Electronic – arduino – Maximum sample rate of Arduino Duemilanove

The minimum sampling rate to enable reconstruction of a bandlimited signal is related to the bandwidth of the signal, not the highest frequency of the signal. If a signal spans the frequency range from near DC to a maximum frequency, say Fmax, then the bandwidth of the signal is equal to the highest frequency and the minimum sampling rate is indeed 2Fmax. But if the signal starts at a frequency other than near DC, then the minimum sampling rate need only be twice the bandwidth. In your case, since the signal spans from 25 kHz to 35 kHz, the bandwidth is only 10 kHz and the minimum sampling rate is only twice that or 20 kHz. The maximum data rate that can be sent over a given bandwidth is determined by 2 factors: one, the bandwidth and two, the signal-to-noise ratio. The relationship is given by the Shannon formula: C = Blog(1+SNR) where C is the maximum bit rate, B is the channel bandwidth, SNR is the power signal-to-noise ratio, and the log is to base 2.

You're doing something wrong with the additional \$2\pi\$. You might just be forgetting that \$cos(x + 2\pi) = cos(x)\$. I assume you are leaving it in and ignoring the order of operations (bad parenthesis).

$$\cos\left(\frac{\pi}{3} n + 2\pi\right) = \cos\left(\frac{\pi}{3} n\right)\require{cancel}$$

So, using your steps and adding the \$2\pi\$.

$$\cos\left(\frac{\pi}{3} n \cancel{+ 2\pi}\right) = \cos(4000\pi nT\cancel{+ 2\pi})$$ $$\cos\left(\frac{\pi}{3} n \right) = \cos(4000\pi nT)$$ $$\frac{\pi}{3} n = 4000\pi nT$$ $$T = \frac{1}{12000}$$

It's a pretty simple mistake, thus rather easy to make.

Edit:

if we shift ONLY x[n] and not x_c(t), then T changes

$$\cos\left(\frac{\pi}{3} n + 2\pi\right) = \cos(4000\pi nT)$$ $$\cos\left(\frac{\pi}{3} n \cancel{+ 2\pi}\right) = \cos(4000\pi nT)$$ $$\cos\left(\frac{\pi}{3} n \right) = \cos(4000\pi nT)$$ $$\frac{\pi}{3} n = 4000\pi nT$$ $$T = \frac{1}{12000}$$

No, it doesn't.

Edit 2:

$$\cos\left(\frac{\pi}{3} n + 2\pi\right) = \cos(4000\pi nT)$$ $$\frac{\pi}{3} n + 2\pi = 4000\pi nT$$

Ok, the problem is you're ignoring the sum angle in your cosine when canceling it. Trigonometry doesn't work that way. You can leave the \$2\pi\$ in, apply the proper identity, and cancel the result. Or, you can simply cross off any multiple of \$2\pi\$ and proceed as you were already doing.