Download presentation

Presentation is loading. Please wait.

Published byShea Lampkins Modified over 2 years ago

1
24 Nov'09Comp30291 DMP Section 51 University of Manchester School of Computer Science Comp 30291 : Digital Media Processing Section 5 z-transforms & IIR-type digital filters

2
24 Nov'09Comp30291 DMP Section 52 Order is maximum of N and M. Recursive if any b j is non-zero. A 2nd order recursive filter has the difference-equation: y[n] = a 0 x[n] + a 1 x[n-1] + a 2 x[n-2] - b 1 y[n-1] - b 2 y[n-2] Introduction General causal digital filter has difference equation:

3
24 Nov'09Comp30291 DMP Section 53 y[n] = a 0 x[n] + a 1 x[n-1] + a 2 x[n-2] - b 1 y[n-1] - b 2 y[n-2] Signal-flow-graph for 2 nd order recursive diff equn z -1 + + + + a0a0 a1a1 a2a2 x[n] -b 2 -b 1 y[n]

4
24 Nov'09Comp30291 DMP Section 54 Derivation of ‘z-transform’ If {x[n]} with x[n] = z n is applied to an LTI system with impulse- response {h[n]}, output would be, by convolution :

5
24 Nov'09Comp30291 DMP Section 55 ‘z-transform’ of {h[n]} If input is {z n }, output is {H(z).z n } z may be any real or complex number for which series converges. For causal & stable IIR, the series converges when |z| 1. Replacing z by e j gives frequency-response:

6
24 Nov'09Comp30291 DMP Section 56 Visualising H(z) On Argand diagram (‘z-plane’), z = e j lies on unit circle. Imaginary part of z Real part of z 1 z = e j

7
24 Nov'09Comp30291 DMP Section 57 Example 5.1 Find H(z) for the non-recursive difference equation: y[n] = x[n] + x[n-1] Solution: {h[n]} = {..., 0, 1, 1, 0,... }, therefore H(z) = 1 + z - 1

8
24 Nov'09Comp30291 DMP Section 58 Example 5.2 Find H(z) for the recursive difference equation: y[n] = a 0 x[n] + a 1 x[n-1] - b 1 y[n-1] Solution: If x[n] = z n then y[n] = H(z) z n, y[n-1] = H(z) z n - 1 Substitute to obtain: H(z) z n = a 0 z n + a 1 z n - 1 - b 1 H(z) z n - 1 H(z) = a 0 + a 1 z - 1 - b 1 H(z) z - 1 (1 + b 1 z -1 ) H(z) = a 0 + a 1 z - 1 When z = -b 1, H(z) =

9
24 Nov'09Comp30291 DMP Section 59 Find H(z) for the 2 nd order difference-equation y[n] = a 0 x[n] + a 1 x[n-1] + a 2 x[n-2] - b 1 y[n-1] - b 2 y[n-2] The same method gives: H(z) z n = a 0 z n + a 1 z n - 1 +a 2 z n - 2 - b 1 H(z) z n - 1 - b 2 H(z)z n-2 a 0 + a 1 z - 1 + a 2 z - 2 H(z) = b 0 + b 1 z - 1 + b 2 z - 2 with b 0 = 1.

10
24 Nov'09Comp30291 DMP Section 510 Consider difference-equation for general digital filter N M y[n] = a i x[n i] b j y[n j] i=0 j=1 The same method gives: This is the ‘system function’ of the digital filter. Also referred to as ‘transfer function’

11
24 Nov'09Comp30291 DMP Section 511 Ratio of 2 polynomials in z -1 Equal to z-transform of impulse-response when this converges. Easily derived from difference-equation & signal-flow graph. Replacing z by e j gives frequency-response System function

12
24 Nov'09Comp30291 DMP Section 512 Example 5.3 Give a signal-flow graph for the system function: Solution: The difference-equation is: y[n] = a 0 x[n] + a 1 x[n-1] + a 2 x[n-2] - b 1 y[n-1] - b 2 y[n-2] Represented by the signal-flow graph below: 2nd order or ‘bi-quadratic’ IIR section in ‘direct form 1’. z -1 + + + + a0a0 a1a1 a2a2 x[n] -b 2 -b 1 y[n]

