Welcome to our community

Be a part of something great, join today!

Current Through A half-wave rectifier

aNxello

New member
Feb 20, 2013
13
Hello!
So I'm stuck in yet another problem, I've been going through some Fourier Series problems (including one that I already asked about around here, and go great help!) and I've managed to get past most, but I got once again stuck on one.

So for this one we got an alternating current of the form
\(\displaystyle i(t) = A\sin(\omega t)\)

that is passed through a half-wave rectifier, which the current when it is flowing in the positive direction only.
So I assumed to do a half-range sine Fourier Series
I approached it by taking the range and re-defining it, realizing that the period was \(\displaystyle 2\frac{\pi }{\omega}\) so from \(\displaystyle -\frac{\pi }{\omega}\) to \(\displaystyle \frac{\pi }{\omega}\)

So since it is only on the positive side I made it so
\(\displaystyle f(t)=\begin{Bmatrix}A\sin(\omega t) & 0 \leq t\leq \frac{\pi}{\omega } \\
0 & -\frac{\pi}{\omega } \leq t \leq 0
\end{Bmatrix}\)

However the problem also told me the answer, but the coefficients I get are nothing like the answer, this is the answer:

\(\displaystyle i_{out} = \frac{A}{\pi } + \frac{A}{2}\sin(\omega t )- \frac{2A}{\pi}\sum \frac{\cos[(n+1)\omega t ]}{n(n+1)}\)


There's also a second part, which is what would happen if the same current went through a full-wave rectifier, and for that the solution is:

\(\displaystyle i_{out} = \frac{2A}{\pi } -\frac{4A}{\pi}\sum \frac{\cos(n \omega t)}{n^{2}-1}\)


I really hope you guys can help, this is driving me crazy (Whew)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Can you show some of your work , to see what goes wrong ?
 

aNxello

New member
Feb 20, 2013
13
img2013031300108.jpg
This is a picture of my work, but if you can't read it I got

\(\displaystyle a_{0} = \frac{4 \omega A}{\pi }\)

\(\displaystyle a_{n} = 0 \)

\(\displaystyle b_{n} = 4n^{2} \int \sin(\omega t)\sin(2nt\omega ) \)
And then that integral keeps repeating :/ (any tips on how to deal with partial integrals that repeat like btw?)

thank you!
 
Last edited:

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
I certainly cannot read that , can you try to write \(\displaystyle A_n , B_n \) to solve them together ?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
Assuming your integral is correct: first of all, your integrand is even. So simplify by saying $\int_{-a}^{a}=2\int_{0}^{a}$. Then try by-parts until you get your original integral back:
\begin{align*}
b_{n}&= 8n^{2} \int_{0}^{ \pi/ \omega} \underbrace{ \sin( \omega t)}_{u} \underbrace{ \sin(2n \omega t) \, dt}_{dv}\\
&= 8n^{2} \left[ \left(- \frac{ \sin( \omega t) \cos(2n \omega t)}{2n \omega} \right) \Bigg|_{0}^{ \pi/ \omega} + \frac{ \omega}{2n \omega} \int_{0}^{ \pi/ \omega} \cos( \omega t) \cos(2n \omega t) \, dt \right]\\
&= 4n \int_{0}^{ \pi/ \omega} \cos( \omega t) \cos(2n \omega t) \, dt.
\end{align*}
Do this again, with $u= \cos( \omega t)$ and $dv= \cos(2n \omega t)\, dt$, until you get back to your original integrand. Then you should be able to solve for $b_{n}$. Done.
 

aNxello

New member
Feb 20, 2013
13
Let me try this, I will update with my results, thank you!
Also the range was already from 0 to Pi/Omega!
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Ok , if you are confused about integration by parts recall :

\(\displaystyle \cos\left( x+y\right) - \cos\left(x-y\right) = -2\sin(x)\sin(y)\)

Then you don't have to do integration by parts
 

aNxello

New member
Feb 20, 2013
13
So guys, I'm still lost, for the second part tho I found the answer,, I just have to do a full fourier series of |sinx|, but for the first time I'm still confused if I should use the half range form or the regular form, and if I have o use the half range, how do I apply it to a function in the range -omega/pi to omega/pi
I'm really confused, and although I went and finished other problems, I still can't finish this one, and it's killing me:(
I just can't get to that answer
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,854
Hello!
So I'm stuck in yet another problem, I've been going through some Fourier Series problems (including one that I already asked about around here, and go great help!) and I've managed to get past most, but I got once again stuck on one.

