If a sequence of +1 and -1 with chip rate ($\frac{1}{0.57 ns}$) is being used for purpose of estimating channel impulse response with delay spread of 6ns.

We decide to have a new rate change to rate $\frac{1}{0.38 ns} $. Then in the digital domain we upsample by 3 the initial signal, filter than downsample by 2 and we have a new sequence with new chip rate.

My question is, given the fact that the to sequence length should be longer than maximum length of channel impulse response to estimate CIF accurately in this case $6 ns$. What is the number of chips that is required for best estimation?

There are two possible answers:

1) Minimum number of chips * 0.57 $\geq 6 $

2) Minimum number of chips * 0.38 $\geq 6 $

I think the correct answer is 2, what do you think?



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