# Nyquist frequency , sampling distance

I have few questions I tried to solve regarding nyquist theorem, and I would like to see your opinion if I'm doing it correctly?(one I know the answer second one not sure).

1.Let $$f(x)$$ and $$g(x)$$ be functions with max frequencys $$B_1$$ and $$B_2$$, what is the max sampling distance allowed according to nyquist to the function $$h(x)=f(x)g(x)$$.

• My answer: according to the convolution theorem $$H(x)=F(x)*G(x)$$(* for convolution), so $$H(x)=\int_{-B_1}^{B_1}F(u)G(x-u)du$$ so $$u \leq B_1$$ now we knowthat $$x-u \leq B_2$$ so we get $$x\leq B_2+B_1$$ and this $$x$$ is our max frequency, now I can say that our sample rate is $$f_s\geq 2(B_1+B_2)$$ so our sample max distance denote $$T_s$$ is $$T_s \leq \frac{1}{2(B_1+B_2)}$$ is that correct?

2.lets mark function $$f(x,y)=sinx*cosy$$ what is the max sampling rate?

• to be honest here I'm really not sure, I think its zero by intuition but not really sure if I'm correct and if so why?

Please note: those are the notation I saw when I found those questions, thanks

• Thanks! The way I solved the first question itself is also fine? secondly I didnt realy understood your answer for the second question,I'm pretty new to this subject, after all we get 2 deltas in fourier for each at $x=\frac{1}{2\pi}$ and $y=\frac{1}{2\pi}$ no? so could you further explain what happends there? – user2323232 Feb 11 at 10:56
• Your solution to the first is ok. Oh!for the second I have taken your $\sin(x)$ and $\cos(y)$ as two sinusoidals at two different frequencies such as $\sin(2 \pi B_1 t)$ and $\cos(2 \pi B_2 t)$... – Fat32 Feb 11 at 10:59