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CHAP. 6]




SINUSOIDAL FUNCTIONS A sinusoidal voltage vðtÞ is given by vðtÞ ¼ V0 cos ð!t þ Þ

where V0 is the amplitude, ! is the angular velocity, or angular frequency, and  is the phase angle. The angular velocity ! may be expressed in terms of the period T or the frequency f, where f  1=T. The frequency is given in hertz, Hz, or cycles/s. Since cos !t ¼ cos tð!t þ 2Þ, ! and T are related by !T ¼ 2. And since it takes T seconds for vðtÞ to return to its original value, it goes through 1=T cycles in one second. In summary, for sinusoidal functions we have ! ¼ 2=T ¼ 2f

f ¼ 1=T ¼ !=2

T ¼ 1=f ¼ 2=!

EXAMPLE 6.1 Graph each of the following functions and specify period and frequency. ðaÞ v1 ðtÞ ¼ cos t

ðbÞ v2 ðtÞ ¼ sin t

ðcÞ v3 ðtÞ ¼ 2 cos 2t

ðdÞ v4 ðtÞ ¼ 2 cos ðt=4  458Þ ¼ 2 cos ðt=4  =4Þ ¼ 2 cos½ðt  1Þ=4 ðeÞ v5 ðtÞ ¼ 5 cos ð10t þ 608Þ ¼ 5 cos ð10t þ =3Þ ¼ 5 cos 10ðt þ =30Þ (a) See Fig. 6-2(a).

T ¼ 2 ¼ 6:2832 s and f ¼ 0:159 Hz.

(b) See Fig. 6-2(b).

T ¼ 2 ¼ 6:2832 s and f ¼ 0:159 Hz.


T ¼ 1 s and f ¼ 1 Hz.

See Fig. 6-2(c).

(d) See Fig. 6-2(d).

T ¼ 8 s and f ¼ 0:125 Hz.


T ¼ 0:2 ¼ 0:62832 s and f ¼ 1:59 Hz.

See Fig. 6-2(e).

EXAMPLE 6.2 Plot vðtÞ ¼ 5 cos !t versus !t. See Fig. 6.3.



If the function vðtÞ ¼ cos !t is delayed by  seconds, we get vðt  Þ ¼ cos !ðt  Þ ¼ cos ð!t  Þ, where  ¼ !. The delay shifts the graph of vðtÞ to the right by an amount of  seconds, which corresponds to a phase lag of  ¼ ! ¼ 2f . A time shift of  seconds to the left on the graph produces vðt þ Þ, resulting in a leading phase angle called an advance. Conversely, a phase shift of  corresponds to a time shift of . Therefore, for a given phase shift the higher is the frequency, the smaller is the required time shift. EXAMPLE 6.3 Plot vðtÞ ¼ 5 cos ðt=6 þ 308Þ versus t and t=6. Rewrite the given as vðtÞ ¼ 5 cos ðt=6 þ =6Þ ¼ 5 cos½ðt þ 1Þ=6 This is a cosine function with period of 12 s, which is advanced in time by 1 s. the left by 1 s or 308 as shown in Fig. 6-4.

In other words, the graph is shifted to

EXAMPLE 6.4 Consider a linear circuit with the following input-output pair valid for all ! and A: Input:

vi ðtÞ ¼ A cos !t


v0 ðtÞ ¼ A cosð!t  Þ

Given vi ðtÞ ¼ cos !1 t þ cos !2 t, find v0 ðtÞ when (a)  ¼ 106 ! [phase shift is proportional to frequency, Fig. 6-5(a)] (b)  ¼ 106 [phase shift is constant, Fig. 6-5(b)] The output is v0 ðtÞ ¼ cos ð!1 t  1 Þ þ cos ð!2 t  2 Þ.