Signal Reconstruction from Projections

We now know how to project a signal onto other signals. We now need to learn how to reconstruct a signal $x\in{\bf C}^N$ from its projections onto different vectors $\sv_k\in{\bf C}^N$ , $k=0,1,2,\ldots,N-1$ . This will give us the inverse DFT operation (or the inverse of whatever transform we are working with).

As a simple example, consider the projection of a signal $x\in{\bf C}^N$ onto the rectilinear coordinate axes of ${\bf C}^N$ . The coordinates of the projection onto the 0th coordinate axis are simply $(x_0,0,\ldots,0)$ . The projection along coordinate axis has coordinates $(0,x_1,0,\ldots,0)$ , and so on. The original signal is then clearly the vector sum of its projections onto the coordinate axes:

$\begin{eqnarray*} x &=& (x_0,\ldots,x_{N-1})\\ &=& (x_0,0,\ldots,0) + (0,x_1,0,\ldots,0) + \cdots (0,\ldots,0,x_{N-1}) \end{eqnarray*}$

To make sure the previous paragraph is understood, let's look at the details for the case . We want to project an arbitrary two-sample signal onto the coordinate axes in 2D. A coordinate axis can be generated by multiplying any nonzero vector by scalars. The horizontal axis can be represented by any vector of the form $\underline{e}_0=(\alpha,0)$ , $\alpha\neq0$ while the vertical axis can be represented by any vector of the form $\underline{e}_1=(0,\beta)$ , $\beta\neq 0$ . For maximum simplicity, let's choose

$\begin{eqnarray*} \underline{e}_0 &\isdef & [1,0], \\ \underline{e}_1 &\isdef & [0,1]. \end{eqnarray*}$

The projection of onto $\underline{e}_0$ is, by definition,

$\begin{eqnarray*} {\bf P}_{\underline{e}_0}(x) &\isdef & \frac{\left<x,\underlin... ...0}) \underline{e}_0 = x_0 \underline{e}_0\\ [5pt] &=& [x_0,0]. \end{eqnarray*}$

Similarly, the projection of onto $\underline{e}_1$ is

$\begin{eqnarray*} {\bf P}_{\underline{e}_1}(x) &\isdef & \frac{\left<x,\underlin... ...1}) \underline{e}_1 = x_1 \underline{e}_1\\ [5pt] &=& [0,x_1]. \end{eqnarray*}$

The reconstruction of from its projections onto the coordinate axes is then the vector sum of the projections:

$\displaystyle x= {\bf P}_{\underline{e}_0}(x) + {\bf P}_{\underline{e}_1}(x) = ... ..._0 + x_1 \underline{e}_1 \isdef x_0 \cdot [1,0] + x_1 \cdot [0,1] = (x_0,x_1)$

The projection of a vector onto its coordinate axes is in some sense trivial because the very meaning of the coordinates is that they are scalars to be applied to the coordinate vectors $\underline{e}_n$ in order to form an arbitrary vector $x\in{\bf C}^N$ as a linear combination of the coordinate vectors:

$\displaystyle x\isdef x_0 \underline{e}_0 + x_1 \underline{e}_1 + \cdots + x_{N-1} \underline{e}_{N-1}$

Note that the coordinate vectors are orthogonal. Since they are also unit length, $\Vert\underline{e}_n\Vert=1$ , we say that the coordinate vectors $\{\underline{e}_n\}_{n=0}^{N-1}$ are orthonormal.

Changing Coordinates

What's more interesting is when we project a signal onto a set of vectors other than the coordinate set. This can be viewed as a change of coordinates in ${\bf C}^N$ . In the case of the DFT, the new vectors will be chosen to be sampled complex sinusoids.

An Example of Changing Coordinates in 2D

As a simple example, let's pick the following pair of new coordinate vectors in 2D:

$\begin{eqnarray*} \sv_0 &\isdef & [1,1] \\ \sv_1 &\isdef & [1,-1] \end{eqnarray*}$

These happen to be the DFT sinusoids for having frequencies (``dc'') and (half the sampling rate). (The sampled complex sinusoids of the DFT reduce to real numbers only for and .) We already showed in an earlier example that these vectors are orthogonal. However, they are not orthonormal since the norm is $\sqrt{2}$ in each case. Let's try projecting onto these vectors and seeing if we can reconstruct by summing the projections.

