1.5Derivation of Four Equivalent Forms

1.5.1Obtain Form 2 from Form 1

Form 1 represents a simple harmonic motion using a cosine function with amplitude Ac and a sine function with amplitude As as:

Therefore, Form 1 becomes:

xðtÞ ¼ A½ cos ðϕÞ cos ðωtÞ sin ðϕÞ sin ðωtÞ&

Use Math 3, cos(a + b) ¼ cos (a) cos (b) sin (a) sin (b), to transfer Form 1 to Form 2 as:

x(t) ¼ A cos (ωt + ϕ) (Form 2:REP)

Note that in Form 2, A is a combined amplitude of the cosine function, and ϕ is the phase shift of the simple harmonic motion.

1.5.2Obtain Form 3 from Form 2

Form 2 represents simple harmonic motion using one cosine function with a phase shift. Form 3 will represent the same simple harmonic motion using complex exponential functions.

Use Math 2, cos ðθÞ ¼ 12 ejθ þ e jθ as a bridge to transfer Form 2 to Form 3 as: xðtÞ ¼ 12 Ae jðωtþϕÞ þ Ae jðωtþϕÞ (Form 3:CEP)

Note that, in Form 3, the right-hand side of the equation is a real number because

the imaginary part is zero due to the addition of the complex conjugate pair, e j(ωt + ϕ) and e j(ωt + ϕ).

1.5.3Obtain Form 4 from Form 3

In Form 3, the constant phase ϕ is combined with the time variable ωt as one part of the complex exponential function, e j(ωt + ϕ). In Form 4, the constant phase ϕ will be separated from the time variable ωt to become a constant coefficient of e jωt.

Use Math 4, ea + b ¼ ea ∙ eb, to transfer Form 3 to get Form 3.5 as:

Note that the above form is called Form 3.5 because phase ϕ has been separated from the time variable ωt. However, e jϕ and e jϕare still not expressed in Form 4 as [Ac + jAs] and [Ac jAs].

Use Math 1, Euler’s formula, to modify Form 3.5 as:

xðtÞ ¼ 12 A½ cos ðϕÞ þ j sin ðϕÞ&ejωt þ A½ cos ðϕÞ j sin ðϕÞ&e jωt

Introducing Geom 3, Ac ¼ A cos (ϕ), and Geom 4, As ¼ A sin (ϕ), to the above equation yields:

xðtÞ ¼ 12 f½Ac þ jAs&ejωt þ ½Ac jAs&e jωtg (Form 4:CIP)

Form 4 can also be derived from Form 1 using Math 2. The derivation is straightforward and is not shown here.