Chapter 5
Digital Processing of
Continuous-Time Signals
§ 5.1 Digital Processing of
Continuous-Time Signals
Digital processing of a continuous-time
signal involves the following basic steps:
(1) Conversion of the continuous-time
signal into a discrete-time signal,
(2) Processing of the discrete-time signal,
(3) Conversion of the processed discrete-
time signal back into a continuous-time
signal
§ 5.1 Digital Processing of
Continuous-Time Signals
Conversion of a continuous-time
signal into digital form is carried
out by an analog-to-digital (A/D)
converter
The reverse operation of converting
a digital signal into a continuous-
time signal is performed by a
digital-to-analog (D/A) converter
§ 5.1 Digital Processing of
Continuous-Time Signals
Since the A/D conversion takes a finite
amount of time,a sample-and-hold (S/H)
circuit is used to ensure that the analog
signal at the input of the A/D converter
remains constant in amplitude until the
conversion is complete to minimize the
error in its representation
§ 5.1 Digital Processing of
Continuous-Time Signals
To prevent aliasing,an analog anti-
aliasing filter is employed before the
S/H circuit
To smooth the output signal of the
D/A converter,which has a
staircase-like waveform,an analog
reconstruction filter is used
§ 5.1 Digital Processing of
Continuous-Time Signals
Since both the anti-aliasing filter and the
reconstruction filter are analog lowpass filters,
we review first the theory behind the design of
such filters
Also,the most widely used IIR digital filter
design method is based on the conversion of
an analog lowpass prototype
Anti-aliasing
filter
S/H A/D D/A ReconstructionfilterDSP
Complete block-diagram
§ 5.2 Sampling of
Continuous-time Signals
As indicated earlier,discrete-time
signals in many applications are
generated by sampling continuous-time
signals
We have seen earlier that identical
discrete-time signals may result from the
sampling of more than one distinct
continuous-time function
§ 5.2 Sampling of
Continuous-time Signals
In fact,there exists an infinite number of
continuous-time signals,which when
sampled lead to the same discrete-time
signal
However,under certain conditions,it is
possible to relate a unique continuous-
time signal to a given discrete-time
signal
§ 5.2 Sampling of
Continuous-time Signals
If these conditions hold,then it is
possible to recover the original
continuous-time signal from its sampled
values
We next develop this correspondence
and the associated conditions
§ 5.2 Sampling of
Continuous-time Signals
Let ga(t) be a continuous-time signal that
is sampled uniformly at t = nT,
generating the sequence g[n] where
g[n] = ga(nT),-? < n <?
with T being the sampling period
The reciprocal of T is called the
sampling frequency FT,i.e.,FT =1/T
§ 5.2 Sampling of
Continuous-time Signals
Now,the frequency-domain
representation of ga(t) is given by its
continuos-time Fourier transform
(CTFT):
dtetgjG tjaa )()(
n njj engeG ][)(
The frequency-domain representation of
g[n] is given by its discrete-time Fourier
transform (DTFT):
§ 5.3 Effect of Sampling in the
Frequency Domain
To establish the relation between Ga(j?)
and G(ej?),we treat the sampling
operation mathematically as a
multiplication of ga(t) by a periodic
impulse train p(t):
n
nTttp )()(?)(tga )(tgp
)(tp
§ 5.3 Effect of Sampling in the
Frequency Domain
p(t) consists of a train of ideal impulses
with a period T as shown below
n
aap nTtnTgtptgtg )()()()()(
The multiplication operation yields an
impulse train:
§ 5.3 Effect of Sampling in the
Frequency Domain
gp(t) is a continuous-time signal
consisting of a train of uniformly spaced
impulses with the impulse at t = nT
weighted by the sampled value ga(nT) of
ga(t) at that instant t=nT
§ 5.3 Effect of Sampling in the
Frequency Domain
There are two different forms of Gp(j?),
One form is given by the weighted sum of the
CTFTs of?(t-nT):
n nTjap enTgjG )()(
k
ktj
k
ktTj Te
T
e
T
tp 11)( )/2(?
TT /?2where
To derive the second form,we note that p(t) can
be expressed as a Fourier series:
§ 5.3 Effect of Sampling in the
Frequency Domain
The impulse train gp(t) therefore can be
expressed as
)()( 1 tgetg a
k
ktj
Tp
T
From the frequency-shifting property of
the CTFT,the CTFT of ej?Tktga(t) is given
by Ga(j(? - k?T))
§ 5.3 Effect of Sampling in the
Frequency Domain
Hence,an alternative form of the CTFT
of gp(t) is given by
k
TaTp kjGjG )()(
1
Therefore,Gp(j?) is a periodic function
of? consisting of a sum of shifted and
scaled replicas of Ga(j?),shifted by integer
multiples of?T and scaled by 1/T
§ 5.3 Effect of Sampling in the
Frequency Domain
The term on the RHS of the previous
equation for k = 0 is the baseband
portion of Gp(j?),and each of the
remaining terms are the frequency
translated portions of Gp(j?)
The frequency range
22
TT
is called the baseband or Nyquist band
§ 5.3 Effect of Sampling in the
Frequency Domain
Assume ga(t) is a band-limited signal
with a CTFT Ga(j?) as shown below
The spectrum P(j?) of p(t) having a sampling
period T=2?/?T is indicated below
§ 5.3 Effect of Sampling in the
Frequency Domain
Two possible spectra of Gp(j?) are
shown below
§ 5.3 Effect of Sampling in the
Frequency Domain
It is evident from the top figure on the
previous slide that if?T>2?m,there is
no overlap between the shifted replicas
of Ga(j?) generating Gp(j?)
