# Current Through A half-wave rectifier

#### aNxello

##### New member
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

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Can you show some of your work , to see what goes wrong ?

#### aNxello

##### New member

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!

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#### ZaidAlyafey

##### Well-known member
MHB Math Helper
I certainly cannot read that , can you try to write $$\displaystyle A_n , B_n$$ to solve them together ?

#### Ackbach

##### Indicium Physicus
Staff member
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
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
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
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
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
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:
$$\frac {a_0} {2} = \frac A \pi$$

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