Digital Signal Processing ECE-307. Tarun Gulati Assoc. Prof., ECE Deptt. MMEC, MMU, Mullana

Transcription

1 Digital Signal Processing ECE-307 Tarun Gulati Assoc. Prof., ECE Deptt. MMEC, MMU, Mullana

2 The z-transform: Introduction

3 Definition of the z-transform Definition:The z-transform of a discrete-time signal x(n) is defined by where z = re jw is a complex variable. The values of z for which the sum converges define a region in the z-plane referred to as the region of convergence (ROC). Notationally, if x(n) has a z-transform X(z), we write The z-transform may be viewed as the DTFT or an exponentially weighted sequence. Specifically, note that with z = re jw, X(z) can be looked as the DTFT of the sequence r --n x(n) and ROC is determined by the range of values of r of the following right inequation.

4 Complex z-plane z = Re(z)+jIm(z) = re jw Zeros and poles of X(z) ROC & z-plane Many signals have z-transforms that are rational function of z: Factorizing it will give: The roots of the numerator polynomial, β k, are referred to as the zeros (o) and α k are referred to as poles (x). ROC of X(z) will not contain poles.

5 ROC properties ROC is an annulus or disc in the z-plane centred at the origin. i.e. A finite-length sequence has a z-transform with a region of convergence that includes the entire z-plane except, possibly, z = 0 and z =. The point z = will be included if x(n) = 0 for n < 0, and the point z = 0 will be included if x(n) = 0 for n > 0. A right-sided sequence has a z-transform with a region of convergence that is the exterior of a circle: ROC: z >α A left-sided sequence has a z-transform with a region of convergence that is the interior of a circle: ROC: z <β The Fourier Transform of x(n) converges absolutely if and only if ROC of z-transform includes the unit circle

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12 Properties of Z-Transform Linearity If x(n) has a z-transform X(z) with a region of convergence Rx, and if y(n) has a z-transform Y(z) with a region of convergence Ry, and the ROC of W(z) will include the intersection of Rx and Ry, that is, Rw contains. Shifting property If x(n) has a z-transform X(z), Time reversal nw )( nax )( nby )( Z zwax )( zby )( z )( R x R y Z n 0 xnn ( ) zx() 0 z If x(n) has a z-transform X(z) with a region of convergence Rx that is the annulus z, the z-transform of the timereversed sequence x(-n) is xn ( ) X( z) and has a region of convergence z, which is denoted Z by Rx

13 Properties of Z-Transform Multiplication by an exponential If a sequence x(n) is multiplied by a complex exponential α n. Convolution theorm If x(n) has a z-transform X(z) with a region of convergence R x, and if h(n) has a z-transform H(z) with a region of convergence R h, The ROC of Y(z) will include the intersection of R x and R h, that is, R y contains R x R h. With x(n), y(n), and h(n) denoting the input, output, and unit-sample response, respectively, and X(z), Y(x), and H(z) their z-transforms. The z-transform of the unit-sample response is often referred to as the system function. Conjugation If X(z) is the z-transform of x(n), the z-transform of the complex conjugate of x(n) is n x Z () n X( z) ny )( nx )( nh )( Z zy )( )( zhzx x() n Z X( z) )(

14 Properties of Z-Transform Derivative If X(z) is the z-transform of x(n), the z- transform of is dx ( z) nx ( n) Z z dz Initial value theorem If X(z) is the z-transform of x(n) and x(n) is equal to zero for n<0, the initial value, x(0), maybe be found from X(z) as follows: x( )0 lim() Xz z

15 DFT algorithm. Conversion of DFT to FFT algorithm. Implementation of the FFT algorithm.

16 DFT Algorithm The Fourier transform of an analogue signal x(t) is given by: X xte j t dt The Discrete Fourier Transform (DFT) of a discrete-time signal x(nt) is given by: N 2 j nk N Xk xne n 0 Where: k0,, N xntxn

17 DFT Algorithm j N If we let: N 2 then : e W Xk N n 0 xnw N nk 2 Sampled signal Amplitude Sample Frequency Domain 0.8 Magnitude Normalised Frequency

18 DFT Algorithm Xk N n 0 xnw N nk x[n] = input X[k] = frequency bins W = twiddle factors X(0) X() X(k) X(N) = x[0]w N0 + x[]w 0* N + + x[n]w 0*(N) N = x[0]w N0 + x[]w * N + + x[n]w *(N) N : = x[0]w N0 + x[]w k* N + + x[n]w k*(n) N : = x[0]w N0 + x[]w (N)* N + + x[n]w (N)(N) N Note: For N samples of x we have N frequencies representing the signal.

