基于Gram-Schmidt正交归一化基处理给定四个信号的技术咨询

Alright, let's walk through the Gram-Schmidt orthogonalization and normalization process step by step for these four signals. First, let's restate the signals clearly to avoid confusion with their piecewise definitions:

Given Signals #

$$
s_0(t) = \begin{cases}
2, & 0 < t \leq 1 \
-2, & 1 < t \leq 2 \
2, & 2 < t \leq 3
\end{cases}, \quad
s_2(t) = \begin{cases}
1, & 0 < t \leq 1 \
-2, & 1 < t \leq 2 \
0, & 2 < t \leq 3
\end{cases}
$$
$$
s_1(t) = \begin{cases}
-1, & 0 < t \leq 1 \
3, & 1 < t \leq 2 \
1, & 2 < t \leq 3
\end{cases}, \quad
s_3(t) = \begin{cases}
-1, & 0 < t \leq 2 \
-3, & 2 < t \leq 3
\end{cases}
$$

(Note: We assume $s_2(t)$ and $s_3(t)$ are 0 in intervals not explicitly defined, since the other signals span up to $t=3$.)

Step 1: Define the Inner Product #

For continuous-time signals, we use the standard inner product over the interval $[0,3]$:
$$
\langle s_i(t), s_j(t) \rangle = \int_{0}^{3} s_i(t)s_j(t) dt
$$
Normalization of a signal $v(t)$ means scaling it so its norm (square root of the inner product with itself) equals 1:
$$
\phi(t) = \frac{v(t)}{\sqrt{\langle v(t), v(t) \rangle}}
$$

Step 2: Gram-Schmidt Orthogonalization Process #

We'll process the signals in the order $s_0(t) \to s_2(t) \to s_1(t) \to s_3(t)$, checking for linear dependence along the way.

1. First Orthonormal Basis Vector $\phi_0(t)$ #

Start with the first signal as our initial orthogonal vector:
$$
v_0(t) = s_0(t)
$$
Calculate its norm squared:
$$
\langle v_0(t), v_0(t) \rangle = \int_0^1 2^2 dt + \int_1^2 (-2)^2 dt + \int_2^3 2^2 dt = 4 + 4 + 4 = 12
$$
Normalize to get $\phi_0(t)$:
$$
\phi_0(t) = \frac{v_0(t)}{\sqrt{12}} = \frac{1}{2\sqrt{3}} s_0(t) = \begin{cases}
\frac{1}{\sqrt{3}}, & 0 < t \leq 1 \
-\frac{1}{\sqrt{3}}, & 1 < t \leq 2 \
\frac{1}{\sqrt{3}}, & 2 < t \leq 3
\end{cases}
$$

2. Second Orthonormal Basis Vector $\phi_1(t)$ #

Next, orthogonalize $s_2(t)$ against $v_0(t)$:
$$
v_1(t) = s_2(t) - \frac{\langle s_2(t), v_0(t) \rangle}{\langle v_0(t), v_0(t) \rangle} v_0(t)
$$
First compute the inner product $\langle s_2(t), v_0(t) \rangle$:
$$
\langle s_2(t), v_0(t) \rangle = \int_0^1 1 \cdot 2 dt + \int_1^2 (-2) \cdot (-2) dt + \int_2^3 0 \cdot 2 dt = 2 + 4 + 0 = 6
$$
Substitute back to find $v_1(t)$:
$$
v_1(t) = s_2(t) - \frac{6}{12}v_0(t) = s_2(t) - 0.5v_0(t)
$$
Write out the piecewise definition:
$$
v_1(t) = \begin{cases}
1 - 0.5 \cdot 2 = 0, & 0 < t \leq 1 \
-2 - 0.5 \cdot (-2) = -1, & 1 < t \leq 2 \
0 - 0.5 \cdot 2 = -1, & 2 < t \leq 3
\end{cases}
$$
Calculate its norm squared:
$$
\langle v_1(t), v_1(t) \rangle = \int_1^2 (-1)^2 dt + \int_2^3 (-1)^2 dt = 1 + 1 = 2
$$
Normalize to get $\phi_1(t)$:
$$
\phi_1(t) = \frac{v_1(t)}{\sqrt{2}} = \begin{cases}
0, & 0 < t \leq 1 \
-\frac{1}{\sqrt{2}}, & 1 < t \leq 2 \
-\frac{1}{\sqrt{2}}, & 2 < t \leq 3
\end{cases}
$$