On the other hand,as indicated by the
figure on the bottom,if?T<2?m,there
is an overlap of the spectra of the shifted
replicas of Ga(j?) generating Gp(j?)
§ 5.3 Effect of Sampling in the
Frequency Domain
If?T>2?m,ga(t) can be recovered
exactly from gp(t) by passing it through
an ideal lowpass filter Hr(j?) with a gain
T and a cutoff frequency?c greater than
m and less than?T -?m as shown
below
§ 5.3 Effect of Sampling in the
Frequency Domain
The spectra of the filter and pertinent
signals are shown below
§ 5.3 Effect of Sampling in the
Frequency Domain
On the other hand,if?T< 2?m,due to
the overlap of the shifted replicas of
Ga(j?),the spectrum Gp(j?) cannot be
separated by filtering to recover Ga(j?)
because of the distortion caused by a
part of the replicas immediately outside
the baseband folded back or aliased into
the baseband
§ 5.3 Effect of Sampling in the
Frequency Domain
Sampling theorem - Let ga(t) be a band-
limited signal with CTFT Ga(j?)=0 for
|?|>?m
Then ga(t) is uniquely determined by its
samples ga(nT),-nif
T? 2?m
where?T=2?/T
§ 5.3 Effect of Sampling in the
Frequency Domain
The condition?T? 2?m is often
referred to as the Nyquist condition
The frequency?T/2 is usually
referred to as the folding frequency
§ 5.3 Effect of Sampling in the
Frequency Domain
and then passing it through an ideal
lowpass filter Hr(j?) with a gain T and a
cutoff frequency?c satisfying
m <?c < (?T -?m )
n
ap nTtnTgtg )()()(?
Given {ga(nT)},we can recover exactly
ga(t) by generating an impulse train
§ 5.3 Effect of Sampling in the
Frequency Domain
The highest frequency?m contained in
ga(t) is usually called the Nyquist
frequency since it determines the
minimum sampling frequency?T =2?m
that must be used to fully recover ga(t)
from its sampled version
The frequency 2?m is called the Nyquist
rate
§ 5.3 Effect of Sampling in the
Frequency Domain
Oversampling - The sampling frequency
is higher than the Nyquist rate
Undersampling - The sampling
frequency is lower than the Nyquist rate
Critical sampling - The sampling
frequency is equal to the Nyquist rate
Note,A pure sinusoid may not be
recoverable from its critically sampled
version
§ 5.3 Effect of Sampling in the
Frequency Domain
In digital telephony,a 3.4 kHz
signal bandwidth is acceptable for
telephone conversation
Here,a sampling rate of 8 kHz,
which is greater than twice the
signal bandwidth,is used
§ 5.3 Effect of Sampling in the
Frequency Domain
In high-quality analog music signal
processing,a bandwidth of 20 kHz
has been determined to preserve the
fidelity
Hence,in compact disc (CD) music
systems,a sampling rate of 44.1 kHz,
which is slightly higher than twice
the signal bandwidth,is used
§ 5.3 Effect of Sampling in the
Frequency Domain
Example - Consider the three
continuous-time sinusoidal signals:
)6c o s ()(1 ttg
)14c o s ()(2 ttg
)26c o s()(3 ttg
)]26()26([)(3jG
)]14()14([)(2jG
)]6()6([)(1jG
Their corresponding CTFTs are:
§ 5.3 Effect of Sampling in the
Frequency Domain
These three transforms are plotted below
§ 5.3 Effect of Sampling in the
Frequency Domain
These continuous-time signals sampled
at a rate of T = 0.1 sec,i.e.,with a
sampling frequency?T =20?rad/sec
The sampling process generates the
continuous-time impulse trains,g1p(t),
g2p(t),and g3p(t)
Their corresponding CTFTs are given
by
31,)(10)( k Tp kjGjG
§ 5.3 Effect of Sampling in the
Frequency Domain
Plots of the 3 CTFTs are shown below
§ 5.3 Effect of Sampling in the
Frequency Domain
These figures also indicate by dotted
lines the frequency response of an ideal
lowpass filter with a cutoff at
c=?T/2=10?and a gain T=0.1
The CTFTs of the lowpass filter output
are also shown in these three figures
In the case of g1(t),the sampling rate
satisfies the Nyquist condition,hence no
aliasing
§ 5.3 Effect of Sampling in the
Frequency Domain
Moreover,the reconstructed output is
precisely the original continuous-time
signal
In the other two cases,the sampling rate
does not satisfy the Nyquist condition,
resulting in aliasing and the filter
outputs are all equal to cos(6?t)
§ 5.3 Effect of Sampling in the
Frequency Domain
Now,the CTFT Gp(jis a
periodic function of? with a period
T = 2?/T
Because of the mapping,the DTFT
G(ej?) is a periodic function of?
with a period 2?
§ 5.4 Recovery of the Analog Signal
The impulse response hr(t) of the
lowpass reconstruction filter is obtained
by taking the inverse DTFT of Hr(j?)
