737
1 INTRODUCTION
We are talking about a critical sampling of an analog
signal when the sampling frequency used for
performing this operation equals exactly twice the
value of a maximal frequency occurring in the
spectrum of this signal. Note that equivalently we can
talk in this case about a critical sampling frequency
(rate).
It is well known (Marks II R. J. 1991), (Korohoda
P., Borgosz J. 1999), (Osgood B. 2014), (Borys A.,
Korohoda P. 2017) that signal sampling performed at
the critical sampling rate can provide some unwanted
problems when carrying out the inverse operation.
That is in reconstructing (restoring) the analog signal
from its samples.
In this paper, we illustrate the problems
mentioned above on an example of a perfect
recovering a cosinusoidal signal of any phase being
sampled critically. We do this through performing
very detailed analysis of that what really happens
when the cosinusoidal signal is sampled with the
Nyquist rate and afterwards reconstructed from its
samples.
We show in this paper that recovering both the
original cosinusoidal signal amplitude and its phase
is not possible. But, what is possible then? It is
possible to recover one of these quantities under the
assumption that the second one is known. However,
carrying out some additional calculations is also then
needed.
In this paper, we show also that transfer functions
of the recovering filters used in the cases of critical
and non-critical sampling are not identical. Their form
is derived here.
The remainder of the paper is organized as
follows. Section 2 introduces an example of a
cosinusoidal signal of any phase, which is discussed
throughout this paper. Also, in this section, a
thorough analysis of the effects appearing during
recovery of the cosinusoidal signal sampled critically
Impossibility of Perfect Recovering Cosinusoidal Signal
of Any Phase Sampled with Nyquist Rate
A. Borys
Gdynia Maritime University, Gdynia, Poland
P. Korohoda
AGH University of Science and Technology, Kraków, Poland
ABSTRACT: In this paper, a problem of a perfect recovering cosinusoidal signal of any phase being sampled
critically is considered. It is shown that there is no general solution to this problem. Its detailed analysis
presented here shows that recovering both the original cosinusoidal signal amplitude and its phase is not
possible at all. Only one of this quantities can be recovered under the assumption that the second one is known.
And even then, performing some additional calculations is needed. As a byproduct, it is shown here that a
transfer function of the recovering filter that must be used in the case of the critical sampling differs from the
one which is used when a cosinusoidal signal is sampled with the use of a sampling frequency greater than the
Nyquist rate. All the results achieved in this paper are soundly justified by thorough derivations.
http://www.transnav.eu
the
International Journal
on Marine Navigation
and Safety of Sea Transportation
Volume 14
Number 3
September 2020
DOI: 10.12716/1001.1
4.03.28
738
is presented. In Section 3, we consider the
reconstruction formula and the form of a transfer
function occurring in it for the case of occurrence of
Dirac deltas in the signal spectrum together with its
critical sampling. In the next section, complementary
results are presented for the case of non-critical
sampling. Finally, Section 5 concludes the paper.
2 A SIMPLE EXAMPLE ILLUSTRATING THE
PROBLEM
Let us start consideration of the problem we are
discussing in this paper with an example. And, to this
end, let us take into account the sampling of a co-
sinusoidal signal of the form
( )
( ) cos 2
m
xt ft
πϕ
= +
, (1)
where
m
f
and
ϕ
are its frequency and phase,
respectively. For simplicity, the amplitude of this
signal is assumed here to be equal to one, and t in
(1) means a continuous time.
So, sampling of (1) will be critical, when the
sampling rate,
s
f
, being the inverse of the distance
between samples, T, is equal to
12
sm
f Tf= =
. (2)
Moreover, see that the Fourier transform of the
cosinusoidal signal given by (1) has the following
form:
( )
(
) (
)
(
)
1
exp
2
mm m
Xf f f f f jff
δδ ϕ
= ++ −
, (3)
where
( )
δ
⋅
means the so-called Dirac delta impulse
(Dirac P. A. M. 1947), (Marks II R. J. 1991), (Osgood B.
2014), which is also called the Dirac delta function
(improperly) or the Dirac distribution (properly) in
the literature.
Next, using the sifting property of the Dirac delta
impulse in (3), we get
( ) ( ) ( )
( ) ( )
1
exp
2
exp .
m
m
Xf f f j
ff j
δϕ
δϕ
= + −+
+−
(4)
Note further that by applying the Euler formula to
( )
exp j
ϕ
−
and
( )
exp j
ϕ
in (4) we arrive at an
equivalent form of the latter, i.e.
