2. 2
Fourier series discussions assume that the signal of interest is periodic.
However, a majority of signals we encounter in signal processing are not
periodic. Furthermore, we have many signals that are bunch of random bits with
no pretense of periodicity. This is the real world of signals and Fourier series
comes up short for these types of signals. This types of signals are called
aperiodic signals.
Applying Fourier Series to Aperiodic Signals
Extending the period to infinity
Fig.16: Can we compute the Fourier series coefficients of this aperiodic signal?
Let us say that the little signal, as shown in Fig.16, has been collected and the
data show no periodicity. To compute its spectrum using Fourier analysis, we
have to make this periodic.
3. 3
We can pretend that the signal in Fig.16 is actually a periodic signal, but we
see one period, the length of which is longer than the length of the data at
hand. We surmise/guess that if the length of the signal is 4s, then it looks
like the top row of Fig.17 with signal repeating with a period of 5s.
Fig.17: Going from periodic to aperiodic signal by extending the period.
4. 4
If we increase the period to infinitely long, then it will be the signal with
zeros extending to infinity on each side. Now we declare that this is now a
periodic signal but with a period extending to . We have turned an
aperiodic signal into a periodic signal with this assumption.
We can apply the Fourier analysis to this extended signal because it is
ostensibly periodic with a period, T extending to infinity. The single piece of
the signal is then one period of a periodic signal, the other periods of which
we can not see. With this assumption, Fourier series coefficients (FSC) of
this signal can be computed by setting its period to, T = .
5. 5
We can apply the Fourier analysis to this extended signal because it is
ostensibly periodic with a period, T extending to infinity. The single
piece of the signal is then one period of a periodic signal, the other
periods of which we can not see.
But if the period is infinitely long, then the fundamental frequency
defined as the inverse of the period becomes infinitely small. The
harmonics are still integer multiples of this infinitely small
fundamental frequency but they are so very close to each other that they
approach a continuous function of frequency. So a key result of this
assumption is that the spectrum of an aperiodic signal becomes a
continuous function of the frequency and is no longer discrete as are
the FSC.
Continuous-Time Fourier Transform (CTFT) for Aperiodic Signals
6. 6
Fig.18(a) shows a pulse train with period T0. As the pulses move further apart in
Fig.18(b and c), the spectral lines or the harmonics are moving closer together.
Fig.18: Take the pulse train in (a), as we increase its period, i.e., allow more
time between the pulses, the fundamental frequency gets smaller, which
makes the spectral lines move closer together as in (c). In the limiting case,
where the period goes to1, the spectrum would become continuous.
7. 7
The Fourier series coefficients (FSC) of a continuous-time signal is:
(26)
To apply this to an aperiodic case, we let T0 go to . In Eq.(26) as the
period gets longer, so there is problem of division by infinity. Putting
the period in the form of frequency can be avoided this problem.
Thereafter, we only have to worry about multiplication by zero. We
write the period as a function of the frequency:
(27)
If T0 is allowed to go to infinity, then 0 is becoming tiny. In this case,
we write frequency 0 as instead, to show that it is becoming
infinitesimally smaller. Now we write the period in the limit as
2
1
1
2
Or,
2
o
o
o
o
o
o
T
T
F
8. 8
(28)
We rewrite Eq. (26) by substituting Eq. (28).
(29)
But now as T0 goes to infinity, approaches zero, and the whole expression
goes to zero. To get around this problem, we start with the time-domain Fourier
series representation of x(t), as given by
(30)
Now substitute Eq.(29) into Eq.(30) for the value of Ck to write
(31)
9. 9
We now change the limits of the period, T0, from a finite number to . We
also change to d, and k0 to just , the continuous frequency and the
summation in Eq.(31) now becomes an integral. Furthermore, the factor 1/2 is
moved outside. Now, we rewrite Eq.(31) incorporating these ideas as:
…(32)
The underlined part is calling the Fourier transform and refer to it by the
expression, X(). Now the Eq.(32) is called the Inverse Fourier Transform
and is equivalent to the Fourier series representation or the synthesis equation.
