Sinusoidal signals are perhaps the most important type of signal that we will encounter in signal processing. There are two basic types of signals, the cosine:
x
(
t
)
=
cos
(
Ω
t
)
x
(
t
)
=
cos
(
Ω
t
)
(1)
and the sine:
x
(
t
)
=
sin
(
Ω
t
)
x
(
t
)
=
sin
(
Ω
t
)
(2)
Plots of the sine and cosine signals are shown in Figure 1. Sinusoidal signals are periodic signals. The period of the cosine and sine signals shown above is given by T=2π/ΩT=2π/Ω. The frequency of the signals is Ω=2π/TΩ=2π/T which has units of rad/sec . Equivalently, the frequency can be expressed as 1/T1/T, which has units of sec-1sec-1, cycles/sec, or Hz. The quantity ΩtΩt has units of radians and is often called the phase of the sinusoid. Recalling the effect of a time shift on the appearance of a signal, we can observe from Figure 1 that the sine signal is obtained by shifting the cosine signal by T/4T/4 seconds, i.e.
sin
(
Ω
t
)
=
cos
(
Ω
(
t
-
T
/
4
)
)
sin
(
Ω
t
)
=
cos
(
Ω
(
t
-
T
/
4
)
)
(3)
and since T=2π/ΩT=2π/Ω, we have
sin
(
Ω
t
)
=
cos
(
Ω
t
-
π
/
2
)
)
sin
(
Ω
t
)
=
cos
(
Ω
t
-
π
/
2
)
)
(4)
Similarly, we have
cos
(
Ω
t
)
=
sin
(
Ω
t
+
π
/
2
)
)
cos
(
Ω
t
)
=
sin
(
Ω
t
+
π
/
2
)
)
(5)
Using Euler's Identity, we can also write:
cos
(
Ω
t
)
=
1
2
e
j
Ω
t
+
e
-
j
Ω
t
cos
(
Ω
t
)
=
1
2
e
j
Ω
t
+
e
-
j
Ω
t
(6)
and
sin
(
Ω
t
)
=
1
2
j
e
j
Ω
t
-
e
-
j
Ω
t
sin
(
Ω
t
)
=
1
2
j
e
j
Ω
t
-
e
-
j
Ω
t
(7)
The quantity ejΩtejΩt is called a complex sinusoid and can be expressed as
e
±
j
Ω
t
=
cos
j
Ω
t
±
j
sin
j
Ω
t
e
±
j
Ω
t
=
cos
j
Ω
t
±
j
sin
j
Ω
t
(8)
There are a number of trigonometric identities which are sometimes useful. These are shown in Table 1. Table 2 shows some basic calculus operations on sine and cosine signals.
Table 1: Useful trigonometric identities.
|
sin
(
θ
)
=
cos
(
θ
-
π
/
2
)
sin
(
θ
)
=
cos
(
θ
-
π
/
2
)
|
|
cos
(
θ
)
=
sin
(
θ
+
π
/
2
)
cos
(
θ
)
=
sin
(
θ
+
π
/
2
)
|
|
sin
(
θ
1
)
sin
(
θ
2
)
=
1
2
cos
(
θ
1
-
θ
2
)
-
cos
(
θ
1
+
θ
2
)
sin
(
θ
1
)
sin
(
θ
2
)
=
1
2
cos
(
θ
1
-
θ
2
)
-
cos
(
θ
1
+
θ
2
)
|
|
sin
(
θ
1
)
cos
(
θ
2
)
=
1
2
sin
(
θ
1
-
θ
2
)
-
sin
(
θ
1
+
θ
2
)
sin
(
θ
1
)
cos
(
θ
2
)
=
1
2
sin
(
θ
1
-
θ
2
)
-
sin
(
θ
1
+
θ
2
)
|
|
cos
(
θ
1
)
cos
(
θ
2
)
=
1
2
cos
(
θ
1
-
θ
2
)
+
cos
(
θ
1
+
θ
2
)
cos
(
θ
1
)
cos
(
θ
2
)
=
1
2
cos
(
θ
1
-
θ
2
)
+
cos
(
θ
1
+
θ
2
)
|
|
a
cos
(
θ
)
+
b
sin
(
θ
)
=
a
2
+
b
2
c
o
s
θ
-
tan
-
1
b
a
a
cos
(
θ
)
+
b
sin
(
θ
)
=
a
2
+
b
2
c
o
s
θ
-
tan
-
1
b
a
|
|
cos
(
θ
1
±
θ
2
)
=
cos
(
θ
1
)
cos
(
θ
2
)
∓
sin
(
θ
1
)
sin
(
θ
2
)
cos
(
θ
1
±
θ
2
)
=
cos
(
θ
1
)
cos
(
θ
2
)
∓
sin
(
θ
1
)
sin
(
θ
2
)
|
|
sin
(
θ
1
±
θ
2
)
=
sin
(
θ
1
)
cos
(
θ
2
)
±
sin
(
θ
1
)
cos
(
θ
2
)
sin
(
θ
1
±
θ
2
)
=
sin
(
θ
1
)
cos
(
θ
2
)
±
sin
(
θ
1
)
cos
(
θ
2
)
|
Table 2: Derivatives and integrals of sinusoidal signals.
