Hi,

Consider Vi (represented in black) a triangular wave with 100 Hz,10 V and average value=0V.
Note that Vo(represented in red is overlapped with Vi.

How can i determine the average output voltage Vomed?

I know that generically \( Vomed(wt)=(1/period)*Integral(amplitude*waveformfunction) d(wt)\)

I don't know the expression of the ouput waveform so can i compute this calculating the area of the trapezoid?

Vomed=(((5ms)(1,6ms))/(2))*(5V)

Thanks

 

Hi,

Consider Vi (represented in black) a triangular wave with 100 Hz,10 V and average value=0V.
Note that Vo(represented in red is overlapped with Vi.

How can i determine the average output voltage Vomed?

I know that generically \( Vomed(wt)=(1/period)*\integral(amplitude*waveformfunction) d(wt)\)

You need to preview your posts to be sure that your tex block render as intended.

Close, but "amplitude*waveformfunction" is pretty meaningless for most waveforms.

\(
V_{avg}=\frac{1}{T}\int_{T_0}^{T_0+T}v(t)dt
\)

I don't know the expression of the ouput waveform

Then you need to learn how to figure it out, especially for waveforms that are piecewise linear like this one.

so can i compute this calculating the area of the trapezoid?

That's part of it. But you aren't looking for the area, you are looking for the average height. Also, don't forget that the areas add algebraically.

Vomed=(((5ms)(1,6ms))/(2))*(5V)

Do these units work out?

What have I been saying to you over and over about checking your units!

Where does the 1.6ms come from?

What would you estimate the average voltage to be? You don't need anything precise, just place as tight a bound on the result as you can in a quick way.

For instance, will the average voltage be positive or negative? What would the average voltage be if the waveform were two positive triangles (i.e., the first half of the waveform was repeated for the second half)? What would the average become if you take away the second half completely? Will adding the actual second half make that go up or go down?

Just doing the above you should be able to come up with an estimate that is bounded by about 1V on either side of the actual answer. A little further thinking and you can probably make a good estimate of where within that range the answer should lie.

 

Then you need to learn how to figure it out, especially for waveforms that are piecewise linear like this one.

The function of the waveform would be y=x untill half of the period and for the other half y=-x

For instance, will the average voltage be positive or negative?

It will be positive since the area in the positive region > area in the negative region

What would the average voltage be if the waveform were two positive triangles (i.e., the first half of the waveform was repeated for the second half)?

It would be half of the amplitude of the triangular wave

What would the average become if you take away the second half completely? Will adding the actual second half make that go up or go down?

I think it would be the same.

Just doing the above you should be able to come up with an estimate that is bounded by about 1V on either side of the actual answer. A little further thinking and you can probably make a good estimate of where within that range the answer should lie.

I know vomed will be > 0V,since the original waveform as an average value of 0V.


Untill t=5ms \( (Vomed=5V) \)

From T=5ms \((Vomed =-2,5 V)\)

\(Vomed =(5 V +(-2.5 V)/(2))=1.25 V \)

Thanks

 

The function of the waveform would be y=x untill half of the period and for the other half y=-x

Will it? Plot y=-x (ignoring the units issue) and see if it lies on top of the second half of the waveform.

It will be positive since the area in the positive region > area in the negative region

Correct.

It would be half of the amplitude of the triangular wave

Correct.

I think it would be the same.

How can removing half of the waveform (all of which is on the same side) leave the average unchanged?

I know vomed will be > 0V,since the original waveform as an average value of 0V.

Correct.

Untill t=5ms \( (Vomed=5V) \)

Correct.

From T=5ms \((Vomed =-2,5 V)\)

Try again. What is the area of that trapezoid.

 

Will it? Try again. What is the area of that trapezoid.

\( ATrapezoid=((B+b)/(2))*h
ATrapezoid=((5+2.5)/(2)*5=37.5


\)

I left the units out because as you mentioned they don't seem to make sense,considering we are calcuting an area...

 

The "area" has units of volt-seconds because we have a "box" that has units of volts in one dimension and units of time in the other. Once you have the total area, you will divide by the total time to get the average voltage. Vs/s = V.

And, again, I ask where this value of 1.6 is coming from?

At what time does the downward sloping portion of the waveform reach -5V?

At what time does the flat portion of the waveform start rampling back up?

 

And, again, I ask where this value of 1.6 is coming from?

I had already changed it in my previous post.

At what time does the downward sloping portion of the waveform reach -5V?

At t=6,25ms

At what time does the flat portion of the waveform start rampling back up?

