我只是采取 KIS（保持简单）方法作为第一步。

将您的（未知）信号定义为 然后只定义错误： 然后你的估计和被选为：

$$s(t) = k A(t) + m B(t) + n(t)$$ $$e(\tilde{k}, \tilde{m}) = \sum_{\forall t} \left| s(t) - \tilde{k} A(t) - \tilde{m} B(t) \right|^2$$ $\hat{k}$$\hat{m}$ $$(\hat{k}, \hat{m}) = \arg \min e(\tilde{k}, \tilde{m})$$

下面的 R 代码执行此操作，并且输出似乎做了正确的事情。这里是均值为零且标准差为的高斯噪声。$n(t)$$0.1$

实际的和值在绿色圆圈处，估计值是它下方的蓝色对角线（在最右边的图中）。$k$$m$

这似乎适用于和的任何值，包括零。您也可以将其扩展为更多信号、等。$k$$m$$C(t)$$D(t)$

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下面的R代码

# 30067

prototype_length <- 40

A <- rep(0, prototype_length)
A[(prototype_length/4+1):(3*prototype_length/4)] <- sin(2*pi*seq(0,prototype_length/2-1)/prototype_length)


B <- rep(0, prototype_length)
B[(prototype_length/2+1):(prototype_length)] <- `^`(sin(2*pi*seq(0,prototype_length/2-1)/prototype_length),12)*3/8

par(mfrow = c(1,3), pty="s")
plot(A, type="l", col="blue", lwd=5)
lines(B, col="red", lwd=5)
title('Original Signals')

k <- sample(0:10, 1)
m <- sample(0:10, 1)

data <- k*A + m*B + 0.1 * rnorm(prototype_length, 0, 0.1)
plot(data, type="l")
title('Scaled & Added Signals')

err <- array(0,c(11,11))
kvals <- array(0,c(11,11))
mvals <- array(0,c(11,11))
for (khat in 0:10)
{
  for (mhat in 0:10)
  {
    err[khat+1,mhat+1] <- sum('^'(abs(data - khat*A - mhat*B), 2))
    kvals[khat+1,mhat+1] <- khat
    mvals[khat+1,mhat+1] <- mhat
  }
}

image(0:10,0:10,err, xlab="k hat", ylab = "m hat")
ix <- which.min(err)
lines(kvals[ix] + c(-0.2,0.2), mvals[ix]+ c(-0.2,0.2), col="blue", lwd = 10)
points(k, m, col="green", lwd = 10)
title('Error, actual values, and minimum')