[B.2 SINUSOIDS](#page-6-0)

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B.2 SINUSOIDS

Consider the sinusoid

$$x(t) = C\cos(2\pi f_0 t + \theta)$$

(B.13)

We know that

$$\cos \varphi = \cos (\varphi + 2n\pi)$$

$n = 0, \pm 1, \pm 2, \pm 3, ...$

Therefore, cos ϕ repeats itself for every change of 2π in the angle ϕ. For the sinusoid in Eq. (B.13), the angle 2πf0t+θ changes by 2π when t changes by 1/f0. Clearly, this sinusoid repeats every 1/f⁰ seconds. As a result, there are f⁰ repetitions per second. This is the frequency of the sinusoid, and the repetition interval T⁰ given by

$$T_0 = \frac{1}{f_0}$$

(B.14)

is the period. For the sinusoid in Eq. (B.13), C is the amplitude, f⁰ is the frequency (in hertz), and θ is the phase. Let us consider two special cases of this sinusoid when θ = 0 and θ = −π/2 as follows:

$$x(t) = C\cos 2\pi f_0 t \qquad (\theta = 0)$$

and

$$x(t) = C\cos(2\pi f_0 t - \pi/2) = C\sin 2\pi f_0 t \qquad (\theta = -\pi/2)$$

The angle or phase can be expressed in units of degrees or radians. Although the radian is the proper unit, in this book we shall often use the degree unit because students generally have a better feel for the relative magnitudes of angles expressed in degrees rather than in radians. For example, we relate better to the angle 24◦ than to 0.419 radian. Remember, however, when in doubt, use the radian unit and, above all, be consistent. In other words, in a given problem or an expression, do not mix the two units.

It is convenient to use the variable ω⁰ (radian frequency) to express 2πf0:

$$\omega_0 = 2\pi f_0 \tag{B.15}$$

With this notation, the sinusoid in Eq. (B.13) can be expressed as

$$x(t) = C\cos{(\omega_0 t + \theta)}$$

in which the period T⁰ and frequency ω⁰ are given by [see Eqs. (B.14) and (B.15)]

$$T_0 = \frac{1}{\omega_0/2\pi} = \frac{2\pi}{\omega_0} \quad \text{and} \quad \omega_0 = \frac{2\pi}{T_0}$$

Although we shall often refer to ω⁰ as the frequency of the signal cos(ω0t+θ ), it should be clearly understood that ω⁰ is the radian frequency; the hertzian frequency of this sinusoid is f⁰ = ω0/2π ).

The signals Ccos ω0t and Csin ω0t are illustrated in Figs. B.6a and B.6b, respectively. A general sinusoid Ccos(ω0t+θ ) can be readily sketched by shifting the signal Ccos ω0t in Fig. B.6a by the appropriate amount. Consider, for example,

$$x(t) = C\cos{(\omega_0 t - 60^\circ)}$$

Figure B.6 Sketching a sinusoid.

This signal can be obtained by shifting (delaying) the signal Ccos ω0t (Fig. B.6a) to the right by a phase (angle) of 60◦. We know that a sinusoid undergoes a 360◦ change of phase (or angle) in one cycle. A quarter-cycle segment corresponds to a 90◦ change of angle. We therefore shift (delay) the signal in Fig. B.6a by two-thirds of a quarter-cycle segment to obtain Ccos(ω0t − 60◦), as shown in Fig. B.6c.

Observe that if we delay Ccos ω0t in Fig. B.6a by a quarter-cycle (angle of 90◦ or π/2 radians), we obtain the signal Csin ω0t, depicted in Fig. B.6b. This verifies the well-known trigonometric identity

$$C\cos{(\omega_0 t - \pi/2)} = C\sin{\omega_0 t}$$

18 CHAPTER B BACKGROUND

Alternatively, if we advance Csin ω0t by a quarter-cycle, we obtain Ccos ω0t. Therefore,

$$C\sin(\omega_0 t + \pi/2) = C\cos\omega_0 t$$

These observations mean that sin ω0t lags cos ω0t by 90◦(π/2 radians) and that cos ω0t leads sin ω0t by 90◦.

B.2-1 Addition of Sinusoids

Two sinusoids having the same frequency but different phases add to form a single sinusoid of the same frequency. This fact is readily seen from the well-known trigonometric identity

Ccos θ cos ω0t −Csin θ sin ω0t = Ccos(ω0t +θ )

Setting a = Ccos θ and b = −Csin θ, we see that

$$a\cos\omega_0 t + b\sin\omega_0 t = C\cos(\omega_0 t + \theta)$$

(B.16)

From trigonometry, we know that

$$C = \sqrt{a^2 + b^2} \qquad \text{and} \qquad \theta = \tan^{-1}\left(\frac{-b}{a}\right) \tag{B.17}$$

Equation (B.17) shows that C and θ are the magnitude and angle, respectively, of a complex number a − jb. In other words, a − jb = Cej^θ . Hence, to find C and θ, we convert a − jb to polar form and the magnitude and the angle of the resulting polar number are C and θ, respectively.