The projection of onto $\sv_0$ is, by definition,^5.12

$\begin{eqnarray*} {\bf P}_{\sv_0}(x) &\isdef & \frac{\left<x,\sv_0\right>}{\Vert... ...+ x_1 \cdot \overline{1})}{2} \sv_0 = \frac{x_0 + x_1}{2}\sv_0. \end{eqnarray*}$

Similarly, the projection of onto $\sv_1$ is

$\begin{eqnarray*} {\bf P}_{\sv_1}(x) &\isdef & \frac{\left<x,\sv_1\right>}{\Vert... ...- x_1 \cdot \overline{1})}{2} \sv_1 = \frac{x_0 - x_1}{2}\sv_1. \end{eqnarray*}$

The sum of these projections is then

$\begin{eqnarray*} {\bf P}_{\sv_0}(x) + {\bf P}_{\sv_1}(x) &=& \frac{x_0 + x_1}... ...} - \frac{x_0 - x_1}{2}\right) \\ [5pt] &=& (x_0,x_1) \isdef x. \end{eqnarray*}$

It worked!

Projection onto Linearly Dependent Vectors

Now consider another example:

$\begin{eqnarray*} \sv_0 &\isdef & [1,1], \\ \sv_1 &\isdef & [-1,-1]. \end{eqnarray*}$

The projections of onto these vectors are

$\begin{eqnarray*} {\bf P}_{\sv_0}(x) &=& \frac{x_0 + x_1}{2}\sv_0, \\ {\bf P}_{\sv_1}(x) &=& -\frac{x_0 + x_1}{2}\sv_1. \end{eqnarray*}$

The sum of the projections is

$\begin{eqnarray*} {\bf P}_{\sv_0}(x) + {\bf P}_{\sv_1}(x) &=& \frac{x_0 + x_1}... ... + x_1}{2} (-1,-1) \\ &=& \left(x_0+x_1,x_0+x_1\right) \neq x. \end{eqnarray*}$

Something went wrong, but what? It turns out that a set of vectors can be used to reconstruct an arbitrary vector in ${\bf C}^N$ from its projections only if they are linearly independent. In general, a set of vectors is linearly independent if none of them can be expressed as a linear combination of the others in the set. What this means intuitively is that they must ``point in different directions'' in -space. In this example so that they lie along the same line in -space. As a result, they are linearly dependent: one is a linear combination of the other ( ).

Projection onto Non-Orthogonal Vectors

Consider this example:

$\begin{eqnarray*} \sv_0 &\isdef & [1,1] \\ \sv_1 &\isdef & [0,1] \end{eqnarray*}$

These point in different directions, but they are not orthogonal. What happens now? The projections are

$\begin{eqnarray*} {\bf P}_{\sv_0}(x) &=& \frac{x_0 + x_1}{2}\sv_0 \\ {\bf P}_{\sv_1}(x) &=& x_1\sv_1. \end{eqnarray*}$

The sum of the projections is

$\begin{eqnarray*} {\bf P}_{\sv_0}(x) + {\bf P}_{\sv_1}(x) &=& \frac{x_0 + x_1}... ...\frac{x_0 + x_1}{2}, \frac{x_0 + 3x_1}{2}\right) \\ &\neq& x. \end{eqnarray*}$

So, even though the vectors are linearly independent, the sum of projections onto them does not reconstruct the original vector. Since the sum of projections worked in the orthogonal case, and since orthogonality implies linear independence, we might conjecture at this point that the sum of projections onto a set of vectors will reconstruct the original vector only when the vector set is orthogonal, and this is true, as we will show.

It turns out that one can apply an orthogonalizing process, called Gram-Schmidt orthogonalization to any linearly independent vectors in ${\bf C}^N$ so as to form an orthogonal set which will always work. This will be derived in Section 5.10.4.

