Calculating Nyquist sampling rate and interval

[Evo8]

Homework Statement

Determine the Nyquist sampling rate and the Nyquist sampling interval for this signal.

sinc(2100\pit)

Homework Equations

N/A

The Attempt at a Solution

Ok I know that the Nyquist sampling rate is double or 2 times the bandwidth of a bandlimited signal. So I would assume the procedure for solving is find the bandwidth and multiply by 2. However I don't know where to start with finding the bandwidth of this signal. Also I am unsure of how to treat the "sinc" function. Any help would be greatly appreciated!

Thanks,

 

[I like Serena]

Homework Statement

Determine the Nyquist sampling rate and the Nyquist sampling interval for this signal.

sinc(2100\pit)

Homework Equations

N/A

The Attempt at a Solution

Ok I know that the Nyquist sampling rate is double or 2 times the bandwidth of a bandlimited signal. So I would assume the procedure for solving is find the bandwidth and multiply by 2. However I don't know where to start with finding the bandwidth of this signal. Also I am unsure of how to treat the "sinc" function. Any help would be greatly appreciated!

Thanks,

Hey Evo8! Long time no see!

More specifically the Nyquist sampling rate is double the highest relevant frequency $f$.

Your highest relevant frequency is the $f$ in $\text{sinc}(2\pi f t) = {\sin(2\pi f t) \over 2\pi f t}$.
The Nyquist sampling interval is the inverse of this doubled frequency.

 

[rude man]

I think you need to look at the spectrum of sinc(2πf₀t).
That spectrum is derived by taking the Fourier integral of sinc(2πf₀t) → |1/2πf₀| rect(f/2πf₀)
(this is a rectangular function between f = -πf₀ and +πf₀ and height = 1).

So the spectrum is flat from zero to f = πf₀
and here 2πf₀ = 2100π or f₀ = 1050 Hz. So the spectrum goes from zero Hz to 1050π Hz and the Nyquist sampling rate is 2100π Hz.

So that makes the Nyquist rate 3.14... times as high as if the sampled wave were a simple sine wave of f₀ = 1050 Hz.

If I did it right ... comments welcomed.

 

[Evo8]

Hi I Like Serena! good to hear from you!

Im a little confused now...

I understand what ILS is saying. So if thef is the highest relevant frequency then my Nyquist sampling rate would then be 2f.

And if my sampling rate is simply the inverse then that would turn out \frac{1}{2f} right? So this means that the 2100 in the sinc function literally has no effect on the nyquist sampling rate or interval..

I have a hard time seeing where rude man gets f_{0}=1050Hz Further more how its determined that the rate is 2100n Hz...

Thanks again for the help. It will take a few min until the mathematical portion of my brain comes out of hibernation. I simply don't use this type of math in my actual work practice so its gets uh let's say dusty...

EDIT:
I took another look at the question and I am even more confused. Where did the f come from? The original signal was sinc(2100\pi t) So is the 2100 my highest relevant frequency?

Thanks again for the help

 

[I like Serena]

After reading rude man's comment, I agree that the Nyquist rate will be $2100\pi \text{ Hz} \approx 6597 \text{ Hz}$.

To get the proper sampling, Nyquist stated that you need to sample at a frequency that is double the highest frequency that can occur, which is $1050 \cdot \pi$ in your case.
(A typical trick is to pass the signal through a low pass filter to remove unwanted high frequencies.)

He got 1050 Hz, since $2\pi f t = 2100\pi t$.
Dividing left and right by $2\pi t$ yields $f = 1050 Hz$.

 

[Evo8]

Ok i think i understand.

Heres one more related question then. Say I have the following signal $5*sinc^{2}(200 \pi t)$

Would the Nyquist sampling rate then be simply be $2*100 \pi = 200 \pi Hz$?

Thanks for braking this down for me. I do appreciate it!

 

[rude man]

Ok i think i understand.

Heres one more related question then. Say I have the following signal $5*sinc^{2}(200 \pi t)$

Would the Nyquist sampling rate then be simply be $2*100 \pi = 200 \pi Hz$?

Thanks for braking this down for me. I do appreciate it!

Once again, I would take the Fourier integral of that function of time to determine the spectrum.

I don't have a table of transforms that includes that function. I supose the choice would be to either do a convolution of F{sinc(2πf₀t)} with itself, or try to perform the actual integral. In the latter case I would try sin(x) = (1/2j)[exp(jx) - exp(-jx)] but it still looks like a bear to perform with the x² in the denominator ... and with the infinite limits it might be necessary to invoke the delta function .. blah blah .. think I'll sit this one out.

 

[I like Serena]

Ok i think i understand.

Heres one more related question then. Say I have the following signal $5*sinc^{2}(200 \pi t)$

Would the Nyquist sampling rate then be simply be $2*100 \pi = 200 \pi Hz$?

Thanks for braking this down for me. I do appreciate it!

Once again, I would take the Fourier integral of that function of time to determine the spectrum.

I don't have a table of transforms that includes that function. I supose the choice would be to either do a convolution of F{sinc(2πf₀t)} with itself, or try to perform the actual integral. In the latter case I would try sin(x) = (1/2j)[exp(jx) - exp(-jx)] but it still looks like a bear to perform with the x² in the denominator ... and with the infinite limits it might be necessary to invoke the delta function .. blah blah .. think I'll sit this one out.

