# digital modulation

Discussion in 'Mathematics and Physics' started by PG1995, Sep 4, 2013.

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Hi

Regards
PG

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2. ### steveBWell-Known MemberMost Helpful Member

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PG,

I dont' think so. With 50% modulation you have 50% of the amplitude in the off state and 100% in the on state. With 100% modulation you have 0% of the amplitude in the off state and 100% in the on state.

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3. ### PG1995Active Member

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Thank you.

Could you please elaborate a little why this is so? Thanks a lot.

Regards
PG

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5. ### steveBWell-Known MemberMost Helpful Member

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I believe it is just a definition. I could be wrong because I'm not an expert on this. Let's see if someone else can confirm it.

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6. ### JimBSuper ModeratorMost Helpful Member

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The depth of modulation of an amplitude modulated signal is defined as:

((P - Q)/(P+Q)) x 100 percent

Where P is the maximum value of the modulated signal
and Q is the minimum value of the modulated signal.

JimB

Last edited: Sep 4, 2013
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7. ### steveBWell-Known MemberMost Helpful Member

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PG,

Now that I think about it, I think you are correct in your original comment. JimB provided the mathematical definition and it seems to correspond to what you said.

I think you are correct that the diagrams are just representative, which is why they show the amplitudes the same. However, if they drew to scale, the 100 % modulation case would be 33% larger in the A region of the 100% plot referenced to the A region of the 50% plot.

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8. ### JimBSuper ModeratorMost Helpful Member

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Another case of over analysis of a simplified drawing.

Case 2 could also be considered as simple on/off keying. Which by definition cannot help but be 100% modulated.

Where a carrier which is correctly amplitude modulated by (say) a sinewave, the peak votage of the modulated waveform will be twice the amplitude of the unmodulated carrier.

JimB

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9. ### steveBWell-Known MemberMost Helpful Member

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I think PG knew this beforehand, according to his own words. I assume his goal is to verify that his conceptualization of what is happening is correct. I think it is since he got it right.

10. ### JimBSuper ModeratorMost Helpful Member

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SteveB, I was just emphasising.

An Exercise for PG
Anyway, It occurs to me that it would be a good exercise for PG to work out what the spectrum of such a signal (RF carrier, AM modulated with an audio frequency square wave) would look like.
It would tie together one or two topics which we have discussed recently.

JimB

11. ### PG1995Active Member

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Thanks a lot, Steve, JimB.

This text puts the modulation index formula provided by JimB in a proper context.

Could you please tell me how you get this "33%"? This is how I was thinking about it. Thanks.

@JimB: If you don't mind, this is the best I can do at the moment. If you want me to do it myself, I can do it later. Thanks.

Regards
PG

1: http://www.electro-tech-online.com/custompdfs/2013/09/modulate-1.pdf
3: http://www.electro-tech-online.com/custompdfs/2013/09/AM20fundamentals-1.pdf

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12. ### JimBSuper ModeratorMost Helpful Member

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Yes thats it.
I think that the important thing to notice is how the square wave spectrum is mirrored on each side of the carrier, each sideband carries the same information, and each sideband can reproduce the original waveform when received correctly.

I notice that the text goes on to illustrate keying the carrier with Morse code characters.
A potential problem when transmitting Morse code, the rise and fall times of the keying envelope must not be too fast.
Notice with the square wave how the harmonics spread far and wide on both sides of the carrier, on/off keying with Morse code will do the same and the transmission will take up much more spectrum space than is necessary.
This condition is known as "key clicks", and it will make you very unpopular with your RF neighbours!

JimB

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13. ### steveBWell-Known MemberMost Helpful Member

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I hope I didn't screw it up, but my thinking was as follows. A carrier of amplitude 1 modulated 100% will go from zero amplitude to an amplitude of 2. A carrier of amplitude 1 modulated 50% will go from 0.5 amplitude to an amplitude of 1.5. Hence, the former case has region "A" with an amplitude of 2, and the latter case has region "A" with an amplitude of 1.5. That's all that really matters for your understanding, and I should have just said that.

However, the 33% number was obtained by using the 1.5 amplitude as a reference and saying that 2 is 33% larger than 1.5. The actual number was not my point, but my point was that you got the concept right and I got it wrong and misled you.

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14. ### PG1995Active Member

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Hi

Q1:
It can be said that modulation (plus multiplexing) is basically used:
i: to transmit more than one channel over the same link
ii: to decrease antenna size

Do I have it correct? Would you like to add something?

Q2:
a: In many cases a wired link, such as the one established by a coaxial cable, could give us more bandwidth but still a radio link is preferred. Because a radio link is easy to implement, maintain and could be comparatively cost effective. Do you agree?

b: On the other hand, sometimes a wired link is preferred for several reasons. Some of the main reasons could be that it wouldn't require you to buy bandwidth from relevant authorities, and as a result no licensing costs. It would also provide you with better data rates. Do you agree?

Q3:
a: Why wouldn't it be a good idea to send digital data directly over a wired link?

This is what I think. We can assign voltage levels to digital data, i.e. "1" and "0".

i: It would limit us to only one half duplex or simplex channel for a single wired link. Each channel would require a separate wired link.
ii: Sending "0's" and "1's" constitutes an almost rectangular or pulse signal which consists of harmonics at high frequencies. A wired link won't be able to carry these high frequency harmonics faithfully and this would lead to errors because as attenuation increases so does the chances of error.

b: Why wouldn't it be a good idea to send digital data directly over a radio link?

