Page 114 - Schaum's Outline of Theory and Problems of Electric Circuits

CHAP. 6]
6.3 SINUSOIDAL FUNCTIONS WAVEFORMS AND SIGNALS 103

A sinusoidal voltage vðtÞ is given by
vðtÞ¼ V 0 cos ð!t þ Þ
where V 0 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 2 Þ, ! 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 2 f 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 v 1 ðtÞ¼ cos t ðbÞ v 2 ðtÞ¼ sin t ðcÞ v 3 ðtÞ¼ 2 cos 2 t
ðdÞ v 4 ðtÞ¼ 2 cos ð t=4 458Þ¼ 2 cos ð t=4 =4Þ¼ 2 cos½ ðt 1Þ=4Š
ðeÞ v 5 ðtÞ¼ 5 cos ð10t þ 608Þ¼ 5 cos ð10t þ =3Þ¼ 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.
(c) See Fig. 6-2(c). T ¼ 1 s and f ¼ 1 Hz.
(d) See Fig. 6-2(d). T ¼ 8 s and f ¼ 0:125 Hz.
(e) See Fig. 6-2(e). T ¼ 0:2 ¼ 0:62832 s and f ¼ 1:59 Hz.

EXAMPLE 6.2 Plot vðtÞ¼ 5 cos !t versus !t.
See Fig. 6.3.

6.4 TIME SHIFT AND PHASE SHIFT
If the function vðtÞ¼ cos !t is delayed by seconds, we get vðt Þ¼ cos !ðt Þ¼ 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 2 f . 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 vðtÞ¼ 5 cos ð t=6 þ 308Þ versus t and t=6.
Rewrite the given as

vðtÞ¼ 5 cos ð t=6 þ =6Þ¼ 5 cos½ ðt þ 1Þ=6Š
This is a cosine function with period of 12 s, which is advanced in time by 1 s. In other words, the graph is shifted to
the left by 1 s or 308 as shown in Fig. 6-4.

EXAMPLE 6.4 Consider a linear circuit with the following input-output pair valid for all ! and A:
Input: v i ðtÞ¼ A cos !t Output: v 0 ðtÞ¼ A cosð!t Þ

Given v i ðtÞ¼ cos ! 1 t þ cos ! 2 t, find v 0 ðtÞ when
6
(a) ¼ 10 ! [phase shift is proportional to frequency, Fig. 6-5(a)]
(b) ¼ 10 6 [phase shift is constant, Fig. 6-5(b)]
The output is v 0 ðtÞ¼ cos ð! 1 t 1 Þþ cos ð! 2 t 2 Þ.