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Chapter 12

Chapter 12. Case Studies in Digital Signal Processing. Case Studies. Dual-Tone Multifrequency (DTMF) Signaling A Software Impedance Bridge Details of JPEG Compression FBI Fingerprint Image Compression Using Wavelets. Case I: Dual-Tone Multifrequency Signaling.

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Chapter 12

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  1. Chapter 12 Case Studies in Digital Signal Processing

  2. Case Studies • Dual-Tone Multifrequency (DTMF) Signaling • A Software Impedance Bridge • Details of JPEG Compression • FBI Fingerprint Image Compression Using Wavelets

  3. Case I: Dual-Tone Multifrequency Signaling • Used to communicate Touchtone® pad information via telephone links • Based on CCITT Q.23 and Q.24 standards

  4. DTMF Frequency Set 4 Low Frequencies 3 High Frequencies No harmonic relationships Example: “5” is the sum of 770 Hz and 1336 Hz

  5. Operational Specifications for DTMF Signaling

  6. Tone Generation The transfer function is a 4-pole resonator with frequencies at one set of the tone pairs (ΩL and ΩH)

  7. >> f1=852; >> f2=1209; >> fs=8000; >> bL=sin(2*pi*f1/fs); >> bH=sin(2*pi*f2/fs); >> aL=2*cos(2*pi*f1/fs); >> aH=2*cos(2*pi*f2/fs); >> b=[0,bL+bH,-(aL*bH+aH*bL),bL+bH]; >> a=[1,-(aL+aH),aL*aH+2,-(aL+aH),1]; >> x=[1,zeros(1,2000)]; %This is an impulse signal to start the oscillator >> y=filter(b,a,x); >> dtft_demof(y,0,1400,1024,8000); %The sampling frequency is 8 kHz >> figure, zplane(b,a) % Examine the poles and zeros Example: Digit “7” (852 and 1209 Hz)

  8. Tone Detection and Validation • Rejection of dial-tone • Estimation of spectral content (Goertzel Algorithm) • Estimation of second-harmonic power • Determination of forward and reverse twist ratios • Tone duration check

  9. >> fs=8000; % The system sampling frequency is 8 Hz >> f350=2*pi*350/fs; % Compute the digital frequencies for each notch >> f440=2*pi*440/fs; >> deltaf=2*pi*5/fs; % Let the -3 dB width of each notch be 5 Hz >> r=1-deltaf/2; % Compute the pole radius >> g350=abs(1-2*r*cos(f350)+r^2)/(2*abs(1-cos(f350))); % Compute the “g” factors >> g440=abs(1-2*r*cos(f440)+r^2)/(2*abs(1-cos(f440))); >> b350=g350*[1,-2*cos(f350),1]; % Compute the b and a coefficients for each filter >> b440=g440*[1,-2*cos(f440),1]; >> a350=[1,-2*r*cos(f350),r^2]; >> a440=[1,-2*r*cos(f440),r^2]; >> b2=conv(b350,b440); % Convolve the coefficient vectors >> a2=conv(a350,a440); >> x=analog([350,440,697,1209],[1,1,1,1],1000,8000); >> y=filter(b2,a2,x); >> subplot(2,1,1),dtft_demof(x,0,1400,2048,8000); title('Digit 1 with Dial-Tone') >> subplot(2,1,2),dtft_demof(y,0,1400,2048,8000); title('Digit 1 with Dial-Tone Removed') Rejection of Dial Tone (350 Hz and 440 Hz )Cascaded Notch Filter

  10. Determination of Tone Frequencies • Tone frequencies are determined by the Goertzel algorithm • The Goertzel algorithm is based on computing the DFT at just the expected tone frequencies and determining which frequencies show the largest response • It can be shown that the Goertzel algorithm is more efficient than the FFT if M < log2N where M is the number of frequencies to be computed by the DFT and N is the signal length. • If M = 7, then N = 136, then the Goertzel algorithm can operate on 136 samples of the MF tone signal, • At a sampling frequency of 8 kHz, tone detection requires a minimum of 136*.125 = 17 ms.

