Frequency

Thread Starter

magnet

Joined Apr 25, 2005
16
I would like to calculate the the frequency of a carrier signal from a graph of amplitude-modulated waveform. Can anyone supply the equation please.

Thanks
 

hgmjr

Joined Jan 28, 2005
9,029
Originally posted by magnet@Jun 11 2005, 03:24 AM
I would like to calculate the the frequency of a carrier signal from a graph of amplitude-modulated waveform. Can anyone supply the equation please.

Thanks
[post=8406]Quoted post[/post]​
Can you arrange to post an image of the graph you mentioned above?
 

Dave

Joined Nov 17, 2003
6,970
This is a little difficult to do without actually seeing the waveform its self.

If you are referring to a standard AM signal, you can usually read off the time period of the highest frequency component in the modulated signal, this is the case since the process of AM impresses the modulating signal onto a higher frequnecy carrier. The frequency is merely the reciprocal of this time. This process will work for both DSB-SC, DSBTC and SSB amplitude modulation, as long as the modulted signal is periodic.

Like I said before this is particularly difficult to explain without actually seeing the waveform you are refering to. If you care to post a copy of the modulated signal I can help by refering to it.
 

n9xv

Joined Jan 18, 2005
329
You need to know the scale of the graph as in units of time per division like an oscilloscope graph. If the carrier is modulated it could be difficult to discern the uninterupted parts of the carrier. Can you provide more info and why your wanting to measure it on paper as opposed to any other method?
 

Brandon

Joined Dec 14, 2004
306
Originally posted by magnet@Jun 11 2005, 03:24 AM
I would like to calculate the the frequency of a carrier signal from a graph of amplitude-modulated waveform. Can anyone supply the equation please.

Thanks
[post=8406]Quoted post[/post]​

Very easy to answer. Formula is

1/(time between 2 peaks)

Since it AM mod, it easy to see. Just look for the AM mod. You will see high peaks and low peaks, one after another, this is the AM mod. The contour of the peaks is the actual signal carried by the AM mod.
 

Thread Starter

magnet

Joined Apr 25, 2005
16
Originally posted by Brandon@Jun 11 2005, 05:52 PM
Very easy to answer. Formula is

1/(time between 2 peaks)

Since it AM mod, it easy to see. Just look for the AM mod. You will see high peaks and low peaks, one after another, this is the AM mod. The contour of the peaks is the actual signal carried by the AM mod.
[post=8418]Quoted post[/post]​
The time base of the oscilloscope is set to 100us per division and the verticle axis to 1 v per division
 

n9xv

Joined Jan 18, 2005
329
Since frequency = 1 / time

The formula for frequency becomes;

1 / (number of divisions between peaks X time base setting)

So, for example if there were 2-1/2 divisions between peaks and the time base setting was 100-uS, the total time or "period" of the waveform would be 250-uS and 1 / 250-uS = 4,000 or 4-KHz.

But you really need to look at the unmodulated carrier to make an acurate measurment.
 

Dave

Joined Nov 17, 2003
6,970
If you zoom into the first image you should be able to see the general form of the modulated signal - from this it should be straight forward enough to observe the carrier (if you knoow what your looking for). Sadly this is not really a very accurate method, but is certainly the simplest.
 
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