hilbert transform

2. Analysis of Signals

Figure 2.45.: Approximate FTs of two bandlimited signals

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The Hilbert transform of a function is by definition,
H {x(t)} = xh (t) =

∞

x(τ ) dτ t −τ
−∞

1 π (2.171)

which is the convolution of x(t) with 1/π t,
H {x(t)} = xh (t) = x(t) ∗

1 πt (2.172)

if we take the FT of this convolution,
Xh (ω ) = X (ω ) × F

1 πt (2.173)

From Example 2.24,
F {sgn(t)} =

2 jω (2.174)

and using duality from Equation (2.126) and Example 2.26, j ⇔ sgn(ω ) πt therefore
F

(2.175)

1 πt (2.176)

= − jsgn(ω )

87

2. Analysis of Signals

(a)

(b)

Figure 2.46.: −90◦ Phase-shifter therefore Xh (ω ) = − jX (ω )sgn(ω )

jXh (ω ) = X (ω )sgn(ω )

(2.177)

From the definition of the Hilbert transform and its FT, we may infer the following properties.
1. The function and its Hilbert transform have the same magnitude spectrum. This property may inferred from the following facts. x(t) ⇔ X (ω )

xh (t) ⇔ − jsgn(ω )X (ω )

(2.178)

therefore,
|Xh (ω )| = |− jsgn(ω )X (ω )|

= | − j||sgn(ω )||X (ω )|

= |X (ω )|

2. For a real time function, if we pass it through a −π /2 phase shifter, the output of the filter is the Hilbert transform of the input. Referring to Figure 2.46 (a), x(t) is the input to the phase shifter whose output is y(t). The phase shifter only has an effect on the phase, while the magnitude is not changed. We know that if we give all the positive frequencies a phase shift of −π /2, for a realisable filter, the negative frequencies will suffer a phase shift of
+π /2. Based on these ideas, the frequency response of the phase shifter is shown in the
(b) part of the figure. If we examine the filter in more detail, the filter response shows that this is the same as the operation of the Hilbert transform on the input function.
3. The Hilbert transform applied twice gives us the same function with a negative sign. The
Hilbert