Orthogonal Functions Question

[O000]

It is quite common to check wether two functions are orthogonal using this definition :

For Functions : F(x) and G(x) & (H(x) = F(x) * G(x))

If
Integration of [H(x)] = 0 , in an interval (A to B)
then F(x) and G(x) are orthogonal

And of course :

sine(wt + Angle1) and sine(wt + Angle2) are orthogonal whenever Abs[(Angle2 - Angle1)] = 90

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Now here's the question :

For interval : A = 0 & B = 2*pi

consider G(x) = 1
and F1(x) = sine(wt + 30)

Using the "Integral definition" above we find that F1 and G are orthogonal.

consider G(x) = 1
and F2(x) = sine(wt + 40)

we find that G and F2 here are also orthogonal.

>So how can G be orthogonal to both F1 and F2, when F1 and F2 are clearly not 180 degrees apart.<

And as most of you have noticed, using the same method, all pure sinusoidal functions (regardless of theta and frequency) with no harmonics, are orthogonal to G.


My hunch is telling me that there's something quite simple behind this question, but I can't find it.
Any help is appreciated,
Thanks.

[Surg]

Whether or not to vectors(in this case functions) are orthogonal depends entirely on your choice of inner product. In the space defined by that inner product those two vectors actually are orthogonal.

Orthogonality isn't necessarily transitive.

[Talith]

(0,1) is orthogonal to (1,0) and (1/2, 0) but (1,0) is not orthogonal to (1/2, 0). You can't compare vectors transitively to check for orthogonality; it isn't an equivalence relation.

[Surg]

(0,1) is orthogonal to (1,0) and (1/2, 0) but (1,0) is not orthogonal to (1/2, 0). You can't compare vectors transitively to check for orthogonality; it isn't an equivalence relation.

The first part of what you said is right; orthogonality isn't a transitive relation. The second part of what you said is misleading in that there are a lot of relations that aren't equivalence relations but are also transitive(like total orderings).

[Talith]

of course, I only mentioned it wasn't an equivalence relation because that's the proposition I guessed that the OP was assuming when he noticed his 'contradiction'. Hopefully I didn't cause any confusion.

[O000]

Oh there was a significant typo in the first post :


>So how can G be orthogonal to both F1 and F2, when F1 and F2 are clearly not orthogonal.<
should be :
>So how can G be orthogonal to both F1 and F2, when F1 and F2 are clearly not 180 apart.<

Thanks Talith & Surg, you're absolutely right. Orthogonality is not transitive.

Now let me elaborate on what I said, to avoid any more confusion :

The way I see it, G can be orthogonal to both F1 and F2 in two cases only :

1 - F1 and F2 are represented by two vectors in the complex(phasor) domain, on the same plane, and separated by 180 degrees.
for example :
F1 = cos(wx) & F2 = cos(wt + 180) = (-1)*cos(wt)
G can be orthogonal to F1 and F2 if G(x) is considered as :
G(x) = 1 = [sin(wx)]^0

but this is too fishy and presents a whole bunch of other confusions itself. And sine^0 is bound to go through [0^0) at some point, which is undefined.

Also, F1 and F2 do need to be both 180 degrees apart to be both orthogonal to G. As in the case of the example in the first post.

2 - In the phasor domain. G(x) exists on another dimension, as a vector normal to a plane which holds both F1 and F2. just like (z) is orthogonal to all vectors in the (xy-plane), regardless of the angle between those vectors.
This could work, but I'm not comfortable about altering the phasor vector space and adding a new dimension to it, since I've never seen this before.(namely, a 3d phasor space)

@Surg : Thanks for the reply.
"In the space defined by that inner product those two vectors actually are orthogonal."
I heard that before, but could you please elaborate on the space of the inner product. How do you define this space ?

[PerchloricAcid]

Take a look at the definition section on the wiki article on vector spaces. An inner product vector space would be a vector space that has another structure - the inner product. Thanks to inner products, you can talk about length or angles - without them, you can't speak of those terms. In the Euclidean* vector space (which you've most likely mostly dealt with), the inner product is called a dot product.
When you can speak of angles, you can speak of orthogonality. If and only if the inner product of two vectors is equal to zero, they are orthogonal.
But, keep in mind that there is more than one way of defining the inner product (space), as has been stated above. Orthogonality in one space doesn't mean orthogonality in other spaces. It depends on the definition of the inner product.

