hilbert

The Hilbert Transform Frank R. Kschischang The Edward S. Rogers Sr. Department of Electrical and Computer Engineering University of Toronto October 22, 2006 1 Definition The Hilbert transform H[g(t)] of a signal g(t) is defined as H[g(t)] =g(t)∗1 πt=1 π/integraldisplay∞ −∞g(τ) t−τdτ=1 π/integraldisplay∞ −∞g(t−τ) τdτ. (1) The Hilbert transform of g(t) is the convolution of g(t) with the signal 1 /πt. It is the response tog(t) of a linear time-invariant filter (called a Hilbert transformer) having impulse response 1/πt. The Hilbert transform H[g(t)] is often denoted as ˆ g(t) or as [ g(t)]∧. A technicality arises immediately. The alert reader will already be concerned with the definition (1) as the integral is improper: the integrand has a singularity and the limits of integration are infinite. In fact, the Hilbert transform is properly defined as the Cauchy principal value of the integral in (1), whenever this value exists. The Cauchy principal value is defined—for the first integral in (1)—as H[g(t)] =1 πlim /epsilon1→0+/parenleftBigg/integraldisplayt−/epsilon1 t−1//epsilon1g(τ) t−τdτ+/integraldisplayt+1//epsilon1 t+/epsilon1g(τ) t−τdτ/parenrightBigg . (2) We see that the Cauchy principal value is obtained by considering a finite range of inte- gration that is symmetric about the point of singularity, but which excludes a symmetric subinterval, taking the limit of the integral as the length of the interval approaches ∞while, simultaneously, the length of the excluded interval approaches zero. Henceforth, whenever we write an integral as in (1), we will mean the Cauchy principal value of that integral (when it exists). 1 2 Some Basic Properties Some obvious properties of the Hilbert transform follow directly from the definition. Clearly the Hilbert transform of a time-domain signal g(t) is another time-domain signal ˆ g(t). If g(t) is real-valued, then so is ˆ g(t). Linearity: The Hilbert transform is linear, i.e., if a1anda2are arbitrary (complex) scalars, andg1(t) and g2(t) are signals, then [a1g1(t) +a2g2(t)]∧=a1ˆg1(t) +a2ˆg2(t). (This follows immediately from the fact that the Hilbert transform is the output of a linear system.) The Hilbert transform of a constant signal: Note that, for any constant c, the Hilbert transform of the constant signal g(t) =cis ˆg(t) = ˆc= 0. (See Exercise 2.) From linearity it follows that H[g(t) +c] = ˆg(t) + ˆc= ˆg(t). Thus, like an ideal differentiator, a Hilbert transformer “loses” dc offsets. Later we will define an inverse Hilbert transform which can recover the original signal up to an additive constant (in the same way that integration can undo differentiation only up to an additive constant). Time-shifting and time-dilation: Ifg(t) has Hilbert transform ˆ g(t), then g(t−t0) has Hilbert transform ˆ g(t−t0), and g(at) has Hilbert transform sgn(a)ˆg(at) (assuming a/negationslash= 0). Convolution: The Hilbert transform behaves nicely with respect to convolution, since [g1(t)∗g2(t)]∧= ˆg1(t)∗g2(t) =g1(t)∗ˆg2(t). To see this, observe from the associative and commutative properties of convolution that [g1(t)∗g2(t)]∗1 πtcan be written as [ g1(t)∗1 πt]∗g2(t) or as g1(t)∗[g2(t)∗1 πt]. Time-derivative: The Hilbert transform of the derivative of a signal is the derivative of the Hilbert transform, i.e., H[d dtg(t)] =d dtH[g(t)]. To see this, recall Leibniz’s Integral Rule, which states that d dc/integraldisplayb(c) a(c)f(x, c)dx=/integraldisplayb(c) a(c)∂ ∂cf(x, c)dx+f(b, c)d dcb(c)−f(a, c)d dca(c). 