2-1

2. Bandpass Signal Theory

• Throughout this course, we will adopt the baseband representation of abandpass signal.

• References :

1. J.G. Proakis, Digital Communications, McGraw Hill, 3rd Edition, 1995, Section 4.1

2.1 The Complex Envelop

2.1.1 Introduction

• By definition, a bandpass signal have null at and near DC.

• Any bandpass signal can be written in the form

( ) ( )cos(2 ) ( )sin(2 )

( ) cos(2 ( ))c c

c

s t x t f t y t f t

A t f t t

π ππ φ

= −= +

Information is contained in the envelop (amplitude and phase), not thecarrier. Distinguish between them with a concise notation :

{ }2( ) Re ( ) cj f ts t z t e π= where

( )( ) ( ) ( ) ( ) j tz t x t jy t A t e φ= + =

2-2

is the complex envelop, or the baseband equivalent of s(t). It is like a time-varying phasor

• The complex representation of a bandpass signal is not necessarily unique.Consider a bandpass signal with the spectrum shown below.

{ }

2

1

2

1

2 2

2 2

2

( ) ( ) 2Re ( )

Re 2 ( )

Re ( )

c

c

c

c

f Wj ft j ft

f W

Wj f t j t

c

W

j f t

s t S f e df S f e df

e S f e d

e z t

π π

π πβ

π

β β

+∞

−∞ −

−

= =

= +

=

∫ ∫

∫

where ( ) 2 ( )cZ f S f f+= + and 2

1

2( ) ( )W

j ft

W

z t Z f e dfπ

−

= ∫ , and subscript + means

+ve frequency only.

Notes

1. choice of “center” frequency is not unique

( )S f

fcf 2cf W+1cf W−cf− 1cf W− +

2cf W− −

( )Z f

f2W1W−

2-3

2. transform is not necessarily symmetric about cf ,3. ( ) and ( )A t tφ are amplitude and phase with respect to the unmodulated

carrier 2 cj f te π .4. transform Z(f) twice as large as either +ve or –ve component of S(f)

2.1.2 Generation of Bandpass Signals by Complex Envelop

• The band pass signal ( )x t can be generated using the quadrature modulatorshown in the diagram below

• Note that

1. x(t) and y(t) are generated by a modem

2. has to be balanced

LocalOscillator

( )x t

( )y t

cos(2 )cf tπ

( )s tsin(2 )cf tπ−

( )z t

2 cj f te π

Re[ ]•( )s t

2-4

a) 90 degree separation or crosstalkb) same gain each branch or distortion (e.g. constant envelop

becomes non constant envelop; SSB generation withy(t)=H[x(t)] has imperfect sideband cancellation).

2.1.3 Recovery of Complex Envelop

• Let the received band pass signal be

{ }2

( ) ( ) cos(2 ) ( )sin(2 )

( ) cos(2 ( ))

Re ( ) c

c c

c

j f t

r t u t f t v t f t

B t f t t

w t e π

π ππ θ

= −= +

=

where ( )( ) ( ) ( ) ( ) j tw t u t jv t B t e θ= + = is the signal we want to recover.

• Consider the multiplication of r(t) by { }22cos(2 ) Re 2 cj f tcf t e ππ = .

From properties of complex numbers, we can easily show that for complex

and α β , { }*12Re{ }Re{ } Reα β αβ αβ= + . So

2 2 412

4

Re{ ( ) }Re{2 } Re{2 ( ) 2 ( ) }

( ) Re{ ( ) }

c c c

c

j f t j f t j f t

j f t

w t e e w t w t e

u t w t e

π π π

π

= +

= +

The first term is a low pass signal and the second term is a doublefrequency term. We can thus recover ( )u t by using a low pass filter (LPF).

2-5

• Similarly, we can recover ( )v t by multiplying r(t) by

{ }22sin(2 ) Re 2 cj f tcf t je ππ− = and then low pass filter the product signal.

• In conclusion, the following structure is known as a product demodulator.

Note that complex envelop phase is with respect to receiver’s localoscillator.