  11. Goertzel Algorithm Example(20 Sample Signal, 250 Hz and 500 Hz)

  12. >> n=1:20; >> f1=2*pi*250/2000; >> f2=2*pi*500/2000; >> x=sin(f1*n)+sin(f2*n); >> dtft_demo(x,0,2*pi,512); % Custom M-file for the DTFT >> hold >> [X,omegax]=dft_demo(x); % Custom M-file for the DFT >> stem(omegax/pi,abs(X), 'k'); >> title('250 and 500 Hz, DTFT (Curve) and DFT (Stem)') >> hold off >> dtft_demo(x,0,2*pi,512); >> hold >> stem([omegax(4),omegax(6)]/pi,[abs(X(4)),abs(X(6))],'k') >> title('DTFT (Curve) and Goertzel Results (Stem)') >> hold off Goertzel Algorithm Example(20 Sample Signal, 250 Hz and 500 Hz)

  13. “goertzel_vector” M-file >> help goertzel_vector G=GOERTZEL_VECTOR(N,FS,F) This function produces a vector of index values, G, to use in the Goertzel algorithm. The index values correspond to the closest Hertzian frequencies given in the vector F, while N = number of sample values in the input signal to the Goertzel command and FS is the sampling frequency of the input signal. >> f=[697,770,852,941,1209,1336,1477]; >> g=goertzel_vector(136,8000,f) g = 13 14 15 17 22 24 26 Goertzel index set for the DTMF tones when N = 136 and Fs = 8 kHz

  14. >> f=[697,770,852,941,1209,1336,1477]; >> g=goertzel_vector(136,8000,f); >> n=0:135; >> f1=2*pi*697/8000; >> f2=2*pi*1209/8000; >> tone1=sin(f1*n)+sin(f2*n); >> X1=goertzel(tone1,g); >> abs(X1); >> stem(f,abs(X1)) >> xlabel('Hz') >> title('Goertzel Algorithm Output for the Tone "1" (697 and 1209 Hz)') Goertzel Algorithm ExampleDetection of Digit “1” Each stem corresponds to one of the 7 tone frequencies. The maximum response decodes the dual tone as 697 Hz and 1209 Hz = digit “1”

  15. >> n=0:135; >> f1off=2*pi*1001/8000; >> f2off=2*pi*1396/8000; >> tone0off=sin(f1off*n)+sin(f2off*n); >> X0off=goertzel(tone0off,g); >> stem(f,abs(X0off)) >> xlabel('Hz') >> title('Goertzel Output for "0" Tones Offset by +60 Hz') Goertzel Algorithm ExampleRejection of Digit “0” The valid tones for “0” are 941 and 1336 Hz. Here the tones have been offset by +60 Hz (3.5%). The upper tone is detected but no lower tone is detected greater than 4 dB below the upper tone. Result: rejection as a valid tone set.

  16. Case II: Identification of Physical Impedances Using an Adaptive Filter The unknown system changes the amplitude and phase of an input sinusoid. The Wiener filter models the response as an FIR system with transfer function H(z). From the frequency response of H(z), the unknown impedance can be estimated.

  17. Analog Impedance Circuit The unknown impedance will change the amplitude and phase of an input sinusoid by A and θ respectively. The resistance Rm is assumed to be known. F(Z) is the “impedance factor” that results in A and θ.

  18. Wiener Filter Estimate of H(z) and the Unknown Impedance In the z-domain: Since both H(Ω) and the impedance factor F(Z) are complex numbers with the same magnitude and phase angle, we can conclude: Using the Wiener filter coefficients:

  19. Example >> Rm=100; >> Rx=50; % This is the “unknown” impedance >> X=-125; >> FZ=(Rx+j*X)/(Rm+Rx+j*X); % This is the complex response of the physical % impedances >> amp=abs(FZ) % This gives the amplitude response amp = 0.6895 >> deg=angle(FZ)*180/pi % This gives the phase response in degrees deg = -28.3930 >> n=0:44099; >> omega=2*pi*200/44100; %The digital frequency for 200 Hz at the %sampling rate of 44.1 kHz. >> x=sin(omega*n); % This is 1 second of the sampled excitation signal >> y=amp*sin(omega*n+angle(FZ)); %This is the digital output signal with an % amplitude and phase that would be % produced by the unknown system % response

  20. Example, Continued >> [err,n_hat,w]=adapt2(6,x,y); %Run the adaptive filter routine >> plot(err); title('Error vs. Samples'); xlabel('Sample') >> b=w'; %Put the estimated impulse response into a row vector >> z=exp(j*omega); %The complex unit circle value of the input frequency >> H=H_evaluate(b,1,z); %Use the custom M-file H_evaluate to compute the % response of the system at the input frequency >> Rm*H/(1-H) % Compute the real and imaginary parts of the unknown % impedance ans = 5.0002e+001 -1.2500e+002 % The estimate of the unknown impedance Rx +jX

  21. Example, Continued The error signal shows the convergence of the Wiener filter to the estimate of the unknown system.