( ) ( ) ( ) ( )
( ) ( ) ( )
1
cos
2
1
sin
2
mm
mm
Xf f f f f
j ff ff
ϕδ δ
ϕδ δ
= ++ − −
− +− −
. (5)
In the literature, the operation of signal sampling
is modeled as a modulation of the so-called Dirac
comb (Marks II R. J. 1991), (Osgood B. 2014), i.e.
( )
comb ( )
T
n
t t nT
δ
∞
=−∞
= −
∑
, (6)
by a given analog signal to be sampled. In other
words, the above operation can be expressed as a
multiplication of the Dirac comb by this signal. That is
by
( ) ( ) ( )
() ()comb
sT
n
x t x t t x nT t nT
δ
∞
=−∞
=⋅= −
∑
, (7)
where
()
s
xt
means a continuous-time sampled
version of the signal
()xt
. So, we get from (7)
( ) ( )
( ) cos 2
sm
n
x t f nT t nT
π ϕδ
∞
=−∞
= +−
∑
. (8)
in the case of (1).
The equivalent of (7) in the frequency domain is
given by
( ) { } { } ( )
{ }
( )
() () comb
,
ss T
ss
n
X f x t xt t
f X f nf
∞
=−∞
==⊗=
= −
∑
(9)
where
{}
⋅
stands for performing the Fourier
transformation of a signal occurring in braces of this
symbol. Moreover, the symbols
⊗
and
s
f
denote
the convolution operation and sampling rate,
respectively.
Looking at the result (9), we see that the Fourier
transform of a sampled signal consists of an infinite
sum of periodically shifted Fourier transforms of its
un-sampled version multiplied by the sampling
frequency
1
s
fT=
. For details regarding derivation
of (9), see, for example, (Marks II R. J. 1991), (Osgood
B. 2014).
Further, applying (5) in (9) leads to
( ) ( )
{
( )
( ) ( ) ( )
( )
}
1
cos
2
sin
.
s s sm
n
sm sm
sm
X f f f nf f
f nf f j f nf f
f nf f
ϕδ
δ ϕδ
δ
∞
=−∞
= −+ +
+ −− − −+ −
− −−
∑
(10)
In the next step, note that the general expression
(10) can be highly simplified in the case of the critical
sampling of the signal (1). That is with the use of
2
sm
ff=
, as given by (2). Then, the following:
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( )
( )
( )
( )
( )
1
2
21 2
22
2
mm
n
m mm
n
m mm
nn
mm
nn
f nf f
f n f ff
f nf f f
f nf f
δ
δ
δ
δ
∞
=−∞
∞
=−∞
∞
′
= − =−∞
∞
′
→ =−∞
− +=
= −− −+=
′
= − −+=
= −−
∑
∑
∑
∑
(11)
holds. Further, application of (11) in (10) gives
( ) ( ) (
)
( )
( ) ( )
( )
2 cos 2 1
2 cos 2 1 .
sm m
n
mm
n
Xf f f n f
f fnf
ϕδ
ϕδ
∞
=−∞
∞
=−∞
= −− =
= −+
∑
∑
(12)
What does it mean the operation of recovering a
signal in the frequency domain? See that it can be
simply formulated as finding a filter possessing the
transfer function, say
(
)
Hf
, which fulfills the
following:
(
) (
) (
)
s
HfX f Xf
=
. (13)
So, applying (9) in (13) gives
( )
( )
( )
( )
( )
s
ss
n
Xf Xf
Hf
Xf
f X f nf
∞
=−∞
= =
−
∑
. (14)
Further, in the next step, one can check that the
function
( )
1
rect
ss
f
Hf
ff
=
, (15)
with an auxiliary function
( )
rect
⋅
defined by
( )
11
rect 1 for and 0 for
22
xx x=≤>
, (16)
is a solution in (14), when the function
( )
Xf
is
well-defined (that is it is a function, not a
distribution). Note also that such a filter as that one
given by (15) is called an interpolation filter (Marks II
R. J. 1991).
Let us now introduce (12) and (15) with
2
sm
ff=
into (13). This leads to
(
)
( )
(
)
(
)
( ) ( ) ( )
1
rect 2 cos
22
21
cos .
rm
mm
m
n
mm
f
Xf f
ff
fnf
ff ff
ϕ
δ
ϕδ δ
∞
=−∞
= ⋅
⋅ −+ =
= ++ −
∑
(17)
Looking at (17), we see that the result obtained
( )
r
Xf
differs from the expected one, that is from
( )
Xf
given by (3) or equivalently by (4). And, for
this reason, it is denoted differently, by
( )
r
Xf
.
