Exploring the Odd Symmetrical Nature of a 1/2 Wave Function

[p75213]

Homework Statement

The book defines a 1/2 wave odd symmetrical function as each 1/2 cycle is a mirror image of the next.
$$\begin{array}{l} {a_0} = 0 \\ {a_n} = {\textstyle{4 \over T}}\int_0^{{\raise0.5ex\hbox{\scriptstyle T} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}} {f(t)\cos n{w_0}t\,dt} {\rm{ - > for n odd}} \\ {a_n} = 0{\rm{ - > for }}n{\rm{ even}} \\ {b_n} = {\textstyle{4 \over T}}\int_0^{{\raise0.5ex\hbox{\scriptstyle T} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}} {f(t)\sin n{w_0}t\,dt} {\rm{ - > for n odd}} \\ {b_n} = 0{\rm{ - > for }}n{\rm{ even}} \\ \end{array}$$


An odd symmetrical function is described as being symmetrical about the vertical axis. f(-t)=-f(t).
$$\begin{array}{l} {a_0} = 0 \\ {a_n} = 0 \\ {b_n} = {\textstyle{4 \over T}}\int_0^{{\raise0.5ex\hbox{\scriptstyle T} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}} {f(t)\sin n{w_0}t\,dt} \\ \end{array}$$

Given this I would therefore categorize the attached periodic function as 1/2 wave odd symmetrical. However the book uses the odd symmetrical category and the corresponding formulas.
Can somebody explain why it is odd symmetrical and not 1/2 wave odd symmetrical?

Homework Equations

The Attempt at a Solution

 

[cepheid]

Although I wouldn't claim to be really sure, based on what I found out from reading here:

http://cnx.org/content/m32877/latest/

It sounds like a signal with half-wave odd symmetry does not necessarily have to have an odd or even symmetry. However, in this case, it DOES have odd symmetry ( f(t) = -f(-t) ), therefore, an additional simplification to the equations for the Fourier series coefficients is possible.

By the way, your choice of LaTeX syntax is...interesting. Why not try something like this?$$a_n = \frac{4}{T}\int_0^{T/2} f(t) \cos(n \omega_0 t)\,dt \longrightarrow \textrm{for}~n~\textrm{odd}$$Right click and then select "Show Math As > TeX Commands" to see the code.

 

[rude man]

Because: take points x = 0 and x = π, then go in the +x direction for both points. If there was half-wave symmetry, as you go in the +x direction, the function would go more negative as you increase above x = π, as it goes more positive as you increase above x = 0. But it doesn't. As you go above π it goes more positive, just as it does as you go above 0.

 

[cepheid]

Because: take points x = 0 and x = π, then go in the +x direction for both points. If there was half-wave symmetry, as you go in the +x direction, the function would go more negative as you increase above x = π, as it goes more positive as you increase above x = 0. But it doesn't. As you go above π it goes more positive, just as it does as you go above 0.

I think this function does have half-wave symmetry, does it not? At least according to the definition given in the link from my first post, the requirement is that x(t) = -x(t + T/2) where T is the period (2π in this instance). Take t = π/2 as an example. x(π/2) = 1.5, and it's clear that x(π/2 + π) = x(3π/2) = -1.5, in accordance with the requirement. It seems obvious that this would be true for any value of t that lies in the first linear ramp (it would be mapped to the equivalent point on the negative of that linear ramp, half a cycle later).

This also fits in with the OP's conception of it as each half cycle being the negative of the preceding one.

EDIT: NO! I see that I'm wrong (particularly with what I said in red). For instance x(0) = 1, whereas x(π) = -2 (not -1). So each point is not mapped to its negative on the next half cycle. The ramp would have to be inverted in slope as well, in order for this to be true (I see that you already stated this). So I was wrong. The function doesn't have half-wave symmetry in this case. My mistake.

 

[rude man]

No sweat, cepheid! You were right in giving us the formal definition of half-wave symmetry, to wit, f(t+T/2) = -f(t).

 

[p75213]

I agree. The surest method is to just go with the formal definition.