)( jH
r
c
cT
,0
,
)(tga^
We now derive the expression for the
output of the ideal lowpass
reconstruction filter Hr(j?) as a function
of the samples g[n]
§ 5.4 Recovery of the Analog Signal
Thus,the impulse response is given by
c c dedejHth tjTtjrr 22 1 )()(
tt t
T
c,
2/
)sin (
np nTtngtg )(][)(
The input to the lowpass filter is the
impulse train gp(t):
§ 5.4 Recovery of the Analog Signal
Substituting hr(t)=sin(?ct)/(?Tt/2) in the
above and assuming for simplicity
c=?T/2=?/T,we get
n
rpra nTthngtgthtg )(][)()()(^ *
)(tga^?
n TnTt
TnTtng
/)(
]/)(s in [][
)(tga^? Therefore,the output of the ideal
lowpass filter is given by:
which is called Poisson sum formula
§ 5.4 Recovery of the Analog Signal
The ideal bandlimited interpolation
process is illustrated below
Illustration of Poisson sum formula
§ 5.5 Implication of the Sampling Process
Consider again the three continuous-
time signals,g1(t)=cos(6?t),
g2(t)=cos(14?t),and g3(t)=cos(26?t)
The plot of the CTFT G1p(j?) of the
sampled version g1p(t) of g1(t) is shown
below
§ 5.5 Implication of the Sampling Process
From the plot,it is apparent that we
can recover any of its frequency-
translated versions cos[(20k?6)?t]
outside the baseband by passing
g1p(t) through an ideal analog
bandpass filter with a passband
centered at? = (20k?6)?
§ 5.5 Implication of the Sampling Process
For example,to recover the signal
cos(34?t),it will be necessary to employ
a bandpass filter with a frequency
response
)( jH
r o th e r w is e,0
)34()34(,1.0
where? is a small number
§ 5.5 Implication of the Sampling Process
Likewise,we can recover the aliased
baseband component cos(6?t) from the
sampled version of either g2p(t) or g3p(t)
by passing it through an ideal lowpass
filter with a frequency response
)( jH
r o th e r w is e,0
)6()6(,1.0
§ 5.5 Implication of the Sampling Process
There is no aliasing distortion unless the
original continuous-time signal also
contains the component cos(6?t)
Similarly,from either g2p(t) or g3p(t) we
can recover any one of the frequency-
translated versions,including the parent
continuous-time signal g2(t) or g3(t) as the
case may be,by employing suitable
filters
§ 5.6 Sampling of Bandpass Signals
The conditions developed earlier for the
unique representation of a continuous-time
signal by the discrete-time signal obtained by
uniform sampling assumed that the
continuous-time signal is bandlimited in the
frequency range from DC to some frequency
T
Such a continuous-time signal is commonly
referred to as a lowpass signal
§ 5.6 Sampling of Bandpass Signals
There are applications where the continuous-
time signal is bandlimited to a higher
frequency range?L? |?|H with?L >0
Such a signal is usually referred to as the
bandpass signal
To prevent aliasing a bandpass signal can of
course be sampled at a rate greater than twice
the highest frequency,i.e,by ensuring
T?2?H
§ 5.6 Sampling of Bandpass Signals
However,due to the bandpass spectrum
of the continuous-time signal,the
spectrum of the discrete-time signal
obtained by sampling will have spectral
gaps with no signal components present
in these gaps
Moreover,if?H is very large,the
sampling rate also has to be very large
which may not be practical in some
situations
§ 5.6 Sampling of Bandpass Signals
A more practical approach is to use
under-sampling
Let=?H -?L define the bandwidth
of the bandpass signal
Assume first that the highest frequency
H contained in the signal is an integer
multiple of the bandwidth,i.e.,
H = M()
§ 5.6 Sampling of Bandpass Signals
We choose the sampling frequency?T to
satisfy the condition
T = 2() = 2?H/M
which is smaller than 2?H,the Nyquist
rate
Substitute the above expression in
k
TaTp kjGjG )()(
1
§ 5.6 Sampling of Bandpass Signals
As before,Gp(j?) consists of a sum of Ga(j?) and
replicas of Gp(j?) shifted by integer multiples
of twice the bandwidthand scaled by 1/T
The amount of shift for each value of k
ensures that there will be no overlap between
all shifted replicas
k kjjGjG aTp )()( 21
This leads to
no aliasing
§ 5.6 Sampling of Bandpass Signals
Figure below illustrate the idea behind
)(?jGa
0LH H?L?
)(?jGp
0
LH H?L?
§ 5.6 Sampling of Bandpass Signals
As can be seen,ga(t) can be recovered from
gp(t) by passing it through an ideal bandpass
filter with a passband given by?L? |?|H
and a gain of T
Note,Any of the replicas in the lower
frequency bands can be retained by passing
through bandpass filters with passbands
L- k()? |?|H - k(),1? k? M-1
providing a translation to lower frequency
ranges
§ 5.7 Analog Lowpass Filter
Specifications
Typical magnitude response |Ha(j?)| of
an analog lowpass filter may be given as
indicated below
§ 5.7 Analog Lowpass Filter
Specifications
In the passband,defined by 0p,we
require
1-?p? |Ha(j?)|? 1+?p,|?|p
i.e.,|Ha(j?)| approximates unity within an
error ofp
In the stopband,defined by?s,we
require
|Ha(j?)|s?s
i.e.,|Ha(j?)| approximates zero within an
error of?s
§ 5.7 Analog Lowpass Filter
Specifications
p - passband edge frequency
s - stopband edge frequency
p - peak ripple value in the passband
s - peak ripple value in the stopband
Peak passband ripple
)(l o g20 10 ss dB
)1(l o g20 10 pp dB
Minimum stopband attenuation
§ 5.7 Analog Lowpass Filter
Specifications
Magnitude specifications may
alternately be given in a normalized
form as indicated below
§ 5.7 Analog Lowpass Filter
Specifications
Here,the maximum value of the magnitude in
the passband assumed to be unity
1 +?(1 +?2) - Maximum passband deviation,
given by the minimum value of the magnitude
in the passband
1/A - Maximum stopband magnitude
§ 5.8 Analog Lowpass Filter
Design
Two additional parameters are defined -
(1) Transition ratio k =?p/?s
For a lowpass filter k<1
(2) Discrimination parameter k1 =?/?(A2 -1)
Usually k1<<1
§ 5.8.1 Butterworth Approximation
First 2N - 1 derivatives of |Ha(j?)|2 at? = 0 are
equal to zero
The Butterworth lowpass filter thus is said to
have a maximally-flat magnitude at? = 0
N
c
a jH 2
2
)/(1
1)(
The magnitude-square response of an N-th
order analog lowpass Butterworth filter is given
by
§ 5.8.1 Butterworth Approximation
Gain in dB is
G(?)=10log10|Ha(j?)|2
As G(0)=0 and
G(?c)=10log10(0.5)=-3.0103?-3 dB
c is called 3-dB cutoff frequency
§ 5.8.1 Butterworth Approximation
Typical magnitude responses with?c =1
0 1 2 3
0
0.2
0.4
0.6
0.8
1
M
a
gni
t
ude
B ut t e r w or t h F i l t e r
N = 2
N = 4
N = 10
§ 5.8.1 Butterworth Approximation
Two parameters completely
characterizing a Butterworth
lowpass filter are?c and N
These are determined from the
specified bandedges?p and?s,and
minimum passband magnitude
1/?(1 +?2),and maximum stopband
ripple 1/A
§ 5.8.2 Chebyshev Approximation
The magnitude-square response of an N-
th order analog lowpass Type 1
Chebyshev filter is given by
)/(1
1)(
22
2
pN
a TsH
1),c o shc o sh (
1),c o sc o s()(
1
1
N
NT
N
where TN(?) is the Chebyshev polynomial
of order N:
§ 5.8.2 Chebyshev Approximation
Typical magnitude response plots of the
analog lowpass Type 1 Chebyshev filter
are shown below
0 1 2 3
0
0.2
0.4
0.6
0.8
1
M
a
gni
t
ude
T ype 1 C he bys he v F i l t e r
N = 2
N = 3
N = 8
§ 5.8.2 Chebyshev Approximation
If at? =?c the magnitude is equal to 1/A,
then
222
2 1
)/(1
1)(
AT
jH
psN
sa
)/1(c o s h
)/1(c o s h
)/(c o s h
)/1(c o s h
1
1
1
1
21
k
kAN
ps
Order N is chosen as the nearest integer greater
than or equal to the above value
Solving the above we get
§ 5.8.2 Chebyshev Approximation
The magnitude-square response of an N-th
order analog lowpass Type 2 Chebyshev (also
called inverse Chebyshev) filter is given by 2
2
2
)/(
)/(
1
1
)(
sN
psN
a
T
T
jH
where TN(?) is the Chebyshev polynomial of
order N
§ 5.8.2 Chebyshev Approximation
Typical magnitude response plots of the
analog lowpass Type 2 Chebyshev filter
are shown below
0 1 2 3
0
0.2
0.4
0.6
0.8
1
M
a
gni
t
ude
T ype 2 C he bys he v F i l t e r
N = 3
N = 5
N = 7
§ 5.8.2 Chebyshev Approximation
The order N of the Type 2 Chebyshev filter is
determined from given?,?s,and A using
)/1(c o s h
)/1(c o s h
)/(c o s h
)/1(c o s h
1
1
1
1
21
k
kAN
ps
6059.2
)/1(c o s h
)/1(c o s h
1
1
1
k
kN
Example - Determine the lowest order of a
Chebyshev lowpass filter with a 1-dB cutoff
frequency at 1 kHz and a minimum attenuation
of 40 dB at 5 kHz -
§ 5.8.3 Elliptic Approximation
where RN(?) is a rational function of order N
satisfying RN(1/?)=1/ RN(?),with the roots of
its numerator lying in the interval 1<?<1 and
the roots of its denominator lying in the
interval 1<? <?
)/(1
1)(
22
2
pN
a RjH
The square-magnitude response of an
elliptic lowpass filter is given by
§ 5.8.3 Elliptic Approximation
For given?p,?s,?,and A,the filter
order can be estimated using
)/1(lo g
)/4(lo g2
10
110
kN?
21' kk
)'1(2
'1
0 k
k
13090500 )(1 5 0)(15)(2
where
§ 5.8.3 Elliptic Approximation
Example - Determine the lowest order of a
elliptic lowpass filter with a 1-dB cutoff
frequency at 1 kHz and a minimum
attenuation of 40 dB at 5 kHz
Note,k = 0.2 and 1/k1=196.5134
Substituting these values we get
k’=0.979796,?0=0.00255135,
=0.0025513525
and hence N = 2.23308
Choose N = 3
§ 5.8.3 Elliptic Approximation
Typical magnitude response plots with
are shown below
0 1 2 3
0
0.2
0.4
0.6
0.8
1
M
a
gni
t
ude
E l l i pt i c F i l t e r
N = 3
N = 4
Homework
Read the textbook from p.299 to 329
Problems
5.2,5.4,5.5,5.8,5.11,5.17,5.25
M5.1,M5.4
Digital Processing of
Continuous-Time Signals
§ 5.1 Digital Processing of
Continuous-Time Signals
Digital processing of a continuous-time
signal involves the following basic steps:
(1) Conversion of the continuous-time
signal into a discrete-time signal,
(2) Processing of the discrete-time signal,
(3) Conversion of the processed discrete-
time signal back into a continuous-time
signal
§ 5.1 Digital Processing of
Continuous-Time Signals
Conversion of a continuous-time
signal into digital form is carried
out by an analog-to-digital (A/D)
converter
The reverse operation of converting
a digital signal into a continuous-
time signal is performed by a
digital-to-analog (D/A) converter
§ 5.1 Digital Processing of
Continuous-Time Signals
Since the A/D conversion takes a finite
amount of time,a sample-and-hold (S/H)
circuit is used to ensure that the analog
signal at the input of the A/D converter
remains constant in amplitude until the
conversion is complete to minimize the
error in its representation
§ 5.1 Digital Processing of
Continuous-Time Signals
To prevent aliasing,an analog anti-
aliasing filter is employed before the
S/H circuit
To smooth the output signal of the
D/A converter,which has a
staircase-like waveform,an analog
reconstruction filter is used
§ 5.1 Digital Processing of
Continuous-Time Signals
Since both the anti-aliasing filter and the
reconstruction filter are analog lowpass filters,
we review first the theory behind the design of
such filters
Also,the most widely used IIR digital filter
design method is based on the conversion of
an analog lowpass prototype
Anti-aliasing
filter
S/H A/D D/A ReconstructionfilterDSP
Complete block-diagram
§ 5.2 Sampling of
Continuous-time Signals
As indicated earlier,discrete-time
signals in many applications are
generated by sampling continuous-time
signals
We have seen earlier that identical
discrete-time signals may result from the
sampling of more than one distinct
continuous-time function
§ 5.2 Sampling of
Continuous-time Signals
In fact,there exists an infinite number of
continuous-time signals,which when
sampled lead to the same discrete-time
signal
However,under certain conditions,it is
possible to relate a unique continuous-
time signal to a given discrete-time
signal
§ 5.2 Sampling of
Continuous-time Signals
If these conditions hold,then it is
possible to recover the original
continuous-time signal from its sampled
values
We next develop this correspondence
and the associated conditions
§ 5.2 Sampling of
Continuous-time Signals
Let ga(t) be a continuous-time signal that
is sampled uniformly at t = nT,
generating the sequence g[n] where
g[n] = ga(nT),-? < n <?
with T being the sampling period
The reciprocal of T is called the
sampling frequency FT,i.e.,FT =1/T
§ 5.2 Sampling of
Continuous-time Signals
Now,the frequency-domain
representation of ga(t) is given by its
continuos-time Fourier transform
(CTFT):
dtetgjG tjaa )()(
n njj engeG ][)(
The frequency-domain representation of
g[n] is given by its discrete-time Fourier
transform (DTFT):
§ 5.3 Effect of Sampling in the
Frequency Domain
To establish the relation between Ga(j?)
and G(ej?),we treat the sampling
operation mathematically as a
multiplication of ga(t) by a periodic
impulse train p(t):
n
nTttp )()(?)(tga )(tgp
)(tp
§ 5.3 Effect of Sampling in the
Frequency Domain
p(t) consists of a train of ideal impulses
with a period T as shown below
n
aap nTtnTgtptgtg )()()()()(
The multiplication operation yields an
impulse train:
§ 5.3 Effect of Sampling in the
Frequency Domain
gp(t) is a continuous-time signal
consisting of a train of uniformly spaced
impulses with the impulse at t = nT
weighted by the sampled value ga(nT) of
ga(t) at that instant t=nT
§ 5.3 Effect of Sampling in the
Frequency Domain
There are two different forms of Gp(j?),
One form is given by the weighted sum of the
CTFTs of?(t-nT):
n nTjap enTgjG )()(
k
ktj
k
ktTj Te
T
e
T
tp 11)( )/2(?
TT /?2where
To derive the second form,we note that p(t) can
be expressed as a Fourier series:
§ 5.3 Effect of Sampling in the
Frequency Domain
The impulse train gp(t) therefore can be
expressed as
)()( 1 tgetg a
k
ktj
Tp
T
From the frequency-shifting property of
the CTFT,the CTFT of ej?Tktga(t) is given
by Ga(j(? - k?T))
§ 5.3 Effect of Sampling in the
Frequency Domain
Hence,an alternative form of the CTFT
of gp(t) is given by
k
TaTp kjGjG )()(
1
Therefore,Gp(j?) is a periodic function
of? consisting of a sum of shifted and
scaled replicas of Ga(j?),shifted by integer
multiples of?T and scaled by 1/T
§ 5.3 Effect of Sampling in the
Frequency Domain
The term on the RHS of the previous
equation for k = 0 is the baseband
portion of Gp(j?),and each of the
remaining terms are the frequency
translated portions of Gp(j?)
The frequency range
22
TT
is called the baseband or Nyquist band
§ 5.3 Effect of Sampling in the
Frequency Domain
Assume ga(t) is a band-limited signal
with a CTFT Ga(j?) as shown below
The spectrum P(j?) of p(t) having a sampling
period T=2?/?T is indicated below
§ 5.3 Effect of Sampling in the
Frequency Domain
Two possible spectra of Gp(j?) are
shown below
§ 5.3 Effect of Sampling in the
Frequency Domain
It is evident from the top figure on the
previous slide that if?T>2?m,there is
no overlap between the shifted replicas
of Ga(j?) generating Gp(j?)
