# Drawing Waveforms or how to draw waveforms

Discussion in 'General Electronics Chat' started by walters, Nov 16, 2005.

1. ### JimBSuper ModeratorMost Helpful Member

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To be quite honest I am surprised that the "regulars" have even bothered to put fingers to keyboard for him.

JimB

2. ### waltersBanned

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Yes u have helped me out in the past and thanks alot for the help this stuff is kinda of hard for me to understand and im slow at learning this but i think im getting it more and more with the help. I just need to go over things a couple of times to really understand it and it make it clear with examples and charts does help me.

3. ### waltersBanned

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Slope= Rate of Change= V/time

When measuring Slope/rate of change its the Voltage/time so

examples:

10V/10us= 1v/us

10V/5ms= 2v/ms

5v/ 60us= .0834v/us

5v/ 400us= .0125v/us

2v/ 40ms= .05v/ms

2v/ 40us= .05/us

5v/100ms= .05v/ms

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5. ### waltersBanned

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Did i do the results right on the slope/rate of change formulas?

6. ### mstechcaNew Member

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Why do you care?
a slope is a diagonal line (with a curve) in it.

Slope can be the Rate of Change.
If you are referring to sinewaves, yes it can mean voltage/time, but if you are referring to product datasheets, then It does not necessarily mean voltage/time.

:?: :?: :?:

When dealing with voltage and time, they are dealt with separately.
For example, an oscillator has an output frequency, and it is measured in Hertz. The frequency is how many times a second something happens. So frequency = 1 / time.
an oscillator also has a voltage, but the time or frequency cannot be determined from voltage alone.

I still think that the above equations you have supplied are useless.

7. ### waltersBanned

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Thanks for the help

These are for Triangle or sawtooth waveforms in my book is says

The Slope of a voltage ramp or slope is V/t

What are the slopes of the voltage ramps or slopes:

10V/10us= 1v/us

If the triangle waveform was 10v and 10us across would it be 1v/us?

10V/5ms= 2v/ms

If the triangle waveform was 10v and 10us across would it be 2v/ms?

5v/ 60us= .0834v/us

If the triangle waveform was 10v and 10us across would it be .0834v/us?

5v/ 400us= .0125v/us

If the triangle waveform was 5v and 400us across would it be .0125v/us?

2v/ 40ms= .05v/ms

If the triangle waveform was 2v and 40ms across would it be .05v/ms?

2v/ 40us= .05/us

If the triangle waveform was 2v and 40us across would it be .05v/us?

5v/100ms= .05v/ms

If the triangle waveform was 5v and 100ms across would it be 1v/us?

8. ### DigiTanNew Member

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It looks like lines 8 and 10 in that last post had typo's, but everything else is correct. These calculations (aka: derivatives) actually have a lot of use. 8)

9. ### waltersBanned

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Thanks for the correction of these calculations

what kind of uses are these used for?

10. ### DigiTanNew Member

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Well, there are tons of everyday uses. Like I mentioned, the slope of a function is the same as the derivative. That lets you find the current through a capacitor...

Icapacitor = C * (dv/dt)

Where "dv/dt" means "change in voltage/change in time." So if there is 0 voltage change per time unit (dv/dt = 0), you know there is 0 current through the cap. Likewise, you can do the same with current to find the voltage on an inductor...

Vinductor = L * (di/dt)

...So we can see that a change in current on an inductor will produce a proportionally large change in the voltage across that inductor. This is why --in alot of our relay threads-- you hear about "inductive kickback" and so on. The inductor always drops a voltage that proportional to the current slope (di/dt).
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You also see this a lot with microcontrollers. Digital devices can tolerate small variations in their supply voltage, but volt variations that are too rapid (caused by power supply ripple, radio interference, motors, neighboring μC's, and other noise-makers) will cause logic errors.

11. ### waltersBanned

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Thanks for the help

the current through a capacitor...

Icapacitor = C * (dv/dt)

So a capacitor has a "Slope" when current goes through a capacitor ?

With current through a Resistor its just linear it just changes the voltage but doesn't change the "Time" or timing of the current or voltage?

With current flowing through a capcaitor it changes the timing of the current and voltage the lead and laging so the Voltage has a slope with time

examples
10V/10us= 1v/us

10V/5ms= 2v/ms

5v/ 60us= .0834v/us

5v/ 400us= .0125v/us

2v/ 40ms= .05v/ms

2v/ 40us= .05/us

5v/100ms= .05v/ms

If these values or derivative are correct
and we use these derivatives to find the Current through a capacitor

Formula::
Icapacitor = C * (dv/dt)

10V/10us= 1v/us
10V/10us * C .01uf = .01 Amps or millamps?

10V/5ms= 2v/ms
10V/5ms * C .01uf = .02 amps or milliamps?

5v/ 60us= .0834v/us
5v/ 60us * C .01uf = 8.33 amps

5v/ 400us= .0125v/us
5v/ 400us * C .01uf = .00125 amps

5v/ 40ms= .05v/ms
5v/ 40ms * C .01uf = .0005 amps

2v/ 40us= .05/us
2v/ 40us * C .01uf = ?

5v/100ms= .05v/ms
5v/100ms * C .01uf = ?

12. ### DigiTanNew Member

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It's the capacitor voltage that has the slope in this case.

This is correct, the resistor currents and voltages are independent of time, so their behavior is always controlled by Vresistor = I*R and Iresistor = V/R

When an AC signal passes through a capacitor, the voltage gets "differentiated" by the (dv/dt) term and causes the currentto lead the voltage by 90 degrees. And inductors will cause the voltage to lead the current by 90 degrees.

For the first one, you use:

Icapacitor = (2/40E-6) * 0.01E-6 = 5E-4 (or 0.5mA, or 0.0005 Amps). Basically, 40E-6 means "40x10^-6." That notation makes things a lot easier when working with prefixes like milli, micro, and so on.

And on the next one:

Icapacitor = (5V/100E-3) * 0.01E-6 = 5E-7 (or 0.5uA or 0.0000005 Amps). The same story goes here. Substitute "E-3" from "milli" and "E-6" from "micro."

13. ### waltersBanned

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thanks alot for the information

What is this differentiated by time?

Inside the capacitor or inductor where is this Differentiated that causes the time to shift lead or lag?

Because inside a resistor there is no Differentiated inside because it doesn't shift the time of the current or voltage lead or lag the current and voltage are insync where a cap or inductor has differentiated why?

14. ### DigiTanNew Member

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Yeah, the dv/dt means we're differentiating with respect to time, so if for example if you had v = sin(wt), the derivative would be w cos(wt) and so on.

The deravitive term is what causes the phase shift (mathmatically at least). Since we know the phase shift by a single capacitor/inductor is always +/- 90 degrees, that makes things a little easier in other areas.

The physical reasoning behind the inductor current lag and capcacitor voltage lag goes back to Gauss's Laws and the Maxwell equations.

15. ### waltersBanned

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Thanks for the help

what are we differentiating? Time VS voltage/current do u mean?
what is the capacitor differentiating? Time VS voltage/current do u mean?
how does a cap or inductor do differentiating?

Whats the different between Deravitive and Differentiating? they both seem to be the same thing or they both do phase shift

Deratvitive inside a capacitor or inductor does phase shifting
Differentiating inside a capacitor or inductor does phase shifting

I sorta of need a Water analogy of what a deravitive would be in this situation? and a water analogy of what a Differentiating would be in this situation?

I can't see in my mind what a deravitive or a Differentiating inside a capacitor or inductor is doing really

16. ### DigiTanNew Member

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Well for now, just remember a derivative describes a function's rate of change.

Without overcomplicating things, I'll just say that when I say "differentiate with respect to time," that means: (1) time is the independent variable (the x-axis), (2) voltage is the dependent variable (the y-axis), and (3) the derivative is the slope. Technically, you can also "differentiate with respect to voltage," but this would be like saying 'hours per mile' instead of 'miles per hour.' ...And only weirdos differentiate the independent variable. :lol:

For sine waves, we've memorized from calculus class that the derivative of sine is cosine. If you plot sine and cosine on a graph, you see that they're identical, except cosine has a 90 degree "head start" on sine. This is the phase shift.

For now, don't worry about how inductors and capacitors do the differention (I don't fully understand Maxwell's equations myself).

17. ### waltersBanned

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Thanks alot this helps me understand it more

so if the Derivative is the Slope what is the Differentiate then?

Are u saying that the Derivative and Differentiate are slopes but at different times? like 90 degrees apart

18. ### DigiTanNew Member

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Nah, "differentiate" is just a verb. It means finding the derivative.

19. ### waltersBanned

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Thanks to make it more clear

So are u saying rate of change is a derivative or many derivatives?

Because "rate of change" changes with time or frequency

examples:
inputs a high frequency like 1K into a capacitor or inductor
the rate of change is a derivative

Now lower the frequency like 800hz into a capacitor or inductor the rate of change is Different derivative based on the inputs freq.

20. ### waltersBanned

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I think the Rate of change is based on the capacitors charging and discharging how fast is can charge and discharge thats what makes the rate of change timing because high frequencys make the capacitor charge faster and discharge faster than lower frequencys

21. ### DigiTanNew Member

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Well, not exactly. For AC sine waves, we always use "sin(x)" or "cos(x)" so we take the deriviates of those. For, the derivative of a 1KHz sine voltage would look like this...

The original voltage:
v(t) = sin(1000t)
The derivative of v(t) is:
dv/dt = 1000*cos(1000t)

[For the other members: Don't give me any static about not using w = 2*pi*frequency. This is just an example.]

So in general, if you have:
v(t) = sin(some_constant*t)
The derivative will be:
dv/dt = some_constant*cos(some_constant*t)

When you differentiate a trigonometric term, you always get another trig term back. [More info here]
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Anyway, it's impossible to find a equations for any AC circuit without using trigonometry and calculus in general, which is why it's not taught much before college or highschool calculus I.