19 Performance of the DFT Algorithm The DFT requires N 2 (NxN) complex multiplications: Each X(k) requires N complex multiplications. Therefore to evaluate all the values of the DFT ( X(0) to X(N) ) N 2 multiplications are required. The DFT also requires (N)*N complex additions: Each X(k) requires N additions. Therefore to evaluate all the values of the DFT (N)*N additions are required.

20 Performance of the DFT Algorithm Number of Multiplications Number of Additions Number of Samples Number of Samples Can the number of computations required be reduced?

21 DFT FFT A large amount of work has been devoted to reducing the computation time of a DFT. This has led to efficient algorithms which are known as the Fast Fourier Transform (FFT) algorithms.

22 DFT FFT x[n] = x[0], x[],, x[n] Lets divide the sequence x[n] into even and odd sequences: x[2n] x[2n+] = x[0], x[2],, x[n-2] = x[], x[3],, x[n]

23 DFT FFT Equation can be rewritten as: Since: N 2 kx 2 N 2 2 2nk xwn N xn n 0 n 0 W 2 nk N [2] W 2nk N e 2 j 2nk N W nk N 2 e 2 j nk N2 W 2 n k k nk N W W N N 2 Then: nk k x 2Wn 2 Xk N 2 N n 0 2 k N k YkWZ N 2 nk N N n 0 2 W xn W

24 Or: N Xk 2 N 2 nk k xnw W Xk DFT FFT Periodicity and symmetry of W can be exploited to simplify the DFT further: N n 0 2 N 2 n 0 xnw N nk 2 N 2 N 2 nk N xnw 2 N n 0 2 W N N2 k N 2 n 0 2 xnw N 2 2N 2 2 k j kj j k j k 2 NN2 N j N Nee ee e W N nk 2 N 2 : Symmetry W k N [3] And: N k 2 N 2 We 2 2N j kj N2 N22 e 2 j k N2 e W k N 2 : Periodicity

25 DFT FFT Symmetry and periodicity: W 8 6 W 8 4 W 8 5 W 8 3 W 8 7 W 8 0 = W 8 8 W 8 = W 8 9 W k+n/2 N = - W k N W k+n/2 N/2 = W k N/2 W k+4 8 = - W k 8 W k+8 8 = W k 8 W 8 2

26 DFT FFT Finally by exploiting the symmetry and periodicity, Equation 3 can be written as: N Xk 2 N 2 nk xwn N n 0 2 k k YkWZ N N 2 nk N 2 N n 0 2 k WxWn [4]

27 DFT FFT k N Xk YkWZkk N;,0 2 N k N Xk YkWZkk N;,0 2 2 Y(k) and W N k Z(k) only need to be calculated once and used for both equations. Note: the calculation is reduced from 0 to N to 0 to (N/2 - ).

28 DFT FFT k N Xk YkWZkk N;,0 2 N k N Xk YkWZkk N;,0 2 2 Y(k) and Z(k) can also be divided into N/4 point DFTs using the same process shown above: k k Uk WVk Zk Pk WQk N Yk N 2 N k Yk Uk WVk N N k Zk Pk WQk N 4 2 The process continues until we reach 2 point DFTs.

29 DFT FFT x[0] x[2] x[4] N/2 point DFT y[0] y[] y[2] X[0] = y[0]+ z[0] X[] = y[]+w z[] x[n-2] y[n-2] x[] x[3] x[5] N/2 point DFT z[0] z[] z[2] W X[N/2] = y[0]- z[0] X[N/2+] = y[]-w z[] x[n] z[n/2] Illustration of the first decimation in time FFT.

30 FFT Implementation Calculation of the output of a butterfly : Y(k) = U r +ju i U =U r +ju i = X(k) W Nk Z(k) = (L r +jl i )(W r +jw i ) L =L r +jl i = X(k+N/2) Key: U = Upper r = real L = Lower i = imaginary Different methods are available for calculating the outputs U and L. The best method is the one with the least number of multiplications and additions.

31 FFT Implementation Calculation of the output of a butterfly : U r +ju i U =U r +ju i (L r +jl i )(W r +jw i ) L =L r +jl i L jl WjW WLjL WjL WLW r i U r i rr ri ir ii rr ii WLWL ri ir WLWLj r i rr ii UWLWL r ri ir UWLWLj i UjU r i rr ii ri ir WLWLj UWL r rr ii WL Uj i ri i WLWL r LUjUWL WL

32 FFT Implementation Calculation of the output of a butterfly : U r +ju i U =U r +ju i = (L r W r - L i W i + U r )+j(l r W i + L i W r + U i ) (L r +jl i )(W r +jw i ) L =L r +jl i = (U r - L r W r + L i W i )+j(u i - L r W i - L i W r ) To further minimise the number of operations (* and +), the following are calculated only once: temp = L r W r temp2 = L i W i temp3 = L r W i temp4 = L i W r temp_2 = temp - temp2 temp3_4 = temp3 + temp4 U r = temp - temp2 + U r = temp_2 + U r U i = temp3 + temp4 + U i = temp3_4 + U i L r = U r - temp + temp2 = U r - temp_2 L i = U i - temp3 - temp4 = U i - temp3_4

33 FFT Implementation To efficiently implement the FFT algorithm a few observations are made: Each stage has the same number of butterflies (number of butterflies = N/2, N is number of points). The number of DFT groups per stage is equal to (N/2 stage ). The difference between the upper and lower leg is equal to 2 stage. The number of butterflies in the group is equal to 2 stage.

34 FFT Implementation Example: 8 point FFT W 2 W W 2 W 2 W 3 Decimation in time FFT: Number of stages = log 2 N Number of blocks/stage = N/2 stage Number of butterflies/block = 2 stage

35 FFT Implementation Example: 8 point FFT () Number of stages: W 2 W W 2 W 2 W 3 Decimation in time FFT: Number of stages = log 2 N Number of blocks/stage = N/2 stage Number of butterflies/block = 2 stage

36 Stage FFT Implementation Example: 8 point FFT () Number of stages: N stages = W 2 W W 2 W 2 W 3 Decimation in time FFT: Number of stages = log 2 N Number of blocks/stage = N/2 stage Number of butterflies/block = 2 stage

37 Stage FFT Implementation Stage 2 Example: 8 point FFT () Number of stages: N stages = 2 W 2 W W 2 W 2 W 3 Decimation in time FFT: Number of stages = log 2 N Number of blocks/stage = N/2 stage Number of butterflies/block = 2 stage

38 Stage FFT Implementation Stage 2 Stage 3 Example: 8 point FFT () Number of stages: N stages = 3 W 2 W W 2 W 2 W 3 Decimation in time FFT: Number of stages = log 2 N Number of blocks/stage = N/2 stage Number of butterflies/block = 2 stage

39 Stage FFT Implementation Stage 2 Stage 3 W 2 Example: 8 point FFT () Number of stages: N stages = 3 (2) Blocks/stage: Stage : W W 2 W 2 W 3 Decimation in time FFT: Number of stages = log 2 N Number of blocks/stage = N/2 stage Number of butterflies/block = 2 stage

40 lock Stage FFT Implementation Stage 2 Stage 3 W 2 Example: 8 point FFT () Number of stages: N stages = 3 (2) Blocks/stage: Stage : N blocks = W W 2 W 2 W 3 Decimation in time FFT: Number of stages = log 2 N Number of blocks/stage = N/2 stage Number of butterflies/block = 2 stage

41 lock lock 2 Stage FFT Implementation Stage 2 Stage 3 W 2 Example: 8 point FFT () Number of stages: N stages = 3 (2) Blocks/stage: Stage : N blocks = 2 W W 2 W 2 W 3 Decimation in time FFT: Number of stages = log 2 N Number of blocks/stage = N/2 stage Number of butterflies/block = 2 stage

42 lock lock 2 Stage FFT Implementation Stage 2 Stage 3 W 2 Example: 8 point FFT () Number of stages: N stages = 3 (2) Blocks/stage: Stage : N blocks = 3 lock 3 W W 2 W 2 W 3 Decimation in time FFT: Number of stages = log 2 N Number of blocks/stage = N/2 stage Number of butterflies/block = 2 stage

43 lock lock 2 Stage FFT Implementation Stage 2 Stage 3 W 2 Example: 8 point FFT () Number of stages: N stages = 3 (2) Blocks/stage: Stage : N blocks = 4 lock 3 W lock 4 W 2 W 2 W 3 Decimation in time FFT: Number of stages = log 2 N Number of blocks/stage = N/2 stage Number of butterflies/block = 2 stage

44 lock Stage FFT Implementation Stage 2 Stage 3 W 2 Example: 8 point FFT () Number of stages: N stages = 3 (2) Blocks/stage: Stage : N blocks = 4 Stage 2: N blocks = W W 2 W 2 W 3 Decimation in time FFT: Number of stages = log 2 N Number of blocks/stage = N/2 stage Number of butterflies/block = 2 stage

45 lock Stage FFT Implementation Stage 2 Stage 3 W 2 Example: 8 point FFT () Number of stages: N stages = 3 (2) Blocks/stage: Stage : N blocks = 4 Stage 2: N blocks = 2 lock 2 W W 2 W 2 W 3 Decimation in time FFT: Number of stages = log 2 N Number of blocks/stage = N/2 stage Number of butterflies/block = 2 stage

46 lock Stage FFT Implementation Stage 2 Stage 3 W 2 W Example: 8 point FFT () Number of stages: N stages = 3 (2) Blocks/stage: Stage : N blocks = 4 Stage 2: N blocks = 2 Stage 3: N blocks = W 2 W 2 W 3 Decimation in time FFT: Number of stages = log 2 N Number of blocks/stage = N/2 stage Number of butterflies/block = 2 stage

47 Stage FFT Implementation Stage 2 Stage 3 W 2 W W 2 Example: 8 point FFT () Number of stages: N stages = 3 (2) Blocks/stage: Stage : N blocks = 4 Stage 2: N blocks = 2 Stage 3: N blocks = (3) B flies/block: Stage : W 2 W 3 Decimation in time FFT: Number of stages = log 2 N Number of blocks/stage = N/2 stage Number of butterflies/block = 2 stage

48 Stage FFT Implementation Stage 2 Stage 3 W 2 W W 2 Example: 8 point FFT () Number of stages: N stages = 3 (2) Blocks/stage: Stage : N blocks = 4 Stage 2: N blocks = 2 Stage 3: N blocks = (3) B flies/block: Stage : N btf = W 2 W 3 Decimation in time FFT: Number of stages = log 2 N Number of blocks/stage = N/2 stage Number of butterflies/block = 2 stage

49 Stage FFT Implementation Stage 2 Stage 3 W 2 W W 2 W 2 W 3 Example: 8 point FFT () Number of stages: N stages = 3 (2) Blocks/stage: Stage : N blocks = 4 Stage 2: N blocks = 2 Stage 3: N blocks = (3) B flies/block: Stage : N btf = Stage 2: N btf = Decimation in time FFT: Number of stages = log 2 N Number of blocks/stage = N/2 stage Number of butterflies/block = 2 stage

50 Stage FFT Implementation Stage 2 Stage 3 W 2 W W 2 W 2 W 3 Example: 8 point FFT () Number of stages: N stages = 3 (2) Blocks/stage: Stage : N blocks = 4 Stage 2: N blocks = 2 Stage 3: N blocks = (3) B flies/block: Stage : N btf = Stage 2: N btf = 2 Decimation in time FFT: Number of stages = log 2 N Number of blocks/stage = N/2 stage Number of butterflies/block = 2 stage

51 Stage FFT Implementation Stage 2 Stage 3 W 2 W W 2 W 3 Decimation in time FFT: W 2 Example: 8 point FFT () Number of stages: N stages = 3 (2) Blocks/stage: Stage : N blocks = 4 Stage 2: N blocks = 2 Stage 3: N blocks = (3) B flies/block: Stage : N btf = Stage 2: N btf = 2 Stage 3: N btf = Number of stages = log 2 N Number of blocks/stage = N/2 stage Number of butterflies/block = 2 stage

52 Stage FFT Implementation Stage 2 Stage 3 W 2 W W 2 W 3 Decimation in time FFT: W 2 Example: 8 point FFT () Number of stages: N stages = 3 (2) Blocks/stage: Stage : N blocks = 4 Stage 2: N blocks = 2 Stage 3: N blocks = (3) B flies/block: Stage : N btf = Stage 2: N btf = 2 Stage 3: N btf = 2 Number of stages = log 2 N Number of blocks/stage = N/2 stage Number of butterflies/block = 2 stage

53 Stage FFT Implementation Stage 2 Stage 3 W 2 W W 2 W 3 Decimation in time FFT: W 2 Example: 8 point FFT () Number of stages: N stages = 3 (2) Blocks/stage: Stage : N blocks = 4 Stage 2: N blocks = 2 Stage 3: N blocks = (3) B flies/block: Stage : N btf = Stage 2: N btf = 2 Stage 3: N btf = 3 Number of stages = log 2 N Number of blocks/stage = N/2 stage Number of butterflies/block = 2 stage

54 Stage FFT Implementation Stage 2 Stage 3 W 2 W W 2 W 3 Decimation in time FFT: W 2 Example: 8 point FFT () Number of stages: N stages = 3 (2) Blocks/stage: Stage : N blocks = 4 Stage 2: N blocks = 2 Stage 3: N blocks = (3) B flies/block: Stage : N btf = Stage 2: N btf = 2 Stage 3: N btf = 4 Number of stages = log 2 N Number of blocks/stage = N/2 stage Number of butterflies/block = 2 stage

55 FFT Implementation Stage Stage 2 Stage 3 W 2 W W 2 W 2 W 3 Start Index Input Index 2 4 Twiddle Factor Index N/2 = 4

56 FFT Implementation Stage Stage 2 Stage 3 W 2 W W 2 W 2 W 3 Start Index Input Index 2 4 Twiddle Factor Index N/2 = 4 4/2 = 2

57 FFT Implementation Stage Stage 2 Stage 3 W 2 W W 2 W 2 W 3 Start Index Input Index 2 4 Twiddle Factor Index N/2 = 4 4/2 = 2 2/2 =

58 FFT Implementation Stage Stage 2 Stage 3 W 2 W W 2 W 2 W 3 Start Index 0 Input Index Twiddle Factor Index N/2 = 4 4/2 = 2 2/2 = Indicies Used W 2 W W 2 W 3

59 Structures for Discrete-Time Systems

60 Example Block diagram representation of nyanya2 nxb ny 2 0

61 Block Diagram Representation LTI systems with rational system function can be represented as constant-coefficient difference equation The implementation of difference equations requires delayed values of the input output

62 Direct Form I General form of difference equation N M k nyâ k k nxbˆ k k0 k0 N M ny knyak knxb k 0k Alternative equivalent form k

63 Direct Form I Transfer function can be M written as k bz k Hz k0 N k az k k Hz Direct Form I Represents k HzXz b z Xz Vz H zh z 2 M k0 k N k az k k M k0 b z k k M bxn k k vn N ayn k k v n yn k0 k

64 Alternative Representation Replace order of cascade LTI systems Hz H zxz Xz Wz k HzWz b z Wz Yz H zh z 2 2 M k0 M k0 b z k N k k k az k k N k az k k N awn k k xn wn k M bwn k k yn k0

65 Alternative Block Diagram We can change the order of the cascade systems N awn k k xn wn k M bwn k k yn k0

66 Direct Form II No need to store the same data twice in previous system So we can collapse the delay elements into one chain This is called Direct Form II or the Canonical Form Theoretically no difference between Direct Form I and II

67 Signal Flow Graph Example representation of a difference Representation Similar to block diagram representation Notational differences A network of directed branches connected at nodes

68 Example Representation of Direct Form II with signal flow graphs w n awn 4 xn w n bw 0 2n bw 4n w2 n wn w n 2 w n 3 w n 4 yn 3 nwaw nnx ny nwb nwb 0

69 Determination of System Function from Flow Graph w n w n w4 n xn w n w2 n xn w3 n w n w n w n 2 3 w n 4 yn 2 4 W z W z 2 W z 3 W z 4 W4 z Xz W z W2 z Xz W3 zz Wz Wz Yz 2 4 W z 2 W z 4 Wz Wz Yz Xz z z Xzz z 2 4 Yz z zh Xz z n n nh nu nu

70 Basic Structures for IIR Systems: Direct Form I

71 Basic Structures for IIR Systems: Direct Form II

72 Basic Structures for IIR Systems: Cascade Form General form Mfor cascade implementation M2 fz k zg k zg k zha k N N2 zc zd zd k k k k k0 zh M k More practical form in 2 nd order systems k b zb bz za az k k k 2 k2 2 k2

73 Example 2 z2z zz 2 25 z z z z5.0.0z25 zh.0z75.0 z5.0.0z25 Cascade of Direct Form I subsections Cascade of Direct Form II subsections

74 Basic Structures for IIR Systems: Parallel Form Represent system function using partial NP fraction expansion N P NP k Ak kkzeb zhzc k 0k k zc k zdzd k k k N P k k zh zc S 0k N k eze k0 k 2 k za k2 za Or by pairing the real poles

75 Example Partial Fraction Expansion zh.0z75 2 z2z 8.0z z5.0 z2.0 7z8.0z75 25 z zh 8 2 Combine poles to get

76 Transposed Forms Linear signal flow graph property: Transposing doesn t change the input-output relation Transposing: Reverse directions H z of all branches az Interchange input and output nodes Example: Reverse directions of branches and interchange

77 Example Transpose 2nxbnxbnxb2nyanyany Both have the same system function or difference equation

78 Basic Structures for FIR Systems: Direct Form Special cases of IIR direct form structures Transpose of direct form I gives direct form II Both forms are equal for FIR systems

79 Basic Structures for FIR Systems: Cascade Form M M n 2 zh znh S bzb k0 k k2 zb 0n k Obtained by factoring the polynomial system function

80 Structures for Linear-Phase FIR Systems Structure for even M Structure for odd M

81 IIR Filters

82 The general form for z-transform transfer functions for a digital filter is m b0 zb bz m zh ). n a za za ( 0 n The inverse of this z-transform transfer function h[n] is the impulse response for this digital filter.

83 As an example if Hz) z (. we have baabb 0, 0,, 2a,0 a2 3.0

84 The inverse z-transform of this transfer function (the impulse response) is n hn [] ( ) u[]. n h[n] n

85 This impulse response function goes on forever. That is, the impulse response is infinite in duration. Such filters are referred to as infinite impulse response or IIR filters.

86 Another example would be Hz ( ) z. The corresponding impulse response is h[ n] [] n [ n ].

87 h[n] n This impulse response is finite in duration. Such filters are referred to as finite impulse response or FIR filters.

88 Given a transfer function, how do we know whether this transfer function corresponds to an impulse response that is infinite or finite in duration? In our previous two examples we had Hz) z (. (IIR) Hz ( ) z. (FIR)

89 In general, any filter whose transfer function has a denominator (that does not factor-out) will have an impulse response that is infinite in duration corresponding to an IIR filter. Any filter whose transfer function does not have a denominator will have an impulse response that is finite in duration corresponding to an FIR filter.

90 The distinction between IIR filter transfer functions and FIR filter transfer functions becomes more clear if we look at the corresponding difference equations. In our examples the corresponding difference equations are y[ n] yn [ ] x []. n yn [] xn [] x [ n ]. (IIR) (FIR)

91 The IIR filter difference equation is recursive in nature: the current output depends upon the previous output y[ n] yn [ ] x []. n Since the current output depends upon the previous output and the previous output depends upon its previous output, the output depends upon the infinite past.

92 The FIR filter difference equation is depends only upon the input yn [] xn [] x [ n ]. If the input is finite in duration (such as an impulse) then the output is finite in duration. The output depends upon the finite past.

93 The difference between IIR and FIR filters can also be seen by looking at the transfer functions and noting that the IIR transfer function can be expanded using a geometric series that is infinite: 2 3 () zh zz z. z The FIR filter is already in a finite series form. Hz ( ) z.

94 Most analog filters have an impulse response which is infinite in duration. IIR filters are generally designed by emulating an analog prototype filter. There are two methods for doing this analog filter emulation: () the matched z-transform or impulse invariant transform (2) the bilinear transformation. In both cases, we are given an analog transfer function H(s), and we transform this function into a digital transfer function H(z).

95 The Matched z-transform In the matched z-transform digital filter design method we try to match the impulse response of the analog filter with that of the digital filter being designed. To match the impulse responses, we take the inverse Laplace transform of the analog filter H(s)h(t), then sample the impulse response h(t)h[n], then take the z-transform of the sampled impulse response to get the z-transform transfer function h[n]h(z).

96 Analog Prototype L H (s) Digital Filter H (z) Z sample h (t) h[n] Once we have our z-transform transfer function H(z), we apply the definition of the transfer function to write our digital filter equations: Yz ( ) H( z). Xz ( )

97 Example: Use the matched filter design method to design the digital equivalent of an integrator. Solution: The analog transfer function is The inverse Laplace transform is H( s). s h( t) u( t).

98 We then sample the impulse response to get h[n]: h[ n] u[ n]. Finally, we take the z-transform of the impulse response to get the digital filter transfer function. Hz () z.

99 Finally, we apply the definition of the z-transform transfer function to get the relationship between the input of the digital filter x[n] and the output of the digital filter y[n]. Yz () Hz (). Xz () z Yz ()[ z ] Xz (). []. y[ n] yn [ ] x n

100 y[ n] yn [ ] x []. n Applying this formula to an arbitrary input x[n], we have n x[n] y[n] 0 x[0] x[0] x[] x[0]+x[] 2 x[2] x[0]+x[]+x[2] 3 x[3] x[0]+x[]+x[2]+x[3]

101 We see that the output is the summation of the input. Thus the digital filter accurately represents the analog filter. The digital filter does not always accurately represent the analog filter as will be seen in the next example.

102 Example: Use the matched filter design method to design the following transfer function: Hs () s c, where c = s /4, and s is the sampling frequency. Solution: First, we find the impulse response ht () e c t c.

103 Then we sample the impulse response: [] cnt nh ce ce 2/n. Then we take the z-transform of the (discrete-time) impulse response: Hz ) c e 2 z ( /.

104 We then find the frequency response of the filter e j H(e j ) H(e j ) e 0.26 je e /2 /2 j 0.98 / /2

105 2 Frequency Response H(e j ) , x

106 As we can see, it is not much of a low-pass filter: the frequency rolloff is not very great. The reason for this small rolloff is aliasing error: the frequency response of H(z) is composed of copies of the frequency response of H(s) at 0, s, 2 s, etc. The copy at 2 s overlaps the copy at 0. Aliasing error is an inherent problem in matched filter design. Unless the cutoff frequency is very low compared to the sampling frequency, we will get substantial error due to aliasing. This aliasing problem is solved using the bilinear transformation.

107 The Bilinear Transformation The bilinear transformation is a fairly direct method of converting H(s) to H(z). Rather than map the analog frequencies =0 to s /2 to the digital frequencies =0 to = (as we had done with the matched z- transform), we will map the analog frequencies =0 to to the digital frequencies =0 to = Matched z-transform: =(0, s /2) =(0, Bilinear Transformation: =(0,) =(0,

108 What kind of function maps =(0,) =(0,? How about this: tan. 2 As we can see from the graph on the following slide, this function does perform the necessary mapping.

109 20 Frequency Warping 5 0 = tan (/2) , x

110 We know the relationship between and ; what is the relationship between s and z? ) ( ) ( cos sin 2 tan 2 / 2 / 2 2 / 2 / j j j j j e e e e

111 .. j j j j e e j e e j Since s=j and z=e j, we have. s z z

112 Because of the warped nature of our transformation, it is necessary to pre-warp our analog prototype critical frequencies so as to coincide with the critical frequencies of the corresponding digital filter. T ' tan tan c. c 2 2

113 Example: Use the bilinear transformation method to find the digital equivalent to the following transfer function: Hs () s c, where c = s /4, and s is the sampling frequency. Solution: First, we must pre-warp the analog frequency: T tan tan s 4 c T tan ' c

114 We then substitute our pre-warped frequency,, ( ) Hs s and apply the bilinear (sz) transformation:. 2 )( () z zh sh z z z z s

115 The frequency response of the filter is e j H(e j ) H(e j ) j 2 /2 j

116 2 Frequency Response H(e j ) , x

117 As can be seen, the frequency response is much improved. At =0, the response is the same as the analog filter at =0, and at =, the frequency response is the same as the analog filter at =.

118 Example: Repeat the previous example Hs () s c, where c = s /6. Solution: Our prewarping is slightly different: T tan tan s 6 c T tan ' c

119 Substituting our pre-warped frequency, we have Hs (). 3 s

120 Applying the bilinear transformation, we have. ) 3 ( 3 3 () () z z Hs Hz z z z z s The frequency plot is given on the following slide.

121 2 Frequency Response H(e j ) , x

122 Example: find the digital equivalent of a second-order Butterworth filter using the bilinear transformation. Let c = s /4. Solution: The second-order Butterworth filter has the following form: Hs (), () s2 (2) s c c (Enter [b a] = butter(2,,'s') in MATLAB.)

123 The prewarping is the same as in the first example: T tan tan s 4 c T tan ' c So, Hs (). s 2 2 s

124 Applying the bilinear transformation we have ) ( ) )( 2( ) ( ) ( 2 () () z z z z z Hs Hz z z z z z z s

125 2 2 z z 2( )2 2( )2 z. 2 The frequency response is on the following slide.

126 2 Frequency Response H(e j ) , x

127 Design of FIR Filters by Windowing FIR filters are designed based on directly approximating the desired frequency response of the discrete-time system. Most techniques for approximating the magnitude response of an FIR system assume a linear phase constraint.

128 Window Method An ideal desired frequency response jw He d hn d n hne d d 2 jw H e w c 0 jwn Heedw jwjwn w c, ww jw c He lp,0 c ww lp sin wn c hn n Many idealized systems are defined by piecewise-constant frequency response with discontinuities at the boundaries. As a result, these systems have impulse responses that are noncausal and infinitely long.