The process of adding two sinusoids with the same frequency can be clarified by using phasors to represent sinusoids. We represent the sinusoid Ccos(ω0t+θ ) by a phasor of length C at an angle θ with the horizontal axis. Clearly, the sinusoid acos ω0t is represented by a horizontal phasor of length a(θ = 0), while bsin ω0t = bcos(ω0t −π/2) is represented by a vertical phasor of length b at an angle −π/2 with the horizontal (Fig. B.7). Adding these two phasors results in a phasor of length C at an angle θ, as depicted in Fig. B.7. From this figure, we verify the values of C and θ found in Eq. (B.17). Proper care should be exercised in computing θ, as explained on page 8 (“A Warning About Computing Angles with Calculators”).

Figure B.7 Phasor addition of sinusoids.

EXAMPLE B.6 Addition of Sinusoids

In the following cases, express x(t) as a single sinusoid:

(a) ^x(t) ⁼ cos ^ω0*^t* ^{− √}3 sin ^ω0*^t*

(b) x(t) = −3 cos ω0t +4 sin ω0t

(a) In this case, ^{a =} 1 and ^b = −√3. Using Eq. (B.17) yields

$$C = \sqrt{1^2 + (\sqrt{3})^2} = 2$$

and $\theta = \tan^{-1}(\frac{\sqrt{3}}{1}) = 60^\circ$

Therefore,

$$x(t) = 2\cos{(\omega_0 t + 60^\circ)}$$

We can verify this result by drawing phasors corresponding to the two sinusoids. The sinusoid cos ω0t is represented by a phasor of unit length at a zero angle with the horizontal. The phasor sin ω0t is represented by a unit phasor at an angle of −90◦ with the horizontal. Therefore, − ^√3 sin ^ω0*^t* is represented by a phasor of length ^√3 at 90◦ with the horizontal, as depicted in Fig. B.8a. The two phasors added yield a phasor of length 2 at 60◦ with the horizontal (also shown in Fig. B.8a).

Figure B.8 Phasor addition of sinusoids.

Alternately, we note that a−jb = 1+j ^{√3 = 2}ejπ/3. Hence, ^{C =} 2 and ^{θ =} π/3. Observe that a phase shift of ±π amounts to multiplication by −1. Therefore, x(t) can also be expressed alternatively as

$$x(t) = -2\cos(\omega_0 t + 60^\circ \pm 180^\circ) = -2\cos(\omega_0 t - 120^\circ) = -2\cos(\omega_0 t + 240^\circ)$$

In practice, the principal value, that is, −120◦, is preferred.

(b) In this case, a = −3 and b = 4. Using Eq. (B.17) yields

$$C = \sqrt{(-3)^2 + 4^2} = 5$$

and $\theta = \tan^{-1}\left(\frac{-4}{-3}\right) = -126.9^{\circ}$

Observe that

$$\tan^{-1}\left(\frac{-4}{-3}\right) \neq \tan^{-1}\left(\frac{4}{3}\right) = 53.1^{\circ}$$

Therefore,

$$x(t) = 5\cos\left(\omega_0 t - 126.9^\circ\right)$$

This result is readily verified in the phasor diagram in Fig. B.8b. Alternately, a−jb = −3−j4 = 5e−j126.9◦ , a fact readily confirmed using MATLAB.

C = abs(-3+4j) C=5 >> theta = angle(-3+4j)*180/pi theta = 126.8699

Hence, C = 5 and θ = −126.8699◦.

We can also perform the reverse operation, expressing Ccos(ω0t +θ ) in terms of cos ω0t and sin ω0t by again using the trigonometric identity

Ccos(ω0t +θ ) = Ccos θ cos ω0t −Csin θ sin ω0t

For example,

$$10\cos\left(\omega_0 t - 60^\circ\right) = 5\cos\omega_0 t + 5\sqrt{3}\sin\omega_0 t$$

B.2-2 Sinusoids in Terms of Exponentials

From Eq. (B.3), we know that ej^ϕ = cos ϕ + jsin ϕ and e−j^ϕ = cos ϕ − jsin ϕ. Adding these two expressions and dividing by 2 provide an expression for cosine in terms of complex exponentials, while subtracting and scaling by 2j provide an expression for sine. That is,

$$\cos \varphi = \frac{1}{2} (e^{j\varphi} + e^{-j\varphi})$$

and $\sin \varphi = \frac{1}{2j} (e^{j\varphi} - e^{-j\varphi})$ (B.18)