• Exercise : Let the transmitted complex envelop be ( ) ( ) ( )z t x t jy t= + and

the received complex envelop be ( ) ( ) jw t z t e ω= . Determine the output ofthe quadrature demodulator shown above.

LocalOscillator

( )r t

2cos(2 )cf tπ( )u t

2sin(2 )cf tπ−

LPF

LPF

( )v t

( )r t

2 cj f te π−

LPF( )w t

2-6

2.1.4 Bandpass filtering

• Adopt the notation that ( )s t! (or ( )S f! in the frequency domain) is a realbandpass signal and ( )s t (or ( )S f in the frequency domain) its complexenvelop

• Let ( )s t! be the input to a bandpass filter with an impulse response ( )h t! .

The output of the filter is denoted by ( )r t! .

• Clearly ( ) ( ) ( )R f H f S f= !! ! and hence ( ) ( ) ( )R f H f S f+ + += !! !

So

( )h t!( )s t! ( )r t!

( )S f!

f

( )H f!

f

( )R f!

fcf 2cf W+1cf W−cf− 1cf W− +

2cf W− −

2-7

( ) ( ) ( )R f H f S f=

where ( ) 2 ( )cS f S f f+= +! , ( ) ( )cH f H f f+= +! , and ( ) 2 ( )cR f R f f+= +! . In otherword, convolution of real bandpass signals is equivalent to convolution ofthe corresponding complex envelopes.

Notes : no factor of 2 in ( ) ( )cH f H f f+= +! .

• Exercise : Use the time domain approach to prove the above property.

• Example : In digital communications, the most common filter used is amatched filter. Let ( ) ( ) cos(2 ) ( )sin(2 )c cs t x t f t y t f tπ π= −! be a bandpass

pulse. Its matched filter has an impulse response equal to ( ) 2 ( )h t s t= −! ! .Determine the output of the matched filter and sketch its implementationstructure.

Solution

- By definition, ( ) 2 ( )cos(2 ) 2 ( )sin(2 )c ch t x t f t y t f tπ π= − + −! . This means

the equivalent low pass response is ( ) ( ) ( )h t x t jy t= − − − . The

complex envelop of the input bandpass signal is ( ) ( ) ( )s t x t jy t= +

- The output of the bandpass matched filter is

[ ]{ }{ }

2

2

( ) Re ( ) ( )

Re ( )

c

c

j f t

j f t

r t h t s t e

r t e

π

π

= ⊗

=

!

where the complex envelop of the output signal is

2-8

[ ] [ ][ ] [ ]

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( )

r t x t jy t x t jy t

x t x t y t y t j y t x t x t y t

a t jb t

= + ⊗ − − −

= ⊗ − + ⊗ − + ⊗ − − ⊗ −= +

- The above equations suggest the following implementation structure

where the filter bank computes a(t) and b(t) from x(t) and y(t)according to the last equation.

- in practice, there is no need to do the remodulation because we areonly interested in the matched filter output at time t kT= , where T isthe bit or symbol duration, and k an integer. In this case, thequadrature modulator in the above diagram can be replaced by theoperation (which implicitly involves sampling)

[ ]{ }2Re ( ) ( ) cj f kTa kT jb kT e π+

Futher simplication can be obtained if cf T is an integer. In this case,the output of the bandpass filter is simply

( ) ( ) ( ) ( ) ( )a t x t x t y t y t= ⊗ − + ⊗ − . In other word, the filter bankonly has the filters x(-t) and y(-t). There is no need to duplicate thesefilters.

2.1.5 Fourier Symmetry Relations

( )r t

QuadratureDemodulator

QuadratureModulator

Real FilterBank

( )s t( )x t

( )y t

( )a t

( )b t

2-9

• Now that we’re committed to complex time functions, better reviewFourier symmetries. They will become handy later on.

