Generating I/Q samples from a real signal

Question

My understanding is that if I have a real signal, it can be called as the 'I' component and its Hilbert transform can be called the 'Q' component. Due to the property of Hilbert transform, 'I' and 'Q' will have 90 degrees phase difference.

Now consider the case of a pure sine wave with constant amplitude. Let us say we sample it at 4 times its frequency, hence taking 4 samples per cycle. Consequently, each sample will have 90 degrees phase difference with respect to the previous sample. In this case, is it correct to say that "the incoming signal delayed by one sample" is the Hilbert transform, and hence the 'Q' component?

Also, now let us say we add random noise to this pure sine wave, thus making it non-periodic. Can we still say that "the incoming signal delayed by one sample" is off by 90 degrees?

I hope my question makes sense! Thanks!

Answer 1

Let us say we sample it at 4 times its frequency, hence taking 4 samples per cycle. Consequently, each sample will have 90 degrees phase difference with respect to the previous sample. In this case, is it correct to say that "the incoming signal delayed by one sample" is the Hilbert transform, and hence the 'Q' component?

It's true that the "delay by one sample" and "Hilbert transform" give the same result signal when they are applied to this particular signal, but I wouldn't say they're the same — they're still different operators.

Also, now let us say we add random noise to this pure sine wave, thus making it non-periodic. Can we still say that "the incoming signal delayed by one sample" is off by 90 degrees?

No, because the noise contains other frequencies, and for those frequencies a one-sample delay will correspond to a phase shift which is not 90 degrees.

There's one particular signal which is the Hilbert transform of your particular sine-plus-noise, and this is not it. "Random noise" is not a featureless blob which stays the same under phase shifts or delays — it's the opposite, in fact.