Integration of below expression

[Hina Gohar]

Homework Statement

it is capacitor charging expression..how to find its integration

Homework Equations

VL(t) = ∫_(T/2)^T▒〖Vme^(-T/2RC) 〗 dt

The Attempt at a Solution

result is 0.5...but how

 

[Simon Bridge]

Homework Statement

it is capacitor charging expression..how to find its integration

Homework Equations

VL(t) = ∫_(T/2)^T▒〖Vme^(-T/2RC) 〗 dt

The Attempt at a Solution

result is 0.5...but how

The expression you write inside the integral is a constant ... perhaps you mean that to be a t not a T in the exponential?

It's hard to make out, but it does not look like you have the expression for a charging capacitor there.

Are you trying to do $$V_L(t)=V_{m}\int_{\tau/2}^\tau(1-e^{-t/\tau})\;dt \qquad \tau=2RC$$

The symbols suggest you are looking for the voltage across an inductor though.
How about relate the complete problem so we know what you are trying to acheive?
... and do show us your working ;)

 

[Hina Gohar]

The expression you write inside the integral is a constant ... perhaps you mean that to be a t not a T in the exponential?

It's hard to make out, but it does not look like you have the expression for a charging capacitor there.

Are you trying to do $$V_L(t)=V_{m}\int_{\tau/2}^\tau(1-e^{-t/\tau})\;dt \qquad \tau=2RC$$

The symbols suggest you are looking for the voltage across an inductor though.
How about relate the complete problem so we know what you are trying to acheive?
... and do show us your working ;)

No it is T not t that's why I am facing problem..it is voltage across load in voltage doubler circuit.I am attaching paper in which this equation is mentioned by (5) and desired one is (6)..in eq(5) from 0 to T/2 I solved but from T/2 to T facing problem in integration.
kindly help

 

[Simon Bridge]

Well, as near as I can make out, you wrote: $$V_L(t)=\int_{T/2}^T V_me^{-T/2RC}\;dt$$... I am sure that is not what you mean. I looked at the paper - and the integrand is only a small part of one case in eq(5)... which is:

$$v_L(t)=\begin{cases}V_m &: t=0\\ \\
V_me^{-t/R_LC_2} &: 0<t<T/2\\ \\
V_m\left[e^{-T/2R_LC_2} +\left(1-e^{(t-(T/2))/rC_2}\right)\left(1-e^{-T/2R_LC_2}\right)\right] &: T/2\leq t \leq T
\end{cases}$$

The case that $T/2 \leq t \leq T$ can be put in form: $A+Be^{\lambda t}$ ... please show your working. Remember it is the lower case $t$ that is being integrated against.
Don't forget to follow the instructions to substitute it into $$V_{L,ave}=\frac{1}{T}\int_{t_1}^{t_1+T} v_L(t)\;dt$$
... which you'll see is somewhat different from what you wrote.

 

[Hina Gohar]

Well, as near as I can make out, you wrote: $$V_L(t)=\int_{T/2}^T V_me^{-T/2RC}\;dt$$... I am sure that is not what you mean. I looked at the paper - and the integrand is only a small part of one case in eq(5)... which is:

$$V_L(t)=\begin{cases}V_m &: t=0\\ \\
V_me^{t/R_LC_2} &: 0<t<T/2\\ \\
V_m\left[e^{T/2R_LC_2} +\left(1-e^{(t-(T/2))/rC_2}\right)\left(1-e^{T/2R_LC_2}\right)\right] &: T/2\leq t \leq T
\end{cases}$$

The case that $T/2 \leq t \leq T$ can be put in form: $A+Be^{\lambda t}$ ... please show your working. Remember it is the lower case $t$ that is being integrated against.
Don't forget to follow the instructions to substitute it into $$V_{L,ave}=\frac{1}{T}\int_{t_1}^{t_1+T} V_L(t)\;dt$$
... which you'll see is somewhat different from what you wrote.

yes it is small part of eq(5)...2nd part I solved which is below while in 3rd part not getting how to integrate especially exponential with T instead of t..
from 0 to T/2 part solution is attached

 

[Simon Bridge]

... in 3rd part not getting how to integrate especially exponential with T instead of t.

... why would you want to? If the text suggests doing that I missed it.

[edit]
I managed to open your Word attachment[*]

In the attachment you wrote:

$$V_L(t)=\int_0^{T/2}V_m e^{-t/RC}\;dt$$ ... (dropping subscripts)

This equation does not appear to follow from anything in the attachment to post #3.
In that attachment, eq(5) gives you $V_L(t)$, you then substitute eq(5) into eq(1) to get $V_{L,ave}$.

I an guessing that this is what you intended to do.
Please be consistent in your notation.