  22. Case III: JPEG Compression • Joint Photographic Experts Group standard adopted in 1994 as ISO 10918-1 (International Organization for Standards). • JPEG is an example of transform compression using the Discrete Cosine Transform. • Because some of the transform information is deleted in the algorithm, JPEG is termed lossy compression. • Suitable for photographs but less so for line drawings and textual graphics.

  23. The Concept of Transform Compression The Fourier series approximations to a triangle wave only require a few low frequency components for an adequate representation. The 10-term signal is a 10-to-1 “compression” of the 100-term signal.

  24. Discrete Cosine Transform • The DCT is a variation on the discrete Fourier Transform (DFT) • Provides a real-valued frequency representation in the range Ω = [0,π].

  25. >> n=0:20; >> f=pi/4; >> x=sin(f*n); >> yf=fft(x); >> yc=dct(x); >> omegaf=0:2*pi/20:2*pi; >> omegac=0:pi/20:pi; >> subplot(1,2,1),stem(omegaf/pi,abs(yf)),title('Discrete Fourier Transform') >> xlabel('units of pi') >> subplot(1,2,2),stem(omegac/pi,yc),title('Discrete Cosine Transform') >> xlabel('units of pi') Comparison of DFT and DCT for aΩ = π/4 Sinusoid The DCT has only real coefficients and is defined only in the interval [0,π] and is therefore a more “efficient” transform of the sinusoid.

  26. Two-Dimensional DCT

  27. Steps in JPEG Compression • Convert the image matrix pixel values to the correct numerical storage class for arithmetic computations. After this step the pixel values (matrix values) will range from 0 to 1 in grayscale. • Level-shift the matrix values so that the middle grayscale value is zero. This would mean that 0.5 is subtracted from each pixel value. This basically eliminates any “DC bias” in the values and makes the DCT more sensitive to frequency variations. • Compute the 2-dimensional DCT on the each 8-by-8 block of the level-shifted image matrix. • Multiply the DCT matrix block by an 8-by-8 “mask” matrix, element-by-element, to select a variable number of DCT coefficients, starting with the lowest frequency coefficients. This is the step that actually produces the compression of the image. • Convert the masked DCT block to a vector by means of a “zigzag” reading of the matrix values and discard trailing zeros. The block vectors form the compressed file for the image. • Reconstruct the image by reversing the above steps. The key step is taking the inverse DCT of the masked coefficient blocks.

  28. Step 1 – Create the 8-by-8 block and convert to double precision >> I=imread('pout.tif'); >> size(I) ans = 291 240 >> block=I(101:108,101:108) block = 120 123 134 142 144 146 149 146 119 123 136 144 147 144 148 145 117 122 133 139 147 144 147 143 117 121 128 133 144 144 146 144 112 119 127 130 138 142 145 145 106 112 121 123 128 133 139 142 95 101 109 112 114 117 125 130 91 91 102 107 108 108 108 113 >> b2=im2double(block) % Convert the matrix values to storage class “double” b2 = 0.4706 0.4824 0.5255 0.5569 0.5647 0.5725 0.5843 0.5725 0.4667 0.4824 0.5333 0.5647 0.5765 0.5647 0.5804 0.5686 0.4588 0.4784 0.5216 0.5451 0.5765 0.5647 0.5765 0.5608 0.4588 0.4745 0.5020 0.5216 0.5647 0.5647 0.5725 0.5647 0.4392 0.4667 0.4980 0.5098 0.5412 0.5569 0.5686 0.5686 0.4157 0.4392 0.4745 0.4824 0.5020 0.5216 0.5451 0.5569 0.3725 0.3961 0.4275 0.4392 0.4471 0.4588 0.4902 0.5098 0.3569 0.3569 0.4000 0.4196 0.4235 0.4235 0.4235 0.4431

  29. Step 2 – Level-shift the matrix values >> b3=b2-.5 % Subtract 0.5 from each matrix value to create +/- 1 pixel values b3 = -0.0294 -0.0176 0.0255 0.0569 0.0647 0.0725 0.0843 0.0725 -0.0333 -0.0176 0.0333 0.0647 0.0765 0.0647 0.0804 0.0686 -0.0412 -0.0216 0.0216 0.0451 0.0765 0.0647 0.0765 0.0608 -0.0412 -0.0255 0.0020 0.0216 0.0647 0.0647 0.0725 0.0647 -0.0608 -0.0333 -0.0020 0.0098 0.0412 0.0569 0.0686 0.0686 -0.0843 -0.0608 -0.0255 -0.0176 0.0020 0.0216 0.0451 0.0569 -0.1275 -0.1039 -0.0725 -0.0608 -0.0529 -0.0412 -0.0098 0.0098 -0.1431 -0.1431 -0.1000 -0.0804 -0.0765 -0.0765 -0.0765 -0.0569

