9.2 Sinusoids

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9.2 Sinusoids

Consider the sinusoidal voltage

$$v(t) = V_m \sin \omega t \tag{9.1}$$

where

Vm = the amplitude of the sinusoid

ω = the angular frequency in radians/s

ωt = the argument of the sinusoid

The sinusoid is shown in Fig. 9.1(a) as a function of its argument and in Fig. 9.1(b) as a function of time. It is e vident that the sinusoid repeats itself every T seconds; thus, T is called the period of the sinusoid. From the two plots in Fig. 9.1, we observe that ωT = 2π,

$$T = \frac{2\pi}{\omega} \tag{9.2}$$

Historical

© Hulton Archives/Getty Images

Heinrich Rudorf Hertz (1857–1894), a German experimental physicist, demonstrated that electromagnetic waves obey the same fundamental laws as light. His work confirmed James Clerk Maxwell’s celebrated 1864 theory and prediction that such waves existed.

Hertz w as born into a prosperous f amily in Hamb urg, German y. He attended the Uni versity of Berlin and did his doctorate under the prominent physicist Hermann von Helmholtz. He became a professor at Karlsruhe, where he be gan his quest for electromagnetic w aves. Hertz successfully generated and detected electromagnetic w aves; he was the first to show that light is electromagnetic ener gy. In 1887, Hertz noted for the first time the photoelectric effect of electrons in a molecular structure. Although Hertz only lived to the age of 37, his discovery of electromagnetic waves paved the way for the practical use of such waves in radio, television, and other communication systems. The unit of frequency, the hertz, bears his name.

The fact that v(t) repeats itself every T seconds is shown by replacing t by t + T in Eq. (9.1). We get

$$v(t+T) = V_m \sin \omega(t+T) = V_m \sin \omega \left(t + \frac{2\pi}{\omega}\right)$$

= $V_m \sin (\omega t + 2\pi) = V_m \sin \omega t = v(t)$ (9.3)

Hence,

$$v(t+T) = v(t) \tag{9.4}$$

that is, v has the same value at t + T as it does at t and v(t) is said to be periodic. In general,

A periodic function is one that satisfies ^f (t) = ^f (t + nT ), for all t and for all integers n.

As mentioned, the period T of the periodic function is the time of one complete cycle or the number of seconds per c ycle. The reciprocal of this quantity is the number of c ycles per second, kno wn as the cyclic frequency f of the sinusoid. Thus,

$$f = \frac{1}{T}$$

(9.5)

From Eqs. (9.2) and (9.5), it is clear that

$$\omega = 2 \pi f \tag{9.6}$$

While ω is in radians per second (rad/s), f is in hertz (Hz).

Let us now consider a more general expression for the sinusoid,

$$v(t) = V_m \sin(\omega t + \phi)$$

\n(9.7)

where (ωt + ϕ) is the argument and ϕ is the phase. Both argument and phase can be in radians or degrees.

Let us examine the two sinusoids

$$v_1(t) = V_m \sin \omega t \qquad \text{and} \qquad v_2(t) = V_m \sin(\omega t + \phi) \tag{9.8}$$

shown in Fig. 9.2. The starting point of v2 in Fig. 9.2 occurs first in time. Therefore, we say that v² leads v1 by ϕ or that v¹ lags v2 by ϕ. If ϕ ≠ 0, we also say that v1 and v2 are out of phase. If ϕ = 0, then v1 and v2 are said to be in phase; they reach their minima and maxima at e xactly the same time. We can compare v1 and v2 in this manner because they operate at the same frequency; they do not need to have the same amplitude.

A sinusoid can be e xpressed in either sine or cosine form. When comparing two sinusoids, it is expedient to express both as either sine or cosine with positive amplitudes. This is achieved by using the following trigonometric identities:

$$sin(A \pm B) = sin A cos B \pm cos A sin B$$

\n

$$cos(A \pm B) = cos A cos B \mp sin A sin B$$

\n(9.9)

With these identities, it is easy to show that

$$sin(\omega t \pm 180^\circ) = -sin \omega t$$

\n

$$cos(\omega t \pm 180^\circ) = -cos \omega t$$

\n

$$sin(\omega t \pm 90^\circ) = \pm cos \omega t$$

\n

$$cos(\omega t \pm 90^\circ) = \mp sin \omega t$$

\n(9.10)

Using these relationships, we can transform a sinusoid from sine form to cosine form or vice versa.

The unit of f is named after the German physicist Heinrich R. Hertz (1857–1894).

Figure 9.3

A graphical means of relating cosine and sine: (a) cos(ωt − 90°) = sin ωt, (b) sin(ωt + 180°) = −sin ωt.

A graphical approach may be used to relate or compare sinusoids as an alternati ve to using the trigonometric identities in Eqs. (9.9) and (9.10). Consider the set of axes shown in Fig. 9.3(a). The horizontal axis represents the magnitude of cosine, while the vertical axis (pointing down) denotes the magnitude of sine. Angles are measured positi vely counterclockwise from the horizontal, as usual in polar coordinates. This graphical technique can be used to relate tw o sinusoids. F or example, we see in Fig. 9.3(a) that subtracting 90° from the ar gument of cos ωt gives sin ωt, o r cos(ωt − 90°) = sin ωt. Similarly, adding 1 80° to the argument of sinωt gives −sin ωt, or sin(ωt + 180°) = −sin ωt, as shown in Fig. 9.3(b).

The graphical technique can also be used to add tw o sinusoids of the same frequency when one is in sine form and the other is in cosine form. To add A cos ωt and B sin ωt, we note that A is the magnitude of cos ωt while B is the magnitude of sinωt, as shown in Fig. 9.4(a). The magnitude and argument of the resultant sinusoid in cosine form is readily obtained from the triangle. Thus,

$$A\cos\omega t + B\sin\omega t = C\cos(\omega t - \theta)$$

\n(9.11)

where

$$C = \sqrt{A^2 + B^2}$$

, $\theta = \tan^{-1} \frac{B}{A}$ (9.12)

For example, we may add 3 cos ωt and −4 sinωt as shown in Fig. 9.4(b) and obtain

$$3\cos \omega t - 4\sin \omega t = 5\cos(\omega t + 53.1^{\circ})$$

(9.13)

Compared with the trigonometric identities in Eqs. (9.9) and (9.10), the graphical approach eliminates memorization. Ho wever, we must not confuse the sine and cosine ax es with the ax es for comple x numbers to be discussed in the ne xt section. Something else to note in Figs. 9.3 and 9.4 is that although the natural tendenc y is to ha ve the vertical axis point up, the positive direction of the sine function is down in the present case.

(a) Adding A cos ωt and B sin ωt, (b) adding 3 cos ωt and −4 sin ωt.

v(t) = 12 cos(50t + 10°) V.

Solution:

The amplitude is Vm = 12 V. The phase is ϕ = 10°. The angular frequency is ω = 50 rad/s. The period T = ___ ^2π ω = ___ ^2π 50 = 0.1257 s. The frequency is f = __1 T = 7.958 Hz.

Given the sinusoid 45 cos(5 πt + 36°), calculate its amplitude, phase, angular frequency, period, and frequency.

Answer: 45, 36°, 15.708 rad/s, 400 ms, 2.5 Hz.

Calculate the phase angle between v1 = −10 cos(ωt + 50°) and v2 = Example 9.2 12 sin(ωt − 10°). State which sinusoid is leading.

Solution:

Let us calculate the phase in three ways. The first two methods use trigonometric identities, while the third method uses the graphical approach.

■ METHOD 1 In order to compare v1 and v2, we must e xpress them in the same form. If we e xpress them in cosine form with posi tive amplitudes,

v1 = −10 cos(ωt + 50°) = 10 cos(ωt + 50° − 180°) v1 = 10 cos(ωt − 130°) or v1 = 10 cos(ωt + 230°) (9.2.1) and v2 = 12 sin(ωt − 10°) = 12 cos(ωt − 10° − 90°) v2 = 12 cos(ωt − 100°) (9.2.2)

It can be deduced from Eqs. (9.2.1) and (9.2.2) that the phase difference between v1 and v2 is 30°. We can write v2 as

v2 = 12 cos(ωt − 130° + 30°) or v2 = 12 cos(ωt + 260°) (9.2.3)

Comparing Eqs. (9.2.1) and (9.2.3) shows clearly that v2 leads v1 by 30°.

■ METHOD 2 Alternatively, we may express v1 in sine form:

v1 = −10 cos(ωt + 50°) = 10 sin(ωt + 50° − 90°) = 10 sin(ωt − 40°) = 10 sin(ωt − 10° − 30°)

Practice Problem 9.1

But v2 = 12 sin(ωt − 10°). Comparing the tw o shows that v1 lags v² by 30°. This is the same as saying that v2 leads v1 by 30°.

■ METHOD 3 We may regard v1 as simply −10 cosωt with a phase shift of +50°. Hence, v1 is as shown in Fig. 9.5. Similarly, v2 is 12 sinωt with a phase shift of −10°, as shown in Fig. 9.5. It is easy to see from Fig. 9.5 that v2 leads v1 by 30°, that is, 90° − 50° − 10°.

Find the phase angle between

i1 = −4 sin(377t + 55°) and i2 = 5 cos(377t − 65°)

Does i1 lead or lag i2?

Answer: 210°, i1 leads i2.