Given $$x[n] = \cos\left(\frac{\pi}{3} n\right)$$ That was obtained by sampling $$x_c(t) = \cos(4000\pi t)$$

At \$T\$ samples per second, find \$T\$. Is \$T\$ unique? Why or why not?

$$x[n] = x_c(nT)$$ $$\cos\left(\frac{\pi}{3} n\right) = \cos(4000\pi nT)$$ $$\frac{\pi}{3} n = 4000\pi nT$$ $$T = \frac{1}{12000}$$

So far so good. Is \$T\$ unique? No, because either of these periodic signals can be phase shifted by \$2\pi\$ and remain essentially unchanged. E.g.

$$x[n] = \cos\left(\frac{\pi}{3} n + 2\pi\right)$$

EDIT TO SHOW STEPS $$\cos\left(\frac{\pi}{3} n\right) = \cos\left(\frac{\pi}{3} n + 2\pi\right)$$ So, $$\cos\left(\frac{\pi}{3} n + 2\pi\right) = \cos(4000\pi nT)$$ $$\frac{\pi}{3} n + 2\pi = 4000\pi nT$$ $$\frac{1}{3} + 2 = 4000 T$$ $$T = \frac{7}{12000}$$ END EDIT

Using this new \$x[n]\$, which is equivalent to the original, and the same relationship as defined above, we find that \$T = \frac{7}{12000}\$. While the math checks out (unless I've seriously screwed up), intuitively this makes no sense to me.

How would lowering the sample rate simply shift the resulting discrete time signal? By this logic, lowering the sample rate won't change the output, which is fundamentally wrong (isn't it??)

I must be missing something here, please enlighten me!

  • \$\begingroup\$ Can you show your steps using the shifted x[n]? It seems that's where your mistake is hiding. \$\endgroup\$
    – Samuel
    Dec 4 '15 at 19:17
  • \$\begingroup\$ @Samuel I just edited to correct my mistake (see the 2nd to last paragraph). However, the question remains. \$\endgroup\$
    – Kyle G.
    Dec 4 '15 at 19:18
  • \$\begingroup\$ Yes, there it is. Review your trig for the step where you "cancel" your cosine. \$\endgroup\$
    – Samuel
    Dec 4 '15 at 20:00

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.


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.

  • \$\begingroup\$ Could you elaborate? The fact that \$\cos\left(\frac{\pi}{3} n + 2\pi\right) = \cos\left(\frac{\pi}{3} n\right)\$ is where \$T\$ not being unique comes from. \$\endgroup\$
    – Kyle G.
    Dec 4 '15 at 19:34
  • \$\begingroup\$ @KyleG. I don't see how it does. \$T\$ is the same if you shift by \$2\pi\$. \$\endgroup\$
    – Samuel
    Dec 4 '15 at 19:42
  • \$\begingroup\$ Ah ok, I see what you're doing. But if we shift ONLY x[n] and not x_c(t), then T changes. However since x[n] = x[n + 2 pi], we've broken no rules. \$\endgroup\$
    – Kyle G.
    Dec 4 '15 at 19:45
  • \$\begingroup\$ I'll edit to show my math \$\endgroup\$
    – Kyle G.
    Dec 4 '15 at 19:46
  • 1
    \$\begingroup\$ @KyleG. The diplomatic thing may be to not bring it up :) Cheers. \$\endgroup\$
    – Samuel
    Dec 4 '15 at 20:19

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