# Nyquist rate and oversampling

Assuming we have time domain signal $$x(t)\,\,\,\,\,\,\, 0\leq t< T$$ and we decide to upsample the signal by factor $L\geq1$.

Then the samples of this oversampled time domain signal are represent as

$$x(kT/L)$$

I have read in paper that setting $L=1$ corresponding to the case of Nyquist rate sampling, does anyone know why this is case

If $L=4$ how many zeros are padded to obtained the upsampled signal?

I provided the text that made me ask the question in first place.

• Your question doesn't make much sense. Signal $x(t)$ seems to be time-limited, so its bandwidth is infinite. Also, it is continuous in time, so it can't be upsampled. It also looks like $T$ defines the signal duration, but it's also the sampling period? Please clarify.
– MBaz
May 25 '15 at 22:51
• Thanks for replying, l will update and provide the example as is from the paper I read. @MBaz Hope I can have your opinion then, May 26 '15 at 2:44

As stated in the paper citation, $(L-1)N$ zeros are padded in frequency domain to achieve an $L$ times oversampling, where $N$ is the number of subcarriers.
Example: Let $N=256$ and $L=4$. For the PAPR simulations use a 1024-IFFT. The 256 "center" inputs of the IFFT are the original OFDM frequency domain signal, the rest is zero. Thus $(L-1)N = 768\,$ zeros have been inserted.
"Nyquist-rate sampling" here means that all IDFT input bins are used and are in general unequal to zero, as for $L=1$ the number of zero padded subcarriers is 0. Since the IDFT input can be interpreted as a frequency domain signal, this means that the complete available bandwidth from $-f_\mathrm s/2$ up to $f_\mathrm s/2$ carries signal energy. In contrast, some of the available bandwidth is unused if zero padding in the frequency domain is applied. This is referred to as oversampling in analogy to sampling an analog signal at a higher rate than twice its maximum frequency. In the context of this question it is worth noting that the oversampling described in the article is only used for simulation purposes and that the signal will usually be analyzed at the IDFT output. A transmission of the oversampled signal is usually not done as the purpose of oversampling here is only PAPR analysis.
• thanks! Why is it that with $L=1$ we obtain Nyquist sampling? May 26 '15 at 13:45