…(33)
The Continuous Time Fourier transform (CTFT) is defined as the underlined
part in Eq.(32) and is equal to
…(34)
10. 10
Energy Density and Power Density Spectrum for Continuous Time
Aperiodic Signal
A continuous time aperiodic signal has infinite energy which is given as
The is represented the energy density spectrum of continuous time
aperiodic signal.
It is an Energy signal. In this case, we can’t calculate power density spectrum
of aperiodic signal due to unknown period.
Example 34-a. Determine the formula for Fourier transform and energy
density spectrum of the unit rectangular pulse of :
T/2
t
T/2
-
1
T/2
t
T/2
-
0
)
(
t
x
11. 11
2
sinc
2
2
sin
2
sin
2
2
sin
2
2
2
1
1
.
1
.
0
)
(
.
0
)
(
)
(
)
(
)
(
)
(
)
2
/
(
)
2
/
(
)
2
/
(
)
2
/
(
)
2
/
(
)
2
/
(
2
/
2
/
2
/
2
/
2
/
2
/
2
/
2
/
2
/
2
/
2
/
2
/
T
T
T
T
T
T
T
T
T
j
e
e
e
e
j
e
e
j
j
e
dt
e
dt
e
dt
e
t
x
dt
e
dt
e
t
x
dt
e
t
x
dt
e
t
x
dt
e
t
x
X
T
j
T
j
T
j
T
j
T
j
T
j
T
T
t
j
T
T
t
j
T
t
j
T
T
t
j
T t
j
T
t
j
T
T
t
j
T t
j
t
j
Answer: We know,
12. 12
Fig.34-a: Energy density spectrum (EDS) for given rectangular pulse.
The graphical representation of energy density spectrum for given rectangular
pulse as shown in Fig.34-a.
So, energy density spectrum for given rectangular pulse is
2
2
2
sinc
)
(
T
T
X
Sinc Function
DESMOS website:
https://www.desmos.com/
calculator/tjl420nxgp
13. 13
Example 4.1: Compute the CTFT of x(t) = (t) (delta function) and show its
inverse CTFT. Explain in detail the representation of the delta function.
Solution: This is the most important function in signal processing. The delta
function can be considered a continuous (Dirac delta function) or a discrete
function (Kronecker delta function), but here we treat it as a continuous
function. We use the CTFT equation, Eq. (34) and substitute delta function for
function x(t). We compute its CTFT as follows.
)
(
)
(
).
(
:
)
(
of
Properties
origin
At the
1
1 t
x
dt
t
t
δ
t
x
t
1
)
0
(
)
(
)
(
)
(
0
.
j
t
j
t
j
t
j
e
dt
e
t
dt
e
t
dt
e
t
x
X
CTFT of An Impulse Function:
14. 14
Since CTFT is constant for all frequencies, so we get a flat line from - to
for the spectrum of the delta function. It encompasses much depth.
The inverse CTFT is:
Substituting in the second step, the definition of the delta function from
Eq. (33), we get the function back, a perfect round trip. The CTFT of a
delta function is 1 in frequency-domain, and the inverse CTFT of 1 in
frequency-domain is the delta function in time-domain.
)
(
.
1
2
1
}
1
{
)
( 1
t
d
e
t
x
t
j
15. 15
The delta function was defined by Dirac as a summation of an infinite
number of exponentials.
The same equation in frequency domain is given by:
The general version of the Dirac delta function with a shift for time and
frequency is given as:
(35)
2
1
)
(
d
e
t t
j
(36)
2
1
)
( dt
e t
j
(38)
2
1
)
(
(37)
2
1
)
(
)
(
0
)
(
0
0
0
dt
e
d
e
t
t
t
j
t
t
j
16. 16
Example 4.2: What is the Fourier transform of the time-domain signal,
x(t) = 1.
From Eq. (36) we get:
Solution: This case is different from Ex.4.1. Here, the time-domain
signal is a constant and not a delta function. It continues forever in time
and is not limited to one single time instant as is the first case of a
single delta function at the origin. Using Eq.(34), we write the CTFT as:
dt
e t
j
2
1
)
(
)
(
2
.
1
2
1
.
2
.
