# Thread: Intgration of a function.

1. ## Intgration of a function.

actually this is not a homework, I found this explanation in a journal paper but I could not understand it. Can someone give me an explanation or possibly a proof that:

if:

$\frac{\mathrm{d}&space;V(t)}{\mathrm{d}&space;x}=\sqrt{2}\sum_{h=1}^{H}h\omega&space;V_{h}cos(h\omega&space;t+\frac{\pi&space;}{2})$

then why integration over whole period is:

$\frac{1}{T}\int_{0}^{T}\left&space;(&space;\frac{\mathrm{d}&space;V(t)}{\mathrm{d}&space;t}&space;\right&space;)^{2}dt=\omega&space;\sum_{h=1}^{H}h^{2}V_{h}^{2}$

I have problem with the power of omega, my solution returns w with power 2, while the power of omega in answer is one, Can someone help me for the reason?

2. What is T?

Isn't T given by: $\displaystyle T={{2\pi}\over{h\omega}}\ ,$ where h=1 ?

… and shouldn't that derivative be: $\displaystyle {{dV(t)}\over{dt}}$ not dx ?

3. Originally Posted by lordcx
actually this is not a homework, I found this explanation in a journal paper but I could not understand it. Can someone give me an explanation or possibly a proof that:

if:

$\frac{\mathrm{d}&space;V(t)}{\mathrm{d}&space;x}=\sqrt{2}\sum_{h=1}^{H}h\omega&space;V_{h}cos(h\omega&space;t+\frac{\pi&space;}{2})$

then why integration over whole period is:

$\frac{1}{T}\int_{0}^{T}\left&space;(&space;\frac{\mathrm{d}&space;V(t)}{\mathrm{d}&space;t}&space;\right&space;)^{2}dt=\omega&space;\sum_{h=1}^{H}h^{2}V_{h}^{2}$

I have problem with the power of omega, my solution returns w with power 2, while the power of omega in answer is one, Can someone help me for the reason?
Frankly I wasn't able to do the summation, but I can address the problem with the $\omega$. The integral becomes:
$\frac{1}{T}\int_0^T \left ( \frac{dV}{dt} \right ) ^2 ~ dt = \frac{2 \omega ^2}{T} \displaystyle\sum_{h = 1}^H \displaystyle\sum_{h' = 1}^H h h' V_h V_{h'} \int_0^T sin(h \omega t)~sin(h' \omega t)~dt$

The integral is trivial and can easily be seen to be proportional to $\frac{1}{ \omega }$ leaving only one $\omega$.

-Dan

4. Originally Posted by topsquark
Frankly I wasn't able to do the summation, but I can address the problem with the $\omega$. The integral becomes:
$\frac{1}{T}\int_0^T \left ( \frac{dV}{dt} \right ) ^2 ~ dt = \frac{2 \omega ^2}{T} \displaystyle\sum_{h = 1}^H \displaystyle\sum_{h' = 1}^H h h' V_h V_{h'} \int_0^T sin(h \omega t)~sin(h' \omega t)~dt$

The integral is trivial and can easily be seen to be proportional to $\frac{1}{ \omega }$ leaving only one $\omega$.

-Dan

Thank you, May I know please which theory did you use to get $\frac{1}{T}\int_0^T \left ( \frac{dV}{dt} \right ) ^2 ~ dt = \frac{2 \omega ^2}{T} \displaystyle\sum_{h = 1}^H \displaystyle\sum_{h' = 1}^H h h' V_h V_{h'} \int_0^T sin(h \omega t)~sin(h' \omega t)~dt$

5. Originally Posted by lordcx
Thank you, May I know please which theory did you use to get $\frac{1}{T}\int_0^T \left ( \frac{dV}{dt} \right ) ^2 ~ dt = \frac{2 \omega ^2}{T} \displaystyle\sum_{h = 1}^H \displaystyle\sum_{h' = 1}^H h h' V_h V_{h'} \int_0^T sin(h \omega t)~sin(h' \omega t)~dt$
$\frac{dV}{dt} = \sqrt{2}~\displaystyle\sum_{h = 1}^{H} h \omega V_h~cos \left (h \omega t + \frac{ \pi }{2} \right )$

Thus
$\left ( \frac{dV}{dt} \right )^2 = \left [ \sqrt{2}~\displaystyle\sum_{h = 1}^{H} h \omega V_h~cos \left (h \omega t + \frac{ \pi }{2} \right ) \right ] \cdot \left [ \sqrt{2}~\displaystyle\sum_{h' = 1}^{H} h' \omega V_{h'}~cos \left (h' \omega t + \frac{ \pi }{2} \right ) \right ]$

Simplifying
$\left ( \frac{dV}{dt} \right )^2 = 2 \omega ^2 \displaystyle\sum_{h = 1}^{H} \displaystyle\sum_{h' = 1}^{H}~hh'V_h V_{h'}~cos \left (h \omega t + \frac{ \pi }{2} \right )~cos \left (h' \omega t + \frac{ \pi }{2} \right )$

Now expand the cosines out in the following format:
$cos(a + b) = cos(a)~cos(b) - sin(a)~sin(b)$

-Dan

PS: For the rest of the problem there is probably a way to write the double summation as a single sum. I'm not good at that kind of thing.