13
24 Nov'09Comp30291 DMP Section 513 Label ‘z -1 ’ box inputs & outputs as shown: To implement ‘direct form 1’ biquad as a program z -1 + + + + a0a0 a1a1 a2a2 X -b 2 -b 1 Y X1 X2 Y1 Y2

14
24 Nov'09Comp30291 DMP Section 514 X1=0; X2=0; Y1=0; Y2=0; while 1 X = input(‘X=’); Y = a0*X + a1*X1 + a2*X2 - b1*Y1 - b2*Y2; disp(sprintf(‘Y = %f’, Y) ) ; % output Y X2 = X1; X1 = X ; Y2 = Y1; Y1 = Y ; end; In MATLAB using floating point arithmetic

15
24 Nov'09Comp30291 DMP Section 515 y = filter([a0 a1 a2], [b0 b1 b2], x ); In MATLAB using Signal Processing toolbox :

16
24 Nov'09Comp30291 DMP Section 516 K=1024; A0=round(a0*K); A1=round(a1*K); A2=round(a2*K); B1=round(b1*K); B2=round(b2*K); X1=0; X2=0; Y1=0; Y2=0; while 1 X = input(‘X=’) ; Y = A0*X + A1*X1 + A2*X2 - A1*Y1 - A2*Y2 ; Y = round(Y/K); % Divide by arith right shift disp(sprint(‘Y=%f’,Y)); % Output Y X2 = X1; X1 = X ; % Prepare for next time Y2 = Y1; Y1 = Y ; end; In MATLAB using fixed point arith & shifting

17
24 Nov'09Comp30291 DMP Section 517 Look again at ‘Direct Form 1’ signal-flow-graph It may be thought of as two signal-flow-graphs: z -1 + + + + a0a0 a1a1 a2a2 x[n] -b 2 -b 1 y[n] Non-recursive part Recursive part x[n] y[n]

18
24 Nov'09Comp30291 DMP Section 518 Re-ordering LTI systems It may be shown than if we have two LTI systems as shown: L1L1 L2L2 x[n]y[n] L2L2 L1L1 x[n]y[n] then re-ordering L 1 & L 2 does not change the behaviour of the overall system. Only guaranteed to work for LTI systems

19
24 Nov'09Comp30291 DMP Section 519 Alternative signal-flow-graph Look again at ‘Direct Form 1’: Re-order the two ‘halves’ & then simplify to ‘direct form 2’: x[n] z -1 + + + + a0a0 a1a1 a2a2 -b 2 -b 1 y[n] a2a2 x[n] z -1 + + a0a0 a1a1 -b 2 z -1 + + -b 1 y[n] a2a2 z -1 + + a0a0 a1a1 y[n] x[n] -b 2 + + -b 1

20
24 Nov'09Comp30291 DMP Section 520 ‘Direct Form II’ signal-flow-graph Its difference equation is: y[n] = a 0 x[n] + a 1 x[n-1] + a 2 x[n-2] - b 1 y[n-1] - b 2 y[n-2] i.e. exactly the same as Direct Form 1 It is a 2nd order (bi-quad) section whose system function is: a2a2 z -1 + + a0a0 a1a1 y[n] x[n] -b 2 + + -b 1 W W1W1 W2W2 Direct form II economises on ‘delay boxes’. Notice labels: W,W1 & W2

21
24 Nov'09Comp30291 DMP Section 521 W1 = 0; W2 = 0; %For delay boxes while 1 X = input(‘X=’) ; % Input to X W =X - b1*W1 - b2*W2; % Recursive part Y = W*a0 + W1*a1 + W2*a2; % Non-rec. part W2 = W1; W1 = W; % For next time disp(sprintf(‘Y=%f’,Y)); % Output Y using disp end; % Back for next sample Program to implement Direct Form II using normal arithmetic

22
24 Nov'09Comp30291 DMP Section 522 K=1024; A0=round(a0*K); A1=round(a1*K); A2=round(a2*K); B1=round(b1*K); B2=round(b2*K); W1 = 0; W2 = 0; %For delay boxes while 1 X = input(‘X=’) ; % Assign X to input W =K*X - B1*W1 - B2*W2; % Recursive part W =round( W / K); % By arith right-shift Y = W*A0+W1*A1+W2*A2; % Non-rec. part W2 = W1; W1 = W; %For next time Y = round(Y/K); %By arith right-shift disp(sprintf( ‘Y=%f’, Y)); %Output Y using disp end; % Back for next sample Direct Form II in fixed point arithmetic with shifting