Obviously, there must be at least vectors in the set. Otherwise, there would be too few degrees of freedom to represent an arbitrary $x\in{\bf C}^N$ . That is, given the coordinates $\{u(n)\}_{n=0}^{N-1}$ of (which are scale factors relative to the coordinate vectors $\underline{e}_n$ in ${\bf C}^N$ ), we have to find at least coefficients of projection (which we may think of as coordinates relative to new coordinate vectors $\sv_k$ ). If we compute only coefficients, then we would be mapping a set of complex numbers to numbers. Such a mapping cannot be invertible in general. It also turns out linearly independent vectors is always sufficient. The next section will summarize the general results along these lines.

General Conditions

This section summarizes and extends the above derivations in a more formal manner (following portions of chapter 4 of $\cite{Noble}$ ). In particular, we establish that the sum of projections of $x\in{\bf C}^N$ onto vectors $\sv_k\in{\bf C}^N$ will give back the original vector whenever the set $\{\sv_k\}_0^{N-1}$ is an orthogonal basis for ${\bf C}^N$ .

Definition: A set of vectors is said to form a vector space if, given any two members and from the set, the vectors and are also in the set, where $c\in{\bf C}^N$ is any scalar.

Definition: The set of all -dimensional complex vectors is denoted ${\bf C}^N$ . That is, ${\bf C}^N$ consists of all vectors $\underline{x}= (x_0,\ldots,x_{N-1})$ defined as a list of complex numbers $x_i\in{\bf C}$ .

Theorem: ${\bf C}^N$ is a vector space under elementwise addition and multiplication by complex scalars.

Proof: This is a special case of the following more general theorem.

Theorem: Let $M\le N$ be an integer greater than 0. Then the set of all linear combinations of vectors from ${\bf C}^N$ forms a vector space under elementwise addition and multiplication by complex scalars.

Proof: Let the original set of vectors be denoted $\sv_0,\ldots,\sv_{M-1}$ . Form

$\displaystyle x_0 = \alpha_0 \sv_0 + \cdots + \alpha_{M-1}\sv_{M-1}$

as a particular linear combination of $\{\sv_i\}_{i=0}^{M-1}$ . Then

$\displaystyle cx_0 = c\alpha_0 \sv_0 + \cdots + c\alpha_{M-1}\sv_{M-1}$

is also a linear combination of $\{\sv_i\}_{i=0}^{M-1}$ , since complex numbers are closed under multiplication ( $c\alpha_i\in{\bf C}$ for each

). Now, given any second linear combination of $\{\sv_i\}_{i=0}^{M-1}$ ,

$\displaystyle x_1 = \beta_0 \sv_0 + \cdots + \beta_{M-1}\sv_{M-1},$

the sum is

$\begin{eqnarray*} x_0 + x_1 &=& (\alpha_0 \sv_0 + \cdots \alpha_{M-1}\sv_{M-1}) ... ... \beta_0) \sv_0 + \cdots + (\alpha_{M-1} + \beta_{M-1})\sv_{M-1} \end{eqnarray*}$

which is yet another linear combination of the original vectors (since complex numbers are closed under addition). Since we have shown that scalar multiples and vector sums of linear combinations of the original vectors from ${\bf C}^N$ are also linear combinations of those same original vectors from ${\bf C}^N$ , we have that the defining properties of a vector space are satisfied. $\Box$

Note that the closure of vector addition and scalar multiplication are ``inherited'' from the closure of complex numbers under addition and multiplication.

Corollary: The set of all linear combinations of real vectors $x\in{\bf R}^N$ , using real scalars $\alpha_i\in{\bf R}$ , form a vector space.

Definition: The set of all linear combinations of a set of complex vectors from ${\bf C}^N$ , using complex scalars, is called a complex vector space of dimension .

Definition: The set of all linear combinations of a set of real vectors from ${\bf R}^N$ , using real scalars, is called a real vector space of dimension .

Definition: If a vector space consists of the set of all linear combinations of a finite set of vectors $\sv_0,\ldots,\sv_{M-1}$ , then those vectors are said to span the space.