  30. Step 3 – Compute the 2-D DCT of the level-shifted matrix >> c1=dct2(b3) c1 = 0.0059 -0.3088 -0.0867 -0.0315 -0.0010 0.0087 0.0009 0.0203 0.3411 -0.0007 -0.0428 0.0175 -0.0007 0.0093 -0.0100 0.0059 -0.1531 0.0294 -0.0058 -0.0161 0.0146 0.0154 0.0056 -0.0076 0.0645 -0.0277 0.0242 -0.0099 -0.0132 0.0017 0.0003 -0.0064 -0.0186 0.0098 -0.0030 0.0160 0.0020 -0.0026 0.0055 -0.0037 -0.0089 -0.0089 0.0143 0.0024 -0.0024 -0.0025 -0.0042 0.0001 0.0019 0.0027 -0.0042 0.0051 -0.0007 0.0004 0.0039 0.0005 -0.0031 -0.0031 -0.0009 -0.0037 -0.0046 -0.0005 -0.0019 -0.0006

  31. Step 4 – Select only low index values of the DCT matrix with a mask. >> mask=mask8(4) % This is a custom M-file for mask generation. % It will select a triangle-shaped portion of the low index % values of the DCT matrix. mask = 1 1 1 1 0 0 0 0 1 1 1 0 0 0 0 0 1 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 >> c2=c1.*mask % Multiply the DCT matrix by the mask element-by-element c2 = 0.0059 -0.3088 -0.0867 -0.0315 0 0 0 0 0.3411 -0.0007 -0.0428 0 0 0 0 0 -0.1531 0.0294 0 0 0 0 0 0 0.0645 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

  32. Step 5 – Read out the retained matrix values into a vector using a “zig-zag” >> c2 c2 = 0.0059 -0.3088 -0.0867 -0.0315 0 0 0 0 0.3411 -0.0007 -0.0428 0 0 0 0 0 -0.1531 0.0294 0 0 0 0 0 0 0.0645 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 >> c2vector=zz(c2); >> length(c2vector) ans = 64 >> c2vector(1:10)' ans = 0.0059 -0.3088 0.3411 -0.1531 -0.0007 -0.0867 -0.0315 -0.0428 0.0294 0.0645 The vector of the retained matrix represents the “compressed” information for the .jpg file

  33. Step 6a – Reconstruct the matrix by computing the inverse DCT of the masked coefficient matrix • >> c2=dezz(c2vector) % This command “de-zig-zags” the stored zig-zag vector • ans = • 0.0059 -0.3088 -0.0867 -0.0315 0 0 0 0 • 0.3411 -0.0007 -0.0428 0 0 0 0 0 • -0.1531 0.0294 0 0 0 0 0 0 • 0.0645 0 0 0 0 0 0 0 • 0 0 0 0 0 0 0 0 • 0 0 0 0 0 0 0 0 • 0 0 0 0 0 0 0 0 • 0 0 0 0 0 0 0 0 • >> r1=idct2(c2) • r1 = • -0.0312 -0.0043 0.0331 0.0619 0.0745 0.0754 0.0733 0.0722 • -0.0396 -0.0130 0.0242 0.0536 0.0677 0.0709 0.0711 0.0715 • -0.0473 -0.0214 0.0151 0.0449 0.0612 0.0680 0.0720 0.0748 • -0.0498 -0.0255 0.0092 0.0384 0.0561 0.0665 0.0745 0.0800 • -0.0567 -0.0347 -0.0032 0.0237 0.0414 0.0540 0.0652 0.0729 • -0.0807 -0.0614 -0.0340 -0.0106 0.0055 0.0187 0.0317 0.0409 • -0.1187 -0.1018 -0.0783 -0.0586 -0.0446 -0.0319 -0.0182 -0.0082 • -0.1487 -0.1332 -0.1120 -0.0946 -0.0822 -0.0701 -0.0562 -0.0460