• Given ( ) ( )s t S f↔ , we have

1. 2 2( ) ( ) ; s( ) ( )j ft j ftS f s t e dt t S f e dfπ π∞ ∞

−

−∞ −∞

= =∫ ∫

2. time/freqency reversal : ( ) ( )s t S f− ↔ −

3. * *( ) *( ); ( ) *( )s t S f S f s t↔ − ↔ −

4. conjugate symmetry : s(t) real ! *( ) ( )S f S f− ↔

5. conjugate odd symmetry : s(t) imag ! *( ) ( )S f S f− ↔ −

6. s(t) real and even ! S(f) real and even

7. If s(t) = x(t) + jy(t), x(t) and y(t) both real, then

*12

*12

Re{ ( )} ( ) ( ) ( )

Im{ ( )} ( ) ( ) ( )j

s t X f S f S f

s t Y f S f S f

↔ = + − ↔ = − −

8. complex frequency shift : 2( ) ( )oj f t

os t e S f fπ ↔ −

9. convolution : * * *( ) ( ) ( ) ( ) ( ) ( )s t r t s r t d S f R fα α α⊗ = − ↔ −∫

10. cross correlation : * * *( ) ( ) ( ) ( ) ( ) ( )s t r t s r t d S f R fα α α⊗ − = − ↔∫

11. Parsaval : * *( ) ( ) ( ) ( )s r d S f R f dfα α α =∫ ∫

2-10

• From Properties 5 & 6, we can deduce that the spectrum of a complexenvelop is not symmetrical about f = 0. A non-symmetrical spectrumcarries more information, which simply reflects the fact that the complexenvelop carries two pieces of information.

2.1.6 Bandpass Random Processes

• Suppose we form a bandpass process by up converting a complex lowpassprocess ( ) ( ) ( )z t x t jy t= + , i.e

{ }2( ) Re ( ) ( ) cos(2 ) ( )sin(2 )cj f tc cz t z t e x t f t y t f tπ π π= = −!

How does the second order statistics of ( )z t! depends on those of z(t) ?

• The autocorrelation function of ( )z t! is defined as

[ ]{ } { }

{ }

2 2 ( )

2 2 (2 )*12

( , ) ( ) ( )

Re ( ) Re ( )

Re ( ) ( ) ( ) ( )

c c

c c

z

j f t j f t

j f j f t

R t t E z t z t

E z t e z t e

E z t z t e z t z t e

π π τ

π τ π τ

τ τ

τ

τ τ

−

−

− = −

= − = − + −

! ! !

Note that the first term is wide-sense stationary (WSS) if z(t) is WSS. Thesecond term depends on t.

• Under what conditions does the second term vanish ? Assuming that x(t)and y(t) are WSS, the second term can be expanded as :

2-11

[ ]{ }{ }

2 (2 )

2 (2 )

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

c

c

j f t

j f tx y xy xy

E x t x t y t y t j x t y t y t x t e

R R j R R e

π τ

π τ

τ τ τ τ

τ τ τ τ

−

−

− − − + − + −

= − + + −

So it should be clear that the second term vanishes if

1. ( ) ( ) and ( ) ( )x y xy xyR R R Rτ τ τ τ= = − − , or

2. instead of [ ]E • , we use a time average in the definition of theautocorrelation function ( , )zR t t τ−! , or

3. if carrier has a random phase θ , i.e instead of 2 cj f te π , we have(2 )cj f te π θ+

• From now on, we ignore the second term and the autocorrelation can nowbe rewritten as

{ }{ }

2*12

2

( , ) ( ) Re ( ) ( )

Re ( )

c

c

j fz z

j fz

R t t R E z t z t e

R e

π τ

π τ

τ τ τ

τ

− = = −

=

! !

where

{ }*1

2

12

( ) ( ) ( )

( ) ( ) ( ) ( )

z

x y xy xy

R E z t z t

R R j R R

τ τ

τ τ τ τ

= −

= + + − −

In other word, ( )zR τ is the complex envelop of ( )zR τ! .

• Note that *( ) ( )z zR Rτ τ− =! ! , i.e. even symmetry.

2-12

• The Fourier transform of ( )zR τ is denoted by ( )zS f and it is the power

spectral density (PSD) of z(t). ( )zS f is real but not necessarily even. It is so

if and only if (iff) ( )zR τ is real, i.e.