You seem to be using eq(1) with $t_1=0$ so that:

$$V_{L,ave} = \int_0^{T/2} V_me^{-t/2R_LC_2}\;dt + \int_{T/2}^T V_m\left[e^{-T/2R_LC_2} +\left(1-e^{(t-(T/2))/rC_2}\right)\left(1-e^{-T/2R_LC_2}\right)\right]\;dt$$ ... is that correct?

----------------------------

[*] ... please do not share information in proprietary formats.
I understand that writing equations online is annoying - the best practice is to use LaTeX.
If you must share a file - then PDF is better, since there are FOSS readers for it.

 

[Hina Gohar]

... why would you want to? If the text suggests doing that I missed it.

[edit]
I managed to open your Word attachment[*]

In the attachment you wrote:

$$V_L(t)=\int_0^{T/2}V_m e^{-t/RC}\;dt$$ ... (dropping subscripts)

This equation does not appear to follow from anything in the attachment to post #3.
In that attachment, eq(5) gives you $V_L(t)$, you then substitute eq(5) into eq(1) to get $V_{L,ave}$.

I an guessing that this is what you intended to do.
Please be consistent in your notation.

You seem to be using eq(1) with $t_1=0$ so that:

$$V_{L,ave} = \int_0^{T/2} V_me^{-t/2R_LC_2}\;dt + \int_{T/2}^T V_m\left[e^{-T/2R_LC_2} +\left(1-e^{(t-(T/2))/rC_2}\right)\left(1-e^{-T/2R_LC_2}\right)\right]\;dt$$ ... is that correct?

----------------------------

[*] ... please do not share information in proprietary formats.
I understand that writing equations online is annoying - the best practice is to use LaTeX.
If you must share a file - then PDF is better, since there are FOSS readers for it.

yes it is correct but I actually showed half part if that equation which I solved from 0 to T/2...while in remaining part I can not succeed.

 

[Hina Gohar]

... why would you want to? If the text suggests doing that I missed it.

[edit]
I managed to open your Word attachment[*]

In the attachment you wrote:

$$V_L(t)=\int_0^{T/2}V_m e^{-t/RC}\;dt$$ ... (dropping subscripts)

This equation does not appear to follow from anything in the attachment to post #3.
In that attachment, eq(5) gives you $V_L(t)$, you then substitute eq(5) into eq(1) to get $V_{L,ave}$.

I an guessing that this is what you intended to do.
Please be consistent in your notation.

You seem to be using eq(1) with $t_1=0$ so that:

$$V_{L,ave} = \int_0^{T/2} V_me^{-t/2R_LC_2}\;dt + \int_{T/2}^T V_m\left[e^{-T/2R_LC_2} +\left(1-e^{(t-(T/2))/rC_2}\right)\left(1-e^{-T/2R_LC_2}\right)\right]\;dt$$ ... is that correct?

----------------------------

[*] ... please do not share information in proprietary formats.
I understand that writing equations online is annoying - the best practice is to use LaTeX.
If you must share a file - then PDF is better, since there are FOSS readers for it.

plus eq(5) I have to derive mathematically which are basically shows instantaneous load voltage expression for three different modes of this respective circuit.Mode 1 is simple in which no switch will be on so on output simple dc voltage appears..but for mode 2 T1 will on and for mode 3 T2 will on then how to find this whole expression.I tried so many times

 

[Simon Bridge]

yes it is correct but I actually showed half part if that equation which I solved from 0 to T/2...while in remaining part I can not succeed.

plus eq(5) I have to derive mathematically...

... let's deal with only one problem at a time: you want to evaluate:

$$V_{L,ave} = \frac{1}{T}\int_0^{T/2} V_me^{-t/R_LC_2}\;dt + \frac{1}{T}\int_{T/2}^T V_m\left[e^{-T/2R_LC_2} +\left(1-e^{(t-(T/2))/rC_2}\right)\left(1-e^{-T/2R_LC_2}\right)\right]\;dt \qquad\text{...(1)}$$

There seems to be a communication problem.

In your attachment you told me that: $$\int_0^{T/2} V_me^{-t/R_LC_2}\;dt=\frac{V_m(1-e^{-T/2RC})RC}{T}$$ ... but this is not correct.
The correct working is:$$\begin{align}
\frac{1}{T}\int_0^{T/2} V_me^{-t/R_LC_2}\;dt &=\left.-\frac{V_m R_LC_2}{T}e^{-t/2R_LC_2}\right|_{0}^{T/2}\\
&= \frac{V_mR_LC_2}{T}\left(1-e^{-T/2R_LC_2}\right)
\end{align}$$I suspect that this is what you meant but the trouble is that it is not what you said. In your version the "divide by T" part appears out of nowhere.