On the other hand,as indicated by the
figure on the bottom,if?T<2?m,there
is an overlap of the spectra of the shifted
replicas of Ga(j?) generating Gp(j?)
§ 5.3 Effect of Sampling in the
Frequency Domain
If?T>2?m,ga(t) can be recovered
exactly from gp(t) by passing it through
an ideal lowpass filter Hr(j?) with a gain
T and a cutoff frequency?c greater than
m and less than?T -?m as shown
below
§ 5.3 Effect of Sampling in the
Frequency Domain
The spectra of the filter and pertinent
signals are shown below
§ 5.3 Effect of Sampling in the
Frequency Domain
On the other hand,if?T< 2?m,due to
the overlap of the shifted replicas of
Ga(j?),the spectrum Gp(j?) cannot be
separated by filtering to recover Ga(j?)
because of the distortion caused by a
part of the replicas immediately outside
the baseband folded back or aliased into
the baseband
§ 5.3 Effect of Sampling in the
Frequency Domain
Sampling theorem - Let ga(t) be a band-
limited signal with CTFT Ga(j?)=0 for
|?|>?m
Then ga(t) is uniquely determined by its
samples ga(nT),-nif
T? 2?m
where?T=2?/T
§ 5.3 Effect of Sampling in the
Frequency Domain
The condition?T? 2?m is often
referred to as the Nyquist condition
The frequency?T/2 is usually
referred to as the folding frequency
§ 5.3 Effect of Sampling in the
Frequency Domain
and then passing it through an ideal
lowpass filter Hr(j?) with a gain T and a
cutoff frequency?c satisfying
m <?c < (?T -?m )
n
ap nTtnTgtg )()()(?
Given {ga(nT)},we can recover exactly
ga(t) by generating an impulse train
§ 5.3 Effect of Sampling in the
Frequency Domain
The highest frequency?m contained in
ga(t) is usually called the Nyquist
frequency since it determines the
minimum sampling frequency?T =2?m
that must be used to fully recover ga(t)
from its sampled version
The frequency 2?m is called the Nyquist
rate
§ 5.3 Effect of Sampling in the
Frequency Domain
Oversampling - The sampling frequency
is higher than the Nyquist rate
Undersampling - The sampling
frequency is lower than the Nyquist rate
Critical sampling - The sampling
frequency is equal to the Nyquist rate
Note,A pure sinusoid may not be
recoverable from its critically sampled
version
§ 5.3 Effect of Sampling in the
Frequency Domain
In digital telephony,a 3.4 kHz
signal bandwidth is acceptable for
telephone conversation
Here,a sampling rate of 8 kHz,
which is greater than twice the
signal bandwidth,is used
§ 5.3 Effect of Sampling in the
Frequency Domain
In high-quality analog music signal
processing,a bandwidth of 20 kHz
has been determined to preserve the
fidelity
Hence,in compact disc (CD) music
systems,a sampling rate of 44.1 kHz,
which is slightly higher than twice
the signal bandwidth,is used
§ 5.3 Effect of Sampling in the
Frequency Domain
Example - Consider the three
continuous-time sinusoidal signals:
)6c o s ()(1 ttg
)14c o s ()(2 ttg
)26c o s()(3 ttg
)]26()26([)(3jG
)]14()14([)(2jG
)]6()6([)(1jG
Their corresponding CTFTs are:
§ 5.3 Effect of Sampling in the
Frequency Domain
These three transforms are plotted below
§ 5.3 Effect of Sampling in the
Frequency Domain
These continuous-time signals sampled
at a rate of T = 0.1 sec,i.e.,with a
sampling frequency?T =20?rad/sec
The sampling process generates the
continuous-time impulse trains,g1p(t),
g2p(t),and g3p(t)
Their corresponding CTFTs are given
by
31,)(10)( k Tp kjGjG
§ 5.3 Effect of Sampling in the
Frequency Domain
Plots of the 3 CTFTs are shown below
§ 5.3 Effect of Sampling in the
Frequency Domain
These figures also indicate by dotted
lines the frequency response of an ideal
lowpass filter with a cutoff at
c=?T/2=10?and a gain T=0.1
The CTFTs of the lowpass filter output
are also shown in these three figures
In the case of g1(t),the sampling rate
satisfies the Nyquist condition,hence no
aliasing
§ 5.3 Effect of Sampling in the
Frequency Domain
Moreover,the reconstructed output is
precisely the original continuous-time
signal
In the other two cases,the sampling rate
does not satisfy the Nyquist condition,
resulting in aliasing and the filter
outputs are all equal to cos(6?t)
§ 5.3 Effect of Sampling in the
Frequency Domain
Now,the CTFT Gp(jis a
periodic function of? with a period
T = 2?/T
Because of the mapping,the DTFT
G(ej?) is a periodic function of?
with a period 2?
§ 5.4 Recovery of the Analog Signal
The impulse response hr(t) of the
lowpass reconstruction filter is obtained
by taking the inverse DTFT of Hr(j?)
)( jH
r
c
cT
,0
,
)(tga^
We now derive the expression for the
output of the ideal lowpass
reconstruction filter Hr(j?) as a function
of the samples g[n]
§ 5.4 Recovery of the Analog Signal
Thus,the impulse response is given by
c c dedejHth tjTtjrr 22 1 )()(
tt t
T
c,
2/
)sin (
np nTtngtg )(][)(
The input to the lowpass filter is the
impulse train gp(t):
§ 5.4 Recovery of the Analog Signal
Substituting hr(t)=sin(?ct)/(?Tt/2) in the
above and assuming for simplicity
c=?T/2=?/T,we get
n
rpra nTthngtgthtg )(][)()()(^ *
)(tga^?
n TnTt
TnTtng
/)(
]/)(s in [][
)(tga^? Therefore,the output of the ideal
lowpass filter is given by:
which is called Poisson sum formula
§ 5.4 Recovery of the Analog Signal
The ideal bandlimited interpolation
process is illustrated below
Illustration of Poisson sum formula
§ 5.5 Implication of the Sampling Process
Consider again the three continuous-
time signals,g1(t)=cos(6?t),
g2(t)=cos(14?t),and g3(t)=cos(26?t)
The plot of the CTFT G1p(j?) of the
sampled version g1p(t) of g1(t) is shown
below
§ 5.5 Implication of the Sampling Process
From the plot,it is apparent that we
can recover any of its frequency-
translated versions cos[(20k?6)?t]
outside the baseband by passing
g1p(t) through an ideal analog
bandpass filter with a passband
centered at? = (20k?6)?
§ 5.5 Implication of the Sampling Process
For example,to recover the signal
cos(34?t),it will be necessary to employ
a bandpass filter with a frequency
response
)( jH
r o th e r w is e,0
)34()34(,1.0
where? is a small number
§ 5.5 Implication of the Sampling Process
Likewise,we can recover the aliased
baseband component cos(6?t) from the
sampled version of either g2p(t) or g3p(t)
by passing it through an ideal lowpass
filter with a frequency response
)( jH
r o th e r w is e,0
)6()6(,1.0
§ 5.5 Implication of the Sampling Process
There is no aliasing distortion unless the
original continuous-time signal also
contains the component cos(6?t)
Similarly,from either g2p(t) or g3p(t) we
can recover any one of the frequency-
translated versions,including the parent
continuous-time signal g2(t) or g3(t) as the
case may be,by employing suitable
filters
§ 5.6 Sampling of Bandpass Signals
The conditions developed earlier for the
unique representation of a continuous-time
signal by the discrete-time signal obtained by
uniform sampling assumed that the
continuous-time signal is bandlimited in the
frequency range from DC to some frequency
T
Such a continuous-time signal is commonly
referred to as a lowpass signal
§ 5.6 Sampling of Bandpass Signals
There are applications where the continuous-
time signal is bandlimited to a higher
frequency range?L? |?|H with?L >0
Such a signal is usually referred to as the
bandpass signal
To prevent aliasing a bandpass signal can of
course be sampled at a rate greater than twice
the highest frequency,i.e,by ensuring
T?2?H
§ 5.6 Sampling of Bandpass Signals
However,due to the bandpass spectrum
of the continuous-time signal,the
spectrum of the discrete-time signal
obtained by sampling will have spectral
gaps with no signal components present
in these gaps
Moreover,if?H is very large,the
sampling rate also has to be very large
which may not be practical in some
situations
§ 5.6 Sampling of Bandpass Signals
A more practical approach is to use
under-sampling
Let=?H -?L define the bandwidth
of the bandpass signal
Assume first that the highest frequency
H contained in the signal is an integer
multiple of the bandwidth,i.e.,
H = M()
§ 5.6 Sampling of Bandpass Signals
We choose the sampling frequency?T to
satisfy the condition
T = 2() = 2?H/M
which is smaller than 2?H,the Nyquist
rate
Substitute the above expression in
k
TaTp kjGjG )()(
1
§ 5.6 Sampling of Bandpass Signals
As before,Gp(j?) consists of a sum of Ga(j?) and
replicas of Gp(j?) shifted by integer multiples
of twice the bandwidthand scaled by 1/T
The amount of shift for each value of k
ensures that there will be no overlap between
all shifted replicas
k kjjGjG aTp )()( 21
This leads to
no aliasing
§ 5.6 Sampling of Bandpass Signals
Figure below illustrate the idea behind
)(?jGa
0LH H?L?
)(?jGp
0
LH H?L?