2 In particular, if aandbare definite limits (independent of c), we have d dc/integraldisplayb af(x, c)dx=/integraldisplayb a∂ ∂cf(x, c)dx. Now d dtH[g(t)] =1 πd dt/integraldisplay∞ −∞g(t−τ) τdτ =1 π/integraldisplay∞ −∞g/prime(t−τ) τdτ =H[g/prime(t)], where g/prime(t) =d dtg(t). 3 Interaction with the Fourier Transform The signal 1 /(πt) has Fourier transform −j sgn(f) =  −j,iff >0 0,iff= 0 j,iff <0 Ifg(t) has Fourier transform G(f), then, from the convolution property of the Fourier trans- form, it follows that ˆ g(t) has Fourier transform ˆG(f) =−j sgn(f)G(f). Thus, the Hilbert transform is easier to understand in the frequency domain than in the time domain: the Hilbert transform does not change the magnitude ofG(f), it changes only the phase . Fourier transform values at positive frequencies are multiplied by −j(correspond- ing to a phase change of −π/2) while Fourier transform values at negative frequencies are multiplied by j(corresponding to a phase change of π/2). Stated yet another way, suppose that G(f) =a+bjfor some f. Then ˆG(f) =b−ajiff >0 and ˆG(f) =−b+ajiff <0. Thus the Hilbert transform essentially acts to exchange the real and imaginary parts of G(f) (while changing the sign of one of them). Energy Spectral Density: Suppose that g(t) is an energy signal. Then, since |ˆG(f)|= |G(f)|, both ˆG(f) and G(f) have exactly the same energy spectral density. Thus, for exam- ple, if G(f) is bandlimited to BHz then so is ˆG(f). It also follows that ˆ g(t) has exactly the same energy as g(t). 3 Symmetry Properties: Ifg(t) is real-valued, then G(f) exhibits Hermitian symmetry, i.e.,G(−f) =G∗(f). Of course then ˆG(−f) =−j sgn(−f)G(−f) = [−j sgn(f)G(f)]∗= ˆG(f)∗, so ˆG(f) also exhibits Hermitian symmetry, as expected. Letg(t) be a real-valued signal. Recall that if g(t) is even (so that g(−t) =g(t)) then G(f) is purely real-valued while if g(t) is odd (so that g(−t) =−g(t)) then G(f) is purely imaginary- valued. Now if G(f) is purely real-valued then certainly ˆG(f) is purely imaginary-valued (and vice-versa). Thus if g(t) is even, then ˆ g(t) is odd and if g(t) is odd, then ˆ g(t) is even. Orthogonality: Ifg(t) is a real-valued energy signal, then g(t) and ˆ g(t) are orthogonal. To see this recall that /angbracketleftg(t),ˆg(t)/angbracketright=/integraldisplay∞ −∞g(t)ˆg∗(t)dt =/integraldisplay∞ −∞G(f)ˆG∗(f)d f =/integraldisplay∞ −∞G(f)[−j sgn(f)G(f)]∗d f =/integraldisplay∞ −∞j|G(f)|2sgn(f)d f = 0 , where we have used the property that, since |G(f)|2is an even function of f,|G(f)|2sgn(f) is an odd function of fand hence the value of the integral is zero. Low-pass High-pass Products: Letg(t) be signal