3
\$\begingroup\$

I need to calculate the Nyquist sampling rate of the function \$x(t) = cos(6t)-sin(5t)\$. By definition the Nyquist sampling rate is the minimum value of \$\omega_s\$ that yields no aliasing distortion. This is usually \$2\omega_{max}\$, which is generally the bandwidth.

However, the fourier transform of this function is \$F(\omega) = \pi(\delta(\omega+6) + \delta(\omega-6) + i\delta(\omega-5) - i\delta(\omega+5)\$. Unfortunately the graph for this cannot be shown, but this yields two real impulses at \$\omega = 6,-6\$ an imaginary impulse at \$\omega = 5\$, all with height \$\pi\$, and an imaginary impulse at \$\omega = -5\$ with height \$-\pi\$.

Therefore, I have two options for my nyquist sampling rate... Either \$10 \frac{rad}{sec}\$ or \$2 \frac{rad}{sec}\$. If I do choose the first option, this will allow for the impulses never to cross each other.

However if I do choose the second option, eventually the imaginary portion will cancel each other out due to the aliasing, and the two real impulses will cross each other. However, since the two real impulses are exactly the same amplitude, would this be considered aliasing? Also, isn't it good if the two imaginary impulses cancel each other out? Or would this all be considered aliasing?

\$\endgroup\$
  • \$\begingroup\$ Your Signal is completely real. Therefor your sampling rate is determined by the nyquist theorem and the highest occurung frequency. Your highest frequency is \$ \omega = 6 \$ so the sampling rate is \$ \omega_s = 12 \$ \$\endgroup\$ – Batuu Nov 25 '13 at 8:58
5
\$\begingroup\$

From rad/sec to Hz

Usually we work in the frequency domain with Hz. Let me first make translations:

  1. Signal \$\sin(6t)\$ is a signal of pulsation of \$ 6 rad.s^{-1}\$, which means it has a frequency \$f = \frac{6}{2\pi} \approx0.955 Hz\$
  2. Same translation for \$\sin(5t)\$, it's a \$\approx~0.795Hz\$

Time view

If you subtract them, you would see something like the blue chart below, which is an approximation of the signal for 94.25 (30 \$\pi\$) seconds

Frequency view

The fourier transform will simply show one dirac for the \$6 rad.s^{-1}\$ signal, and another one for \$5 rad.s^{-1}\$. See the fourier transform below. Real part is in green, and imaginary part is in black.

enter image description here

Nyquist frequency must be greater than twice the highest frequency of the signal, so it must be greater than \$2*6 = 12 rad.s^{-1}\$. If you choose \$ 10 rad.s^{-1}\$, you will face a aliasing issue.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ +1. Nice graphics; which tool did you use to create them? \$\endgroup\$ – Brian Drummond Nov 25 '13 at 10:53
  • \$\begingroup\$ I use Pyzo. It's a Python distribution for scientific applications. You could just use matplotlib (python) lib for the graphics. \$\endgroup\$ – RawBean Nov 25 '13 at 10:57
  • \$\begingroup\$ I would suggest that you window the time domain function to be an integer number of periods, then you would see the proper delta function sign, here you have spectral leakage into your FT. \$\endgroup\$ – daaxix Mar 17 '15 at 23:42
  • \$\begingroup\$ Thnkas daaxis for your comment, I've updated my answer accordingly. I hope I still didn't miss a thing. \$\endgroup\$ – RawBean Mar 18 '15 at 9:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.