So for this one we got an alternating current of the form
\(\displaystyle i(t) = A\sin(\omega t)\)

that is passed through a half-wave rectifier, which the current when it is flowing in the positive direction only.
So I assumed to do a half-range sine Fourier Series
I approached it by taking the range and re-defining it, realizing that the period was \(\displaystyle 2\frac{\pi }{\omega}\) so from \(\displaystyle -\frac{\pi }{\omega}\) to \(\displaystyle \frac{\pi }{\omega}\)

So since it is only on the positive side I made it so
\(\displaystyle f(t)=\begin{Bmatrix}A\sin(\omega t) & 0 \leq t\leq \frac{\pi}{\omega } \\
0 & -\frac{\pi}{\omega } \leq t \leq 0
\end{Bmatrix}\)

However the problem also told me the answer, but the coefficients I get are nothing like the answer, this is the answer:

\(\displaystyle i_{out} = \frac{A}{\pi } + \frac{A}{2}\sin(\omega t )- \frac{2A}{\pi}\sum \frac{\cos[(n+1)\omega t ]}{n(n+1)}\)


There's also a second part, which is what would happen if the same current went through a full-wave rectifier, and for that the solution is:

\(\displaystyle i_{out} = \frac{2A}{\pi } -\frac{4A}{\pi}\sum \frac{\cos(n \omega t)}{n^{2}-1}\)


I really hope you guys can help, this is driving me crazy (Whew)
So guys, I'm still lost, for the second part tho I found the answer,, I just have to do a full fourier series of |sinx|, but for the first time I'm still confused if I should use the half range form or the regular form, and if I have o use the half range, how do I apply it to a function in the range -omega/pi to omega/pi
I'm really confused, and although I went and finished other problems, I still can't finish this one, and it's killing me:(
I just can't get to that answer
Hi aNxello! :)

You seem to have mixed up your constants a bit.

From mathworld:
$$f(t) = \frac {a_0}{2} + \sum a_n \cos \frac {n \pi t} {L} + \sum b_n \sin \frac {n \pi t} {L}$$
where
$$\begin{aligned}
a_0 &= \frac 1 L \int_{-L}^L f(t) dt \\
a_n &= \frac 1 L \int_{-L}^L f(t) \cos \frac {n \pi t} {L} dt \\
b_n &= \frac 1 L \int_{-L}^L f(t) \sin\frac {n \pi t} {L} dt \\
\end{aligned}$$

Since you have $L=\dfrac \omega \pi$ and $f(t)=A\sin \omega t$ on the interval from $0$ to $L$, you get:
$$\begin{aligned}
a_0 &= \frac \omega \pi \int_0^{\pi/\omega} A \sin \omega t dt \\
a_n &= \frac \omega \pi \int_0^{\pi/\omega} A \sin \omega t \cos n \omega t dt \\
b_n &= \frac \omega \pi \int_0^{\pi/\omega} A \sin \omega t \sin n \omega t dt \\
\end{aligned}$$

Let's pick $a_0$. This one yields:
$$\begin{aligned}
a_0 &= \frac \omega \pi \int_0^{\pi/\omega} A \sin \omega t dt \\
&= \left. \frac \omega \pi (-\frac A \omega \cos \omega t) \right|_0^{\pi/\omega} \\
&= \frac A \pi (-\cos \pi - -\cos 0) \\
&= \frac {2A} \pi \\
\end{aligned}$$

So the first term is:
$$\frac {a_0} {2} = \frac A \pi$$
 
Last edited: