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8-1b we see a collection of sine wave harmonics as deÞned in Fig.. These sine wave harmonics are the Fourier series constituents of the symmetrical square wave in Fig.. These − cosine wa

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0 16 32 48 64 80 96 112 128

0

−1

1

x(n)

n

(a)

N := 128 n := 0, 1 N

k := 0, 1 N

x(n) := 0

−1 if n >

1 if n> 0

0 if n = N

0 if n=

N

2

N

2

(c)

XH(k) := −j⋅X(k) if k < N

2

N

2

0 if k=

N 2

j⋅X(k) if k >

−1

−0.5

0

0.5

1

Im(X(k))

k

(b)

Imaginary

X(k) := 1 ∑nN−1= 0 N

n N x(n) ⋅exp −j⋅2⋅π⋅ ⋅k

Figure 8-1 Example of the Hilbert transform.

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0 16 32 48 64 80 96 112 128

−1

−0.5

0

0.5

1

Re(XH(k))

k

Real

(d )

−4

−3

−2

−10

1

2

3

4

(xh(n))

x(n)

n

(f )

xh1(n) : = 0.25⋅xh(n − 1) + 0.5⋅xh(n) + 0.25⋅xh(n + 1) xh2(n) : = 0.25⋅xh1(n − 1) + 0.5⋅xh1(n) + 0.25⋅xh1(n + 1)

(g)

−4

−3

−2

−10

1

2

3

4

xh2(n)

x(n)

n

(h)

(e)

N (XH(k)) ⋅exp j⋅2⋅π⋅ ⋅n

Figure 8-1 (continued)

Trang 3

chapter we will continue to use DFT and IDFT and stay focused on the main objective, understanding the Hilbert transform

Why do the samples in Fig 8-1f and h bunch up at the two ends and

in the center to produce the large peaks? The answer can be seen by comparing Fig 8-1b and d In Fig 8-1b we see a collection of (sine) wave harmonics as deÞned in Fig 2-2c These sine wave harmonics are the Fourier series constituents of the symmetrical square wave in Fig 8-1a In Fig 8-1d we see a collection of (− cosine) waves as deÞned

in Fig 2-2b These (− cosine) wave harmonic amplitudes accumulate at

the endpoints and the center exactly as Fig 8-1f and h verify As the harmonics are attenuated, the peaks are softened The smoothing also tends to equalize adjacent amplitudes slightly The peaks in Fig 8-1h rise about 8 dB above the square-wave amplitude, which is almost always too much There are various ways to deal with this One factor is that the square-wave input is unusually abrupt at the ends and center Smoothing

(equivalent to lowpass Þltering) of the input signal x(n), is a very useful

approach as described in Chapter 4 This method is usually preferred in circuit design

It is useful to keep in mind, especially when working with the HT,

that the quadrature of θ, which is θ ± 90◦, is not always the same thing

as the conjugate of θ If the angle is +30◦, its conjugate is−30◦, but its quadrature is +30◦± 90◦= +120◦ or −60◦ The HT uses θ◦± 90◦ For example, Fig 8-1b shows+j0.5 at k = 127 The HT multiplies this by + j

to get a real value of− 0.5 in part (d) at k = 127 This is a quadrature positive phase shift at negative frequency k= 127 A similar event occurs

at k = 1 Also, at k = 0 and N the phase jump is 180◦ from± j to ∓j, and the same, although barely noticeable, at N /2 (Use a highly magniÞed

vertical scale in Mathcad.)

A different example is shown in Fig 8-2 The baseband signal in part (b) is triangular in shape, and this makes a difference The abrupt changes in the square wave are gone, the baseband spectrum (d) contains only real cosine-wave harmonics, and the Hilbert transform (f) contains

only sine-wave harmonics The sharp peaks in xh(n) that we see in

Fig 8-1g disappear in Fig 8-1h However, for equal peak-to-peak ampli-tude, the square wave of Fig 8-1 has 4.8 dB more average power than the triangular wave of Fig 8-2 It is not clear what practical advantage

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0 16 32 48 64

n

−4

−2

0 2

x(n)

−0.5

0.5

0

1 1.5

Re(X(k))

x(n):=

N:= 128 n := 0, 1 N k := 0, 1 N

3 if n= 0

3 if n = N

(a)

(b)

k

(d )

(c )

∑N−1

n = 0

X(k) := 1

N

−j⋅2⋅π⋅ ⋅k

12·n N

−9 + 12·n if n >

N N

2

Figure 8-2 Hilbert transform using a triangular waveform.

the triangular wave would have The importance of peak amplitude limits and peak power limits in circuit design must always be kept in mind

BASIC PRINCIPLES OF THE HILBERT TRANSFORM

There are many types of transforms that are useful in electronics work The DFT and IDFT are well known in this book because they transform back

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xH(k) := −j ⋅ X(k) if k <

0 if k=

1

0.5

0

(e)

N

2

j ⋅ X(k) if k > N

2

N

2

−1

−0.5

4

−4

−3

−2

−10

1

2

3

0 16 32 48 64

k

(f )

(g)

80 96 112 128

0 20 40

n (h)

60 80 100 120

Im(XH(k))

x(n)

Re(xh(n))

∑N−1

n = 0

xh(n) := (XH(k)⋅exp k

N

j ⋅2⋅π⋅ ⋅k

Figure 8-2 (continued)

and forth between the discrete time-domain signal x(n) and the discrete frequency-domain spectrum X (k).

Another very popular transform converts a linear differential equation into a linear algebraic equation For example, consider the differential

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