1
)
(
)
(
dt
e
dt
e
dt
e
t
x
X
t
j
t
j
t
j
CTFT of A Constant:
17. 17
CTFT of A Sinusoid:
Example 4.3: Since a sinusoid is a periodic function, we will select only one
period of it to make it aperiodic. Here, we have just a piece of a sinusoid. We
make no assumption about what happens outside the selected time frame.
Solution: We can compute the CTFT of this little piece of cosine as:
We use Equation (38) to write this result as:
(39) dt
e t
j
)
(
0
0
2
1
)
(
By similarity, the Fourier transform of a sine is given by
(40)
18. 18
CTFT of A Complex Exponential:
Example 4.4: Now we calculate the CTFT of a very important function,
the complex exponential.
Solution: We have already calculated the CTFT of a sine and a cosine
given by:
(41)
(42)
Figure: The CTFT of a
complex exponential. The
Real part is a cosine,
hence the spectrum looks
like Fig.(a) and the
Imaginary part is a sine,
and hence this plot is
exactly the same as
Fig.(b).
19. 19
By the linearity principle, we write the Fourier transform of the CE
keeping the sine and cosine separate.
The result is a single delta function located at o. In Fig. we see why it is
one sided.
(43)
20. 20
Example 4.5: Compute the iCTFT of a single impulse located at
frequency 1.
Solution: We take the iCTFT per Eq. (33).
The result is a complex exponential of frequency 1 in time-domain.
The result is a complex exponential of
frequency 1 in time-domain.
Because this is a complex signal, it
has a non-symmetrical frequency
response that consists of just one
impulse located at the CE’s frequency.
In Fig. 4.9, we see why it is one sided.
)
(
)
(
).
(
:
)
(
of
Properties
origin
At the
1
1 t
x
dt
t
t
δ
t
x
t
t
j
t
j
e
d
e
t
x
1
1
1
2
1
).
(
2
1
)}
(
{
)
( 1
21. 21
Here, we write the two important CTFT pairs. The CTFT of a CE is
one-sided, an impulse at its frequency. (The CTFT of all complex
functions are asymmetrical.)
(44)
(45)
22. 22
Continuous-Time Fourier Transform (CTFT) for Periodic Signals
The Fourier transform came about so that the Fourier series could be
made rigorously applicable to aperiodic signals. The signals we
examined so far in this chapter are all aperiodic, even the cosine wave,
which we limited to one period. Can we use the CTFT for a periodic
signal? Our intuition says that this should be the same as the Fourier
series. Let us see if that is the case.
Take a periodic signal x(t) with fundamental frequency of o = 2/T0
and write its FS representation.
Taking the CTFT of both sides of this equation, we get
23. 23
We can move the coefficients out of the CTFT because they are not function
of frequency. They are just numbers.
The Fourier transform of the complex exponential ejot is a delta function
located at the frequency o . Making the substitution, we get
(46)
This equation says that the CTFT of a periodic signal is a sampled version of
the FSC. The FSC are being sampled at frequency of the signal, o, with k the
index of repetition. However, the FSC are already discrete! Thus, the only
thing the Fourier transform does is change the scale. The magnitude of the
CTFT of a periodic signal is 2 times bigger than that computed with FSC as
seen by the factor 2 front of Eq. (46).
Important observation: The CTFT of an aperiodic signal is aperiodic and
continuous whereas the CTFT of a periodic signal is aperiodic but discrete.
dt
t
j
t
dt
e t
j
)
sin
(cos
2
1
2
1
)
(
DESMOS
https://www.des
mos.com/calcula
tor/xc5znoelkq
}
{
)
( t
jk
k
k
o
e
X C
)
(
2
)
( o
k
k
C
X
24. 24
Example **: Example **: Examine the CTFT of the periodic square pulse
of 1 Volt that lasts seconds and repeats every T seconds:
Solution: First we have to calculate the FSC of this periodic pulse train as
cycle
duty
e,
wher
);
(
sinc
)
/
(
sinc
/
/
sin(
/
)
/
sin(
1
/
1
2
1
)
/
(
1
2
1
/
1
2
1
)
/
2
2
/
2
/
1
.
