I know that why we use Carrier of high frequency to send message signal over a long distance, when the frequency is low , energy will be obviously low. To increase the energy of the signal we need to increase the frequency. This is achieved by multiplying the message signal with the carrier signal (with high frequency).


Carrier signal frequency = 2800KHz
message signal frequency = 3KHz

the two generated sidebands will be ,

2800 + 3 = 2803 KHz
2800 - 3 = 2797 KHz

and eventually, the bandwidth of the signal is,

BW = 2803 - 2797 = 6KHz

this is when the carrier frequency is higher , and we noticed that bandwidth is just twice of the highest frequency of modulating signal. and the signal will be easily demodulated at the receiver end.

but what will happen if it is reversed ?


Message signal frequency = 2800KHz
Carrie signal frequency = 3KHz

please explain would happen here ?

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    $\begingroup$ In simple terms, you can combine any two waveforms you want. However, some combinations are more useful than others. In particular, if you want to, at some remote location, separate the "signal" from the "carrier", then it's useful to not have the "carrier" in the same frequency band as the "signal". (Remember, your 2800KHz "signal" is not a single frequency but a BAND.) Additionally, there's no advantage to an "upside down" carrier/signal arrangement -- it doesn't help the signal be transmitted, but only introduces interference that must be somehow separated. $\endgroup$ – Daniel R Hicks May 29 '12 at 3:21
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    $\begingroup$ The use of "high" frequencies has nothing to do with energy. The main purpose of using a "carrier" is to transmit multiple signals of smaller bandwidth over a channel that has much larger bandwidth. $\endgroup$ – Hilmar May 29 '12 at 11:02
  • $\begingroup$ @Hilmar: indeed, though the fact that higher frequencies have shorter wavelengths and therefore can be transmitted more effectively with short antennas is also important for many applications. $\endgroup$ – leftaroundabout May 29 '12 at 14:18
  • $\begingroup$ In Modulation and antenna propagation, the adsorption of the wave makes you to need more energy to transmit. $\endgroup$ – user1434 May 29 '12 at 15:58
  • $\begingroup$ @Hilmar: also due to the practicalities of antenna design and atmospheric absorption windows en.wikipedia.org/wiki/Radio_window $\endgroup$ – endolith May 30 '12 at 11:57
suppose, Carrier signal frequency = 2800KHz message signal frequency = 3KHz

Then you will get a signal that looks like this in the frequency plane. enter image description here

Obviously this is not to scale, but you get the idea.

but what will happen if it is reversed ?

i.e Message signal frequency = 2800KHz Carrie signal frequency = 3KHz

please explain would happen here ?

Then you will get the following. enter image description here

The positive and negative sides of the signal will almost completely overlap, and thus distort each other to the point that they likely cannot be recovered. Due to the +/- 3 kHz carrier frequency there will be 6 kHz that is non-overlapping (though they will still be distorted by the other signal's roll-off), but the rest will overlap.

To increase the energy of the signal we need to increase the frequency.

Signal power has nothing to do with increasing the signal frequency. I think that you are thinking of light photons, which do increase in energy as their frequency goes up. Communication signals, though they are (at a really, really, low-level) composed of photons, it is a non-issue because adding power just means adding more photons.

Anyway, suffice to say that you don't need to increase the frequency to increase the signal power. We increase the carrier frequency for the following reasons:

  • Spectrum Availability. You send your signal where the FCC (or whatever governing body applies) says you can send it.
  • Increased bandwidth. As the answer to your question demonstrated, to send wide-band signals you need higher carrier frequencies.
  • Practicalities of antennas. It is difficult (impossible?) to make a good wide-band antenna. An antenna's bandwidth, though, is proportional to its center frequency, so increasing the carrier frequency makes it much easier to make good antennas with wide passbands.
  • Channel characteristics. Different frequencies behave in different ways. Some frequencies get absorbed by rain and some resist that. Some frequencies bounce off the ionosphere and so can travel farther than "line of sight".
  • $\begingroup$ @JimClay Nice answer - how did you create those diagrams btw? $\endgroup$ – Spacey May 30 '12 at 17:21
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    $\begingroup$ @Mohammad Created in Powerpoint, then I screen capture it and open in "Paint" to crop the image and save it in PNG format. $\endgroup$ – Jim Clay May 30 '12 at 18:10

I suppose it should be noted that there are a few applications where a "carrier" may be "beat" against a higher-frequency signal to lower it's frequency.

For instance, a sensor may return a signal that is a frequency varying between, say, 999KHz and 1.001MHz. Even though the bandwidth (2KHz) isn't particularly broad, this is an awkward signal to transmit any distance -- requiring coax cable. So one might "beat" it against, say, a 998KHz signal to produce a 1-3KHz signal than can easily be transmitted over ordinary audio/phone channels.

