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Here is an image of the question.

enter image description here

I am stuck at part a, which asks to find the DC Offset based of a modulating index of 0.7, or 70%. So essentially, you will transform the maximum and minimum, yes? How do you know the minimum and maximum values?

I know that m (modulating index) = (max-min)/(max+min). Should the Vpp value be preserved and stay the same as original after the transformation?

How do you do this? Would you do trial and error? Or am I using the wrong formula?

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Here's what 50% AM looks like: -

enter image description here

Now imagine the slowish sinewave shape (in red) is your signal so roughly when your signal bottoms out at about -0.033 "units", this has to correspond to the -70% level as implied in my picture. Yes I know my pictures shows +50% and -50% but you can see that there will also be a point that equals + and -70%.

Your signal has to fit in those limits exactly and, because AM can be regarded as simple 2 quadrant multiplication, there needs to be a DC offset applied to your signal to get it to fit as I've prescribed i.e. the "centre line" of your signal needs to be moved to the nominal level in my picture.

Note that the above sentence previously said "the average level of your signal needs to be moved,,,," BUT, because the signal is asymmetric about the average level this was misleading.

Now I have no idea what the graph in your question is actually shwoing in terms of volts because the axis aren't labelled and believe you me I'm not going to start plotting out the formula BUT, I can see that the graph does not bear much resemblance to the formula in terms of amplitude so you are on your own on this.

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you have the correct formula -- look like your extremes are +0.045 and -0.032. When you add a DC offset, it will add to each of these values. Select the offset that meets the modulation index requirement.

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I have this same question in my homework (we are probably in the same class). You have the right formula, (modulating index) = (max-min)/(max+min). What I did is first find the max and min of the signal. This is hard to do from the graph, so I used Mathematica to find the actual values from the given equation. You can use MATLAB, your calculator, whatever, just find the max and min. I get Vmax = 45 mV and Vmin = -32.6 mV, but you should check that to make sure. Then I used the equation, adding in Vdc because that is what you are adding on and solving for. For my calculations, I put all my values in Volts. So we have

.7 = [(Vdc + .045) - (Vdc - .0326)] / [(Vdc + .045) + (Vdc - .0326) ].

Solve for Vdc.

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  • \$\begingroup\$ Thanks! But don't we use the envelope minimum value instead of -0.0326? \$\endgroup\$ – MURICA Nov 28 '15 at 0:24

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