§ 5.6 Sampling of Bandpass Signals
As can be seen,ga(t) can be recovered from
gp(t) by passing it through an ideal bandpass
filter with a passband given by?L? |?|H
and a gain of T
Note,Any of the replicas in the lower
frequency bands can be retained by passing
through bandpass filters with passbands
L- k()? |?|H - k(),1? k? M-1
providing a translation to lower frequency
ranges
§ 5.7 Analog Lowpass Filter
Specifications
Typical magnitude response |Ha(j?)| of
an analog lowpass filter may be given as
indicated below
§ 5.7 Analog Lowpass Filter
Specifications
In the passband,defined by 0p,we
require
1-?p? |Ha(j?)|? 1+?p,|?|p
i.e.,|Ha(j?)| approximates unity within an
error ofp
In the stopband,defined by?s,we
require
|Ha(j?)|s?s
i.e.,|Ha(j?)| approximates zero within an
error of?s
§ 5.7 Analog Lowpass Filter
Specifications
p - passband edge frequency
s - stopband edge frequency
p - peak ripple value in the passband
s - peak ripple value in the stopband
Peak passband ripple
)(l o g20 10 ss dB
)1(l o g20 10 pp dB
Minimum stopband attenuation
§ 5.7 Analog Lowpass Filter
Specifications
Magnitude specifications may
alternately be given in a normalized
form as indicated below
§ 5.7 Analog Lowpass Filter
Specifications
Here,the maximum value of the magnitude in
the passband assumed to be unity
1 +?(1 +?2) - Maximum passband deviation,
given by the minimum value of the magnitude
in the passband
1/A - Maximum stopband magnitude
§ 5.8 Analog Lowpass Filter
Design
Two additional parameters are defined -
(1) Transition ratio k =?p/?s
For a lowpass filter k<1
(2) Discrimination parameter k1 =?/?(A2 -1)
Usually k1<<1
§ 5.8.1 Butterworth Approximation
First 2N - 1 derivatives of |Ha(j?)|2 at? = 0 are
equal to zero
The Butterworth lowpass filter thus is said to
have a maximally-flat magnitude at? = 0
N
c
a jH 2
2
)/(1
1)(
The magnitude-square response of an N-th
order analog lowpass Butterworth filter is given
by
§ 5.8.1 Butterworth Approximation
Gain in dB is
G(?)=10log10|Ha(j?)|2
As G(0)=0 and
G(?c)=10log10(0.5)=-3.0103?-3 dB
c is called 3-dB cutoff frequency
§ 5.8.1 Butterworth Approximation
Typical magnitude responses with?c =1
0 1 2 3
0
0.2
0.4
0.6
0.8
1
M
a
gni
t
ude
B ut t e r w or t h F i l t e r
N = 2
N = 4
N = 10
§ 5.8.1 Butterworth Approximation
Two parameters completely
characterizing a Butterworth
lowpass filter are?c and N
These are determined from the
specified bandedges?p and?s,and
minimum passband magnitude
1/?(1 +?2),and maximum stopband
ripple 1/A
§ 5.8.2 Chebyshev Approximation
The magnitude-square response of an N-
th order analog lowpass Type 1
Chebyshev filter is given by
)/(1
1)(
22
2
pN
a TsH
1),c o shc o sh (
1),c o sc o s()(
1
1
N
NT
N
where TN(?) is the Chebyshev polynomial
of order N:
§ 5.8.2 Chebyshev Approximation
Typical magnitude response plots of the
analog lowpass Type 1 Chebyshev filter
are shown below
0 1 2 3
0
0.2
0.4
0.6
0.8
1
M
a
gni
t
ude
T ype 1 C he bys he v F i l t e r
N = 2
N = 3
N = 8
§ 5.8.2 Chebyshev Approximation
If at? =?c the magnitude is equal to 1/A,
then
222
2 1
)/(1
1)(
AT
jH
psN
sa
)/1(c o s h
)/1(c o s h
)/(c o s h
)/1(c o s h
1
1
1
1
21
k
kAN
ps
Order N is chosen as the nearest integer greater
than or equal to the above value
Solving the above we get
§ 5.8.2 Chebyshev Approximation
The magnitude-square response of an N-th
order analog lowpass Type 2 Chebyshev (also
called inverse Chebyshev) filter is given by 2
2
2
)/(
)/(
1
1
)(
sN
psN
a
T
T
jH
where TN(?) is the Chebyshev polynomial of
order N
§ 5.8.2 Chebyshev Approximation
Typical magnitude response plots of the
analog lowpass Type 2 Chebyshev filter
are shown below
0 1 2 3
0
0.2
0.4
0.6
0.8
1
M
a
gni
t
ude
T ype 2 C he bys he v F i l t e r
N = 3
N = 5
N = 7
§ 5.8.2 Chebyshev Approximation
The order N of the Type 2 Chebyshev filter is
determined from given?,?s,and A using
)/1(c o s h
)/1(c o s h
)/(c o s h
)/1(c o s h
1
1
1
1
21
k
kAN
ps
6059.2
)/1(c o s h
)/1(c o s h
1
1
1
k
kN
Example - Determine the lowest order of a
Chebyshev lowpass filter with a 1-dB cutoff
frequency at 1 kHz and a minimum attenuation
of 40 dB at 5 kHz -
§ 5.8.3 Elliptic Approximation
where RN(?) is a rational function of order N
satisfying RN(1/?)=1/ RN(?),with the roots of
its numerator lying in the interval 1<?<1 and
the roots of its denominator lying in the
interval 1<? <?
)/(1
1)(
22
2
pN
a RjH
The square-magnitude response of an
elliptic lowpass filter is given by
§ 5.8.3 Elliptic Approximation
For given?p,?s,?,and A,the filter
order can be estimated using
)/1(lo g
)/4(lo g2
10
110
kN?
21' kk
)'1(2
'1
0 k
k
13090500 )(1 5 0)(15)(2
where
§ 5.8.3 Elliptic Approximation
Example - Determine the lowest order of a
elliptic lowpass filter with a 1-dB cutoff
frequency at 1 kHz and a minimum
attenuation of 40 dB at 5 kHz
Note,k = 0.2 and 1/k1=196.5134
Substituting these values we get
k’=0.979796,?0=0.00255135,
=0.0025513525
and hence N = 2.23308
Choose N = 3
§ 5.8.3 Elliptic Approximation
Typical magnitude response plots with
are shown below
0 1 2 3
0
0.2
0.4
0.6
0.8
1
M
a
gni
t
ude
E l l i pt i c F i l t e r
N = 3
N = 4
Homework
Read the textbook from p.299 to 329
Problems
5.2,5.4,5.5,5.8,5.11,5.17,5.25
M5.1,M5.4