  34. Step 6b – Continue reconstruction by restoring the pixel levels >> r2=r1+.5 r2 = 0.4688 0.4957 0.5331 0.5619 0.5745 0.5754 0.5733 0.5722 0.4604 0.4870 0.5242 0.5536 0.5677 0.5709 0.5711 0.5715 0.4527 0.4786 0.5151 0.5449 0.5612 0.5680 0.5720 0.5748 0.4502 0.4745 0.5092 0.5384 0.5561 0.5665 0.5745 0.5800 0.4433 0.4653 0.4968 0.5237 0.5414 0.5540 0.5652 0.5729 0.4193 0.4386 0.4660 0.4894 0.5055 0.5187 0.5317 0.5409 0.3813 0.3982 0.4217 0.4414 0.4554 0.4681 0.4818 0.4918 0.3513 0.3668 0.3880 0.4054 0.4178 0.4299 0.4438 0.4540

  35. Step 6c – Compare to the original image block matrix >> b2-r2 ans = 0.0018 -0.0133 -0.0076 -0.0051 -0.0098 -0.0029 0.0110 0.0004 0.0063 -0.0046 0.0091 0.0111 0.0088 -0.0062 0.0093 -0.0029 0.0061 -0.0001 0.0065 0.0002 0.0153 -0.0033 0.0045 -0.0140 0.0086 -0.0000 -0.0072 -0.0168 0.0086 -0.0018 -0.0019 -0.0152 -0.0041 0.0013 0.0013 -0.0139 -0.0002 0.0029 0.0035 -0.0043 -0.0036 0.0006 0.0085 -0.0070 -0.0036 0.0028 0.0134 0.0159 -0.0087 -0.0021 0.0057 -0.0022 -0.0083 -0.0093 0.0084 0.0180 0.0056 -0.0099 0.0120 0.0142 0.0058 -0.0064 -0.0202 -0.0108 Analysis - Compute the average percent error between the original and reconstructed block: >> mean2(((b2-r2)./b2)*100) ans = -0.0314

  36. Full Image JPEG: The Block Processing Command >> help blkproc BLKPROC Implement distinct block processing for image. B = BLKPROC(A,[M N],FUN) processes the image A by applying the function FUN to each distinct M-by-N block of A, padding A with zeros if necessary. FUN is a function that accepts an M-by-N matrix, X, and returns a matrix, vector, or scalar Y: Y = FUN(X) BLKPROC does not require that Y be the same size as X. However, B is the same size as A only if Y is the same size as X. B = BLKPROC(A,[M N],FUN,P1,P2,...) passes the additional parameters P1,P2,..., to FUN.

  37. Full Image JPEG:The Compression M-File >> type cjpeg1 function cx = cjpeg1(x,mask) % function cx = cjpeg1(x,mask) % % Basic jpeg-like transform coder. % MASK is an 8 by 8 matrix that sets the DCT coefficients % to be saved (i.e., compression amount), generated by the M-file MASK8.M. % <x> is the input image. % The retained coefficients are returned in <cx>. % The DC coefficient should be top left of each 8x8 block % This command uses DCT2 for computing the block transform. x=im2double(x); %convert image to double precision for the inline commands x = x-.5; %level-adjust the image around the mid-range grayscale fun=@dct2; %Define the 2-D DCT function B = blkproc(x,[8 8],fun); %Block process the image to create DCT blocks cx=blkproc(B, [8 8], 'x.*P1',mask); %Multiply the blocks by the mask %end of cjpeg1

  38. Full Image JPEG:M-Files cjpeg1 – This function carries out an 8-by-8 sub-block computation of the DCT on a level-adjusted black-and-white image and then multiplies each block by an 8-by-8 mask generated by mask8. mask8 – This file generates an 8-by-8 logical mask matrix that will select the DCT coefficients from the lowest to the highest index values. When multiplied by an 8-by-8 DCT block, this selects just a subset of the coefficients in the block, setting the others to zero. This effects the compression of the block. djpeg1 – This function reconstructs (decompresses) an image from the compressed transform file generated by cjpeg. jpeg_demo – This combines the compression and reconstruction routines (cjpeg1 and djpeg1) and generates an absolute-value error image between the original image and its reconstruction. It also computes the compression ratio by determining the fraction of DCT coefficients retained after the block masking operation.

