1
$\begingroup$

So the question is to determine if this statement is true or false. The question can be seen below.

I know that two signals are orthogonal if their inner product is equal to 0. I know how this calculation is done with vectors but not with signals. Would be really helpful i someone could explain this to me. Is it also possible to just directly see it in the two graphs if they are orthogonal to each other? enter image description here

$\endgroup$
  • $\begingroup$ You can convert the signals into vectors where for each time until you take the corresponding amplitude. For example, $\mathbf{s}_1 = [+A, +A, -A, +A, -A, -A]$ where $A$ is the amplitude. $\endgroup$ – BlackMath Aug 12 at 7:29
  • $\begingroup$ Do you have enough information to validate an answer? $\endgroup$ – Laurent Duval Oct 3 at 20:26
2
$\begingroup$

I would draw a new function which is the multiplication of both functions at each point and afterwards compute the "area"(:= Integral) under the graph. This process describes the formula in an intuitive way.

enter image description here

$\endgroup$
2
$\begingroup$

The inner product of two continuous-time signals is defined by

$$\langle\mathbf{s}_1,\mathbf{s}_2\rangle=\int_{-\infty}^{\infty}s_1(t)s^*_2(t)dt\tag{1}$$

where $*$ denotes complex conjugation. In your example the conjugation is irrelevant because both signals are real-valued. So you just need to multiply the signals and determine the area of the result.

With $(1)$ it should be easy for you to show that the two given signals are indeed orthogonal.

$\endgroup$
2
$\begingroup$

Signals that are plus or minus constants on unit intervals (here $\pm 2$) are great example, because they help you understand some of the concepts of orthogonality, like a combination of amplitude and support, and you can almost evaluate it visually, without much computations. Plus, you don't have to deal with complex conjugates (as Matt did).

First, orthogonality does not depend on amplitude, so you can divide your signals by $2$, to have $\pm 1$ amplitudes. With this, you'll only have to multiply $1 \times 1$, $-1 \times -1$ or $1 \times -1$, which save mistakes. On a graph paper, you can draw $s_1$, $s_2$ and their product:

graph of Hadamard-like functions

Each square of the product is thus of "unit surface", either positive or negative:

  • positive if $s_1$ and $s_2$ have the same sign;
  • negative if $s_1$ and $s_2$ have opposite signs;

Signals will be orthogonal if there is the same number of positive and negative unit squares, so they sum up to zero surface. This is the layman version of the integral or sum definition of orthogonality for such simple signals, by counting areas multiplying support (unit $x$) and amplitude. So as there are three green bars up, and three down, they are orthogonal. Here is the same graph, showing same ($s$) and opposite ($o$) signs.

graph of Hadamard-like functions: positive, negative, orthogonality

When you are used to this, to can do it more quickly, by cancelling unit squares of opposite signs, sequentially. For the first two $x$-units (say $1$ and $2$), $s_1$ and $s_2$ are of same-then-opposite signs, so they cancel each other. Then, $s_1$ and $s_2$ are of same signs on squares $5$ and $6$, but opposite on $3$ and $4$. Once again, those cancel each other.

Comment:

  • such sequences are important in mathematics and were in signal processing. They are related to (orthogonal) Walsh, Paley or Hadamard sequences (and Haar), that can be used as orthogonal transformations with only adds and subtracts, that can save time and power. One source: Transmission of Information by Orthogonal Functions, Harmuth, Henning F.
  • This reasoning might seem quite simple, but it is the base of integration: cutting either the time or amplitude axes in chunks, and summing them.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.