It is important to say what you mean in maths if you are not to get confused and if you don't want to confuse other people.

So much for the first integral in the expression (1).

Like I already told you, the second integral can be written in form: $$\int_{T/2}^T A+Be^{\lambda t}\; dt$$ ... I do not understand why you are having trouble with this integral.
Please find the values of A, B, and λ; and show me your working for the integration.

If you do not follow suggestions, I cannot help you.
If you do not provide needed information, I cannot help you.
Please bear in mind that it is against the rules for me to do your work for you.

 

[Hina Gohar]

... let's deal with only one problem at a time: you want to evaluate:

$$V_{L,ave} = \frac{1}{T}\int_0^{T/2} V_me^{-t/R_LC_2}\;dt + \frac{1}{T}\int_{T/2}^T V_m\left[e^{-T/2R_LC_2} +\left(1-e^{(t-(T/2))/rC_2}\right)\left(1-e^{-T/2R_LC_2}\right)\right]\;dt \qquad\text{...(1)}$$

There seems to be a communication problem.

In your attachment you told me that: $$\int_0^{T/2} V_me^{-t/R_LC_2}\;dt=\frac{V_m(1-e^{-T/2RC})RC}{T}$$ ... but this is not correct.
The correct working is:$$\begin{align}
\frac{1}{T}\int_0^{T/2} V_me^{-t/R_LC_2}\;dt &=\left.-\frac{V_m R_LC_2}{T}e^{-t/2R_LC_2}\right|_{0}^{T/2}\\
&= \frac{V_mR_LC_2}{T}\left(1-e^{-T/2R_LC_2}\right)
\end{align}$$I suspect that this is what you meant but the trouble is that it is not what you said. In your version the "divide by T" part appears out of nowhere.

It is important to say what you mean in maths if you are not to get confused and if you don't want to confuse other people.

So much for the first integral in the expression (1).

Like I already told you, the second integral can be written in form: $$\int_{T/2}^T A+Be^{\lambda t}\; dt$$ ... I do not understand why you are having trouble with this integral.
Please find the values of A, B, and λ; and show me your working for the integration.

If you do not follow suggestions, I cannot help you.
If you do not provide needed information, I cannot help you.
Please bear in mind that it is against the rules for me to do your work for you.

I also did the same thing which I showed in word file..main problem I thing is in equations writing.
Problem I have in 2nd part is integration of exponential power of 'T'.This is T not t so it should be treated as constant?

 

[Simon Bridge]

Problem I have in 2nd part is integration of exponential power of 'T'.This is T not t so it should be treated as constant?

... do not get confused between T and t in the expression.
The "dt" at the end of the expression means that everything that is not a lower case "t" is treated as a constant.
So "T" is a constant, "t" is a variable.

This is how it should have worked out when you followed the suggestion:

Please find the values of A, B, and λ; and show me your working for the integration.​

... in post #9 which was repeated from earlier.
If you will not follow suggestions I cannot help you.

 

[Hina Gohar]

... do not get confused between T and t in the expression.
The "dt" at the end of the expression means that everything that is not a lower case "t" is treated as a constant.
So "T" is a constant, "t" is a variable.

This is how it should have worked out when you followed the suggestion:

Please find the values of A, B, and λ; and show me your working for the integration.​

... in post #9 which was repeated from earlier.
If you will not follow suggestions I cannot help you.

how to write eq(5) last expression in Latex.. I tried this but error comes
V_{L}(t)=int_0^\T/2V_{m} \mathrm{e}^{-\frac{t}{R_{L}C_{2}}}\,\mathrm{d}t

 

[Simon Bridge]

how to write eq(5) last expression in Latex.. I tried this but error comes
V_{L}(t)=int_0^\T/2V_{m} \mathrm{e}^{-\frac{t}{R_{L}C_{2}}}\,\mathrm{d}t

... that last line is not an error message.
If you click the "quote" button on one of my posts, you will see how I did it.

That last line renders as:
$$V_{L}(t)=int_0^\T/2V_{m} \mathrm{e}^{-\frac{t}{R_{L}C_{2}}}\,\mathrm{d}t$$
You need a backslash before the "int" and curly-brackets around the upper limit.
But that won't produce an error.$$V_{L}(t)=\int_0^{T/2}V_{m} \mathrm{e}^{-\frac{t}{R_{L}C_{2}}}\,\mathrm{d}t$$... better. But it usually renders better if you don't use "frac" in an exponent. If you must, then use \exp{} instead of e^{}

 

[Hina Gohar]

... that last line is not an error message.
If you click the "quote" button on one of my posts, you will see how I did it.

