Let's calculate the spectrum of the periodic pulse signal
shown here. The pulse width is
Δ
Δ
, the period
T
T
, and the amplitude
A
A.
The complex Fourier spectrum of this signal is given by
c
k
=AT∫0Δⅇ-ⅈ2πkΔTdt=-Aⅈ2πkⅇ-ⅈ2πkΔT−1
c
k
A
T
t
0
Δ
2
k
Δ
T
A
2
k
2
k
Δ
T
1
At this point, simplifying this expression requires knowing an
interesting property.
1−ⅇ-ⅈθ=ⅇ-ⅈθ2ⅇ+ⅈθ2−ⅇ-ⅈθ2=ⅇ-ⅈθ22ⅈsinθ2
1
θ
θ
2
θ
2
θ
2
θ
2
2
θ
2
(1)
Armed with this result, we can simply express the Fourier
series coefficients for our pulse sequence.
c
k
=Aⅇ-ⅈπkΔTsinπkΔTπk
c
k
A
k
Δ
T
k
Δ
T
k
(2)
Because this signal is real-valued, we find that the
coefficients do indeed have conjugate symmetry:
c
k
=
c
−
k
¯
c
k
c
−
k
.
The periodic pulse signal has neither even nor odd symmetry;
consequently, no additional symmetry exists in the
spectrum. Because the spectrum is complex valued, to plot it
we need to calculate its magnitude and phase.
|
c
k
|=A|sinπkΔTπk|
c
k
A
k
Δ
T
k
(3)
∠
c
k
=-πkΔT+πnegsinπkΔTπksignk
c
k
k
Δ
T
neg
k
Δ
T
k
sign
k
(4)
What is the value of
c
0
c
0
?
Recalling that this spectral coefficient corresponds to
the signal's average value, does your answer make sense?
c
0
=AΔT
c
0
A
Δ
T
. This quantity clearly corresponds to the periodic pulse
signal's average value.
The somewhat complicated expression for the phase results because
the sine term can be negative; magnitudes must be positive, leaving
the occasional negative values to be accounted for as a phase shift of
π
. The
neg·
neg
·
equals -1 if its argument is negative and zero otherwise.
Also note the presence of a linear phase term (the first term in
∠
c
k
c
k
is proportional to frequency
kT
k
T
. Comparing this term with that predicted from
delaying a signal, a delay of
Δ2
Δ
2
is present in our signal. Advancing the signal by this amount
centers the pulse about the origin, leaving an even signal,
which in turn means that its spectrum is real-valued. Thus,
our calculated spectrum is consistent with the properties of
the Fourier spectrum.
Investigate the half-wave rectified sine wave's spectrum for
a linear phase term. If one is present, show how to delay or
advance the signal to create an even or odd signal. If one
is not present, convince yourself that no delay would yield
a signal having even or odd symmetry.
A half-wave rectified sine wave occurs when
Δ=T2
Δ
T
2
.
Thus,
c
k
¯=-πk2−πnegsinπk2πk
c
k
k
2
neg
k
2
k
, which corresponds to a linear function of
k
k.
The linear phase can be removed by delaying the signal by
14
1
4
of a period.
The phase plot shown in Figure 2
requires some explanation as it does not seem to agree with what
Equation 4 suggests. There, the phase has
a linear component, with a jump of
π
every time the sinusoidal term changes sign. We must realize that
any integer multiple of
2π
2
can be added to a phase at each frequency without
affecting the value of the complex spectrum. We see
that at frequency index 4 the phase is nearly
-π
The phase at index 5 is undefined because the magnitude is zero
in this example. At index 6, the formula suggests that the
phase of the linear term should be less than (more negative)
than
-π
. In addition, we expect a shift of
-π
in the phase between indices 4 and 6. Thus, the phase value
predicted by the formula is a little less than
-2π
2
. Because we can add
2π
2
without affecting the value of the spectrum at index 6, the
result is a slightly negative number as shown. Thus, the formula
and the plot do agree. In phase calculations like those made in
MATLAB, values are usually confined to the range
-ππ
by adding some (possibly negative) multiple of
2π
2
to each phase value.