Btw, I'm not studying in English, so I might have used wrong names for some things. If that's the case, I'd like somebody to correct me.

______
*That's a real coordinate space with real number coordinates (in 1D, it's a real line, in 2D it's the Cartesian plane, in 3D it's a coordinate space with 3 real number coordinates, in nD it's a coordinate space with n real number coordinates). You have other types of vector spaces, such as unitary (complex) vector spaces.

[jestingrabbit]

@O000: Consider A=(1,0,0), B=(0,1,0), C=(0,1, 2), D=(0,0,1). Notice that if we have the usual inner product on these vectors, then A is perpendicular to the other three, and the other three have a variety of angles between them: a right angle, arctan(1/2) and arctan(2). There is no constraint on the angles between vectors that are perpendicular to a third in high dimensional spaces. In fact, in three dimensional space, given two vectors that aren't on the same line through the origin, the cross product of the vectors is guaranteed to be perpendicular to both.

You are right in 2 dimensions. There, vectors that are perpendicular to the same vector all lie on the same line through the origin. But in 3 dimensions, the set of vectors perpendicular to a particular vector is a plane through the origin, and in general, the set of vectors that are perpendicular to some particular vector form a subspace with 1 less dimension that the vectors. In your particular expample, the space of functions is infinite dimensional, so there is no problem with functions that are perpendicular to some third function having any angle at all between them.

[O000]

OK Great, Thank you Jest and Perchloric.

So to clarify : The problem I was having lies in the fact that I was only considering 2D space as the Inner Product Space.

But in 3 dimensions, the set of vectors perpendicular to a particular vector is a plane through the origin, and in general, the set of vectors that are perpendicular to some particular vector form a subspace with 1 less dimension that the vectors. In your particular expample, the space of functions is infinite dimensional, so there is no problem with functions that are perpendicular to some third function having any angle at all between them.

This is synonymus to :

2 - In the phasor domain. G(x) exists on another dimension, as a vector normal to a plane which holds both F1 and F2. just like (z) is orthogonal to all vectors in the (xy-plane), regardless of the angle between those vectors.
This could work, but I'm not comfortable about altering the phasor vector space and adding a new dimension to it, since I've never seen this before.(namely, a 3d phasor space)

It all makes golly sense now, cheers..

[jestingrabbit]

OK Great, Thank you Jest and Perchloric.

So to clarify : The problem I was having lies in the fact that I was only considering 2D space as the Inner Product Space.

<snip>

This is synonymus to :

2 - In the phasor domain. G(x) exists on another dimension, as a vector normal to a plane which holds both F1 and F2. just like (z) is orthogonal to all vectors in the (xy-plane), regardless of the angle between those vectors.
This could work, but I'm not comfortable about altering the phasor vector space and adding a new dimension to it, since I've never seen this before.(namely, a 3d phasor space)

It all makes golly sense now, cheers..

Yes, that's right. Basically, a phasor domain can only deal with functions that are the same frequency and sinusoidal. Now, that's useful for analysing what a solenoid will do to an ac-voltage, but its useless for dealing with multiple frequencies. The constant function can be thought of as a different frequency, or a different shape of wave entirely. So, whatever you've learnt about the phasor domain won't be applicable to circumstances where there are other kinds of electrical signal.

[eney09]

Show that the given function are orthogonal on the indicated interval

f1(x) = e^x, f2(x) = sin(x); [π/4, 5π/4]

if someone could please explain how to do this question and what the answer is in a neat, orderly form that would be much appreciated

i've tried, but still cannot get the correct answer..

[Meem1029]

Where are you having problems? Do you know what the definition of 2 functions being orthogonal on a given interval is?

[Ben-oni]

Show that the given function are orthogonal on the indicated interval

f1(x) = e^x, f2(x) = sin(x); [π/4, 5π/4]

if someone could please explain how to do this question and what the answer is in a neat, orderly form that would be much appreciated

i've tried, but still cannot get the correct answer..

What, exactly, have you tried?