We can assign voltage levels to digital data, i.e. "1" and "0", and then 'radiate' these voltage levels using a transmitter.

i: As said above, it would limit us to only one half duplex or simplex channel for a single wired link. Each channel would require a separate wired link.
ii: Sending "0's" and "1's" constitute an almost rectangular wave which consists of harmonics at high frequencies. Dominant harmonics would lie at lower frequencies and to capture these important low frequency harmonics would require a large antenna.

Note to self: The frequency determines the wavelength and hence the physical size of antenna we require. A minimum length for an efficient antenna is a half wavelength; c=fλ.

Q4:
In general terms, the more complex a signal is, the more complex its spectrum would be and it will many harmonics to represent it. The complexity of a signal is related to number of abrupt changes in a signal.

Regards
PG

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15. ### JimBSuper ModeratorMost Helpful Member

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JimBs replies in GREEN

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16. ### PG1995Active Member

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Thank you, JimB.

I don't know why you find it so silly, at least from theoretical point of view. Suppose, you want to transmit baseband signal which extends from 0Hz to 5000 Hz. Unless, you modulate it using a high frequency carrier, you would need very large antennae.

Agreed! It's still helpful for a learner to confirm this from an expert.

Okay! That was just a note to myself. Many a time I intentionally make the text of my posts redundant because it helps me when I refer back to my posts after some time and I hope this also helps anyone like me who stumbles onto any of my posts.

Q4:
In general terms, the more complex a signal is, the more complex its spectrum would be and it will many harmonics to represent it. The complexity of a signal is related to number of abrupt changes in a signal. For example, a triangular signal requires less number of harmonics for faithful representation compared to a square wave signal because comparatively square wave is a complex signal.

It is said that B-ASK requires more bandwidth compared to 4-ASK although 4-ASK looks, at least to me, more complex than B-ASK. Here , it shows B-ASK and 4-ASK. In other words, 4-ASK has more abrupt changes compared to B-ASK and hence it should require more bandwidth (or, in words number of harmonics). Why is it so? (If it's still not clearer, please let me know.) Thank you.

Regards
PG

17. ### JimBSuper ModeratorMost Helpful Member

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Because no sane person with a vague understanding of radio would even contemplate trying it.

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18. ### PG1995Active Member

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Thank you.

You are right because transmitting baseband signal of 0Hz to 5000Hz without modulation would require impractically large antenna. So, in my humble opinion, I don't think that saying that modulation is used to make the antenna size practical (among other things) that wrong.

If possible, please give Q4 from my previous post another look. If it's still not clear then please let me know so that I can try to rephrase it. Thanks.

Regards
PG

Last edited: Mar 12, 2015
19. ### JimBSuper ModeratorMost Helpful Member

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You could if you wanted to look foolish in front of a room full of radio engineers.
While the expression is true, it is as logical as saying that road vehicles have round wheels because a square wheel and tyre would be mode difficult to make.

.
Sorry, that is something with which I am not familiar. Sorry I cannot help.

JimB

Last edited: Mar 12, 2015
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20. ### steveBWell-Known MemberMost Helpful Member

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PG,

I think the problem with your Q4 is three-fold.

First you are being very vague, redundant and confusing by using words like "complex" and abrupt.

Second, you are saying incorrect things like "triangle-wave has less harmonics than square wave". The (infinite) number of harmonics is the same, but the amplitudes of the harmonics are different.

Third, you are trying to intuitively determine the spectral content of modulation methods, when this is something that has to be calculated with mathematics. I'm also not familiar enough with the modulation schemes you mentioned to say much, other than the fact that (if I cared to know) the spectrums, I could look up the details of what the method entails and then calculate the spectrums.

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21. ### PG1995Active Member

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Thank you, JimB, Steve.

Agreed and already confessed to this in my original post. I didn't know that how to put my query more clearly.

"And I'm sorry about making too many generalized statements above."

I tried to be little precise with this. A square wave signal has sharper edges compared to a triangular wave signal.

"The complexity of a signal is related to number of abrupt changes in a signal. For example, a triangular signal requires less number of harmonics for faithful representation compared to a square wave signal because comparatively square wave is a complex signal."

Let me try to elaborate on "abrupt changes". You can see here that a square wave signal has sharper rise and fall times compared to both triangle and sawtooth signals and hence it will require more number of harmonics for 'faithful' representation. Further, a sawtooth signal has sharper fall time compared to triangle signal and hence more number of harmonics are required for almost accurate representation. This signal involves more abrupt changes compared to square, triangle or sawtooth signals and hence require more number of harmonics for almost accurate or 'faithful' representation.

Along the same lines, a B-ASK signal looks less complex (i.e. involves less abrupt changes) compared to 4-ASK signal as shown here. But still 4-ASK signal requires less bandwidth compared to B-ASK signal.

I do agree with you because, again, I would say that I didn't know that how to phrase the query in any clearer manner.

Please have a look here on the highlighted text. It might be helpful. Here, it shows B-PSK and 4-PSK signals. Thank you.

Regards
PG

References:
1: http://www.dspguide.com/graphics/F_13_10.gif
2: Telecommunications Demystified by Carl Nassar (page #147, ebook)
3: http://lpsa.swarthmore.edu/Fourier/Series/ExFS.html

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