# Confused! - Max frequency versus fundamental frequency

1. Jul 20, 2011

### sajib333

Confused! -- Max frequency versus fundamental frequency

Hello

I understand the following question is very silly, however I am not sure about the answer.
Lets consider we are saying there is a signal with a frequency of 4 Hz (for example). In such a statement, does it mean the maximum frequency is 4 Hz for this signal? If it does, why we refer it as the maximum frequency? What are other frequencies of this signal then? Or, should we say the max frequency is 4 Hz and the min frequency is 0 Hz for this signal, hence the baseband bandwidth is equal to 4 Hz if it is considered as a baseband signal.

In addition, what would be the fundamental frequency (the lowest frequency component based on what the harmonics are determined) of such signal if we say a signal exists with frequency of 4 Hz?
So, how do you identify the fundamental frequency and the maximum frequency?

2. Jul 20, 2011

### Drakkith

Staff Emeritus
Re: Confused!

Are you referring to a single wave source, or something like a carrier wave?

3. Jul 20, 2011

### sajib333

Re: Confused!

A single wave source, just like a audio signal, for instance.

4. Jul 20, 2011

### Oscuro93

Re: Confused!

As far as I understand, the 4 Hz in an audio signal would refer to its "mean" frequency, so to speak. It is practically impossible to to produce a signal (of any kind, if I´m not wrong) with a perfectly constant frequency. So the 4 Hz would be the frequency that you perceive, theoretically (the human ear can only hear from 20 Hz to 20000 Hz). And, to answer the second question, the fundamental frequency would be exactly that, the 4 Hz of the signal.

5. Jul 20, 2011

6. Sep 6, 2011

### sajib333

Re: Confused!

Thanks! Thanks a lot, well, we know frequency= Light speed/ Wavelength, f= C/Lemda. Accordingly I found its written in one website " If we like to transmit a low frequency signal directly, for instance an 1 KHz audio signal, from the equation F= C/ Lemda, we get the corresponding wavelength value lemda is almost 300 Km. " My question is the audio signal of 1 KHz referred in this example is the bandwidth of the signal OR the constant frequency of a single signal (in such a case, may be this will be known as fundamental freq.) ?
I appreciate your time spending regarding this.

7. Sep 7, 2011

### sajib333

Re: Confused!

Re: Confused!
As far as I understand, the 4 Hz in an audio signal would refer to its "mean" frequency, so to speak. It is practically impossible to to produce a signal (of any kind, if I´m not wrong) with a perfectly constant frequency. So the 4 Hz would be the frequency that you perceive, theoretically (the human ear can only hear from 20 Hz to 20000 Hz). And, to answer the second question, the fundamental frequency would be exactly that, the 4 Hz of the signal.

Thanks! Thanks a lot, well, we know frequency= Light speed/ Wavelength, f= C/Lemda. Accordingly I found its written in one website " If we like to transmit a low frequency signal directly, for instance an 1 KHz audio signal, from the equation F= C/ Lemda, we get the corresponding wavelength value lemda is almost 300 Km. " My question is the audio signal of 1 KHz referred in this example is the bandwidth of the signal OR the constant frequency of a single signal (in such a case, may be this will be known as fundamental freq.) ?
I appreciate your time spending regarding this.

8. Sep 7, 2011

### Drakkith

Staff Emeritus
Re: Confused!

The f=c/wavelength is only for electromagnetic waves, not sound waves. I think you are mixing up a transmitted audio signal, like to a radio, with an actual sound wave that goes through the air.

9. Sep 7, 2011

### sajib333

Re: Confused!

10. Sep 7, 2011

### Mordred

Re: Confused!

Not my area mainly but I'll take a stab at this by using a rope analogy. If you shake a rope and the other end is fixed a continous wave will travel down to the fixed end and be reflected back. As you continue to shake the rope the frequencies will be travelling in both directions and the wave coming back will interfere with the wave going forward. If you shake the rope at just the right frequency the sum of the two waves will interfere in such a way to that that a large standing wave will be produced (4htz in your case) because it does not appear to be moving the points of destructive interence is called nodes and the constructive (not destructive) interference called antinodes. Standing waves occur at more than one frequency. but are always multiples of the lowest standing wave. The frequencies at which standing waves are produced are the natural frequencies or resonant frequencies of the rope. the lowest frequency called the fundamental frequency corresponds to one antinode. The other natural frequencies are called overtones when they are integral multiples of the fundamental frequency they are called harmonics. with the first fundamental referrd to as the first harmonic, the next loop after the fundamental is called second harmonic or first overtone.

I'm really not sure what you refer to as max frequency of a signal or min frequency of a signal given at a specific frequency. if you have a frequency however the min frequency cannot be zero the harmonic frequencies can continuosly increase into larger multiples of the fundamental frequency so max freqency is also a confusing term.

Last edited: Sep 7, 2011
11. Sep 7, 2011

### Drakkith

Staff Emeritus
Re: Confused!

