对两个信号进行卷积
计算以下信号的卷积（您的答案将采用方程的形式）：
h[n] = δ[n-1] + δ[n+1]，x[n] = δ[na] + δ[n+b]
我不知道自己在做什么用 h 和 x 做。我只是将它们相乘吗？ h[n]*x[n]？我用几种类型的模糊和边缘检测器编写了卷积程序，但我不知道如何将这些知识转化为这个问题。请帮忙！
Calculate the convolution of the following signals (your answer will be in the form of an equation):
h[n] = δ[n-1] + δ[n+1], x[n] = δ[n-a] + δ[n+b]
I'm lost as to what I do with h and x. Do I simply multiply them? h[n]*x[n]? I programmed convolution with several types of blurs and edge detectors, but I don't see how to translate that knowledge to this problem. Please help!
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卷积是一种与乘法不同的运算。如果
h[n] = delta[na]表示 n=a 处的脉冲，则 h 与任意函数f[n]的卷积等于conv(h,f) = f[na]，并且您应该能够通过叠加确定问题的答案，因为卷积和加法都是线性运算符。Convolution is an operation distinct from multiplication. If
h[n] = delta[n-a]represents an impulse at n=a, then the convolution of h and any functionf[n]is equal toconv(h,f) = f[n-a], and you should be able to determine the answer to your question through superposition, since convolution and addition are both linear operators.h 和 x (hx) 的卷积类似于 hx = SUM h[nq]*x[n]，其中根据 h 的大小对所有可能的 q 执行求和。如果 h 和 x 无穷大，则总和从 -INF 到 +INF。
The convolution of h and x (hx) will be something like hx = SUM h[n-q]*x[n], where the sum is performed over all possible q's depending on the size of h. If h and x are infinite then the sum goes from -INF to +INF.
在这种情况下（如果您不确定卷积，则更是如此），进行 卷积 会很方便以图形方式，掌握基础知识并将其与分析公式联系起来。
In this case, (and more so if you are unsure about convolution), it is convenient to do the convolution graphically, to grap the basics and relate that to the analytical formula.