(The net of this is that "beating" signals together is done for various reasons, and you must understand those reasons to know why some combos are more "popular" than others.)


Simply put, the relativly higher frequency usually has the ability to move from point A to point B, unlike the message signal. Also the carrier has the ability to span over the messgae signal, but not the otherwise. This is why it is called carrier, it is like the bus (carrier) carying people (data or message) and moving them from point A to point B. Under normal conditions and when the message can move from my mouth to your ears directly then there is no need to embedd the signal on top of a carrier.


What is the bandwidth of the message signal? If the bandwidth is narrow enough and the center frequency is high enough and suitable for the channel to be used, then you don't even need a carrier (it could be 0 Hz for some modulation schemes).

  • $\begingroup$ no, i am actually asking why ? What would happen if a 1khz signal amplitude modulate a 1khz carrier signal? $\endgroup$ – Sufiyan Ghori May 28 '12 at 18:26
  • $\begingroup$ If a 1KHz signal "modulates" a 1KHz carrier then you get a 1KHz signal with less amplitude variation. $\endgroup$ – Daniel R Hicks May 28 '12 at 19:48

...to send message signal over a long distance, when the frequency is low , energy will be obviously low.

Yes, for the case of electromagnetic radiation . If you were to apply the exact same comms techniques (modulation for example) in the case of an underwater channel you would be dealing with mechanical waves where you can "jump" from one frequency to the next without having to spend more energy to produce it. (Effectively though, high frequency waves do not propagate as far as low frequency waves and you would have to "boost them" but that's something to do with the characteristics of the propagation of the wave in that medium (not the physics of producing a mechanical wave at some frequency as it happens with producing an electromagnetic wave).)

To increase the energy of the signal we need to increase the frequency. This is achieved by multiplying the message signal with the carrier signal (with high frequency).

Not necessarily (for the first part...as above).

The reason for modulation is to match a given baseband signal as best as possible to the characteristics of a channel. Amplitude modulation (which is what you describe in your question) is not the only form of modulation.

what will happen if it is reversed ?

Modulation, is a commutative operation. It doesn't matter if it is carried out A*B or B*A. Therefore, it doesn't matter if the frequency of signal A is higher than the frequency of signal B. The end-result will be the same.

If you are wondering what happens in the case of $F_A-F_B<0$: You will be getting "negative frequencies". Frequency is the rate of phase change. Negative frequency would result in a reversal of this phase change which would result to something similar to aliasing. (This is hinted to but not exactly clarified in the Wikipedia page about AM. You might want to first read through Spectrum and then Mod. Index for this effect)

  • $\begingroup$ -1. Lower frequency does NOT mean that the energy is lower. Photon quanta has nothing to do with this. $\endgroup$ – Jim Clay May 29 '12 at 14:26
  • $\begingroup$ Thanks, it could be possible that that part of my answer is wrong but would you mind elaborating a bit more on why do you think it is? $\endgroup$ – A_A May 29 '12 at 15:10
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    $\begingroup$ Photon quanta energy only matters when certain interactions- like causing an electron to jump from one energy level to another- depends on that one individual photon having a certain amount of energy. For macro processes like signals, where trillions of photons are involved, all that matters is the total energy, which is the sum of the individual photons' energy. It is more useful to think of the transmitter energy in terms of the power amp specifications, not at the photon level. $\endgroup$ – Jim Clay May 29 '12 at 15:47
  • $\begingroup$ Thank you, i will be looking this up to clarify where a missconception might have stemmed from and get back. At the moment, i am not convinced that this is entirely wrong. In a nutshell, if photons of energy E are emitted by an antenna, they have obtained this energy from "something". If that required E is linked to f, then that link should also be reflected to that "something" which would validate that "...when the frequency is low energy will be obviously low" (at least for Electromagnetic Radiation which is what the article i cite refers to). $\endgroup$ – A_A May 30 '12 at 0:20
  • $\begingroup$ The right way to approach it is to calculate the energy of two different sinusoids with equal amplitudes but different frequencies. If you average the energy over time, making sure that you do it over an integer number of cycles, you will see that the energy is exactly equal. $\endgroup$ – Jim Clay May 30 '12 at 0:55

Antenna height also has got something to do with carrier frequency, i believe. To receive an EM wave of wave length 'l', the antenna height should be atleast 'l/10'. As u all know, wavelength of an EM wave can be calculated as 'c/frequency in Hz'. Where c is the speed of light which is equal to 3E8 m/s. Results u can just calculate and see. So, thinking logically, I believe there is no point in modulating a lower frequency carrier using a high frequency message signal.


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