23
24 Nov'09Comp30291 DMP Section 523 Re-express as: Poles & zeros of H(z) For a discrete time filter: Now factorise numerator & denominator:

24
24 Nov'09Comp30291 DMP Section 524 Poles & zeros of H(z) continued z 1, z 2,..., z N, are ‘zeros’. p 1,p 2,..., p N, are ‘poles’. H(z) generally infinite with z equal to a pole. H(z) generally zero with z equal to a zero. For a causal stable system all poles must satisfy p i < 1. i.e. on Argand diagram: poles must lie inside unit circle. No restriction on the positions of zeros.

25
24 Nov'09Comp30291 DMP Section 525 Design of IIR ‘notch’ filter Design a 4th order 'notch' filter to eliminate an unwanted sinusoid at 800 Hz without severely affecting rest of signal. The sampling rate is FS = 10 kHz. One way is to use the MATLAB function ‘butter’ as follows: FL = 800 – 25 ; FU = 800+25; [a b] = butter(2, [FL FU]/(FS/2),’stop’); a = [0.98 -3.43 4.96 -3.43 0.98] b= [ 1 -3.47 4.96 -3.39 0.96] freqz(a, b); freqz(a, b, 512, FS); % Better graph axis([0 FS/2 -50 5]); % Scales axes Notch has -3 dB frequency band: 25 + 25 = 50 Hz. MATLAB response

26
24 Nov'09Comp30291 DMP Section 526 Gain/phase response of notch filter

27
24 Nov'09Comp30291 DMP Section 527 Gain/phase responses of notch filter

28
24 Nov'09Comp30291 DMP Section 528 How sharp is the notch? Can answer this question by specifying notch’s -3 dB bandwidth. Have just designed what Barry calls a 4th order band-stop filter. MATLAB calls it a 2nd order band-stop filter. For a sharper notch, decrease -3 dB bandwidth But this will decrease its ‘depth’, i.e. the attenuation at the ‘notch’ frequency. If necessary, increase the order to 6 (3) or 8 (4). Details

29
24 Nov'09Comp30291 DMP Section 529 Sketch the same gain-response The -3dB frequencies are at ( 800 + ) and ( 800 - ) Hz. Given (= 25 Hz say) can sketch gain-response: 1 FS/2 F Gain 0.5 0 800 800 + 800 - 0 dB -3 dB

30
24 Nov'09Comp30291 DMP Section 530 Implement the 4 th order ‘notch’ filter MATLAB function gave us: a = [0.98 -3.43 4.96 -3.43 0.98] b= [ 1 -3.47 4.96 -3.39 0.96] Transfer (System) Function is:

31
24 Nov'09Comp30291 DMP Section 531 z -1 + + + + + + + + 0.98 3.47 -4.96 3.39 -0.96 0.0.98 x[n] y[n] -3.43 4.96 -3.43 A direct Form 2 implementation of the 4th order IIR notch filter

32
24 Nov'09Comp30291 DMP Section 532 Problems with ‘direct form’ IIR implementations Implementation on previous slide works fine in MATLAB. But ‘direct form’ IIR implementations of order >2 are rarely used. Sensitivity to round-off error in coeff values will be high. Also range of ‘intermediate’ signals in z -1 boxes is high. High wordlength floating point arithmetic hides this problem But in fixed point arithmetic, great difficulty occurs. Instead we use ‘cascaded biquad sections’

33
24 Nov'09Comp30291 DMP Section 533 Using biquad (2 nd order) IIR sections H(z) x[n]y[n] Given 4 th order H(z), instead of: we prefer to arrange two biquad sectns as follows: H 1 (z)H 2 (z) x[n]y[n] G

34
24 Nov'09Comp30291 DMP Section 534 Converting to the new implementation Get a & b for 4 th order H(z) as before: [a b] = butter(2, [FL FU]/(FS/2),’stop’); Then execute: [SOS G] = tf2sos(a,b) MATLAB responds with: SOS = 1 -1.753 1 1 -1.722 0.9776 1 -1.753 1 1 -1.744 0.9785 G = 0.978 Transfer function to 2 nd order sectns First sectn 2 nd sectn