Example: The coordinate vectors in ${\bf C}^N$ span ${\bf C}^N$ since every vector $x\in{\bf C}^N$ can be expressed as a linear combination of the coordinate vectors as

$\displaystyle x= x_0 \underline{e}_0 + x_1 \underline{e}_1 + \cdots + x_{N-1}\underline{e}_{N-1}$

where $x_i\in{\bf C}$ , and

$\begin{eqnarray*} \underline{e}_0 &=& (1,0,0,\ldots,0),\\ \underline{e}_1 &=& (... ...x{, and so on up to }\\ \underline{e}_{N-1} &=& (0,\ldots,0,1). \end{eqnarray*}$

Definition: The vector space spanned by a set of vectors from ${\bf C}^N$ is called an -dimensional subspace of ${\bf C}^N$ .

Definition: A vector $\underline{s}\in{\bf C}^N$ is said to be linearly dependent on a set of $M\le N$ vectors $\sv_i\in{\bf C}^N$ , $m=0,\ldots,M-1$ , if $\underline{s}$ can be expressed as a linear combination of those vectors.

Thus, $\underline{s}$ is linearly dependent on $\{\sv_i\}_{i=0}^{M-1}$ if there exist scalars $\{\alpha_i\}_{i=0}^{M-1}$ such that $\underline{s}= \alpha_0\sv_0 + \alpha_1\sv_1 + \cdots + \alpha_{M-1}\sv_{M-1}$ . Note that the zero vector is linearly dependent on every collection of vectors.

Theorem: (i) If $\sv_0,\ldots,\sv_{M-1}$ span a vector space, and if one of them, say $\sv_m$ , is linearly dependent on the others, then the same vector space is spanned by the set obtained by omitting $\sv_m$ from the original set. (ii) If $\sv_0,\ldots,\sv_{M-1}$ span a vector space, we can always select from these a linearly independent set that spans the same space.

Proof: Any in the space can be represented as a linear combination of the vectors $\sv_0,\ldots,\sv_{M-1}$ . By expressing $\sv_m$ as a linear combination of the other vectors in the set, the linear combination for becomes a linear combination of vectors other than $\sv_m$ . Thus, $\sv_m$ can be eliminated from the set, proving (i). To prove (ii), we can define a procedure for forming the required subset of the original vectors: First, assign $\sv_0$ to the set. Next, check to see if $\sv_0$ and $\sv_1$ are linearly dependent. If so (i.e., $\sv_1$ is a scalar times $\sv_0$ ), then discard $\sv_1$ ; otherwise assign it also to the new set. Next, check to see if $\sv_2$ is linearly dependent on the vectors in the new set. If it is (i.e., $\sv_2$ is some linear combination of $\sv_0$ and $\sv_1$ ) then discard it; otherwise assign it also to the new set. When this procedure terminates after processing $\sv_{M-1}$ , the new set will contain only linearly independent vectors which span the original space.

Definition: A set of linearly independent vectors which spans a vector space is called a basis for that vector space.

Definition: The set of coordinate vectors in ${\bf C}^N$ is called the natural basis for ${\bf C}^N$ , where the th basis vector is

$\displaystyle \underline{e}_n = [\;0\;\;\cdots\;\;0\;\underbrace{1}_{\mbox{$n$th}}\;\;0\;\;\cdots\;\;0].$

Theorem: The linear combination expressing a vector in terms of basis vectors for a vector space is unique.

Proof: Suppose a vector $x\in{\bf C}^N$ can be expressed in two different ways as a linear combination of basis vectors $\sv_0,\ldots,\sv_{N-1}$ :

$\begin{eqnarray*} x&=& \alpha_0 \sv_0 + \cdots \alpha_{N-1}\sv_{N-1} \\ &=& \beta_0 \sv_0 + \cdots \beta_{N-1}\sv_{N-1} \end{eqnarray*}$

where $\alpha_i\neq\beta_i$ for at least one value of $i\in[0,N-1]$ . Subtracting the two representations gives

$\displaystyle \underline{0}= (\alpha_0 - \beta_0)\sv_0 + \cdots + (\alpha_{N-1} - \beta_{N-1})\sv_{N-1}.$

Since the vectors are linearly independent, it is not possible to cancel the nonzero vector $(\alpha_i-\beta_i)\sv_i$ using some linear combination of the other vectors in the sum. Hence, $\alpha_i=\beta_i$ for all $i=0,1,2,\ldots,N-1$ .