It turns out not to be so bad.

According to wiki the Fourier transform of $\text{sinc}^2(ax)$ is ${1 \over \sqrt{2\pi a^2}} \text{tri}({\omega \over 2\pi a})$, where $\text{tri}$ is the triangular function that has its last non-zero value at 1.

In our case we have $~~a = 2\pi f_0$ and $\omega_{highest} = 2\pi f_{highest}$.
So $f_0 = 100 \text{ Hz}$.

The highest frequency is given by:
$${\omega_{highest} \over 2\pi a} = 1$$
$${2\pi f_{highest} \over 2\pi \cdot 2\pi f_0} = 1$$
$$f_{highest} = 2\pi f_0 = 2\pi \cdot 100 \text{ Hz}$$
Therefore the Nyquist sample rate is:
$$f_{Nyquist} = 2 f_{highest} = 4\pi\cdot 100 \approx 1257 \text{ Hz}$$

 

[rude man]

Thanks ILS.

 

[Evo8]

Ok I am hanging on here... barely.

I understand how you applied transform 203 from the table and the basic algebra to get the $f_{highest}=2 \pi f_{0}$ then finding the sample rate is even more streight forward.

The next signal I need to evaluate is something like this. $sinc(2100 \pi t)+sinc^{2}(200 \pi t)$

Soo looking at the tables on the wiki I don't see a transform that matches. Or a theorem to add two transforms. I can't image its as simple as adding the two highest frequencies together right?

Giving me $(2100 \pi)+(4 \pi 100)=7854Hz$ for a nyquist rate?

 

[I like Serena]

And if my sampling rate is simply the inverse then that would turn out \frac{1}{2}f right?

I hope you intended T_{Nyquist} = \frac{1}{2f_{highest}}.

 

[I like Serena]

Ok I am hanging on here... barely.

I understand how you applied transform 203 from the table and the basic algebra to get the $f_{highest}=2 \pi f_{0}$ then finding the sample rate is even more streight forward.

The next signal I need to evaluate is something like this. $sinc(2100 \pi t)+sinc^{2}(200 \pi t)$

Soo looking at the tables on the wiki I don't see a transform that matches. Or a theorem to add two transforms. I can't image its as simple as adding the two highest frequencies together right?

Giving me $(2100 \pi)+(4 \pi 100)=7854Hz$ for a nyquist rate?

Transform 101 tells you what happens if you add: you just add the two transforms.
So no, you don't add the two highest frequencies together.
The frequency spectrums are simply added together.
The highest resulting frequency is the maximum of the 2 frequencies.

 

[Evo8]

I hope you intended T_{Nyquist} = \frac{1}{2f_{highest}}.

Yes! Thanks for catching that. I wrote one thing down in my notebook and typed another!

 

[Evo8]

Ok for some reason I thought 101 looked like it wasn't exactly what I had. On second inspection i see it.

Now If i add them I get something like $(\frac{1}{2100 \pi}* rect(\frac{f_{highest}}{2100 \pi})+(\frac{1}{200 \pi}*tri \frac{f_{highest}}{200 \pi})$

However I am nor sure I understand this comment.

So no, you don't add the two highest frequencies together.
The frequency spectrums are simply added together.
The highest resulting frequency is the maximum of the 2 frequencies.

Soo I have the two highest frequencies of $100 \pi$ and $1050 \pi$ So the highest frequency would be $1050 \pi$ therefore the sampling rate is $2100 \pi$? I have a feeling I am not understanding fully.. right?

 

[I like Serena]

Ok for some reason I thought 101 looked like it wasn't exactly what I had. On second inspection i see it.

Now If i add them I get something like $(\frac{1}{2100 \pi}* rect(\frac{f_{highest}}{2100 \pi})+(\frac{1}{200 \pi}*tri \frac{f_{highest}}{200 \pi})$

However I am nor sure I understand this comment.

Soo I have the two highest frequencies of $100 \pi$ and $1050 \pi$ So the highest frequency would be $1050 \pi$ therefore the sampling rate is $2100 \pi$? I have a feeling I am not understanding fully.. right?

You'd get something like $\text{rect}(\frac{f}{2100 \pi})+\text{tri}(\frac{f}{200 \pi})$.
I'm leaving out the amplitudes for convenience.
In particular this is a function of the frequency f and not of f_highest.
See W|A.
Their frequency spectrums overlap.
The highest frequency where this function is non-zero is determined by the rectangle.

 

[rude man]

Ok I am hanging on here... barely.

I understand how you applied transform 203 from the table and the basic algebra to get the $f_{highest}=2 \pi f_{0}$ then finding the sample rate is even more streight forward.

The next signal I need to evaluate is something like this. $sinc(2100 \pi t)+sinc^{2}(200 \pi t)$

Soo looking at the tables on the wiki I don't see a transform that matches. Or a theorem to add two transforms. I can't image its as simple as adding the two highest frequencies together right?

?

Better than that, you just pick the higher of the two sampling rates!