35
24 Nov'09Comp30291 DMP Section 535 H(z) may now be realised as:

36
24 Nov'09Comp30291 DMP Section 536 Example A digital filter with a sampling rate of 200 Hz is required to eliminate an unwanted 50 Hz sinusoidal component of an input signal without affecting the magnitudes of other components too severely. Design a 4th order "notch" filter for this purpose whose 3dB bandwidth is not greater than 3.2 Hz. Solution method: FS=200; FL=50-1.6; FU=50+1.6; [a b]=butter(2,[FL,FU]/(FS/2), ‘stop’); [SOS G] = tf2sos(a,b)

37
24 Nov'09Comp30291 DMP Section 537 Many design techniques for IIR digital filters have adopted ideas of analogue filters. Can transform analogue ‘prototype’ transfer function H a (s) into H(z) for an IIR digital filter. Analogue filters have infinite impulse-responses. Many gain-response approximations exist which are realisable by analogue filters e.g. Butterworth low-pass approximation which can be transformed to high-pass, band-pass & band-stop. IIR digital filter design by bilinear transformation

38
24 Nov'09Comp30291 DMP Section 538 Butterworth low-pass gain approximation of order n At C, Gain 0.71 i.e -3 dB

39
24 Nov'09Comp30291 DMP Section 539 Can transform H a (s), with gain-response G a ( ), to H(z) for an IIR digital filter with similar gain-response G( ). Many ways exist. Most famous is ‘bilinear transformation’. Replaces s by 2(z-1)/(z+1) to transform H a (s) to H(z). Fortunately MATLAB does all this for us. Transformations from H a (s) to H(z) for IIR digital filter

40
24 Nov'09Comp30291 DMP Section 540 Properties of bilinear transformation (i) Order of H(z) = order of H a (s) (ii) If H a (s) is causal & stable, so is H(z). (iii) G( ) = G a ( ) where = 2 tan( /2) So gain of analog filter at radians/second becomes gain of digital filter at radians/sample where = 2tan( /2).

41
24 Nov'09Comp30291 DMP Section 541 Frequency warping By (iii), from - to mapped to in range - to .

42
24 Nov'09Comp30291 DMP Section 542 Shape of G( ) will change under the transformation. If G a ( ) is Butterworth, G( ) will not have exactly the same shape, but we still call it Butterworth. Mapping approx linear for in the range -2 to 2. As increases above 2, a given increase in produces smaller and smaller increases in . Frequency warping (cont)

43
24 Nov'09Comp30291 DMP Section 543 G( ) becomes more and more compressed as . Illustrate for an analog gain-response with ripples: (a): Analogue gain response (b): Effect of bilinear transformation Ga()Ga() G( ) Comparing G a ( ) with G( )

44
24 Nov'09Comp30291 DMP Section 544 ‘Prototype’ analogue transfer function Although the shape changes, we would like G( ) at its cut off C to the same as G a ( ) at its cut-off frequency. If G a ( ) is Butterworth, it is -3dB at its cut-off freq So we would like G( ) to be -3 dB at its cut-off C. Achieved if analogue prototype is designed to have its cut- off frequency at C = 2 tan( C /2). C is the ‘pre-warped’ cut-off frequency. Designing analog prototype with cut-off freq 2 tan( C /2) guarantees that the digital filter will have its cut-off at C.

45
24 Nov'09Comp30291 DMP Section 545 Let required cut-off frequency C = /4 radians/sample. Need prototype transfer fn H a (s) for 2nd order Butt low-pass filter with 3 dB cut-off at 2tan( C /2) = 2 tan( /8) radians/second. C = 2 tan( /8) = 0.828 I happen to remember that the transfer fn for a 2nd order Butt low- pass filter with cut-off C is: If you don’t believe me, check that replacing s by j and taking the modulus gives G( ) = 1/ [1+( / C ) 2n ] with n=2. Set C = 0.828 in this formula, then replace s by 2(z-1)/(z+1). Gives us H(z) for an IIR digital filter. That’s it! Design 2nd order IIR lowpass digital filter by bilinear transfm

46
24 Nov'09Comp30291 DMP Section 546 Resulting IIR digital filter x[n] y[n] 0.098 2 0.94 - -0.33