|
d
d
t
cos
(
Ω
t
)
=
-
Ω
sin
(
Ω
t
)
d
d
t
cos
(
Ω
t
)
=
-
Ω
sin
(
Ω
t
)
|
|
d
d
t
sin
(
Ω
t
)
=
Ω
cos
(
Ω
t
)
d
d
t
sin
(
Ω
t
)
=
Ω
cos
(
Ω
t
)
|
|
∫
cos
(
Ω
t
)
d
t
=
1
Ω
sin
(
Ω
t
)
∫
cos
(
Ω
t
)
d
t
=
1
Ω
sin
(
Ω
t
)
|
|
∫
sin
(
Ω
t
)
d
t
=
-
1
Ω
cos
(
Ω
t
)
∫
sin
(
Ω
t
)
d
t
=
-
1
Ω
cos
(
Ω
t
)
|
|
∫
0
T
sin
(
k
Ω
o
t
)
cos
(
n
Ω
o
t
)
d
t
=
0
∫
0
T
sin
(
k
Ω
o
t
)
cos
(
n
Ω
o
t
)
d
t
=
0
|
|
∫
0
T
sin
(
k
Ω
o
t
)
sin
(
n
Ω
o
t
)
d
t
=
0
,
k
≠
n
∫
0
T
sin
(
k
Ω
o
t
)
sin
(
n
Ω
o
t
)
d
t
=
0
,
k
≠
n
|
|
∫
0
T
cos
(
k
Ω
o
t
)
cos
(
n
Ω
o
t
)
d
t
=
0
,
k
≠
n
∫
0
T
cos
(
k
Ω
o
t
)
cos
(
n
Ω
o
t
)
d
t
=
0
,
k
≠
n
|
|
∫
0
T
sin
2
(
n
Ω
o
t
)
d
t
=
T
/
2
∫
0
T
sin
2
(
n
Ω
o
t
)
d
t
=
T
/
2
|
|
∫
0
T
cos
2
(
n
Ω
o
t
)
d
t
=
T
/
2
∫
0
T
cos
2
(
n
Ω
o
t
)
d
t
=
T
/
2
|
Now suppose that we have a sum of two sinusoids, say
x
(
t
)
=
cos
(
Ω
1
t
)
+
cos
(
Ω
2
t
)
x
(
t
)
=
cos
(
Ω
1
t
)
+
cos
(
Ω
2
t
)
(9)
It is of interest to know what the period TT of the sum of 2 sinusoids is. We must have
x
(
t
-
T
)
=
cos
(
Ω
1
(
t
-
T
)
)
+
cos
(
Ω
2
(
t
-
T
)
)
=
cos
(
Ω
1
t
-
Ω
1
T
)
+
cos
(
Ω
2
t
-
Ω
2
T
)
x
(
t
-
T
)
=
cos
(
Ω
1
(
t
-
T
)
)
+
cos
(
Ω
2
(
t
-
T
)
)
=
cos
(
Ω
1
t
-
Ω
1
T
)
+
cos
(
Ω
2
t
-
Ω
2
T
)
(10)
It follows that Ω1T=2πkΩ1T=2πk and Ω2T=2πlΩ2T=2πl, where kk and ll are integers. Solving these two equations for TT gives T=2πk/Ω1=2πl/Ω2T=2πk/Ω1=2πl/Ω2. We wish to select the shortest possible period, since any integer multiple of the period is also a period. To do this we note that since 2πk/Ω1=2πl/Ω22πk/Ω1=2πl/Ω2, we can write
Ω
1
Ω
2
=
k
l
Ω
1
Ω
2
=
k
l
(11)
so we seek the smallest integers kk and ll that satisfy Equation 11. This can be done by finding the greatest common divisor between kk and ll. For example if Ω1=10πΩ1=10π and Ω2=15πΩ2=15π, we have k=2k=2 and l=3l=3, after dividing out 5, the greatest common divisor between 10 and 15. So the period is T=2πk/Ω1=0.4T=2πk/Ω1=0.4 sec. On the other hand, if Ω1=10πΩ1=10π and Ω2=10.1πΩ2=10.1π, we find that k=100k=100 and l=101l=101 and the period increases to T=2πk/Ω1=20T=2πk/Ω1=20 sec. Notice also that if the ratio of Ω1Ω1 and Ω2Ω2 is not a rational number, then x(t)x(t) is not periodic!
If there are more than two sinusoids, it is probably easiest to find the period of one pair of sinusoids at a time, using the two lowest frequencies (which will have a longer period). Once the frequency of the first two sinusoids has been found, replace them with a single sinusoid at the composite frequency corresponding to the first two sinusoids and compare it with the third sinusoid, and so on.