Further, note also that the inverse Fourier transform
of
( )
r
Xf
is given by
( ) ( )
{
}
( ) ( )
1
2cos cos 2
rr m
x t X f ft
ϕπ
−
= =
, (18)
where
{ }
1−
⋅
stands for performing the inverse
Fourier transformation of a Fourier transform in
braces of this symbol.
Comparison of (18) with (1) shows that the
reconstructed signal
( )
r
xt
evidently differs from the
original analog one,
( )
xt
. In what follows, we will
look for the cause of this. So, to this end, let us first
check whether the transfer function given by (15),
which was derived from (13) and (14), is a correct one
in our case.
3 TRANSFER FUNCTION IN RECONSTRUCTION
FORMULA IN CASE OF OCCURRENCE OF
DIRAC DELTAS IN SIGNAL SPECTRUM AND
CRITICAL SAMPLING
Consider now in more detail (13), which is the
reconstruction formula expressed in the frequency
domain, when both the signals
( )
s
Xf
and
( )
Xf
in it contain Dirac deltas.
Further, we continue in this section analysis of the
example introduced in the previous section about a
critical sampling of an analog cosinusoidal signal.
And, note now that because of a critical character of
the latter operation it is highly advisable to check
correctness of calculations at edges of the
characteristics of
( )
Hf
. That is for the left-hand
side edge occurring at
2
sm
ff f=−=−
and for
its right-hand side counterpart at
2
sm
ff f= =
. For
this purpose, we modify slightly
( )
Hf
given by
(15) by introducing there a coefficient c, which we
assume to be unknown at first instance. So, we
rewrite then (15) as
( )
rect
c
ss
cf
Hf
ff
=
, (19)
where c means a real number and
( )
c
Hf
stands for
a modified
( )
Hf
.
Substituting
( )
s
Xf
given by (12),
( )
c
Hf
given by (19), and
( )
Xf
given by (5) into (13), with
2
sm
ff=
, leads to
740
( )
( )
( )
( ) ( ) ( )
( ) ( ) ( )
rect 2 cos
22
21
1
cos
2
1
sin .
2
m
mm
m
n
mm
mm
cf
f
ff
fnf
ff ff
j ff ff
ϕ
δ
ϕδ δ
ϕδ δ
∞
=−∞
⋅
⋅ −+ =
= ++ − −
− +− −
∑
(20)
In the next step, note that performing the
operations indicated in the expression on the right-
hand side of (20) results in
( )
(
)
( )
( ) ( ) ( )
rect 2 cos
22
21
cos .
m
mm
m
n
mm
cf
f
ff
fnf
c ff ff
ϕ
δ
ϕδ δ
∞
=−∞
⋅
⋅ −+ =
=⋅ ++ +
∑
(21)
Next, introducing (21) into (20), we get finally
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
cos
1
cos
2
1
sin .
2
mm
mm
mm
c ff ff
ff ff
j ff ff
ϕδ δ
ϕδ δ
ϕδ δ
⋅ ++ + =
= ++ − −
− +− −
(22)
Observe now that there is no real number c that
fulfills equation (22), in general, for all phases
ϕ
of
the cosinusoidal signal given by (1). However, for a
particular value of
0
ϕ
=
radians, equation (22)
simplifies to
( ) ( )
( )
( )
1
,
2
mm
mm
c ff ff
ff ff
δδ
δδ
⋅ ++ + =
= ++ −
(23)
and the resulting equation (23) is identically satisfied
for
12c =
.
Observe also that our previous results (17) and (18)
will be slightly modified if we take into account the
coefficient
12c =
in (19) instead of
1c =
; the latter
follows from (14) and (15). Then, we will have
( ) ( ) ( ) ( )
1
cos
2
rc m m
Xf ff ff
ϕδ δ
= ++ −
(24)
and
( ) ( )
{ }
( ) ( )
( )
1
cos cos 2
cos 2
rc rc m
m
x t X f ft
A ft
ϕ
ϕπ
π
−
= = =
=
, (25)
respectively, where
( )
rc
Xf
means a modified
version of
( )
r
Xf
and
( )
rc
xt
is the inverse
Fourier transform of the former one. Moreover,
A
ϕ
means an amplitude of the cosinusoidal signal in (25).
Now, see finally that for
0
ϕ
=
radians the
original signal
( )
xt
given by (1) and its recovered
version
( )
rc
xt
given by (25) are identical. That is in
this particular case we have to do with a perfect
reconstruction. And, this is a very strong argument
for the use of the transfer function (19) with
12c
=
instead of the one given by (15) in our further
considerations.
