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The expression $F(t)=\frac{4}{\pi}[cos(2\pi ft)-\frac{1}{3}cos(6 \pi ft)+\frac{1}{5}cos(10 \pi ft )]$ is an approximation to a unit square-wave of frequency f at time t. Find the phase $\phi_i$ for $each$ $cosine$ $term.$

The mark scheme gives the following solutions:

Denote $F(t)$ by $$F(t)=\frac{4}{\pi}[F_1(t)-\frac{1}{3}F_2(t)+\frac{1}{5}F_3(t) )]$$

then $$\phi_1=0, \phi_2=\pm \pi, \phi_3=0.$$

But I have difficulties understanding the the meaning of $phase$, let alone interpreting the solution.

If the phase of a periodic function is defined by $$\phi(t)=2\pi\left[\left[\frac{t-t_1}{T} \right]\right] $$ where $\left[\left[x \right]\right]=x-\lfloor x \rfloor$ (i.e. the fractional part of any real number),

then how can the $\phi_i$ be constant numbers?

Isn't the phase of a function defined with initial conditions like $$\phi=sin^{-1}\left(\frac{y(0)}{A}\right)$$ ($y(0)$=initial displacement)?

Can someone please explain where my conceptual errors/over-complications lie?

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As you correctly noted the argument of the cosine (sine or exponential) function $\cos(\Phi(t))$ is called (time dependent) phase. Therefore, your solution is technically correct. However, when considering a cosine (sine or exponential) one often is also interested in the initial phase at time $t=0$. Thus, we often write $$ \cos(\Phi(t)) = \cos(\omega t + \phi_0) \tag1 $$ where $\phi_0$ is the phase at time $t=0$. This is what the given solution describes.

Furthermore, please note that the use definition of the initial phase depends on the choice to use the cosine function in eq. (1). If the functions are all cosine functions, the definition used above seams "natural". However, we could define the initial phase by $$ \sin(\Phi(t)) = \sin(\omega t + \phi_0) $$ which shifts each phase by $-\pi/2$. Hence, the initial phase is not a well-defined quantity by itself.

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  • $\begingroup$ Thanks. But what makes $F_2(t)=cos(6 \pi ft)$ special that its initial phase is $\pm \pi$? Shouldn't it be zero as well? $\endgroup$ Jun 26 '20 at 9:36
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    $\begingroup$ No, it should not. Check what you obtain, if you use $\phi_0 = \pi$. $\endgroup$
    – Semoi
    Jun 26 '20 at 12:28

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