1
1
)
(
1
)
/
(
)
/
(
)
/
(
)
/
(
2
/
)
/
2
(
2
/
)
/
2
(
)
/
2
(
2
2
/
2
/
2
2
/
2
/
T
τ
r
r
k
r
T
k
T
T
k
T
k
T
T
k
T
k
T
T
k
j
e
e
T
T
k
j
e
e
T
T
k
j
e
e
T
T
jk
e
T
dt
e
T
dt
e
t
x
T
C
T
jk
T
jk
T
jk
T
jk
T
jk
T
jk
t
T
jk
t
f
jk
t
f
jk
T
T
k o
o
This is the sinc function. It comes up so often in signal processing that it is probably the second most
important equation in DSP after the Euler’s equation.
25. 25
If the Duty cycle (= /T) is 1/2, then we can write
)
2
1
(
sinc
2
1
k
Ck
Figure: (a) The periodic square wave
with duty cycle of 0.5. (b) Its FSC and
(c) its CTFT. Only the scale is different.
We plot these FSC in Fig.(b). To compute CTFT, we set 0 = 1 and now
we can write the CTFT expression as:
The result is the sampled version
of the FSC scaled by 2 which
are of course themselves discrete
Fig.(c).
27. 27
We consider a piece of a discrete and ostensibly aperiodic signal as shown in
Fig. 20(a). Conceptually, we can extend its period. This signal x[n] is just 5
samples, but we can pretend that the signal is periodic with period N0 >> 5.
This period may be very long, even infinitely long. So if we extend the period
of this signal to , we basically get back the original signal, x[n] of 5 samples,
which is now surrounded by zeros.
Fig.20: An aperiodic discrete-time
signal can be considered periodic if
period is assumed to be infinitely long.
(47)
Discrete-time Fourier Transform (DTFT) For Aperiodic Signal
28. 28
As we increase N0, so it can now be considered a periodic signal, although
only in a mathematical sense. We can’t see any of the periods. They are too
far apart. Now since the signal is periodic, we can use the discrete-time
Fourier series (DTFS) to write its frequency representation in terms of its
complex coefficients as
…(48)
The fundamental digital frequency of a discrete periodic signal is 0=2/N0,
with N0 as period of the signal in samples. As N0 goes to infinity, the
fundamental frequency goes to zeros as well. We can think of the
fundamental frequency as the resolution of the spectrum; so if this
number is zero, then the frequency becomes continuous and k, the
harmonic identifier drops out entirely. Hence, there are no unique
harmonics. Now the signal of Eq.(47) can be written as a periodic signal x[n].
29. 29
Here, since x[n] is discrete-time aperiodic signal, so the Summation sign
has not been replaced by the Integration sign. The expression does not
contain any reference to the harmonic index, k.
…..(50)
Now we define a new transform called the Discrete-time Fourier Transform
(DTFT) for a discrete aperiodic signal, assuming that N0= as
…..(49)
Comparison to the CTFT:
The CTFT frequency is termed whereas the digital
frequency for discrete signals is given by . The
CTFT, X() is continuous in frequency. The DTFT or
X() is also continuous in frequency for the same
reason: due to the extension of the period to . Both
X() and X() are continuous functions, hence we
have written them with round brackets.
The inverse DTFT is similarly given by this expression.
30. 30
DTFT is Continuous and Periodic with Period 2 :
Unlike continuous frequency, the whole range of digital frequency is
limited to 2. The spectrum computed thereof is also limited to that range and
is called the principal alias. Because of this condition, the coefficients for
harmonic frequencies outside 0 to 2 are just copies. Hence there is no need
to compute X() outside this range. We can ignore all these “replicated
spectrum” as they are identical to the principal alias. We write this property as
So first we say that the period of a signal is assumed to be infinitely long and
now we are saying that the DTFT is periodic. How can that be? Yes, in time
domain, we are assuming that the period is infinitely long. But in frequency
domain the spectrum is periodic and X() repeats with 2. This talk of a
frequency that is measured in multiples of can be confusing. But we must
accept the fact that the DTFT is defined in terms of the digital frequency
which is special type of frequency. The signal consists of discrete values and
in order to make the analysis independent of real physical time, i.e. the time
between the samples, the DTFT is defined in terms of the digital frequency.