By the way, note that using another method it has
been shown in (Borys A., Korohoda P. 2017) that
reasonable results of analysis of the example
discussed in this paper are achieved only when the
value of the coefficient
12
c =
in (19).
Let us now come back to consideration of (22) for
phases
0
ϕ
≠
radians. It can be easily shown that
there is no such real or complex-valued coefficient c
that satisfies equation (22). Here, we propose to
overcome this problem in a way presented below.
And, afterwards, we check whether results achieved
are reasonable.
We start with ignoring the imaginary component
in (22). That is we ignore the existence of
( )
( ) ( )
1
sin
2
mm
j ff ff
ϕδ δ
− +− −
(26)
in (22). Or, equivalently, we ignore the fact that the
imaginary components on both side of (22) do not
compensate each other. That is we ignore that the
following:
( ) ( ) ( )
1
sin
2
0
mm
j ff ff
ϕδ δ
− +− −
≠
(27)
holds. And, what remains then? It remains consider-
ation of the real parts on both sides of (22) exclusively.
That is consideration of the following
( ) ( ) ( )
( ) ( ) ( )
cos
1
cos ,
2
mm
mm
c ff ff
ff ff
ϕδ δ
ϕδ δ
⋅ ++ + =
= ++ −
(28)
where it is assumed that the coefficient c is real-
valued.
Observe now that for
(
)
cos 0
ϕ
≠
equation (28)
simplifies to (23). Therefore, its solution
12
c =
is
identical with the previous one as well as are the
related solutions (24) and (25) describing the form of
the reconstructed signal.
Let us now take a closer look at the recovered
signal given by (25). Obviously, it is not identical with
the original one given by (1). So, the signal recovery
process in the case of
0
ϕ
≠
and
( )
cos 0
ϕ
≠
at the
same time, as described above, is not perfect.
However, not all is lost here. Namely, proceeding
heuristically as shown below allows us to arrive at the
original form of the signal (1) when we have the result
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(25). Simply, considering the amplitude of the
cosinusoidal signal given by (25), we can write
(
)
arccos
A
ϕ
ϕ
=
. (29)
With this in mind, we can modify the previous
recovery algorithm for the case of
0
ϕ
≠
and
( )
cos 0
ϕ
≠
at the same time - as follows.
1 Perform all the operations to get
( )
rc
xt
as given
by (25). Observe that frequency of the original
signal is correctly recognized in
( )
rc
xt
. However,
signal amplitude there is equal to
(
)
cos
ϕ
instead of 1.
2 Calculate the phase
ϕ
from (29).
3 In (25), replace the amplitude
( )
cos
ϕ
with 1 and
add also there the calculated (in point II above)
value of
ϕ
to the cosine function argument
2
m
ft
π
. It results finally in getting (1).
The algorithm sketched in points I, II, and III
above can be viewed as an improved signal recovery
algorithm for the case of recovery of cosinusoidal
signals of any phase. As seen, its application leads
then to a perfect signal reconstruction.
Note however that all this presented above works
correctly only under the assumption of knowing the
amplitude of a cosinusoidal signal before sampling.
And, in this context, we recall that we assumed for
simplicity in our example the signal given by (1) that
amplitude is 1. Obviously, without this knowledge,
the quantity
A
ϕ
in (25) cannot be fully attributed to
( )
cos
ϕ
. Then, the recovered quantity
A
ϕ
is a
product of an unknown amplitude of the un-sampled
cosinusoidal signal and of a value of the cosine
function of its phase. That is it is a product of two
unknowns. Hence, in this case, nothing can be said
about the amplitude and phase of the original
cosinusoidal signal.
Furthermore, it follows from the detailed analysis
performed in this section that there is no such a
“recovery” transfer function
( )
Hf
in sense of
equation (13), which allows a perfect recovery of a
cosinusoidal signal of any phase sampled with
Nyquist rate. As shown, this fact follows from the
impossibility to satisfy equation (22).
4 TRANSFER FUNCTION IN RECONSTRUCTION
FORMULA IN CASE OF OCCURRENCE OF
DIRAC DELTAS IN SIGNAL SPECTRUM AND
NON-CRITICAL SAMPLING
It is worthy to complete our derivations presented in
the previous section by considering also the case of a
non-critical sampling of the cosinusoidal signal given
by (1). That is we will sample now the signal given by
(1) with a sampling frequency fulfilling the following
inequality:
12
sm
f Tf= >
. (30)
So, using the general formula (9) and (4), we arrive
at
( ) ( )
( ) ( )
( ) ( )
=
exp
2
exp
ss s
n
s
sm
n
sm
X f f X f nf
f
f nf f j
f nf f j
δϕ
δϕ
∞
=−∞
∞
=−∞
= −
= −+ −+
+ −−
∑
∑
(31)
with, do not forget now,
2
sm
ff
>
.