6. Originally Posted by topsquark
$\frac{dV}{dt} = \sqrt{2}~\displaystyle\sum_{h = 1}^{H} h \omega V_h~cos \left (h \omega t + \frac{ \pi }{2} \right )$

Thus
$\left ( \frac{dV}{dt} \right )^2 = \left [ \sqrt{2}~\displaystyle\sum_{h = 1}^{H} h \omega V_h~cos \left (h \omega t + \frac{ \pi }{2} \right ) \right ] \cdot \left [ \sqrt{2}~\displaystyle\sum_{h' = 1}^{H} h' \omega V_{h'}~cos \left (h' \omega t + \frac{ \pi }{2} \right ) \right ]$

Simplifying
$\left ( \frac{dV}{dt} \right )^2 = 2 \omega ^2 \displaystyle\sum_{h = 1}^{H} \displaystyle\sum_{h' = 1}^{H}~hh'V_h V_{h'}~cos \left (h \omega t + \frac{ \pi }{2} \right )~cos \left (h' \omega t + \frac{ \pi }{2} \right )$

Now expand the cosines out in the following format:
$cos(a + b) = cos(a)~cos(b) - sin(a)~sin(b)$

-Dan

PS: For the rest of the problem there is probably a way to write the double summation as a single sum. I'm not good at that kind of thing.
sorry if I take your time but h'=h, and V=V' then we have:

$\frac{2\omega&space;^{2}}{T}\int_{0}^{T}\sum_{h=1}^{H}h^{2}V_{(h)}^{2}sin^{2}(h\omega&space;t)dt$

and over whole period:

$\frac{1}{T}\int_{0}^{T}sin^{2}(hwt)dt=\frac{T}{2}$

then we will have

$\omega&space;^{2}\sum&space;h^{2}V^{2}$ not $\omega&space;\sum&space;h^{2}V^{2}$

am I wrong??

7. Originally Posted by lordcx
sorry if I take your time but h'=h, and V=V' then we have:

$\frac{2\omega&space;^{2}}{T}\int_{0}^{T}\sum_{h=1}^{H}h^{2}V_{(h)}^{2}sin^{2}(h\omega&space;t)dt$

and over whole period:

$\frac{1}{T}\int_{0}^{T}sin^{2}(hwt)dt=\frac{T}{2}$

then we will have

$\omega&space;^{2}\sum&space;h^{2}V^{2}$ not $\omega&space;\sum&space;h^{2}V^{2}$

am I wrong??
Let's take a simpler example.
$\left ( \displaystyle\sum_{h = 1}^5 h \right ) ^2$

I say that this is
$\displaystyle\sum_{h = 1}^5 \displaystyle\sum_{h'=1}^5 h h'$

$= 1 \cdot (1 + 2 + 3 + 4 + 5) + 2 \cdot (1 + 2 + 3 + 4 + 5) + ~...$

$= 1 \cdot 1 + 1 \cdot 2 + 1 \cdot 3 + 1 \cdot 4 +1 \cdot 5$

$+ 2 \cdot 1 + 2 \cdot 2 + 2 \cdot 3 + 2 \cdot 4 +2 \cdot 5 +~...$

This is the same sequence as $(1 + 2 + 3 + 4 + 5)^2 = (1 + 2 + 3 + 4 + 5)(1 + 2 + 3 + 4 + 5)$

$= 1 \cdot (1 + 2 + 3 + 4 + 5) + 2 \cdot (1 + 2 + 3 + 4 + 5) + ~...$

So the h and h' cannot be equal. If they were then we would have

$\left ( \displaystyle\sum_{h = 1}^5 h \right ) ^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2$

which it is not.

-Dan

8. Originally Posted by lordcx
and over whole period:

$\frac{1}{T}\int_{0}^{T}sin^{2}(hwt)dt=\frac{T}{2}$
Also

$\frac{1}{T}~ \int_0^T sin^2(h \omega t)~dt = \frac{1}{2} - \frac{1}{4h \omega T}~sin(2h \omega T)$

-Dan