( ) ( )xy xyR Rτ τ= −

This is satisfied if x(t) and y(t) are uncorrelated (i.e. ( ) 0xyR τ = ) oridentical.

• Conclude : if x(t) and y(t) are uncorrelated, ( )zS f is real and even.

Consequently ( )zS f! , the spectrum of the bandpass signal is symmetricalabout cf .

• We have { }2( ) Re ( ) cj fz zR R e π ττ τ=! . This means ( )1

2z z x yP P P P= = +! , i.e

modulation cuts the power in half.

• Up to this point, we were considering the “modulation” aspect of bandpassprocesses, i.e, given that the low pass processes x(t) and y(t) are WSS, wedetermine the properties of the bandpass process ( )z t! .

Now look at the “demodulation” aspect of bandpass processes. Given that( )z t! is WSS, what are the properties of x(t) and y(t) ?

• The autocorrelation function of ( )z t! is now written as :

[ ]{ } 2 2 (2 )*1

2

( , ) ( ) ( )

Re ( ) ( ) { ( ) ( )}c c

z

j f j f t

R t t E z t z t

E z t z t e E z t z t eπ τ π τ

τ τ

τ τ −

− = −

= − + −

! ! !

2-13

Since ( )z t! is WSS, ( , )zR t t τ−! is a function only of τ . This means

*12 ( ) ( ) ( )zE z t z t Rτ τ − =

and[ ]1

2 ( ) ( ) 0E z t z t τ− =They imply

( ) ( )x yR Rτ τ= and ( ) ( )xy xyR Rτ τ= − −

{ } { }1 12 2( ) ( ) ( ) ( ) ( )

( ) ( )

z x y xy xy

x xy

R R R j R R

R jR

τ τ τ τ ττ τ

= + + − −

= −

The last equation shows even more clearly that, unless ( )zS f! issymmetrical about cf , x(t) and y(t) are correlated.

• Example : If ( )z t! is a bandlimited white noise with PSD equal to / 2oN , i.e.

then

cf /2cf W+/2cf W−/2cf W− +/2cf W− −cf−

( )zS f!

2/oN

f

2/W2/W−

oN

( )zS f

f

2-14

So x(t) and y(t) are uncorrelated (independent if Gaussian), each with PSDoN

2.2 Sampling Theorem

• In the received modem, the received signal is first filtered and thensampled. Recovery of the transmitted data can be achieved by properlyprocessing the samples, using for example a DSP. So lets revisit thesampling theorem.

• Let x(t) be a real signal with a low pass spectrum X(f) shown below. Thebandwidth of the signal is B

Impulse-sampling of x(t) yields

2/W2/W−

oN

( )xS f

f

2/W2/W−

oN

( )yS f

f

sf−

( )X f

2 sf− sfB− Bf

i i i i i i

( )sX f

B− Bf

2 sf

2-15

( ) ( ) ( / )s sk

x t x t t k fδ= ⊗ −∑

where sf is the sampling frequency. Minimum 2sf B= real samples persecond.

• What about complex z(t) ?

Still sample at 2sf B≥ , but take complex samples (2 reals). This reflectsthe extra information in non-symmetric spectrum Z(f).

• Now look at sampling requirements for a bandpass signal ( )z t! withbandwidth B.

Do we need 2( / 2)s cf f B≥ + ? The answer is NO and the proofs are asfollow.

Proof 1 :

Since the carrier cf carries no information, bring the bandpass signal downto baseband using a quadrature demodulator :

QuadratureDemodulator

( )z t!( )z t rate 2(B/2) complex samples/s

or 2B real samples/s

i i ii i i

2 sf− sf− sf

( )Z f

B− Bf

( )sZ f

B− Bf

2 sf

2-16

Proof 2 :

If we take real samples of ( )z t! at rate 2B, we have (for convenience, we set2 / 2cf B B= + ):

So information is preserved with 2B reals/sec.

Can generalize this result to 2 / 2cf Bn B= + , where n an arbitrary integer.