Note that while the linear combination relative to a particular basis is unique, the choice of basis vectors is not. For example, given any basis set in ${\bf C}^N$ , a new basis can be formed by rotating all vectors in ${\bf C}^N$ by the same angle. In this way, an infinite number of basis sets can be generated.

As we will soon show, the DFT can be viewed as a change of coordinates from coordinates relative to the natural basis in ${\bf C}^N$ , $\{\underline{e}_n\}_{n=0}^{N-1}$ , to coordinates relative to the sinusoidal basis for ${\bf C}^N$ , $\{\sv_k\}_{k=0}^{N-1}$ , where $\sv_k(n)= e^{j\omega_k t_n}$ . The sinusoidal basis set for ${\bf C}^N$ consists of length sampled complex sinusoids at frequencies $\omega_k=2\pi k f_s/N, k=0,1,2,\ldots,N-1$ . Any scaling of these vectors in ${\bf C}^N$ by complex scale factors could also be chosen as the sinusoidal basis (i.e., any nonzero amplitude and any phase will do). However, for simplicity, we will only use unit-amplitude, zero-phase complex sinusoids as the Fourier ``frequency-domain'' basis set. To summarize this paragraph, the time-domain samples of a signal are its coordinates relative to the natural basis for ${\bf C}^N$ , while its spectral coefficients are the coordinates of the signal relative to the sinusoidal basis for ${\bf C}^N$ .

Theorem: Any two bases of a vector space contain the same number of vectors.

Proof: Left as an exercise (or see [47]).

Definition: The number of vectors in a basis for a particular space is called the dimension of the space. If the dimension is , the space is said to be an dimensional space, or -space.

In this book, we will only consider finite-dimensional vector spaces in any detail. However, the discrete-time Fourier transform (DTFT) and Fourier transform (FT) both require infinite-dimensional basis sets, because there is an infinite number of points in both the time and frequency domains. (See Appendix B for details regarding the FT and DTFT.)

Signal/Vector Reconstruction from Projections

We now arrive finally at the main desired result for this section:

Theorem: The projections of any vector $x\in{\bf C}^N$ onto any orthogonal basis set for ${\bf C}^N$ can be summed to reconstruct exactly.

Proof: Let $\{\sv_0,\ldots,\sv_{N-1}\}$ denote any orthogonal basis set for ${\bf C}^N$ . Then since is in the space spanned by these vectors, we have

$\displaystyle x= \alpha_0\sv_0 + \alpha_1\sv_1 + \cdots + \alpha_{N-1}\sv_{N-1} \protect$

(5.3)

for some (unique) scalars $\alpha_0,\ldots,\alpha_{N-1}$ . The projection of

onto $\sv_k$ is equal to

$\displaystyle {\bf P}_{\sv_k}(x) = \alpha_0{\bf P}_{\sv_k}(\sv_0) + \alpha_1{\bf P}_{\sv_k}(\sv_1) + \cdots + \alpha_{N-1}{\bf P}_{\sv_k}(\sv_{N-1})$

(using the linearity of the projection operator which follows from linearity of the inner product in its first argument). Since the basis vectors are orthogonal, the projection of $\sv_l$ onto $\sv_k$ is zero for $l\neq k$ :

$\displaystyle {\bf P}_{\sv_k}(\sv_l) \isdef \frac{\left<\sv_l,\sv_k\right>}{\l... ...ll} \underline{0}, & l\neq k \\ [5pt] \sv_k, & l=k. \\ \end{array} \right.$

We therefore obtain

$\displaystyle {\bf P}_{\sv_k}(x) = 0 + \cdots + 0 + \alpha_k{\bf P}_{\sv_k}(\sv_k) + 0 + \cdots + 0 = \alpha_k\sv_k.$

Therefore, the sum of projections onto the vectors $\sv_k$ , $k=0,1,\ldots, N-1$ , is just the linear combination of the $\sv_k$ which forms

$\displaystyle \sum_{k=0}^{N-1} {\bf P}_{\sv_k}(x) = \sum_{k=0}^{N-1} \alpha_k \sv_k = x$

by Eq. $\,$ (5.3). $\Box$

Gram-Schmidt Orthogonalization

Recall from the end of §5.10 above that an orthonormal set of vectors is a set of unit-length vectors that are mutually orthogonal. In other words, orthonormal vector set is just an orthogonal vector set in which each vector $\sv_i$ has been normalized to unit length $\sv_i/ \vert\vert\,\sv_i\,\vert\vert$ .