I did read it. It's just not easy to understand exactly what you want since the discussion has ranged from sound waves to electronic signals. In your quote I think they are talking about the frequency of the wave itself is 1 khz. If it were anything other than the frequency then they wouldn't be talking about the wavelength being 300 km.

12. Jan 29, 2012

### sajib333

Re: Confused!

Hej!

Isnt it the maximum frequency of a signal actually means(or equal to) the highest/largest harmonics of that signal?

13. Sep 10, 2012

### sajib333

Re: Confused!

The following text is not only defined in an weblink (i.e http://www.ni.com/white-paper/3002/en ) , but also written in a number of other sites (with the same menaing).

''" AM radio stations transmit audio signals, with a range from 20 Hz to 20 kHz, using carrier waves with a range from 500 kHz to 1.7 MHz. If they were to transmit audio signals directly they would need an antenna that is around 10,000 km! ""
Where the antenna height is calcualted using the formula, f=c/wavelength

I also support your justification in that a audio signal can not be traveled with velocity of light, but , on the other hand, its not very unlikely that all of these websites are providing wrong justification.
My question is to you, how do you calculate (using which formula) the value of wavelength?
What you will set the velocity of the sound wave, if you like to find out the wavelength value?

Thanks

14. Sep 10, 2012

### Staff: Mentor

Re: Confused!

It means were are referring to a perfectly sinusoidal signal, it comprises a single frequency, 4 Hz. There are no other frequency components.
The fundamental is 4 Hz. There are no other frequency components; it contains no harmonics. (Ideally, of course.)

As for audio wavelengths, if we consider the medium to be air at STP:
Refer to tables to find the speed of sound in rock, steel, or quartz (glass).

15. Sep 10, 2012

### Drakkith

Staff Emeritus
Re: Confused!

Radio stations do not transmit sound waves at all. They encode data and send it out using radio waves, which travel at c. The data is then decoded by the receiving radio and turned into sound waves by the speakers.

As for finding the wavelength, check here: http://en.wikipedia.org/wiki/Wavelength

16. Sep 10, 2012

### sajib333

Re: Confused!

Thanks!

I understand the point. All the low frequency waves need to be superimposed on to a carrier wave, so that the message can be transmitted. But, my question is on the second line of the following paragraph, where they mentioned if a carrier were not used , then the antenna height would have been 10000 Km to transmit a audio signal without any carrier signal (without modulation). Here, 10000 km is found dividing the velocity of light by 20 Hz.
Now, my point is that the way they found 10000 km is not an appropriate calculation, since velocity of light can not be used when we consider a 20 Hz signal.

However, a plenty of sources considered the same value of C as speed of light, when they calculated the wavelength of a low frequency signal.
My question is are they all wrong? Or what it should actually be.
Thanks a lot for your time.

'" AM radio stations transmit audio signals, with a range from 20 Hz to 20 kHz, using carrier waves with a range from 500 kHz to 1.7 MHz. If they were to transmit audio signals directly they would need an antenna that is around 10,000 km! ""

I understnad the antenna height is calcualted using the formula, f=c/wavelength, where c= speed of light is considered.

source:http://www.ni.com/white-paper/3002/en (there are plenty of other sources who did the calculation in a same way)

17. Sep 10, 2012

### sajib333

Re: Confused!

Thanks!
However,please consider the following text again.
'" AM radio stations transmit audio signals, with a range from 20 Hz to 20 kHz, using carrier waves with a range from 500 kHz to 1.7 MHz. If they were to transmit audio signals directly they would need an antenna that is around 10,000 km! ""

I understnad the antenna height is calcualted using the formula, f=c/wavelength, where c= speed of light is considered.
source:http://www.ni.com/white-paper/3002/en (there are plenty of other sources who did the calculation in a same way)
All they tried to explain that one of the purposes of modulation is to reduce the antenna height.

However, according to your opinion there is no dependency of antenna height on Modulation. Right? please clarify .
I would highly appreciate your time.

18. Sep 10, 2012

### Drakkith

Staff Emeritus
Re: Confused! -- Max frequency versus fundamental frequency

They are saying that if you used a 20 Hz - 20 kHz signal instead of the 500 kHz - 1.7 MHz signal, you would need an antenna that large because the frequency is so low it makes the wavelength very very large. Hence why they use a higher frequency, to make the antenna smaller.

19. Sep 10, 2012

### Staff: Mentor

Re: Confused!

The dimensions of the antenna are related to the frequencies that you wish to transmit. If you send an electrical signal of 10 kHz into a copper wire antenna then to have its electromagnetic wave wirelessly reach a distant point you must use a very long antenna.

But, if you first modulate a 1.0 MHz carrier with your 10 KHz wave (in the process producing sinusoidal signals of 0.990 MHz and 1.010 MHz) then these high-frequency signals can be transmitted via a one-hundred times smaller antenna. At the receiving end, your listener will need a radio set to convert that received 0.990 MHz and/or 1.010 MHz signal back to audio of 10 KHz so he hears or understands it.

20. Sep 11, 2012

### sajib333

Re: Confused!

Hello