Good modulation - Why do we need Fp fs?

[seledeur]

Homework Statement

Why do we need Fp>>fs for a good modulation, where Fp is the frequency of the Carrier signal and fs the frequency of the originalsignal

The Attempt at a Solution

If we had Fp≤ fs or less there is no use of modulation, we need a big frequency so waves can go long distances .

 

[seledeur]

Any suggestion please ?

 

[collinsmark]

Hello seledeur,

Welcome to Physics Forums!

Homework Statement

Why do we need Fp>>fs for a good modulation, where Fp is the frequency of the Carrier signal and fs the frequency of the originalsignal

What is the context of the question? What particular area is being covered in the course material? Is the communication necessarily wireless or can it be wired?

Here are some questions to think about when forming an answer (if applicable):

What does the relationship between Fp and fs have when designing the rolloff of the lowpass filter in the receiver (possibly part of the AM detector)?

If the system is wireless, how does the length of the antenna relate?

The Attempt at a Solution

If we had Fp≤ fs or less there is no use of modulation, we need a big frequency so waves can go long distances .

Yes, if modulation frequency is less than the peak, source frequency, that would introduce a whole new realm of problems.

 

[seledeur]

Thank you for the answer .
Actually I am talking about AM modulation . We saw that modulation is 'good' if h≤1 where h is the index of modulation, because the envelope detection in the receiver deletes the negative part , but I don't see a relationship with frequencies ?? or maybe in another step in the demodulation ?

 

[collinsmark]

Thank you for the answer .
Actually I am talking about AM modulation . We saw that modulation is 'good' if h≤1 where h is the index of modulation, because the envelope detection in the receiver deletes the negative part , but I don't see a relationship with frequencies ?? or maybe in another step in the demodulation ?

Although the modulation index is an important concept in its own right, I suspect it is not related to this problem.

Instead, you might want to consider how an AM detector works. It's basically a rectifier (a simple diode can suffice) and lowpass filter. What is the purpose of the filter? What criteria should the filter satisfy relating to the envelope's peak frequency and the lowest frequency of the modulated spectrum?

[Edit: I'm speaking of the filter at the detector itself, after the Radio Frequency (RF) "tuner" circuit, and after the diode].

 

[seledeur]

I think Fp should be close to 1/(2∏√LC)

 

[collinsmark]

Allow me to rephrase my question.

In the AM detector circuit, if you look at the signal at a point immediately after the diode, but before the filter, that signal will not look much at all like the original signal, because, although rectified, it still has the high frequency carrier in it. It's the lowpass filter's job to filter out the high frequency carrier part of the waveform.

What if the cuttoff frequency of the filter is too high? What if it is too low?

What if Fp and fs are close together in frequency?

 

[seledeur]

after the diode

if the cuttoff is too high then nothing will pass through it(or maybe only Fp ?), if it's too low then it will change nothing . if they are close how can the filter differ between Fp and fs ?
EDIT : but still why not fs >> Fp ?

 

[collinsmark]

after the diode

if the cuttoff is too high then nothing will pass through it(or maybe only Fp ?), if it's too low then it will change nothing .

You might have that flipped around backwards. But that's roughly the right approach to think about.

if they are close how can the filter differ between Fp and fs ?
EDIT : but still why not fs >> Fp ?

In the figure shown, a simple RC filter is shown. Single pole, RC filters such as the one shown have a rolloff of -6 dB/octave (-20 dB/decade). So if the Fp and fs are close together, what would need redesign to ensure that the carrier doesn't significantly leak into the final output?

(By the way, I'm using the variables "Fp" and "fs" simply because you used them in the original problem statement.)

(My point of all of the above is that if they are close together, one must consider what that means of the filter design. Perhaps a bigger reason why the carrier frequency is large involves antenna design, but that's a different subject than the AM detector.)

(Also, with modern electronics these days, it might not be a big deal if Fp and fs are closer together. But if you want to make the receiver as simple as possible. ...)