At t=8.75

 

You are getting close.

Using proper units and signs:

What is the area of the triangle in the first half?

What is the area of the trapezoid in the second half?

What is the total area?

What is the average voltage?

I won't keep helping you unless you start using proper units and signs.

 

\( ATrapezoid=((B+b)/(2))*h
ATrapezoid=((5+2.5)/(2)*5=37.5
\)

Remember the other golden rule. Always ask if the answer makes sense.

The area of the first half is 25mVs. Here you are claiming that the area of the second part is -37.5mVs. Does it make sense for the magnitude of the area for the second half to be greater than the first half?

 

Remember the other golden rule. Always ask if the answer makes sense.

The area of the first half is 25mVs. Here you are claiming that the area of the second part is -37.5mVs. Does it make sense for the magnitude of the area for the second half to be greater than the first half?

No you are rigth,now that you pointed that it doesn't make much sense in fact.

You are getting close.

Using proper units and signs:

What is the area of the triangle in the first half?

\( Atriangle=(b*h)/(2)=(5ms*10V)/(2)=5V*ms \)

What is the area of the trapezoid in the second half?

\( Atrapezoid=(B+b)/(2)*h=((5ms+2.5ms)/(2))*(-5 V)=- 18.75 V*ms \)

What is the total area?

\( Total Area=6.25 V*ms \)

What is the average voltage?

\( Average Output Voltage=6.25 V*ms \) rigth?

Thanks

 

Very, very close.

Again, check your units.

What units should AverageOutputVoltage have?

Be more explicit in going from TotalArea to AverageOutputVoltage and you should find to remaining mistake.

 

Very, very close.


What units should AverageOutputVoltage have?

It should be expressed in Volts.

Be more explicit in going from TotalArea to AverageOutputVoltage and you should find to remaining mistake.

Do you mean this :

\( Average Output Voltage=((Atriangle)+(Atrapezoid))/(2)
Average Output Voltage =(25 V*ms)+(-18.75 V*ms)/(2)=3.125V*ms...\)

The units aren't correct wich probably means that i should divide average output voltage by the period to get my result in Volts,right?Or should this division be done with the reference to the total area?

Thanks

 

With reference to the total area.

Think of if you were asked to find the average width of a box. You would find the total area and divide by the length. Same idea.

 

It should be expressed in Volts.



Do you mean this :

\( Average Output Voltage=((Atriangle)+(Atrapezoid))/(2)
Average Output Voltage =(25 V*ms)+(-18.75 V*ms)/(2)=3.125V*ms...\)

Why are you dividing by 2?

The units aren't correct wich probably means that i should divide average output voltage by the period to get my result in Volts,right?Or should this division be done with the reference to the total area?

Thanks

Average output voltage is what you're trying to get. If you already had it you wouldn't need to do anything more, so don't call what you have so far "average output voltage". What you have is the sum of the two areas, and the units of those areas are volt*milliseconds, not volts (or "average output voltage" as you called it).

Add the areas and divide by the period. Remember what you had in post #1:

\( Vomed(wt)=(1/period)*Integral(amplitude*waveformfunction) d(wt)\)

 

Why are you dividing by 2?



Average output voltage is what you're trying to get. If you already had it you wouldn't need to do anything more, so don't call what you have so far "average output voltage". What you have is the sum of the two areas, and the units of those areas are volt*milliseconds, not volts (or "average output voltage" as you called it).

Add the areas and divide by the period. Remember what you had in post #1:

\( Vomed(wt)=(1/period)*Integral(amplitude*waveformfunction) d(wt)\)

\( Total Area=6.25 V*ms

Average Output Voltage=(6.25 V*ms) /(10ms)=0.625 V \) rigth?

Thanks

 

I wondered what you wanted the average for?

 

\( Total Area=6.25 V*ms

Average Output Voltage=(6.25 V*ms) /(10ms)=0.625 V \) rigth?

Thanks

That is correct.

 

\( Total Area=6.25 V*ms

Average Output Voltage=(6.25 V*ms) /(10ms)=0.625 V \) rigth?

Thanks

Correct.

Another way to get the answer is to recognize that if you add the missing triangle from the tip of the bottom portion of the waveform and also add a corresponding triangle from the top that you haven't changed the average but now the main part of the waveform has an average voltage of zero and all you are left with is the positive tip. That triangle has a height of 2.5V and a base of 2.5ms giving an area of 6.25mVs. The average for the entire cycle is thus 6.25mVs/10ms=0.625V