  39. >> I=imread('moon.tif'); >> mask=mask8(4); % 6-to-1 compression >> jpeg_demo(I,mask); >> mask=mask8(2); % 21-to-1 compression >> figure, jpeg_demo(I,mask); Full Image JPEG:Example >> I=imread('moon.tif'); >> mask=mask8(4); % 6-to-1 compression >> jpeg_demo(I,mask); >> mask=mask8(2); % 21-to-1 compression >> figure, jpeg_demo(I,mask); 6 to 1 Compression 21 to 1 Compression

  40. Case IV: FBI Fingerprint Compression Using Wavelets • FBI fingerprint information could require up to 300 terabytes of storage. • Compression of the image data by factors up to 20-to-1 were desired. • JPEG compression is not satisfactory because of the loss of fine image detail. • Wavelet compression gives superior results at high compression ratios. • The method used is similar to the new JPEG 2000 image compression standard.

  41. Uncompressed Fingerprint Image >> load detfingr.mat >> FPbw=X/(max(max(X))); >> imshow(FPbw),title('Uncompressed B/W Intensity Fingerprint Image')

  42. JPEG Compressed Fingerprint Images >> mask_2=mask8(2); %Set a DCT mask to 21-to-1 compression >> mask_3=mask8(3); %Set a DCT mask to 11-to-1 compression >> FPr_11=jpeg_demo(FPbw,mask_3); %Generate a reconstructed image 11-tto-1 >> close %Close the figure created by jpeg_demo >> FPr_21=jpeg_demo(FPbw,mask_2); % Generate a reconstructed image 21-to-1 >> close %Close the figure created by jpeg_demo >> imshow(FPr_11),title('Reconstructed JPEG 11-to-1 Compression') >> figure,imshow(FPr_21),title('Reconstructed JPEG 21-to-1 Compression')

  43. JPEG Compressed Fingerprint Images(21-to-1 Compression) Note the loss of fine detail and the block artifacts in the reconstructed image

  44. Wavelet Compression Using the “wdencmp “ Command (help file) [XC,CXC,LXC,PERF0,PERFL2]=WDENCMP('gbl',X,'wname',N,THR,SORH,KEEPAPP) returns a de-noised or compressed version XC of input signal X (1-D or 2-D) obtained by wavelet coefficients thresholding using global positive threshold THR. Additional output arguments [CXC,LXC] are the wavelet decomposition structure of XC, PERFL2 and PERF0 are L^2 recovery and compression scores in percentages. PERFL2 = 100*(vector-norm of CXC/vector-norm of C)^2 where [C,L] denotes the wavelet decomposition structure of X. Wavelet decomposition is performed at level N and 'wname' is a string containing the wavelet name. SORH ('s' or 'h') is for soft or hard thresholding (see WTHRESH for more details). If KEEPAPP = 1, approximation coefficients cannot be thresholded, otherwise it is possible.

  45. Wavelet Compression Using the “wdencmp” >> wav='bior4.4'; %Set the wavelet family to be used >> level=3; % Set the decomposition level >> sorh='h'; % Set the thresholding method to “hard” >> keepapp=1; % Do not threshold the approximation coefficients >> thr=40; % Set the thresholding level and use globally (‘gbl’) >> [XC,CXC,LXC,PERF0,PERFL2]= wdencmp('gbl',X,wav,level,thr,sorh,keepapp); >> imshow(XC/max(max(XC))),colormap(bone(128)); >> title('20-to-1 Compression, bior4.4 Wavelet, Threshold = 40') >> PERF0 PERF0 = 94.8699

  46. Wavelet Compression Using wdencmpResults for bior4.4 Wavelet Family

  47. Wavelet Compression Using jpeg2000_demoResults for haar Wavelet Family >> load detfingr.mat; % Load the image >> wav='haar'; % Set the wavelet family >> level=3; % Set the decomposition level >> thresh=40; % Set the thresholding for compression >> [XC,cmp_ratio,energy,thr]=jpeg2000_demo(X,wav,level,thresh); >> close % Close the figure created by jpeg2000_demo >> imshow(XC/max(max(XC))),colormap(bone(128)); %Display intensity image >> title('19-to-1 Compression, Haar Wavelet, Threshold =40') >> cmp_ratio % Compression ratio is 19-to-1 for this example cmp_ratio = 19

  48. Wavelet Compression Using jpeg2000_demoResults for haar Wavelet Family

  49. Summary • Dual Tone Multifrequency Signaling • Method by which tone-coded information is exchanged in the telephone network • Digital oscillators used for tone generation • The Goertzel algorithm used for tone identification • DSP Impedance Bridge • Adaptive filter system identification is used to model the magnitude and phase response of the Thevenin equivalent of a physical impedance. • JPEG Compression • A form of block-by-block transform compression of images using the discrete cosine transform (DCT) • Fingerprint Image Compression • The use of wavelets for image compression • Similar to the new JPEG 2000 image compression standard

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