Theorem: Given a set of linearly independent vectors $\sv_0,\ldots,\sv_{N-1}$ from ${\bf C}^N$ , we can construct an orthonormal set $\underline{\tilde{s}}_0,\ldots,\underline{\tilde{s}}_{N-1}$ which are linear combinations of the original set and which span the same space.

Proof: We prove the theorem by constructing the desired orthonormal set $\{\underline{\tilde{s}}_k\}$ sequentially from the original set $\{\sv_k\}$ . This procedure is known as Gram-Schmidt orthogonalization.

First, note that $\sv_k\ne \underline{0}$ for all , since $\underline{0}$ is linearly dependent on every vector. Therefore, $\vert\vert\,\sv_k\,\vert\vert \ne 0$ .

Set $\underline{\tilde{s}}_0 \isdef \frac{\sv_0}{\left\Vert\,\sv_0\,\right\Vert}$ .
Define $\underline{x}_1$ as $\sv_1$ minus the projection of $\sv_1$ onto $\underline{\tilde{s}}_0$ :

$\displaystyle \underline{x}_1 \isdef \sv_1 - {\bf P}_{\underline{\tilde{s}}_0}(... ...) = \sv_1 - \left<\sv_1,\underline{\tilde{s}}_0\right>\underline{\tilde{s}}_0$
The vector $\underline{x}_1$ is orthogonal to $\underline{\tilde{s}}_0$ by construction. (We subtracted out the part of $\sv_1$ that wasn't orthogonal to $\underline{\tilde{s}}_0$ .) Also, since $\sv_1$ and $\sv_0$ are linearly independent, we have $\vert\vert\,\underline{x}_1\,\vert\vert \ne 0$ .
Set $\underline{\tilde{s}}_1 \isdef \frac{\underline{x}_1}{\left\Vert\,\underline{x}_1\,\right\Vert}$ (i.e., normalize the result of the preceding step).
Define $\underline{x}_2$ as $\sv_2$ minus the projection of $\sv_2$ onto $\underline{\tilde{s}}_0$ and $\underline{\tilde{s}}_1$ :

$\displaystyle \underline{x}_2 \;\isdef \; \sv_2 - {\bf P}_{\underline{\tilde{s}... ...ilde{s}}_0 - \left<\sv_2,\underline{\tilde{s}}_1\right>\underline{\tilde{s}}_1$
Normalize: $\underline{\tilde{s}}_2 \isdef \frac{\underline{x}_2}{\left\Vert\,\underline{x}_2\,\right\Vert}$ .
Continue this process until $\underline{\tilde{s}}_{N-1}$ has been defined.

The Gram-Schmidt orthogonalization procedure will construct an orthonormal basis from any set of linearly independent vectors. Obviously, by skipping the normalization step, we could also form simply an orthogonal basis. The key ingredient of this procedure is that each new basis vector is obtained by subtracting out the projection of the next linearly independent vector onto the vectors accepted so far into the set. We may say that each new linearly independent vector $\sv_k$ is projected onto the subspace spanned by the vectors $\{\underline{\tilde{s}}_0,\ldots,\underline{\tilde{s}}_{k-1}\}$ , and any nonzero projection in that subspace is subtracted out of $\sv_k$ to make the new vector orthogonal to the entire subspace. In other words, we retain only that portion of each new vector $\sv_k$ which ``points along'' a new dimension. The first direction is arbitrary and is determined by whatever vector we choose first ( $\sv_0$ here). The next vector is forced to be orthogonal to the first. The second is forced to be orthogonal to the first two (and thus to the 2D subspace spanned by them), and so on.

This chapter can be considered an introduction to some important concepts of linear algebra. The student is invited to pursue further reading in any textbook on linear algebra, such as [47].^5.13

Matlab/Octave examples related to this chapter appear in Appendix I.

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Signal Projection Problems
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The Inner Product