whose Fourier transform satisfies G(f) = 0 for |f| ≥Wand let h(t) be a signal with H(f) = 0 for |f|< W . Then H[g(t)h(t)] =g(t)ˆh(t), i.e., to compute the Hilbert transform of the product of a low-pass signal with a high-pass signal, only the high-pass signal needs to be transformed. To see this, let s(t) =g(t)h(t) so that S(f) =G(f)∗H(f) =G(f)∗[H(f)u(f) +H(f)u(−f)], where udenotes the unit step function. It is easy to see that G(f)∗(H(f)u(f)) is zero if f <0; similarly G(f)∗(H(f)u(−f)) is zero if f >0. Thus ˆS(f) = −j sgn(f)S(f) =−jG(f)∗(H(f)u(f)) + jG(f)∗(H(f)u(−f)) =G(f)∗[−jH(f)u(f) + jH(f)u(−f)] =G(f)∗[−j sgn(f)H(f)] =G(f)∗ˆH(f). 4 An important special case of this arises in the case of QAM modulation. Assuming that mI(t) and mQ(t) are bandlimited to WHz, then, if fc> W , we have H[mI(t)cos(2πfct) +mQ(t)sin(2πfct)] =mI(t)sin(2πfct)−mQ(t)cos(2πfct). (3) Amplitude-modulated Signals: The Hilbert transform of a general amplitude-modulated signal is given by [1] H[g(t)cos(2πfct+θ)] =/bracketleftbigg g(t)∗cos(2πfct) πt/bracketrightbigg cos(2πfct+θ) +/bracketleftbigg g(t)∗sin(2πfct) πt/bracketrightbigg sin(2πfct+θ). To see this, write g(t) asg(t) =gL(t) +gH(t), where gL(t), the “low-pass component” of g(t), is given as gL(t) =g(t)∗2fcsinc(2fct) =g(t)∗sin(2πfct) πt andgH(t), the “high-pass component” of g(t), is given as gH(t) =g(t)−gL(t) =g(t)∗[δ(t)−2fcsinc(2fct)]. Clearly gL(t) contains all frequency components of g(t) from dc up to frequency fc, while gH(t) contains the remaining components at higher frequencies. We have H[g(t)cos(2πfct+θ)] = H[gL(t)cos(2πfct+θ)] +H[gH(t)cos(2πfct+θ)] =gL(t)H[cos(2πfct+θ)] +H[gH(t)]cos(2πfct+θ) where the second equality follows from application of the “low-pass high-pass product” result. Clearly H[cos(2πfct+θ)] = sin(2πfct+θ). It remains to find H[gH(t)]. This is given as H[gH(t)] = g(t)∗[δ(t)−2fcsinc(2fct)]∗1 πt =g(t)∗[1 πt−2fcsinc(2fct)∗1 πt] (a)=g(t)∗[1 πt−2fcsinc(fct)sin(πfct)] =g(t)∗[1 πt(1−2sin2(πfct)] =g(t)∗cos(2πfct) πt, and the main result follows directly. In the equality (a), we have used the fact that H[sinc(t)] = sinc(t/2)sin(πt/2); see Exercise 6. 5 Inverse Hilbert Transform: Note that [ˆG(f)]∧= (−j sgn(f))2G(f) =−sgn2(f)G(f). Except at f= 0, sgn2(f) = 1. Thus, unless G(f) has some sort of singularity (e.g., a delta function) at f= 0, we get H[H[g(t)]] = −g(t). A delta function at f= 0 corresponds to a nonzero dc offset; which, as already remarked upon, is lost by the Hilbert transform. Thus, assuming that g(t) has zero mean1, we may recover g(t) from ˆ g(t): the inverse Hilbert transform is given by applying the Hilbert transform again, and negating the result: g(t) =−H[ˆg(t)] =−ˆg(t)∗1 πt. In general, we have, for some constant c, g(t) =−ˆg(t)∗1 πt+c. Zero-mean signals g(t) and ˆ g(t) are often referred to as a Hilbert transform pair. For every Hilbert transform pair g(t) and ˆ g(t) there is also the dual pair ˆg(t) and −g(t). Table 1 lists some Hilbert transform pairs. Fig. 1 plots rect(t) and H[rect(t)] =1 πln|(2t+ 1)/(2t−1)|. (The rectangular pulse rect(t) is defined as u(t+ 1/2)−u(t−1/2), where u(t) is the unit step.) -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 Figure 1: The function rect(t) and its Hilbert transform1 πln|(2t+ 1)/(2t−1)|. 