Next, let us come to the operation of recovery of
( )
Xf
given by (4) from
( )
s
Xf
expressed by
(31). To this end, we will take into account, as before,
the reconstruction formula in the frequency domain
presented in (13). Further, observe that a good
candidate to play a role of
( )
Hf
in (13) is
(
)
c
Hf
given by (19) because it filters out all the spectral
components outside the range of frequencies
:
s
ff f≤
, as wanted. So, applying
( )
c
Hf
to (31)
gives
( ) ( ) ( ) ( )
( ) ( )
exp
2
exp .
cs m
m
c
H fX f f f j
ff j
δϕ
δϕ
= + −+
+−
(32)
And, substituting (32) into (13) with
( )
Xf
given by (3) results in
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
exp exp
2
1
= exp exp .
2
mm
mm
c
ff j ff j
ff j ff j
δ ϕδ ϕ
δ ϕδ ϕ
+ −+ − =
+ −+ −
(33)
In the next step, grouping Dirac deltas with the
same arguments in (33), we get
( ) ( ) ( )
( ) ( ) ( )
1 exp
1 exp 0 .
m
m
c j ff
c j ff
ϕδ
ϕδ
− − ++
+− − =
(34)
Observe now that (34) will be satisfied if and only
if the coefficients multiplying the Dirac deltas in (34)
are equal zero. This follows from the theory of
distributions (Hoskins R. F. 2009). That is from the
fact that
( )
0a
δ
⋅ ⋅=
if and only if
0a =
. So,
applying this in (34) leads to the following conclusion:
c must be equal to 1 in the case considered in this
section. Further, substituting
1c =
in (19) results in
(
) (
)
1
1
rect
c
ss
f
H f Hf
ff
=
= =
. (35)
So, we can conclude finally that in the case of
occurrence of Dirac deltas in signal spectrum and
non-critical sampling the transfer function of a
reconstruction filter is equal to
( )
Hf
given by (15).
At the end of this section, it is also worthy to draw
attention to the fact that some algebraic operations on
distributions are forbidden (Hoskins R. F. 2009). For
example, the following notation:
742
( )
( )
(
)
( ) ( ) (
) ( )
( ) ( )
( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( )
( )
1
exp exp
2
exp exp
2
exp exp
1 11
1
exp exp
s
mm
s
sm sm
n
mm
s m m ss
Xf
Hf
Xf
ff j ff j
f
f nf f j f nf f j
ff j ff j
f ff j ff j f f
δ ϕδ ϕ
δ ϕδ ϕ
δ ϕδ ϕ
δ ϕδ ϕ
∞
=−∞
= =
+ −+ −
= =
−+ −+ −−
+ −+ −
= = ⋅=
+ −+ −
∑
(36)
is highly incorrect, mainly because of the occurrence
of divisions of Dirac deltas. They are forbidden (or
undefined) in the theory of distributions.
5 CONCLUSIONS
The reasons of impossibility to recover both the
original cosinusoidal signal amplitude and its phase
from samples of this signal sampled critically have
been recognized in this paper. They follow from the
very detailed analyses presented.
Furthermore, it has been shown that only when
one of the aforementioned quantities is known in the
process of signal reconstruction, the value of the
second one can be recovered.
Finally, it has been also shown that a transfer
function of the reconstruction filter that must be used
in the case of a critical sampling differs from the one
which is used when a cosinusoidal signal is not
sampled critically.
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Interpolation Theory, Springer-Verlag, New York.
Korohoda P., Borgosz J. 1999. Explanation of sampling and
reconstruction at critical rate, Proceedings of the 6th
International Conference on Systems, Signals, and Image
Processing (IWSSIP), Bratislava, Slovakia, 157-160.
Osgood B. 2014. The Fourier Transform and Its
Applications, Lecture Notes EE261, Stanford University.
Borys A., Korohoda P. 2017. Analysis of critical sampling
effects revisited, Proceedings of the 21st International
Conference Signal Processing: Algorithms, Architectures,
Arrangements, and Applications SPA2017, Poznań, Poland,
131 – 136.
Dirac P. A. M. 1947. The Principles of Quantum Mechanics,
3rd Ed., Oxford Univ. Press, Oxford.
Hoskins R. F. 2009. Delta Functions: An Introduction to
Generalised Functions, Horwood Pub., Oxford.