3. Check Linear Dependence of $s_1(t)$ #

Now orthogonalize $s_1(t)$ against $v_0(t)$ and $v_1(t)$:
$$
v_2(t) = s_1(t) - \frac{\langle s_1(t), v_0(t) \rangle}{\langle v_0(t), v_0(t) \rangle}v_0(t) - \frac{\langle s_1(t), v_1(t) \rangle}{\langle v_1(t), v_1(t) \rangle}v_1(t)
$$
Compute the required inner products:
$$
\langle s_1(t), v_0(t) \rangle = \int_0^1 (-1) \cdot 2 dt + \int_1^2 3 \cdot (-2) dt + \int_2^3 1 \cdot 2 dt = -2 -6 +2 = -6
$$
$$
\langle s_1(t), v_1(t) \rangle = \int_0^1 (-1) \cdot 0 dt + \int_1^2 3 \cdot (-1) dt + \int_2^3 1 \cdot (-1) dt = 0 -3 -1 = -4
$$
Substitute these values:
$$
v_2(t) = s_1(t) - \frac{-6}{12}v_0(t) - \frac{-4}{2}v_1(t) = s_1(t) + 0.5v_0(t) + 2v_1(t)
$$
Calculating the piecewise values shows $v_2(t) = 0$ for all $t \in (0,3)$. This means $s_1(t)$ is a linear combination of $v_0(t)$ and $v_1(t)$, so it doesn't add any new dimension to our basis.

4. Check Linear Dependence of $s_3(t)$ #

Repeat the process for $s_3(t)$:
$$
v_3(t) = s_3(t) - \frac{\langle s_3(t), v_0(t) \rangle}{\langle v_0(t), v_0(t) \rangle}v_0(t) - \frac{\langle s_3(t), v_1(t) \rangle}{\langle v_1(t), v_1(t) \rangle}v_1(t)
$$
Compute the inner products:
$$
\langle s_3(t), v_0(t) \rangle = \int_0^1 (-1) \cdot 2 dt + \int_1^2 (-1) \cdot (-2) dt + \int_2^3 (-3) \cdot 2 dt = -2 +2 -6 = -6
$$
$$
\langle s_3(t), v_1(t) \rangle = \int_0^1 (-1) \cdot 0 dt + \int_1^2 (-1) \cdot (-1) dt + \int_2^3 (-3) \cdot (-1) dt = 0 +1 +3 = 4
$$
Substitute back:
$$
v_3(t) = s_3(t) - \frac{-6}{12}v_0(t) - \frac{4}{2}v_1(t) = s_3(t) + 0.5v_0(t) - 2v_1(t)
$$
Again, calculating the piecewise values shows $v_3(t) = 0$ for all $t \in (0,3)$. So $s_3(t)$ is also a linear combination of $v_0(t)$ and $v_1(t)$ and doesn't contribute a new basis vector.

Final Orthonormal Basis #

The four signals span a 2-dimensional subspace, so our orthonormal basis consists of:
$$
\phi_0(t) = \begin{cases}
\frac{1}{\sqrt{3}}, & 0 < t \leq 1 \
-\frac{1}{\sqrt{3}}, & 1 < t \leq 2 \
\frac{1}{\sqrt{3}}, & 2 < t \leq 3
\end{cases}, \quad
\phi_1(t) = \begin{cases}
0, & 0 < t \leq 1 \
-\frac{1}{\sqrt{2}}, & 1 < t \leq 2 \
-\frac{1}{\sqrt{2}}, & 2 < t \leq 3
\end{cases}
$$

内容的提问来源于stack exchange，提问作者Kristoffer Jerzy Linder