# Power Spectral Density of a sequence of rnadom binary PAM symbols

Status
Not open for further replies.

#### Usjes

##### Newbie level 2 Hi,

I'd like an explanation of the PSD of a sequence of independent random PAM symbols or amplitude +1 or -1. Now any standard textbook tells us that the PSD is the Fourier Transform of the autocorrelation of the signal, in my case the symbols are INDEPENDENT and => the autocorrelation is non-zero only over the symbol duration 'T', beyond this it goes to zero. Going through the maths I come out with and answer PSD = sinc^2(wT/2), it is basically just the square of the ESD of a rectangular pulse. Again any standard textbook confirms that this is the 'correct' answer. Here is my problem though; sinc^2(wT/2) has a maximum at w = 0, ie it is saying that the most powerful component of the signal is at DC. But if I simply take the mean of my sequence I get zero (the symbols are RANDOM +/-1 and => the average is zero). So, my signal has mean zero (=> zero DC component) yet the accepted 'correct' PSD for this signal implies that its most powerful component is at D.C. Can anyone explain how this can be correct ?

Thanks,

Usjes.

#### KlausST

##### Super Moderator
Staff member Hi,

I'm no expert in this, but you use +/-1. Isn't it possible that others work with 0 and 1?
Then the DC component makes sense.

Klaus

#### Usjes

##### Newbie level 2 Hi,

I'm no expert in this, but you use +/-1. Isn't it possible that others work with 0 and 1?
Then the DC component makes sense.

Klaus

In a word: No. The PSD for a sequence using amplitudes +/-1 will have a maximum at DC despite the signal having mean zero. The same sequence using amplitudes 1,0 will also have PSD max. at DC and while this sequence will have mean = 0.5 the max in the PSD at zero will in fact be smaller than the max in the PSD for the +/-1 sequence despite this sequence having mean 0. It is quite odd.

#### FvM

##### Super Moderator
Staff member Isn't it that the text book chapters about relation of PSD and autocorrelation are presuming a zero mean signal?

Status
Not open for further replies.