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?
Best Answer
From rad/sec to Hz
Usually we work in the frequency domain with Hz. Let me first make translations:
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.
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.