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In discrete-time signal processing, understanding signals as vectors within a vector space allows us to use tools of analysis and linear algebra to examine signal properties. One of the properties we may want to consider is the similarity of (or difference between) two vectors. A mathematical tool that provides insight into this is the inner product .

Transposing vectors

One of the ways to express the inner product of two vectors is through matrix multiplication, so first we must introduce the concept of transposing vectors. In order to multiply two matrices (a vector is simply a matrix in which one of the dimensions is 1), the column dimension of the first must match the row dimension of the second. To make those match for two vectors of the same length, we must transpose one of them. To take the transpose of a matrix, simply turn the rows into columns: the first row will become the first column in the transposed matrix; the second row, the second column, and so on. Here is how that looks for a vector. An $N$-row, single column vector transposed becomes a 1 row, $N$-column vector: $\begin{bmatrix}x[0]\\ x[1] \\ \vdots \\ x[N-1]\end{bmatrix}^T = \begin{bmatrix}x[0]&x[1]&\cdots&x[N-1] \end{bmatrix}$

Now, when it comes to complex valued vectors, we can take a transpose in the same way, but for the purposes of finding an inner product we actually need to take the conjugate, or Hermitian, transpose, which involves taking the transpose and then the complex conjugate: $\begin{bmatrix}x[0]\\ x[1] \\ \vdots \\ x[N-1]\end{bmatrix}^H =\begin{bmatrix}x[0]^*&x[1]^*&\cdots&x[N-1]^* \end{bmatrix}$Of course, for real valued vectors, the regular transpose and Hermitian transpose are identical.

The inner product

The inner product of two complex (or real) valued vectors is defined as: $\langle x, y \rangle = y^H x = \sum_{n=0}^{N-1} x[n]\, y[n]^*$So the inner product operation takes two vectors as inputs and produces a single number. It turns out that the number it produces is related to the angle $\theta$ between the two vectors: $\cos \theta_{x,y} = \frac{\Re{\langle x, y \rangle}}{\|x\|_2 \, \|y\|_2}$This formula works for complex and real vectors, although taking the real part of the inner product is redundant for real valued ones.

For two (or three) dimensional vectors, this angle is exactly what you would expect it to be. Let $x=\begin{bmatrix}1 \\ 2 \end{bmatrix}$, ~ $y=\begin{bmatrix}3 \\ 2 \end{bmatrix}$We have $\|x\|_2^2 = 1^2 + 2^2 = 5$, $\|y\|_2^2 = 3^2 + 2^2 = 13$ ,and $\langle x, y \rangle=(1)(3)+(2)(2)=7$.The angle between them is $\arccos\left(\frac{7}{\sqrt{(5)(13)}}\right) \approx 0.519~{\rm rad} \approx 29.7^\circ$. If you plot the vectors out in the Cartesian plane, you will indeed see an angle between them of about 30 degrees.

For higher dimensional signals the result of the inner product--how it relates to the angle between signals--may not seem as intuitive, but the information it provides is still just as useful, and of course it is computed in the same was as with shorter vectors. Consider the signals below:

Image Image
The inner product of these two signals, computed according to the formula above, is $\langle x, y \rangle ~=~ y^T x ~=~ 5.995$, which corresponds to an angle of $\theta_{x,y} ~=~ 64.9^\circ$.

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Source:  OpenStax, Pdf generation test. OpenStax CNX. Sep 10, 2015 Download for free at http://legacy.cnx.org/content/col11884/1.4
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