129 hn Window Method The most straightforward approach to obtaining a causal FIR approximation is to truncate the ideal impulse response.,0 nm nh d,0 otherwise hnwn hn d, 0nM nw,0 otherwise jw jwjw eh eh ewd d 2

130 Windowing in Frequency Domain Windowed frequency response j eh 2 d j j eh ew d The windowed version is smeared version of desired response

131 Window Method If nw n jwn enw 2 n k jw ew 2 d jw d He HeWedHe jw j jw 2 W e jw 5 wk H e jw w c 0 w c

132 Choice of Window w n is as short as possible in duration. This minimizes computation in the implementation of the filter., 0nM nw,0 otherwise jw W e approximates an impulse. M jw jwn jwn We wne e n n 0 jwm e sin wm 2 jwm 2 jw e e sin2 w 2 M M We 2 M jw

133 Window Method w jw If n is chosen so that W e is concentrated in a narrow band of frequencies around w 0 then jw H e would look like jw H, d e except jw where changes very abruptly. H d e 2 jw jwwj eh ehewd eh M We jw d Hd e jw jw d 2 M 2 M w c 0 w c

134 Rectangular Window jw W e generalized linear phase. for the rectangular window has a wm jw jwm 2 We e sin 2 sin2 w As M increases, the width of the main lobe decreases. 4 M mw While the width of each lobe decreases with M, the peak amplitudes of the main lobe and the side lobes grow such that the area under M each lobe is a constant. M M M M 2 M 2 M

135 H discontinuity. Rectangular Window e jw jw d ew d will oscillate at the The oscillations occur more rapidly, but do not decrease in magnitude as M increases. The Gibbs phenomenon can be moderated through the use of a less abrupt truncation of the Fourier series.

136 Rectangular Window By tapering the window smoothly to zero at each end, the height of the side lobes can be diminished. The expense is a wider main lobe and thus a wider transition at the discontinuity.

137 Design of FIR Filters by Windowing Method To design an ilowpass FIR Filters jw H e, ww jw c He lp,0 c ww lp sin wn c hn n, 0 hn hnwn d eh 2 jw d wn 0, nm otherwise jwjw eh ewd M We jw sin cwnm 2 nm2 w c 0 M M 2 w c M 0 M 2 M M 2 M 2 M

138 Properties of Commonly Used Windows Rectangular nw, 0nM nw,0 otherwise Bartlett (triangular) 2nM, 0 nm2 22 nmm, 2 nm,0 otherwise

139 Properties of Commonly Used Windows Hanning 2 Mn, cos 0 nm nw,0 othe Hamming 2, Mn cos 0nM nw,0 othe

140 7.2. Properties of Commonly Used Windows Blackman nw 2 Mn cos.0 08 cos 4 Mn,,0.0 0 nm other

141 7.2. Properties of Commonly Used Windows

142 Frequency Spectrum of Windows (a) Rectangular, (b) Bartlett, (c) Hanning, (d) Hamming, (e) Blackman, (M=50) (a)-(e) attenuation of sidelobe increases, width of mainlobe increases.

143 Properties of Commonly Used Windows biggest,high oscillations at discontinuity smallest,the sharpest transition

144 Incorporation of Generalized Linear Phase In designing FIR filters, it is desirable to obtain causal systems with a generalized linear phase response. The above five windows are all symmetric about the point M 2,i.e., Mwn, 0 nm nw,0 otherw

145 Incorporation of Generalized Linear Phase Their Fourier transforms are of the form jw e Wee jw jwm 2 e ew jw is a W e real and even func wof nh h d nwn : causal ifhmnhnhnhnwn d d d hmnhngeneralizedlinearphase : jw jw Aee jwm 2 He e M 2 M

146 if Incorporation of Generalized Linear Phase nmh dnh nh dnwnh nmh : d d nhgene lin p jw jaee jwjw M2 He o M 2 M

147 Frequency Domain Representation n ifhmn h d jw jwjw He He e M2 d d jw jw j dee He HW d 2 wn wmn 2jwjwM 2 H ee j jm Wee d 2 e e jw jwm 2 Aee e jw wherea ehwd 2 e e e jw e Wee jw jwm 2 W e hn hnwn ejw ee jw d

148 Properties of Commonly Used Windows biggest,high oscillations at discontinuity smallest,the sharpest transition

149 The Kaiser Window Filter Design Method 2 2 I0 n wn,0 nm I 0 0, otherwise where M2, IuzerordermodifiedBeselfunctionofthefirstkind : 0 twoparameters: lengthm :, Iu 0 r u 2 2 r! r shape paramete : Trade side-lobe amplitude for main-lobe width