DSP Question Nyquist Sample Rate

Status
Not open for further replies.
S

steppermotor

Newbie level 6
Joined
Oct 14, 2010
Messages
12
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,281
Activity points
1,365
Question: A continuous-time signal x(t) = cos(f(t)), where f(t) = w0 + sin(w0t), is to be sampled at a rate fs.

What is the system’s Nyquist sample rate if w0 = 1000/(2pi) r/s?


I believe that w0=2*pi*f --> f= (1000/2*pi)*(1/2*pi) = 25.33 Hz and then multiplying by 2 to get the Nyquist sampling rate of: 50.66 Hz.

Am I on the right track or do I need to take into account the w0+sin(w0t) in some manner. I thought about taking the derivative: w0cos(w0t), but not sure how this helps me.
 

Nyquist rate must be double the highest frequency

since f(t)= w0 + sin(w0t) then MAX[f(t)]= w0 + 1

because the maximum value for sin function is 1

so Fs = (w0 + 1)/pi

I guess
 

flashking said:
Nyquist rate must be double the highest frequency

since f(t)= w0 + sin(w0t) then MAX[f(t)]= w0 + 1

because the maximum value for sin function is 1

so Fs = (w0 + 1)/pi

I guess
Click to expand...

This is completely wrong logic about the nyquist criteria , it simply says "For proper reproducibility of a signal , the sampling rate must be ATLEAST twice the maximum frquency component in the signal" , now w0 term just presents a VARIABLE DC offset at every frequency instant of sampling in time domain , so we do not have to consider it for sampling frq analysis.


I believe that w0=2*pi*f --> f= (1000/2*pi)*(1/2*pi) = 25.33 Hz and then multiplying by 2 to get the Nyquist sampling rate of: 50.66 Hz
Click to expand...

Yep , exactly , thats the minimum frequency of sampling required

---------- Post added at 07:36 ---------- Previous post was at 07:36 ----------

hope it helped !!!
 

If you take a sin(2*pi*fc*t) and sample it at fs=2fc, then you get sin(2*pi*fc*n/fs). fs=2fc. So, sin(2*pi*fc*n/(2*fc)).

On simplifying, you get, sin(n*pi) which is 0 for all n. So, even though you satisfy the nyquist rate, you'll be sampling at all points that have a zero value. In case of cosine signal, you'll get all ones. So, use a value greater than twice the nyquist rate. Regarding the nyquist rate for your question, i'd go with what flashking told.

@ph: the question says sample x(t)= cos( w0 + sin(wo t)).
 

@ph: the question says sample x(t)= cos( w0 + sin(wo t))
Click to expand...

oops , my mistake , flashking , thats true
 

Status
Not open for further replies.

Similar threads

Menu
Log in
Register
Cookies are required to use this site. You must accept them to continue using the site. Learn more…