47
24 Nov'09Comp30291 DMP Section 547 Design of 2nd order IIR low-pass digital filter by bilinear transform using MATLAB If required cut-off freq is /4 radians/sample, type: [a b] = butter(2, 0.25) MATLAB gives us: a = [0.098 0.196 0.098] b = [1 -0.94 0.33] The required expression for H(z) is therefore: To save multipliers it is a good idea to re-express this as:

48
24 Nov'09Comp30291 DMP Section 548 Realise by ‘direct form 2’ signal-flow graph x[n] y[n] 0.098 2 0.94 -0.33

49
24 Nov'09Comp30291 DMP Section 549 Higher order IIR digital filters Example: Design 4th order Butterwth-type IIR low-pass digital filter with 3 dB c/o at f S / 16.. Solution: Relative cut-off frequency is /8. Typing: [a b] = butter(4, 0.125) gives the response: a = 0.0009 0.0037 0.0056 0.0037 0.0009 b = 1 -2.9768 3.4223 -1.7861 0.3556

50
24 Nov'09Comp30291 DMP Section 550 z -1 + + + + + + + + 0.00093 2.977 -3.422 1.79 -0.356 0.00093 x[n] y[n] 0.0037 0.0056 0.0037 A direct Form 2 implementation of 4th order IIR filter

51
24 Nov'09Comp30291 DMP Section 551 0.00093 x[n] z -1 + + + + + + + + 2.977 -3.422 1.79 -0.356 y[n] 4 6 4 Better ‘direct Form 2’ implementation of 4th order IIR filter

52
24 Nov'09Comp30291 DMP Section 552 Higher order IIR digital filters not normally implemented in Direct Form 1 or 2. Instead, implement as cascaded biquad (sos) sections by typing: [a b] = butter(4, 0.125); [sos G] = tf2sos(a,b) MATLAB responds with: sos = 1 2 1 1 -1.365 0.478 1 2 1 1 -1.612 0.745 G = 0.00093 Cascaded bi-quad sections ‘transfer function’ to ‘second order sections’

53
24 Nov'09Comp30291 DMP Section 553 H(z) may be realised as: Fourth order IIR Butterworth filter with cut-off fs/16

54
24 Nov'09Comp30291 DMP Section 554 Better realisation of H(z) At =0, gain of 1st section is (1+2+1)/(1-1.612+0.0745) =30.12 At =0, gain of 2nd section is (1+2+1)/(1-1.365+0.488) = 35.56 Make gain of each section one at =0 by scaling as follows: Fourth order IIR Butterworth filter with cut-off Fs/16

55
24 Nov'09Comp30291 DMP Section 555 Gain-responses for 4th order analog & IIR digital filter

56
24 Nov'09Comp30291 DMP Section 556 Compare gain-response of 4th order Butt low-pass transfer function used as a prototype, with that of derived digital filter. Both are 1 (0 dB) at zero frequency. Both are 0.707 (-3 dB) at the cut-off frequency. Analogue gain approaches 0 as whereas digital filter gain becomes exactly zero at = . Shape of Butt gain response is "warped" by bilinear transfn. For digital filter, cut-off rate becomes sharper as because of the compression as .

57
24 Nov'09Comp30291 DMP Section 557 High-pass band-pass and band-stop IIR filters Example: 2nd (4th)order bandpass filter with L = /4, u = /2. Solution: [a b] = butter(2,[0.25 0.5]) a = 0.098 0 -0.195 0 0.098 b = 1 -1.219 1.333 -0.667 0.33 freqz(a,b); [sos G] = tf2sos(a,b) sos = 1 2 1 1 -0.1665 0.5348 1 -2 1 1 -1.0524 0.6232 G = 0.098 Many people call this 4th order - MATLAB calls it 2nd order.

58
24 Nov'09Comp30291 DMP Section 558 2nd (4th) order IIR bandpass

59
24 Nov'09Comp30291 DMP Section 559 Arranged as 2 biquad sections x[n] 0.1 -2.17 -0.54 1 y[n] 2 1.05 -0.62

60
24 Nov'09Comp30291 DMP Section 560 Higher order band-stop IIR filter Example: 4th (8th)order bandpass filter with L = /4, u = /2. Solution: [a b] = butter(4,[0.25 0.5],’stop’) a = 0.35 -1.15 2.815 -4.24 5.1 -4.24 2.815 -1.15 0.35 b = 1 -2.472 4.309 -4.886 4.477 -2.914 1.519 -0.5 0.12 freqz(a,b); [sos G] = tf2sos(a,b) sos =1 -.828 1 1 -0.351 0.428 1 -.828 1 1 -0.832 0.49 1 -.828 1 1 -0.046 0.724 1 -0.828 1 1 -1.244 0.793 G = 0.347

61
24 Nov'09Comp30291 DMP Section 561 Gain & phase resp of IIR band-stop filter

62
24 Nov'09Comp30291 DMP Section 562 Band-pass as arrangement of 4 ‘sos’ biquads H 1 (z)H 2 (z)H 3 (z)H 4 (z) Careful with scaling for band-pass & high-pass sections Can make gain =1 in pass-band for each section. Not at =0 this time !!!

63
24 Nov'09Comp30291 DMP Section 563. Must use [sos G] = tf2sos ([a 4 a 3 a 2 a 1 a 0 ], [b 4 … b 0 ]) ‘help tf2sos’ to find out abt this function Transfer-function to second order sections: ‘tf2sos’

64
24 Nov'09Comp30291 DMP Section 564 Option of cascading high pass & low-pass digital filters to give band-pass or band-stop filters must be used with care. It is much simpler & avoids the factorisation problem. Then make sure that analogue prototype is wide-band High-pass IIR filters are designed by: [a b] = butter(4,0.125,’high’); Wide-band band-pass & band-stop filters

65
24 Nov'09Comp30291 DMP Section 565 Comparison of IIR and FIR digital filters Advantage of IIR type digital filters: Economical in use of delays, multipliers and adders. Disadvantages: (1) Sensitive to coefficient round-off inaccuracies & effects of overflow in fixed point arith. These effects can lead to instability or serious distortion. (2) An IIR filter cannot be exactly linear phase.

66
24 Nov'09Comp30291 DMP Section 566 Advantages of FIR filters: (1) may be realised by non-recursive structures which are simpler and more convenient for programming especially on devices specifically designed for DSP. (2) FIR structures are always stable. (3) Because there is no recursion, round-off and overflow errors are easily controlled. (4) An FIR filter can be exactly linear phase. Disadvantage of FIR filters: Large orders can be required to perform fairly simple filtering tasks.

67
24 Nov'09Comp30291 DMP Section 567 Problems 1 Find H(z) for the following difference equations (a) y[n] = 2x[n] - 3x[n-1] + 6x[n-4] (b) y[n] = x[n-1] - y[n-1] - 0.5y[n-2] 2 Show that passing {x[n]} thro’ H(z) = z - 1 produces {x[n-1]}. 3 Calculate the impulse-response of the digital filter with 1 H(z) = 1 - 2 z - 1 Draw its signal flow graph, plot its poles and zeros and comment on them.

68
24 Nov'09Comp30291 DMP Section 568 4. If LTI systems L1 & L2, with imp-responses {h1[n]} & {h2[n]} are arranged as below, calculate overall impulse- response. Show that this is affected by interchanging L1 & L2. 5. Low-pass IIR digital filter required with cut-off at f s /4 & stop-band attenuation at least 20 dB for all frequencies above 3f s /8 & below f s /2. Design by bilinear transfn using MATLAB.

69
24 Nov'09Comp30291 DMP Section 569 6. Design a 4th order band-pass IIR digital filter with lower & upper cut-off frequencies at 300 Hz & 3400 Hz when f S = 8 kHz. 7. Design a 4th order band-pass IIR digital filter with lower & upper cut-off frequencies at 2000 Hz & 3000 Hz when f S = 8 kHz. 8. What limits how good a notch filter we can implement on a fixed point DSP processor? In theory we can make notch sharper & sharper by moving the -3dB points closer & closer. What limits us in practice? How sharp a notch could we could get in 16-bit fixed pt arithmetic? 9. What order of FIR low-pass filter would be required to be approx as good as the 2nd order IIR low-pass filter ( /4 cut-off) designed in these notes?

Similar presentations

OK

Unit 9 IIR Filter Design 1. Introduction The ideal filter Constant gain of at least unity in the pass band Constant gain of zero in the stop band The.

Unit 9 IIR Filter Design 1. Introduction The ideal filter Constant gain of at least unity in the pass band Constant gain of zero in the stop band The.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google