• Exercise : Consider the case of B reals/sec. Show that information is lostwith this sampling frequency.

• In conclusion, need 2B real numbers/sec to represent a real bandlimitedsignal. This in independent of whether the signal is lowpass or bandpass.

cf− cf

B( )Z f!

fi i i i i i

Combined

fi i i i i i

-ve images

fi i i i i i

+ve images

f

2-17

2.3 Random Gaussian Vectors

• Reference : Proakis

• Recall multivariate Gaussian probability density function (pdf) for reals :

1 2( , , , )tNx x x=x # - a real (column) vector, N components,

[ ]E =x m , [( )( ) ]tE − − =x m x m Φ

1, −Φ Φ are, real, symmetric, semi-positive definite.

Then ( ) ( )( )1121/ 2/ 2

1( ) exp

(2 )

t

Np

π−= − − −x x x m Φ x m

Φ

• For complex Gaussian :

( )1 2,, ,t

Nz z z=z … - a complex (column) vector, N components,

[ ]E =z m , †12 [( )( ) ]E − − =z m z m Φ , †( )• represents Hermitian

transpose.

Now Φ is Hermitian, †=Φ Φ , so it still has real eigenvalues. Sincecovariance matrix they are all non-negative, i.e. semi-positive definite.

The pdf is

( ) ( )( )† 112

1( ) exp

(2 )Np

π−= − − −z z z m Φ z m

Φ

2-18

• Alternatively, write the complex Gaussian vector z as a real vector oflength 2N,

1 1 2 2( , , , , , , )tN Nx y x y x y=w #

[ ]E=m w , ( )( )tE = − − Φ w m w m , and

( ) ( )( )1121/ 2

1( ) exp

(2 )

t

Np

π−= − − −w w w m Φ w m

Φ

• Let 1 2( , , , )tNω ω ω=ω # . The characteristic function of the complex form is

:

( ) ( )† † †12( ) exp expjM j E e j j = = −

ω zz ω m ω ω Φω

Note that

kk

Mjm

ω =0∂ =∂

zω and ( )zM j j∇ =ω ω m

etc. for other moment generation.

• Exercise : Obtain explicit expressions for the joint pdf and characteristicfunction of two zero mean complex Gaussian random variables 1x and 2x .Let their covariance matrix be

11 12*12 22

φ φφ φ

=

Φ

2-19

• Exercise : For the complex Gaussian random variables 1x and 2x in the lastexercise, determine the conditional pdf

1 2

2 1

1

, 1 22 1

1

( , )( )

( )x x

x xx

p x xp x x

p x=

Properly interprete the results.

• Exercise (Generalization of the result in the last exercise) : Consider twozero mean complex Gaussian vectors ( )1 2, , ,

t

nx x x=x # and ( )1 2, , ,t

my y y=y #with a correlation matrix :

( )† †1

2xx xy

yx yy

E

= =

Φ ΦxΦ x y

Φ Φy

Determine the conditional pdf

, ( , )( )

( )x y

y xx

pp

p=

x yy x

x

Use the following identities ( • denotes the determinant of a matrix) :

1xx yy yx xx xy

−= × −Φ Φ Φ Φ Φ Φ

( )11

11 11

xxxx xy

yx xxyy yx xx xy

−−−

− −−

− = − −

Φ 0 I 0I Φ ΦΦ

Φ Φ I0 I 0 Φ Φ Φ Φ

• Quadratic form of Gaussian RVs :

2-20

†D = z Fz

where †=F F , and ( )1 2,, ,t

Nz z z=z … is a collection of zero mean complex

Gaussian RVs.

• Exercise : Obtain an expression for D for the case of N = 2 and

0 1

1 0

=

F

• Characteristic function of quadratic form :

( ) ( )sD sDD zs E e e p d− − Φ = = ∫

z

z z

• Exercise : Determine the characteristic function of the quadratic form inthe last exercise, assuming the correlation matrix for 1 2( , )tz z=z is

11 12*12 22

φ φφ φ

=

Φ

• Decomposition of the correlation matrix Φ :

The correlation matrix Φ of the complex Gaussian RVs ( )1 2,, ,t

Nz z z=z … is

positive semi-definite and can be written as :

†z z z=Φ U Λ U

2-21

where zU is a unitary matrix whose columns are orthogonal with 1 †z z− =U U ,

and ( ),1 ,2 ,diag , , ,z z z z Nλ λ λ=Λ … is a diagonal matrix containing the eigenvalues

of the correlation matrix Φ .

