I have a question regarding the sampling rate. Suppose that we transmitted a signal $m(t)$ then at receiver side, we received the continuous signal $x(t)$, which is time varying and sampled with a period of $T$ seconds at regular intervals of time. It's multiplied by the sampling function $s(t)$ which is series of impulses, then the resultant signal $x(n)$ which is a discrete-time signal is given by: $$x(n) = x(t)s(t)|_{t=nT}~~~,~~~ \text{for all }~~ n \in Z. $$

The continuous signal is as below:

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The given signals output is shown in figure below:

enter image description here

The given signal output is shown in the Figure above $$x(nT) = [10, 9, 8, 7, 6, 5, 4, 3, 2, 1] ~~~,~~~\text{for } ~~~n = 0,1,2,...,9$$

If we sampled the continuous signal $x(t)$ by sampling period of $T/2$ instead of $T$, so we supposed to have $20$ output samples, which are: $$X(nT/2) = [10 , 9.5 , 9, 8.5, 8, 7.5, 7, 6.5, 6, 5.5, 5, 4.5, 4, 3.5, 3, 2.5 , 2, 1.5, 1, 0.5]$$ $$~~~~,~~~~ \text{for }~~ n= 0,1,2...,19 $$

so we have a signal output of $20$ samples, In case of we took two polyphase components of signal X(nT/2), with length of $10$ samples similar into the $X(nT)$, let's call these two components X1 and X2 .. X1 = 10; 9; ...... 1 X2 = 9.5; 8.5 .....0.5

My question is, Do $X1$ and $X2$ are similar ? I mean Do they give the same performance if we demodulated X1 and X2 and continued with each one of them .. will they give, for example, the same BER

OR .. when changing the period of sampling, we should change the filters parameter or something else ?

thank you

  • $\begingroup$ You talk about demodulation and BER, yet you didn't refer to any communication system or modulation scheme. In which context are these signals being used? $\endgroup$ – Tendero Jul 20 '18 at 14:01
  • $\begingroup$ @Tendero ... Ok, let's consider that we are using OFDM, then when we receive the signal, we upsample it as mentioned in the question, and then downsample to get two polyphase components. SO the question, either using X1 or X2 can will be OK. For example, Multiplying X1 by received signal, then convert S/P and remove the CP then FFT .. will get same results when sampling the received signal with X2 ? .. Hope the question is clear now. $\endgroup$ – New_student Jul 20 '18 at 14:41

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