1The mean of g(t) is defined as lim T→∞1 T/integraltextT/2 −T/2g(t)dt, when this limit exists. 6 Table 1: Hilbert transform pairs. (See also [2].) g(t) ˆg(t) a1g1(t) +a2g2(t);a1, a2∈C a1ˆg1(t) +a2ˆg2(t) h(t−t0) ˆh(t−t0) h(at);a/negationslash= 0 sgn(a)ˆh(at) d dth(t)d dtˆh(t) δ(t)1 πt ejt−jejt e−jtje−jt cos(t) sin(t) rect(t)1 πln|(2t+ 1)/(2t−1)| sinc(t)πt 2sinc2(t/2) = sin(πt/2)sinc(t/2) 1/(1 + t2) t/(1 + t2) 7 4 Single-sideband Modulation For any signal g(t), let g+(t) =1 2[g(t) + jˆg(t)] g−(t) =1 2[g(t)−jˆg(t)] be two complex-valued signals associated with g(t). The significance of these two signals can be seen from their Fourier transforms. We have G+(f) =1 2[G(f) + jˆG(f)] =1 2[G(f)−j2sgn(f)G(f)] =G(f)1 2[1 + sgn(f)] =G(f)u(f) where u(f) is the unit step function, and, similarly, G−(f) =G(f)u(−f). Thus g+(t) has spectral components (equal to those of g(t)) at positive frequencies only , i.e., g+(t) has a right-sided spectrum. Similarly, g−(t) has spectral components (equal to those ofg(t)) at negative frequencies only and hence has a left-sided spectrum. These spectra are illustrated in Fig. 2. W −WG(f) f WfG+(f) −WfG−(f) (a) (b) (c) Figure 2: Signal spectra: (a) G(f), (b) the right-sided spectrum G+(f), (c) the left-sided spectrum G−(f). It is now straightforward to express upper- and lower-sideband signals in terms of g+(t) and g−(t). Let g(t) be the modulating signal, assumed bandlimited to WHz, and let fc> W be the carrier frequency. In the frequency domain, the upper sideband signal is given by SUSB(f) =G+(f−fc) +G−(f+fc), and the lower sideband signal is given by SLSB(f) =G−(f−fc) +G+(f+fc), as sketched in Fig. 3 below. 8 f fc −fcSUSB(f) f fc −fcSLSB(f) (a) (b) Figure 3: Single-sideband spectra: (a) upper-sideband, (b) lower-sideband. It follows from the frequency-shifting property of the Fourier transform that sUSB(f) = g+(t) exp( j2πfct) +g−(t) exp( −j2πfct) =1 2(g(t) + jˆg(t)) exp( j2πfct) +1 2(g(t)−jˆg(t)) exp( −j2πfct) =g(t)1 2[exp( j2πfct) + exp( −j2πfct)] + ˆg(t)1 2[jexp( j2πfct)−jexp(−j2πfct)] =g(t)cos(2πfct)−ˆg(t)sin(2πfct). A similar derivation shows that sLSB(f) =g(t)cos(2πfct) + ˆg(t)sin(2πfct). Thus we see that single-sideband modulation can be regarded and implemented as a form of quadrature amplitude modulation (QAM), with the modulating signal g(t) placed in the in-phase channel and the Hilbert transform of g(t) (or its negative) placed in the quadrature channel. A block diagram illustrating this approach is given in Fig. 4. productmodulatorproductmodulator phaseshiftercos(2πfct) H sin(2πfct)+ + LSB−USB message signal g(t) ˆg(t) Figure 4: Generation of an SSB-modulated signal. 9 5 Exercises 1.Given an expression for the Cauchy principal value of the second integral in (1). 2.Show that ˆ c= 0 for any constant c. 3.Find the Hilbert transform of g(t) = 1 /t. 4.Verify that H[rect(t)] =1 πln|(2t+ 1)/(2t−1)|. 