Added: 2 years ago
From: wahlrite
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  • life is hell

  • Although it's cute and, to some, impressive, to quote positive modulation numbers above 100 % what you're actually producing is reduced carrier AM. At its worst, such signals sound distorted on a typical receiver because of inadequate carrier for the diode detector to demodulate.

    Ideally, refined schemes of compression and levels of carrier that match this denser quality of modulation produce louder audio that sounds best in the majority of receivers.

  • I take it that a less than 100% coverage braid cable, like RG-8 or RG-213 is used to make the 'tap', rather than RG-214 (double shielded silver-plated cable) or a solid-copper shield like Heliax?

    .

  • To figure modulation we use the 'modulation index' formula. In the video the carrier is about one box negative and one box positive. Agreed? The modulation is about three boxes negative and three positive. Agreed?

    modulation equals (in this case) 6(boxes)-2(boxes) divided by 6(boxes)+2(boxes) which equals a fraction: 4/8. Which simplifies to 1/2.

    This is 50%. In order to have 100% the carrier must be equal to Zero.

  • @NIGHTCRAWLER672

    NO, not agreed- when reading modulation on a scope you DO NOT READ OR TAKE ANY MEASUREMENT from below the center line of the screen.

    The negative reading is the darker valley between the lighted sign wave above the center line, and this is why you can NOT read anything more than 100% in the negative direction, unlike reading in the positive direction, each box up is another 100% positive as long as you start out dead keying one box up and one box down.

  • @wahlrite Ok how about this using only positive values above the centerline as you suggested:

    Modulation equals (in this case) 3(boxes)-1(box) divided by 3(boxes)+1(box), which equals a fraction: 2/4, which simplifies as 1/2 equals 0.5 aka 50% MODULATION.

    To achieve 100% Mod. The carrier must be zero, which it was not in the video.

  • @wahlrite Ok how about this using only positive values above the centerline as you suggested:

    Modulation equals (in this case) 3(boxes)-1(box) divided by 3(boxes)+1(box), which equals a fraction: 2/4, which simplifies as 1/2 equals 0.5 aka 50% MODULATION.

    To achieve 100% Mod. The carrier must be zero, which it was not in the video.

  • @NIGHTCRAWLER672

    Nightcrawler, this is a common misconception when reading RF and percent of modulation. almost everyone has a hard time understanding exactly how oscilloscopes actually work. look around on the web their are many good videos that explain how to use a scope to read modulation.

  • Hi there Nightcrawler, the video is showing exactly what it is supposed to . I have now idea how you see 50% after the explanation, if you still don't understand, I really don't know what to say. Maybe you are referring tot he fact that it was not totally pinched off. but even still it was at about 96% negative, and the brightness on the scope made it look like maybe 85% or 90% The positive peaks are clearly way up there.

  • should be said, 200% isn't good depending on what sort of receivers you plan on demodulating the signal. A product detector won't have any problems, but an envelope detector won't decode over 100% worth dick.

  • @Polybun , I aggree. thanks for the coment

  • @Polybun

    Actually a standard Cb radio will demodulate up to about 150% modulation OK, But you are correct, it does start cutting off, and anything more that 150% is not passes to the Rx side of the receiver unless you have replaced the 2IF filter with a larger one similar to the mod we do to the 29 or 148 to allow more thru.

  • Thanks for the tutorial on the RF Tap. I'm definitely picking up a Mauldulator soon. Some killer technology for sure.

  • well said 302 !

    Dont worrie they will come around. notice I was just trying to show how to read the scope, he wanted so bad to say something bad about the mauldulator

  • @wahlrite I have been trying to leave a comment on this site but everytime I go to type a comment, my only choice is to cancel, not post. Is the comments disabled?

  • @NIGHTCRAWLER672

    I'm responding to your post on my video site, no I do not have the comment page turned off.

    Rick

  • @wahlrite I am trying to include some formulas that explain how to compute modulation but I am unable to post them I get error messages. :(

    Basically, Carrier voltage is say one volt or one box, and volts max is 3 box or 3 volt. so mod is 3-1/3+1 times 100 percent which is 50% modulation.

  • @wahlrite The signal at the beginning of the video shows 50 percent modulation not 200 percent.

  • @RFXtreme1 i dont see you having any video's showing that you know how to use a scope

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