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For a signal with amplitude,

$$V = A\sin (\omega t)$$

any idea at what point the maximum slope (gradient, dv/dt) is? I have gone through the method of differentiating which yields

$$A\omega \cos(\omega t) \tag{i}$$

and then doing a second differential, which yields

$$-A \omega^2 \sin (\omega t) \tag{ii}$$

Equating (i) to 0 and then substituting back into (ii) to find which point is where I got stuck. I can do this with a normal equation but trigonometry sometimes confuses me. Any help will be appreciated.

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    \$\begingroup\$ You need to set the second derivative to zero, in order to find the extrema of the first derivative. \$\endgroup\$
    – user110971
    Nov 30 '16 at 16:17
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The maximum slope for a sine wave that has no offset and an ampliutde \$A_0\$ occurs exactly during the zero crossings. Its value is simply \$A_0 \omega\$

The derivation is $$ \frac{d}{dt} A_0 \sin (\omega t) = A_0 \omega \cos(\omega t) $$ which gives the slope of the sine wave.

The maximum of the cosine is 1. Therefore the maximum is \$A_0 \omega\$. $$ \max \left\{ A_0 \omega \cos(\omega t) \right\}= A_0 \omega $$ The results makes sense, since intuitively the slope has to increase with the amplitude as well as the frequency.

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    \$\begingroup\$ I always remember that the maximum slew rate is Pi/2 times that of a sawtooth of the same p-p amplitufe. (i.e. ~1.57* the sawtooth slew rate) \$\endgroup\$ Nov 30 '16 at 20:27
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Perhaps I could not understood well, but I think you done a mistake (or two).

The maximum of the slope is where its (of the slope) derivative is 0, then you should take the slope (the first derivative), derive it and put this second derivative equal to zero.

P.S. You also differentiating wrong:

\$\frac{d}{dx} sin(x) = cos(x) \$

and

\$\frac{d}{dx} cos(x) = -sin(x) \$

Then, ...look at the intersection with x axis... ;-)

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  • \$\begingroup\$ He is not asking for the maximum of the sine wave, he is asking at what point you can find the maximum derivative. \$\endgroup\$
    – DerStrom8
    Nov 30 '16 at 16:36
  • \$\begingroup\$ @DerStrom8, I can't understand your comment and the downvote; I answered exactly how to find the maximum slope, i.e the second derivative, that is the maximum of the prime derivative, and I suggested also how to look for the maximum slope "Then, ...look at the intersection with x axis... ;-)" \$\endgroup\$
    – Antonio
    Nov 30 '16 at 20:14
  • \$\begingroup\$ Just to be clear I didn't downvote you, that must have been someone else. Anyway, I missed the last comment in your answer, but the first part of your answer is a bit confusing. The maximum slope is not where the first derivative is 0. I think you need to specify "The maximum of the slope is where the second derivative is zero, not where the first derivative is zero \$\endgroup\$
    – DerStrom8
    Nov 30 '16 at 22:46
  • \$\begingroup\$ @DerStrom8, no problem for the downvote but thanks for the precisation. :-) Regarding to my answer, I edited for say it better, but I never said that the first derivative of the signal is the maximum slope. I said that a maximum is where the derivative is zero, therefore, I specified immediately after, to find the maximum slope we need to get the derivative, derive it one more time, and put the result equal to zero. It isn't? \$\endgroup\$
    – Antonio
    Dec 1 '16 at 0:03
  • \$\begingroup\$ I think we might be hitting a bit of a language barrier? This quote: "The maximum of the slope is where its (of the slope) derivative is 0" suggests that you are saying the maximum value of the slope (that is, the highest slope) is where the derivative is zero, which is not the case. I think this is what threw me off, I misread what you were trying to say. \$\endgroup\$
    – DerStrom8
    Dec 1 '16 at 2:28

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