5.Verify that H[1/(1 + t2)] =t/(1 + t2). 6.Verify that H[sinc(t)] = sin(πt/2)sinc(t/2). 7.If hL(t) =sin(2πWt) πt is the impulse response of an ideal lowpass filter with cutoff frequency W, what oper- ation does the LTI system with impulse response hH(t) =cos(2πWt) πt perform? Show that hH(t) implements the Hilbert transform of the output of an ideal highpass filter with cutoff frequency W. 6 Solutions to Exercises 1. 1 π/integraldisplay∞ −∞g(t−τ) τdτ= lim /epsilon1→0+/parenleftBigg 1 π/integraldisplay−/epsilon1 −1//epsilon1g(t−τ) τdτ+1 π/integraldisplay1//epsilon1 /epsilon1g(t−τ) τdτ/parenrightBigg . 2.We have ˆc= lim /epsilon1→0+/parenleftBigg 1 π/integraldisplay−/epsilon1 −1//epsilon1c τdτ+1 π/integraldisplay1//epsilon1 /epsilon1c τdτ/parenrightBigg = lim /epsilon1→0+/parenleftBigg −1 π/integraldisplay1//epsilon1 /epsilon1c τdτ+1 π/integraldisplay1//epsilon1 /epsilon1c τdτ/parenrightBigg = lim /epsilon1→0+/parenleftBigg 1 π/integraldisplay1//epsilon1 /epsilon1/parenleftBigc τ−c τ/parenrightBig dτ/parenrightBigg = 0 10 3.We have H[δ(t)] =1 πt, and H[1 πt] =−δ(t). Thus H[1/t] =−πδ(t). 4.Letg(t) =πrect(t). Then ˆg(t) =/integraldisplayt+1/2 t−1/21 τdτ. There are four cases. •In case t >1/2, this integral evaluates to ln((t+ 1/2)/(t−1/2)). •By symmetry, in case t <−1/2, this integral evaluates to −ln((t−1/2)/(t+1/2)), which can be written as ln((t+ 1/2)/(t−1/2)). •In case 0 < t < 1/2, the region of integration spans the singularity at τ= 0. The Cauchy principal value of the integral over the interval [ t−1/2,−t+ 1/2] is zero; therefore what remains is the integral over the interval ( −t+ 1/2, t+ 1/2), which evaluates to ln((t+ 1/2)/(−t+ 1/2)). •By symmetry, in case −1/2< t < 0, we find that the integral over the interval [−t−1/2, t+ 1/2] is zero; therefore what remains is the integral over the interval [t−1/2,−t−1/2], which evaluates to −ln((−t+1/2)/(t+1/2) = ln((t+1/2)/(−t+ 1/2)). These four cases can be combined by writing ˆg(t) = ln|(t+ 1/2)/(t−1/2)|=ln|(2t+ 1)/(2t−1)|. 5.Letg(t) = 2 /(1+t2), so that g(t) =1 1−jt+1 1+ jt. We have F[1/(1−jt)] = 2 πe−2πfu(f). By the time-reversal property of the Fourier transform, we get F[1/(1+ jt)] = 2 πe2πfu(−f), and hence G(f) = 2 πe−2πfu(f) + 2 πe2πfu(−f). Multiplying by −j sgn(f), we get ˆG(f) =−j2πe−2πfu(f) + j2πe2πfu(−f), so ˆg(t) =−j 1−jt+j 1 + jt=2t 1 +t2. 6.From the identity sinc(t) = sinc(t/2)cos(πt/2), apply the low-pass high-pass product result to get H[sinc(t)] = sinc(t/2)sin(πt/2). 7.An ideal highpass filter with cutoff frequency Whas frequency response H(f) = 1 − rect(f/2W), and hence has impulse response δ(t)−2Wsinc(2Wt). Convolving this 11 with 1 /(πt) (to get the Hilbert transform) we get hH(t) =1 πt−2Wsinc(2Wt)∗1 πt =1 πt−2Wsinc(Wt)sin(πWt) =1 πt(1−2sin2(πWt)) =cos(2πWt) πt References [1]S. L. Hahn, “Comments on ‘A Tabulation of Hilbert Transforms for Electrical Engi- neers’,” IEEE Trans. on Commun. , vol. 44, p. 768, July 1996. [2]S. L. Hahn, “Hilbert transforms,” in The Transforms and Applications Handbook (A. Poularakis, Ed.), Boca Raton FL: CRC Press, 1996, ch. 7. 12