DISCRETE-SIGNAL ANALYSIS AND DESIGN- P10 ppt
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SINE, COSINE, AND θ 31
cos
−cos
j cos
−j cos
sin
−sin
j sin
(
h
)
(g)
(f )
(e)
(d )
(c)
− j sin
(a)
(b)
RealImaginary
Figure 2-2 Phasor polarity and type for all possible sine–cosine waves.
Solid lines: real; dashed lines: imaginary.
amplitude 8. The spectrum is positive-sided at 2 and 6.
x(n) = 5cos
2π
n
N
2
− j 8cos
2π
n
N
6
(2-1)
32 DISCRETE-SIGNAL ANALYSIS AND DESIGN
In Fig. 2-3 the (n) step size 0.1 from 0 to N −1 provides smooth curves
in the real and imaginary x(n) graphs. Figure 2-3 shows the following:
•
The real cosine spectrum at k =2 and 14, amplitude 2.5 +2.5 =
+5.0 (see Fig. 2-2a).
•
The −j cosine spectrum at k =6 and 10, amplitude −4 −4 =−8
(see Fig. 2-2f).
•
The phase at k =2 and 14 ( =0
◦
).
•
The phase at k =6 and 10 ( =−90
◦
).
Observe that Mathcad provides the correct two-sided phasors (Fig. 2-2)
that the subsequent two-sided IDFT restores to the input x(n). If x (n)is
viewed from 0 to N −1, the positive frequencies 2 and 6 (number of
cycles per record length) are visible.
In more complicated situations it is a good idea to avoid possible con-
fusion by making sure that all of the Re[X (k)] and Im[X (k )] phasor pairs
are combined into the correct one-sided real and imaginary sine and cosine
positive-frequency constituents, as deÞned in Fig. 2-2. For the example in
Fig. 2-3, the one-sided output is +5 cosine at f =2at0
◦
and −j 8 cosine
at f =6at−90
◦
.
Note the use of the Mathcad function [atan2 (Re(x), Im(x))]·180/π
◦
,
which covers the range ±180
◦
, as compared with atan(x) · 180/π
◦
, which
only covers ±90
◦
.
There are some possibilities for “de-cluttering” the results:
•
If Im(X (k )) < 0.0001, set φ(k ) =0 to avoid clutter in the phase data.
•
If Re(X (k)) < 0.0001, set Re(X (k)) =0.0001.
•
If j Im(X(k )) > j 1000, set j Im(X (k)) =j 1000.
•
If j Im(X(k )) < −j 1000, set j Im(X (k)) =−j 1000.
•
Scale the problem to avoid values of X (k) < 0.001 or > 1000.
In the next example we use just the cosine wave with a phase advance
of +60
◦
=(π/3 radians):
X(k) =
1
N
N−1
n=0
7cos
2π
n
N
4 +
60 ·
π
180
exp
− j2π
n
N
k
(2-2)
SINE, COSINE, AND θ 33
−10
10
−5
0
5
−5
5
k
180
π
24 8 1214
610
0
0
n
24 8 12146100
2
0 if ⏐Im(X(k))⏐ < .0001
atan2 [Re(X(k)), Im((X(k)))]
48
6
10 12 14
−90
−60
−30
30
60
90
φ(k)
k
φ(k) :=
⋅
Re(X(k))
Im(X(k))
Re(x(n))
Im(x(n))
1
N
∑
X(k) :=
x(n) :=
N−1
n = 0
5⋅cos ⋅exp− j⋅8⋅cos2⋅π⋅ ⋅2
n
N
⎛
⎛
⎛
⎛
⎛
⎛
⎛
⎛
⎛
⎛
⎛
⎛
2⋅π⋅ ⋅6
n
N
⎛
⎛
⎛
⎛
−j⋅2⋅π⋅ ⋅k
n
N
5⋅cos − j⋅8⋅cos2⋅π⋅ ⋅2
n
N
⎛
⎛
⎛
⎛
2⋅π⋅ ⋅6
n
N
⎛
⎛
⎛
⎛
⋅
N := 16 k := 0, 1 N − 1 n := 0, 0.1 N − 1
Figure 2-3 Example of two-tone signal and its spectrum components
and phase (deg).
Figure 2-4 shows that a +60
◦
component has been added to the cosine
component, as Fig. 2-2d conÞrms. The amplitude of the cosine seg-
ment is 1.75 +1.75 =3.5V. The amplitude of the −sin θ component is
3.03 +3.03 =6.06V. The phase is shown as +60
◦
.
In these two examples we started with the equations for the signal, then
by examining the phasors in Fig. 2-2 we veriÞed that they coincided with
34 DISCRETE-SIGNAL ANALYSIS AND DESIGN
x(n) :=
7⋅cos 2⋅π⋅ ⋅4 + 60⋅
n
N
π
180
N := 16 n := 0,.01 N − 1 k := 0, 1 N − 1
−10
0
10
n
24 8 12146100
−1
1
2
0
−5
5
0
k
24 8 12146100
k
24 8 12146100
−100
100
50
−50
0
k
24 8 1214166100
Im(x(n))
Im(X(k))
Re(x(n))
Re(X(k))
1
N
X(k) : =
⋅
x(n)⋅exp −j⋅2⋅π⋅
n
N
⋅k
180
π
0 if ⏐Im(X(k))⏐< .001
atan2 (Re(X(
k)), Im((X(k)))
θ(k) :=
θ(k)
⋅
∑
N−1
n = 0
Figure 2-4 Sine wave with 60 phase advance.
SINE, COSINE, AND θ 35
the given signal. In practice we might not know the waveform ahead of
time and we would be asked to examine the phasor amplitudes and phases
and use Fig. 2-2 to determine the signal.
We will Þnd uses for the positive-side techniques described in this
chapter and others. Because of the simplicity and the familiarity with
positive-domain frequency usage in many practical engineering situations,
the sine–cosine–θ approach described here is encouraged.
When setting up a problem to get the line spectrum, be sure that (k ),
the frequency index, is deÞned as an integer array. In Mathcad the assig-
ment statement k : =0, 1 N −1 works. Otherwise, the DFT generates
a lot of complex answers for noninteger values of k and the spectrum
becomes “smeared,” which obscures the desired line spectrum answers.
This general rule is a good idea in discrete wave analysis problems. Oper-
ate at integral values of (n) (time domain) and (k) (frequency domain) if
possible. Procedures for dealing with noninteger values will be covered
in later chapters, especially Chapters 3 and 4.
The calculations in Example 2-1 demonstrate how the two-sided DFT
X (k ) of the two-sided x (n), converted to positive-frequency X (k) only,
can be an excellent spectrum amplitude and phase analyzer for a nonlinear
device, circuit, or system in the real-world positive frequency domain.
Adjustments to signal level(s) of two or more signals and device biasing
parameters can very quickly present a picture of the spectrum response
of the circuit. This is an almost-free Spectrum Analyzer that can be very
accurate and versatile over a very wide frequency range if (a big “if”)
the nonlinear devices used are deÞned correctly. The main requirement is
that we have either an equation or an example data Þle for the device.
One interesting experiment is to look at nonlinear reactions to a two-
tone signal, then vary the amplitude of a strong third signal that is on a
third out-of-band frequency to see the degradation of the in-band two-tone
signal. This can be very illuminating and very practical.
Example 2-1: Nonlinear AmpliÞer Distortion and Square
Law Modulator
To get some hands-on experience, this example will look at the intermodu-
lation (mixing) products of an ampliÞer circuit that is not perfectly linear.
We will use the DFT [Eq. (1-2)] to get the spectrum and IDFT (Eq. 1-8)
