pi,abs(Hhp2),'--k') %This plots the magnitude of Hhp2 versus frequency >> xlabel('Units of Pi') >> title('Comparison of Ideal Window and Sampling Method for HP Design') >> legend('Sampling Method – hhp1','Windowed Method –hhp2')

  40. Design Example 1

  41. Sampling Design 2 • Specifications: • low-pass filter of order 100, • a cut-off frequency 500 Hz • sampling frequency of 3000 Hz • Hamming window.

  42. Sampling Design 2MATLAB Code >> n=0:100; >> fs=3000; >> fc=500; >> omega_cutoff=2*pi*fc/fs; >> [x,f]=selectH_lp(omega_cutoff,length(n)); % Create the response samples >> M=length(x)-1; >> H=exp(-j*f*M/2).*x; % Compute the causal filter frequency response >> h=inv_dft_demo(H); % Compute the impulse response >> h_hamming=h.*hamming(length(n))'; % Window with a Hamming window >> subplot(2,1,1),dtft_demof(h,0,1500,512,3000); % Plot the magnitude response >> title('Sampling Design - Rectangular Window') >> subplot(2,1,2),dtft_demof(h_hamming,0,1500,512,3000); >> title('Sampling Design - Hamming Window')

  43. Sampling Design 2 Results

  44. Using FIR2 for Sampling Designs • >> B=fir2(N,F,A) • N = order • F = Frequency break points, Ω in units of π • A = Amplitudes corresponding to the break points in F F = [0, 0.3, 0.3, 1] A = [1, 1, 0, 0] 0 0.3π π

  45. Sampling Design with FIR2 >> N=100; % Set filter order >> fs = 3000; % Set sampling frequency >> fc=500; % Set cutoff frequency >> F=[0,2*fc/fs,2*fc/fs,1]; % Vector of frequency break points (omega values in units of pi) >> A=[1,1,0,0]; % Amplitudes corresponding to F >> B=fir2(N,F,A); % Compute impulse response >> dtft_demof(B,0,1500,512,3000); >> title('Sampling Design Example with MATLAB FIR2') Note: By default FIR2 uses a Hamming window

  46. Sampling Design with FIR2Results

  47. Optimal Design in MATLAB The values for order, transition width, and ripple cannot be independently specified. Specifying two of the parameters forces a particular value for the third. For a specified ripple and transition width the order is “optimal” in a Parks-McClellan design.

  48. A Parks-McClellan Design for a Low-Pass Filter • Filter specifications: • Order 20 low-pass • Critical frequency π/4 • Transition width 0.2π • By specifying the order and the transition width, the forced parameter is the pass-band and stop-band ripple >> f=[0 .15 .35 1]; % This sets the band edges with normalized frequencies % The transition is between the .15 and .35 values = 0.20 >> a=[1 1 0 0]; % This sets the desired amplitude response in the pass-band and % stop-band by specifying amplitudes at the band edges. >> N=20; % This sets the filter order. The filter length will be N + 1. >> h=firpm(N,f,a); >> fvtool(h,1)

  49. Parks-McClellan Low PassResults To improve the ripple performance, either the order must be increased or the transition width must be increased

  50. Comparison of the 3 Methods >> % Ideal Windowed Design >> n=0:20; >> omega=pi/4; >> hwin=(omega/pi)*sinc(omega*(n-10)/pi).*blackman(21)'; >> % Sample Design >> [H,f]=selectH_lp(pi/4,21); %This is a custom M-file for generating response samples % and frequencies for a low-pass filter >> M=20; >> Hk=exp(-j*f*M/2).*H; >> hs=inv_dft_demo(Hk); >> hsamp=hs.*blackman(21)'; % Blackman window the sampled response >> % Optimal Design >> f=[0,.15,.35,1]; >> a=[1,1,0,0]; >> w=[1,1]; >> N=20; >> hopt=firpm(N,f,a,w);

More Related