This also makes the math, easier.
(51)
31. 31
This comes from the observation that
Every 2, X() is identical. This property simplifies the computation
as we need only integrate over a 2 range of the digital frequency.
Since the area under a periodic signal for one period does not
change no matter where we start the integration, we can generalize
the DTFT equation over any range. We can for example write the
equation for the iDTFT, with integration range written as just 2, and
both are valid.
(52)
)
(
2
1
)
(
2
1
]
[ 2
dΩ
e
Ω
X
dΩ
e
Ω
X
n
x n
j
n
j
33. 33
Example 1: Find the DTFT and Energy Density spectrum of the signal:
otherwise
0
4
n
0
1
]
[n
x
Solution:
4
3
2
4
3
2
0
.
4
.
3
.
2
.
1
.
0
.
4
0
1
0
1
.
1
.
1
.
1
.
1
.
1
]
4
[
]
3
[
]
2
[
]
1
[
]
0
[
]
[
]
[
)
(
j
j
j
j
j
j
j
j
j
j
j
j
j
n
j
n
n
j
N
n
e
e
e
e
e
e
e
e
e
e
x
e
x
e
x
e
x
e
x
e
n
x
e
n
x
X
But, using this expression it is very difficult to calculate the magnitude
and phase, so we should use the easy method.
34. 34
)
2
/
sin(
)
2
Ω
5
sin(
)
(
2
2
/
2
/
2
/
2
/
5
2
/
5
2
/
5
2
/
2
/
2
/
2
/
5
2
/
5
2
/
5
5
.
4
0
4
0
2
2
1
1
.
1
/
X
j
j
j
j
j
j
j
j
j
j
j
j
j
j
j
n
n
j
n
j
n
e
j
e
e
e
j
e
e
e
e
e
e
e
e
e
e
e
e
e
c
c
c
N
N
n
n
1
1
1
0
Here, [sin(5/2)/sin(/2)] is magnitude and -2 is phase. The Energy
spectrum is [sin(5/2)/sin(/2)]2.
35. 35
Discrete-time Fourier Transform (DTFT) For Periodic Signal
Let’s take a periodic, discrete-time signal with a period of N0 samples and
write its discrete Fourier series equation. We did not talk about a period
when discussing DTFT of aperiodic signals, but we will now. Period now
becomes relevant because these signals are periodic, so they have a
period! And whenever, we have a period, the frequency
resolution must be discrete. However to derive a transform for
periodic discrete signals, we have to go back to discrete-time Fourier
Series.
The Fourier series is written in form of Fourier series coefficients for
discrete-time signals as follows:
…(53)
where 0 = 2/N0 is the digital frequency of the discrete signal and N0 is
the period of the signal in samples.
36. 36
The coefficients of the harmonics are given by
…(54)
Since now we have N0 samples of a periodic signal, we can indeed
compute these coefficients. Let’s take the DTFT of Eq. (53).
…(55)
The coefficients in Eq. (55) are not a function of frequency, so they are
pulled out in front. The DTFT of the underlined part, a summation of
complex exponentials is a train of impulses:
dt
e t
j
2
1
)
(
37. 37
Substituting this expression into Eq.(55), we get the equation for the
DTFT of a periodic signal:
…(56)
This equation says that the DTFT of a periodic discrete signal repeats the
DTFS coefficients, Ck at every integer multiple of the digital frequency
which is clear from 2k/No. This formulation is quite different from the
DTFT of an aperiodic signal which we computed by Eq.(49). The
situation here is analogous to the case of the CTFT of periodic signals.
The CTFT of a periodic signal is a discrete version of the CTFSC.
Similarly the DTFT of a periodic signal is also a discrete version of the
DTFSC. The only difference, and a very big one, is that the coefficients
of the DTFT are periodic. This of course has to do with the frequency
ambiguity of discrete signals.
Important observation: The DTFT of both the aperiodic and the periodic
signal repeats but is discrete only for the periodic signals.
38. 38
Example: We want to compute the DTFT of this periodic
signal of period N = 4, x[n] = [0, 1, 2, 1, 0, 1, 2, 1, ...].
