1ENSC327

Communications Systems

6: Hilbert Transform

Jie Liang

School of Engineering Science

Simon Fraser University

2Why Hilbert Transform? Fourier Spectrum of real signals:

Amplitude: even function

Phase: odd function.

Only needs to know half of the spectrum.

1

Original spectrum, DSB, and single sideband (SSB):

Original: lowpass signal

DSB: moved to fc and fc.

SSB: only transmit one sideband:

The power & bandwidth can

be reduced by 50%.

How to get SSB?

USB: Move pos (neg) freq to fc (-fc)

LSB: Move neg (pos) freq to fc (-fc)

How to implement this?

f

f

M(f)

DSB(f)

f

USB(f)

f

LSB(f)

fc-fc

amp

phase

3Roadmap

Implementation of the SSB needs analytic signal and

Hilbert transform

Proof of the Hilbert transform needs sgn(t), sgn(f),

exponential function, and duality property

The SSB signal can be written as

( ) ( )( )tftmtftmA

tscc

c

ssbpipi 2sin)(2cos)(

2)( m=

)( tm is the Hilbert transform of m(t).

4Analytic Signal (Pre-Envelope) Decompose X(f) into positive and negative parts:

( )1

( ) ( ) ( )2

p nX f X f X f= +

Xp(f)2Xn(f)

2X(f)

1

=

0. f ,1

0,f ,0

0,f ,1

)sgn( Use f

xp(t) is called analytic signal or pre-envelope of the positive frequencies.

xn(t) is called analytic signal or pre-envelope of the negative frequencies.

5Derivation of Hilbert Transform To get xp(t) in time domain, we need IFT of sgn(f).

To get IFT of sgn(f), we start from exponential pulse (pp. 24),

because direct calculation is not convergent (Page 47).

=

0. t,0

0, t,5.0

0, t,1

)(u where t

),()( tuetg at=fja

fGpi2

1)(+

=

Proof:

Similarly: )()( = tuetg at

6Derivation of Hilbert Transform

Signum function (pp. 47):

=

0. t,1

0, t,0

0, t,1

sgn(t)

Proof:

{ }fjpi

1sgn(t)F =

Signum function can be viewed as the limit of

0. a when

0. t,

0, t,0

0, t,

g(t)

=

at

at

e

e

:0 a

(pp. 27)

g(t) can also be written as:

Its FT is thus:

7Definition of Hilbert Transform

Signum function: sgn(t)g(t) =fj

fGpi

1)( =

By duality property:

8Definition of Hilbert Transform

The Hilbert Transform (HT) of x(t) is the convolution of x(t) with the filter h(t) = 1 / (pit). The HT of x(t) is denoted as

t

txtx

pi

1)()( =

h(t)x(t) )( tx

)( tx

From the result in the last page, we see that

The HT of x(t) has the following properties:

If x(t) is real, then is also real.

Phases at positive frequencies are shifted by -90o.

Phases at negative frequencies are shifted by 90o.

The amplitudes of FT are not changed.

But DC component of the input is discarded!

)( tx

9Some Hilbert Transform Pairs

The HT of a cos signal is a sin signal.

Proof:

0 ),2cos()(00>= ftftx pi ).2sin()( 0tftx pi=

10

Some Hilbert Transform Pairs

0 ),2sin()(00>= ftftx pi

The HT of a sin signal is -cos signal.

Proof:

).2cos()(0tftx pi=

11

Analytic Signal (Pre-envelope) Now go back to our SSB signal:

=+=

0. f ,0

0,f ),0(

0,f ),(2

)()sgn()()( X

fX

fXffXfXp

Xp(f)2

X(f)1

By the HT notation, this can be written as

Note 1: The analytic signal xp(t) is a complex signal.

Note 2: Although xp(t) only has the positive freq of x(t), it has

all info about x(t), because

Taking IFT

12

Analytic Signal (Pre-envelope)

Example:

Solution:( ) ?

px t =

0 ),2cos()(00>= ftftx pi

This is as expected, since Xp(f) is defined to have twice of

the pos. part of X(f):

13

Analytic Signal of the Neg. Freq.

By the HT notation, this can be written as

Note 1: xn(t) is also a complex signal.

Note 2: xn(t) also has all info about x(t), because

Usually we only use xp(t).

Taking IFT

)()sgn()()( fXffXfXn

=

The analytic signal of the negative frequency is: