Lecture 11 Final Topic on AM:

Vestigial Sideband (VSB)

ELG3175 Introduction to Communication Systems

Motivation

•  For wideband information signals, SSB is difficult to implement.

•  For frequency discrimination, the filter must have a sharp cutoff near the frequency fc so as to be able to eliminate one band without distorting the other.

•  When we use phase discrimination, we require Hilbert transformers which are difficult to implement if the signal m(t) has a large bandwidth.

VSB Modulation

•  When SSB is difficult to implement, we use vestigial sideband (VSB) modulation.

•  VSB is implemented by frequency discrimination but the filtering process does not completely eliminate the unwanted band.

•  In fact, some of the desired band is also partially filtered out.

VSB Modulator

•  A VSB modulator and its filter’s frequency response are shown below.

•  The response of the VSB filter is denoted as HVSB(f). •  We notice a transition band around the frequency fc.

m(t) ×

Accos(2πfct)

HVSB(f) sVSB(t)

f -fc fc

HVSB(f)

fc-x fc+x

m(t) ×

Accos(2πfct)

HVSB(f) sVSB(t)

f -fc fc

HVSB(f)

fc-x fc+x

Spectrum of a VSB signal

•  In the example given, we consider a system which retains the upper sideband as the principal band and the lower sideband contributes the vestigial sideband part. However, either sideband can be used to form the principal band in practice.

M(f)

-Bm Bm f

-fc-Bm –fc -fc+Bm fc-Bm fc fc+Bm f

SDSB-SC(f)

SVSB(f)

-fc-Bm –fc-fc+x fc-x fc fc+Bm f

M(f)

-Bm Bm f

-fc-Bm –fc -fc+Bm fc-Bm fc fc+Bm f

SDSB-SC(f)

SVSB(f)

-fc-Bm –fc-fc+x fc-x fc fc+Bm f

•  In the example given, the filter has a gain of K for all frequencies |f|>fc+x (passband).

•  For the frequencies fc < |f| < fc+x, which reside in the principal band, the gain is less than K, so some loss compared to the passband occurs.

•  For the frequencies fc-x < |f| < fc, the filter’s gain is not 0, therefore some of the other sideband’s frequency components are passed by the filter and sVSB(t) has a vestigial sideband.

•  The bandwidth of sVSB(t) = Bm+x. Generally, since we are trying to reduce the bandwidth compared to DSB-SC, x is smaller than Bm.

SVSB(f)

)()(4

)()(4

)()(4

)()(4

)()(2

)()(2

)()()(

fHffMA

fHffMA

fHffMA

fHffMA

fHffMA

fHffMA

fHfSfS

VSBcc

VSBcc

VSBcc

VSBcc

VSBcc

VSBcc

VSBSCDSBVSB

−−

−+

+−

++

−

++++

−+−=

++−=

=

where

⎩⎨⎧

<

>=+

000)(

)(fffH

fH VSBVSB and

⎩⎨⎧

<

>=−

0)(00

)(ffHf

fHVSB

VSB

We note that due to Hermitian symmetry in the frequency response of real systems.

)()( * fHfH VSBVSB −= +−

Demodulation of VSB

•  We use the same demodulator as DSB-SC

•  If we want z(t) = G m(t), then we need to impose a

constraint on the frequency response of the modulator’s filter HVSB(f).

sVSB(t) SVSB(f)

×

Arcos(2πfct)

x(t) X(f)

LPF z(t) = G m(t) Z(f)

X(f)

)(2

)(2

)( cVSBr

cVSBr ffSAffSAfX ++−=

)()2(8

)()2(8

)()(8

)()(8

)()(8

)()(8

)()2(8

)()2(8

cVSBcrc

cVSBcrc

cVSBrc

cVSBrc

cVSBrc

cVSBrc

cVSBcrc

cVSBcrc

ffHffMAA

ffHffMAA

ffHfMAA

ffHfMAA

ffHfMAA

ffHfMAA

ffHffMAA

ffHffMAA

++++++

++++

−+−+

−−+−−=

−−

−+

+−

++

−−

−+

+−

++

Baseband

Z(f)

)()(8

)()(8

)()(8

)()(8

)(

cVSBrc

cVSBrc

cVSBrc

cVSBrc

ffHfMAAffHfMAA

ffHfMAAffHfMAAfZ

++++

−+−=

+−

++

−−

−+

))()()((8

))()()((8

)(

cVSBcVSBrc

cVSBcVSBrc

ffHffHfMAA

ffHffHfMAAfZ

++−+

++−=

+−−

+−+

We want Z(f) = G M(f), where G is a constant. If we ensure that

KffHffH cVSBcVSB =++− +− )()(

)(4

)(8

)(8

)( fMKAA

fMKAA

fMKAA

fZ rcrcrc =+= −+

•  Therefore z(t) = (AcArK/4)m(t). •  Let us replace and f by Δf and we

get

(This criteria only need be true over the frequency range of the VSB signal).

)(by )( * fHfH VSBVSB −+−

KffHffH cVSBcVSB =Δ++Δ− ++ )()(*

fc

K

fc-Δf fc+Δf

x1

x2

x1*+x2 = K

Possible USB filters

•  Filters with linear transition bands •  Raised cosine filters

fc fc-x fc+x

Example

•  The signal m(t) = 2cos(2π10t)+3cos(2π30t). We wish to transmit this signal using VSB with carrier c(t) = 5cos(2π500t). The VSB filter’s response is shown below. Find sVSB(t) as well as its bandwidth.

480 500 520

1

Solution

•  Since VSB uses frequency discrimination, it is probably best to work in the frequency domain.

•  Let us find M(f) and SDSB-SC(f).

M(f)

-30 -10 10 30

1.5 1

SDSB-SC(f)

470 490 510 530

3.75 2.5

-530 -510 -490 -470

•  Next we find SVSB(f) = SDSB-SC(f)HVSB(f).

SDSB-SC(f)HVSB(f)

470 490 510 530

3.75 2.5

-530 -510 -490 -470

HVSB(470)=0 HVSB(490)=1/4 HVSB(510)=3/4 HVSB(530)=1

SVSB(f)

470 490 510 530

3.75 15/8

-530 -510 -490 -470

5/8 5/8 15/8 3.75

sVSB(t) = 7.5cos(2π530t)+3.75cos(2π510t)+ 1.25cos(2π490t)

40

Example 2

•  Show that we can demodulate sVSB(t) of the previous example to obtain G m(t).

•  sVSB(t)cos(2π500t) = 7.5cos(2π530t) cos(2π500t) +3.75cos(2π510t)cos(2π500t) + 1.25cos(2π490t)cos(2π500t) = 3.75cos(2π30t) + 3.75cos(2π1030t) + 1.875cos(2π10t) + 1.875cos(2π1010t) + 0.625cos(2π10t) + 0.625cos(2π990t) .

•  After lowpass filtering z(t) = 3.75cos(2π30t) + 1.875cos(2π10t) + 0.625cos(2π10t) = 3.75cos(2π30t) + 2.5cos(2π10t) = 1.25m(t).