Note – all eigenvalues are non-negative.

• Basically, what the above formula is saying is that correlated Gaussian RVscan be obtained by mixing independent Gaussian RVs in different ways.

Let ( )1 2,, ,t

Nn n n=n … be N independent complex Gaussian RVs with zero

mean and unit variance. The transformation

1/ 2z z=z U Λ n

generates a set of correlated Gaussian RVs with a correlation matrix

( )( )†† 1/ 2 1/ 21 12 2 z z z zE E = =

zz U Λ n U Λ n Φ

• Decomposition of the quadratic form :

- Useful in bit-error probability (BEP) analysis

- The quadratic form †D = z Fz is equivalent (in a statistical sense) to

2

.1

N

g k kk

D nλ=

=∑

where ,g kλ are the eigenvalues of the matrix

=G ΦF

2-22

and ( )1 2,, ,t

Nn n n=n … are a set of independent and identical distributed

(iid) complex Gaussian RVs with zero mean and unit variance. Theproof is given in Appendix 2A

• Exercise : Show that the characteristic function of the quadratic form†D = z Fz can now be written as

( ) ( ) ( )

1

2

sD sD sDD z ns E e e p d e p d

s

− − − Φ = = =

=+

∫ ∫z n

z z n n

I ΦF

Hints :

1. =G ΦF can be written as g g g=G A Λ B where 1g g

−=B A and gΛ is adiagonal matrix containing the eigenvalues of G

2. For any square matrices A, B, and C, × × = × ×A B C A B C

• Major conclusion of this section :

- know your matrix theory, it can be your friend !

2-23

Appendix 2A : Decomposition of a Gaussian quadratic form

• The original expression for the quadratic form is †D = z Fz .

• From the decomposition of the correlation matrix, we can replace z by1/ 2

z z=z U Λ q . This means the quadratic form can now be written as

( )† 1/ 2 † 1/ 2

† 1/ 2 † 1/ 2

†

( )

z z z z

z z z z

D = ≡

=

=

z Fz U Λ q FU Λ q

q Λ U FU Λ q

q Bq

where1/ 2 † 1/ 2z z z z=B Λ U FU Λ

• Since B is Hermitian, it can be written as

†b b b=B U Λ U

where bU is a unitary matrix whose columns are orthogonal, and

( ),1 ,2 ,diag , , ,b b b b Nλ λ λ=Λ … is a diagonal matrix containing the eigenvalues of

the matrix B. This means the quadratic form can now be written as

( ) ( )† †b b bD = q U Λ U q

• Let†

1 2( , , , )tb Nn n n= =n U q …

2-24

This means the quadratic form can now be rewritten as

2†.

1

N

b b k kk

D nλ=

= =∑n Λ n

Since the covariance matrix of n is

† † †1 12 2

†

n b b

b b

E E = = = =

Φ nn U qq U

U IU I

this means the kn are iid Gaussian RVs with zero mean and unit variance.

• So what are the eigenvalues of the matrix B ? By definition, ,k b k kλ=BV V ,where kV is an eigenvector of B. This means

( )1/ 2 † 1/ 2,z z z z k b k kλ=Λ U FU Λ V V

or

( )( ) ( )1/ 2 1/ 2 † 1/ 2 1/ 2,z z z z z z k b k z z kλ=U Λ Λ U FU Λ V U Λ V

or

( ) ( ) ( )1/ 2 1/ 2 † 1/ 2 1/ 2,z z z z z z k b k z z kλ=U Λ Λ U F U Λ V U Λ V

or

,( ) k b k kλ=ΦF W W

In other word, the eigenvalues of B are the eigenvalues of

=G ΦF