1

I am trying to understand the Complex Fourier series solution for the following function, as printed on "Fundamentals of Electric Circuits" by Alexander & Sadiku:

The solution printed on the solutions manual is:

enter image description here

Please note the highlighted functions inside the red squares.

Take for example the identity inside the first red square (the one on the left). I was under the impression that $e^{jn\pi/2}$ was equal to $jsin(n\pi/2)$ only when 'n' is odd, meaning that:

\begin{array}{l l} cos(n\pi/2) & \quad \text{if $n$ is even}\\ jsin(n\pi/2) & \quad \text{if $n$ is odd} \end{array}

Or in other words: \begin{array}{l l} (-1)^{n/2} & \quad \text{if $n$ is even}\\ j(-1)^{(n-1)/2} & \quad \text{if $n$ is odd} \end{array}

Is any scenario possible in which the identities inside the red squares are true for any n, regardless if n is odd or even?

Anixx
  • 9,119
S.s.
  • 199
  • 1
  • 11

1 Answers1

1

Eulers' identity is $$e^{ix} = \cos x + i\sin x$$ Where I use $i$ for the imaginary unit and $x\in\mathbb R$. This gives $$e^{\pm i \pi \frac n2} = \cos ( \pm \pi \frac n2 ) + i \sin (\pm \pi \frac n2) = \cos(\pi \frac n2) \pm i \sin (\pi \frac n2)$$ By axial symmetry of $\cos$ and point symmetry of $\sin$. Your "thought" isn't quite correct, $$e^{i\pi \frac n2} = (e^{i\frac \pi 2})^n = i^n = \cases{i & $n\equiv 1 \mod 4$ \\ -1 & $n\equiv 2 \mod 4$ \\ -i & $n\equiv 3 \mod 4$ \\ 1 & $n\equiv 0 \mod 4$}$$ This means your values are correct for the expression $e^{i\pi \frac n4}$

AlexR
  • 24,905
  • Agreed, following that reasoning is correct, however it seems odd to me that if if instead of factoring the n, you apply directly Eulers identity, it yields $e^{\pm i \pi \frac n2} = \cos ( \pm \pi \frac n2 ) + i \sin (\pm \pi \frac n2) = \cos(\pi \frac n2) \pm i \sin (\pi \frac n2)$ and in that case, $cos ( \pm \pi \frac n2 ) = (-1)^{n/2}$ instead of zero. – S.s. Dec 21 '14 at 21:19
  • @JoeM You can verify that the two statements are the same (for the $+$ case), $$\cos(\pi \frac n2) + i \sin(\pi \frac n2) = i^n \qquad \forall n\in\mathbb Z$$ – AlexR Dec 21 '14 at 21:21
  • I think we both commented at the same time, please check my comment. Thanks! – S.s. Dec 21 '14 at 21:25
  • Note that $\cos( \pi \frac n2) = 0$ if $n$ is odd, everything is fine as-is... – AlexR Dec 21 '14 at 21:28
  • Yes, I can see your solution is correct, what im trying to say is that I can't see how $e^{\pm i \pi \frac n2} = \cos ( \pm \pi \frac n2 ) + i \sin (\pm \pi \frac n2) = \cos(\pi \frac n2) \pm i \sin (\pi \frac n2)=\pm isin (\pi \frac n2)$ like is stated on the red squares of my question, again $isin (\pi \frac n2) = \pm(i(-1)^{n-1/2})$ when n is even and $0$ when its pair – S.s. Dec 21 '14 at 21:41
  • Again, if the identities in the red squares are correct, then by substituing say n=2 at both sides of the equation, should yield the same result. So again, taking the left side of the equation inside the left red square as an example, if n=2 $cos(2\pi/2) + j sin(2\pi/2)=-1$ this should be exactly the same if whe take the right side of the equation inside the red square, again for n=2 $j sin(2*\pi/2)=0$, which is not the same as the $-1$ on the left side of the equation, so my question remains, is the solution wrong or what is going on? – S.s. Dec 21 '14 at 21:56
  • You missed a bit. $\cos(\pi \frac n2)$ need not (and is not for $n$ even) be $0$. It says that (for $n=2$) $$e^{i\pi} = -1 = \cos(\pi) + i\sin(\pi)$$ Wich is correct. – AlexR Dec 21 '14 at 22:00
  • So you agree that the numerical substitution I did in my previous comment is correct and therefore the solution to the problem I posted is incorrect. Again, im not doubting the identities you posted, what im interested in is finding out how exactly $e^{\pm i \pi \frac n2} = \pm i \sin(\pi \frac n2)$ as the solution to the problem suggests. – S.s. Dec 21 '14 at 22:04
  • 1
    @JoeM Oh, wow I just noticed that equals sign. Yes, we are both correct and the solution incorrectly claims $\cos (\pi \frac n2) = 0$. – AlexR Dec 21 '14 at 22:06
  • Thats the answer I was looking for, thankyou! – S.s. Dec 21 '14 at 22:07