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Textbook by: Carlos E. Davila. E-mail the author

# Sinusoidal Signals

Module by: Carlos E. Davila. E-mail the author

## Sinusoidal Signals

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 ) = A cos ( Ω t ) x ( t ) = A cos ( Ω t )
(1)

and the sine:

x ( t ) = A sin ( Ω t ) x ( t ) = A sin ( Ω t )
(2)

where AA is a real constant. 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:

A cos ( Ω t ) = A 2 e j Ω t + e - j Ω t A cos ( Ω t ) = A 2 e j Ω t + e - j Ω t
(6)

and

A sin ( Ω t ) = A 2 j e j Ω t - e - j Ω t A sin ( Ω t ) = A 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 Ω t ± j sin Ω t e ± j Ω t = cos Ω t ± j sin 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.

 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 )
 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.

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