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Why an Impulse is Not White Noise

Given the test for white noise that its sample autocorrelation must approach an impulse in the limit, one might suppose that the impulse signal $ \delta(n)$ is technically ``white noise'', because its sample autocorrelation function is a perfect impulse. However, the impulse signal fails the test of being stationary. That is, its statistics are not the same at every time instant. Instead, we classify an impulse as a deterministic signal. What is true is that the impulse signal is the deterministic counterpart of white noise. Both signals contain all frequencies in equal amounts. We will see that all approaches to noise spectrum analysis (that we will consider) effectively replace noise by its autocorrelation function, thereby converting it to deterministic form. The impulse signal is already deterministic.

We can modify our white-noise test to exclude obviously nonstationary signals by dividing the signal under analysis into $ K$ blocks and computing the sample autocorrelation in each block. The final sample autocorrelation is defined as the average of the block sample autocorrelations. However, we can also test to see that the blocks are sufficiently ``comparable''. A precise definition of ``comparable'' will need to wait, but intuitively, we expect that the larger the block size (the more averaging of lagged products within each block), the more nearly identical the results for each block should be. For the impulse signal, the first block gives an ideal impulse for the sample autocorrelation, while all other blocks give the zero signal. The impulse will therefore be declared nonstationary under any reasonable definition of what it means to be ``comparable'' from block to block.

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