That last line renders as:
$$V_{L}(t)=int_0^\T/2V_{m} \mathrm{e}^{-\frac{t}{R_{L}C_{2}}}\,\mathrm{d}t$$
You need a backslash before the "int" and curly-brackets around the upper limit.
But that won't produce an error.$$V_{L}(t)=\int_0^{T/2}V_{m} \mathrm{e}^{-\frac{t}{R_{L}C_{2}}}\,\mathrm{d}t$$... better. But it usually renders better if you don't use "frac" in an exponent. If you must, then use \exp{} instead of e^{}

yeh now it is working..if kindly again post eq(5) so I would come to know how to write in latex

 

[Simon Bridge]

It's in post #4 ... Oh OK then:

This is equation (5)
$$v_L(t)=\begin{cases}
V_m &: t=0\\ \\
V_me^{-t/R_LC_2} &: 0<t<T/2\\ \\
V_m\left[e^{-T/2R_LC_2} +\left(1-e^{(t-(T/2))/rC_2}\right)\left(1-e^{-T/2R_LC_2}\right)\right] &: T/2\leq t \leq T
\end{cases}$$
Here's the markup I used:

v_L(t)=\begin{cases}
V_m &: t=0\\ \\
V_me^{-t/R_LC_2} &: 0<t<T/2\\ \\
V_m\left[e^{-T/2R_LC_2} +\left(1-e^{(t-(T/2))/rC_2}\right)\left(1-e^{-T/2R_LC_2}\right)\right] &: T/2\leq t \leq T
\end{cases}

And this is equation (1)
$$V_{L,ave}=\frac{1}{T}\int_{t_1}^{t_1+T} v_L(t)\;\mathrm{d}t$$
... and the markup for that:

V_{L,ave}=\frac{1}{T}\int_{t_1}^{t_1+T} v_L(t)\;\mathrm{d}t

 

[Hina Gohar]

yeh now it is working..if kindly again post eq(5) so I would come to know how to write in latex

∫ [T/2 to T] (A + Be^(−αt)) dt
= (At − B/α e^(−αt)) | [T/2 to T]
= (AT − B/α e^(−αT)) − (AT/2 − B/α e^(−αT/2))
= AT/2 − B/α e^(−αT) + B/α e^(−αT/2)
where A=Vme^{-T/2RLC2}
while B=[1-e^(-t-T/2)/rC2]*[1-e^(-T/2RlC2)]

But this is not required result as in eq(6)

 

[Simon Bridge]

I don't think that's the correct value for B.
The text also says "using a well known method" or something like that right?
This may mean that there is an approximation involved.
Certainly you should not expect to get eq(6) out the end of the integration.

 

[Hina Gohar]

It's in post #4 ... Oh OK then:

This is equation (5)
$$v_L(t)=\begin{cases}
V_m &: t=0\\ \\
V_me^{-t/R_LC_2} &: 0<t<T/2\\ \\
V_m\left[e^{-T/2R_LC_2} +\left(1-e^{(t-(T/2))/rC_2}\right)\left(1-e^{-T/2R_LC_2}\right)\right] &: T/2\leq t \leq T
\end{cases}$$
Here's the markup I used:

v_L(t)=\begin{cases}
V_m &: t=0\\ \\
V_me^{-t/R_LC_2} &: 0<t<T/2\\ \\
V_m\left[e^{-T/2R_LC_2} +\left(1-e^{(t-(T/2))/rC_2}\right)\left(1-e^{-T/2R_LC_2}\right)\right] &: T/2\leq t \leq T
\end{cases}

And this is equation (1)
$$V_{L,ave}=\frac{1}{T}\int_{t_1}^{t_1+T} v_L(t)\;\mathrm{d}t$$
... and the markup for that:

V_{L,ave}=\frac{1}{T}\int_{t_1}^{t_1+T} v_L(t)\;\mathrm{d}t

When I wrote these equations in separate lines in latex after compilation in pdf file they appear on same line then I put $$ before and after these equations to be in separate lines but it shows error.

 

[Simon Bridge]

I am using pdfTeX, Version 3.1415926-1.40.10 (TeX Live 2009/Debian) and it works fine.
You can see that the above code also renders fine in these forums.
I cannot comment further without the text of the error message.

 

[Hina Gohar]

I am using pdfTeX, Version 3.1415926-1.40.10 (TeX Live 2009/Debian) and it works fine.
You can see that the above code also renders fine in these forums.
I cannot comment further without the text of the error message.

That previous circuit of DC-DC converter which I previously sent in pdf file..need some help..there in paper voltage equation mentioned and experimentally I checked that voltage is doubling but its current on output is so low i.e. in mA..I need help that how to determine its load current expression and also for Capacitor values means equation for capacitance from which I can esimate its particular values..

 

[Simon Bridge]

I think that is a separate problem and needs a separate thread.