 

[seledeur]

Actually I'm just a high school student, I don't understand ' -6 dB/octave (-20 dB/decade' , and I still don't get why can't we have fs >> Fp ? or maybeif fs is too big, we don't really need modulation ?

 

[collinsmark]

Actually I'm just a high school student, I don't understand ' -6 dB/octave (-20 dB/decade' , and I still don't get why can't we have fs >> Fp ? or maybeif fs is too big, we don't really need modulation ?

Oh, this is high-school. I'm impressed! Okay, never mind about the -6 dB per octave thing or the details of filter design (unless you've already covered RC circuits).

Back to the definitions in the original problem statement.
Fp is the frequency of the Carrier
fs the frequency of the original signal

At the very least it's obvious that Fp must be greater than fs, by plotting the modulated signal's power or amplitude as a function of frequency (i.e., by plotting its spectrum).

Plot the original signal's unmodulated spectrum, with frequency on the horizontal axis and power spectral density (or singnal amplitude density) on the vertical axis. Don't forget negative frequencies. Zero Hz is right in the middle, with negative frequencies on the left and positive frequencies on the right.

Don't worry if you don't know the true shape of the original signal. Just make something up. Suffice it to say it has some sort of a shape in positive frequencies, with a peak frequency. The negative frequencies are a mirror image of the positive frequencies.

Now draw the power spectral density (or amplitude spectrum density) of the modulated signal, where the original signal is "shifted up the dial" so to speak, by the carrier frequency, Fp. Something like this (with different variable names):

Now ask yourself what would happen if the orignal frequency's peak frequency was greater than the carrier frequency? What would the spectrum of the modulated signal look like then? Would it not overlap with itself?

That explains why the carrier frequency must be greater than the peak frequency of the original signal. But why must it be much, much greater (">>") as opposed to just greater (">")? The only things I can think of are practical reasons I brought up earlier, such as filter rolloff requirements and antenna length. Other than practical concerns, the carrier frequency only needs to be "greater"; not much, much greater; than the original signal's frequency. Other than that, you might want to skim through your coursework again and see if the material hints at anything more specific [I can't think of anything else].

 

[seledeur]

After drawing the original signal I don't see how I can draw the modulated signal

 

[collinsmark]

After drawing the original signal I don't see how I can draw the modulated signal

Do you mean the modulated frequency spectrum?

The modulated spectrum looks essentially identical to the original spectrum, with the following, notable differences:
(1) The center of the spectrum gets shifted up by the carrier frequency; What was once 0 Hz is now the carrier frequency (we're calling it Fp here, although most other books/material call it f_c).
(2) It also gets shifted down in negative frequencies too, making the total result symmetrical.
(3) The center of each image (what was 0 Hz in the original image, now the +/- the carrier frequency in the modulated image) contains an impulse representing the carrier.
(4) The amplitude of the modulated image can be different than the original, unmodulated spectrum amplitude. But that's not important for this problem.

My last post has such an image in it, where the carrier frequency is greater than the signal frequency.

--

All that said, I'm not 100% sure this concept is what you are being asked for in the original problem. Radio modulation and radio communication are really broad topics. Without proper context, the original question can have many answers. That's why I asked more specific questions about the context. The proper answer to the question likely depends on the specific material covered in the coursework/textbook immediately before the question was asked.

 

[NascentOxygen]

Homework Statement

Why do we need Fp>>fs for a good modulation, where Fp is the frequency of the Carrier signal and fs the frequency of the originalsignal

The Attempt at a Solution

If we had Fp≤ fs or less there is no use of modulation, we need a big frequency so waves can go long distances .

Apart from the technical benefits you have alluded to (Mhz electro-magnetic frequencies propagate farther than kHz, circuits & antennae are of more practiceable sizes, etc.), translating each signal into a different high frequency region allows transmission & reception of many signals smultaneously without interference. Using AM allows you to create multiple separate independent channels, each centred on its respective f_c while sharing a common medium. The higher these carrier frequencies, the more such channels you can have.

Once you delve into the technical minutae, you can discover other advantages, too, as collinsmark explains.