Current Through A half-wave rectifier

In summary: I'm really confused, and although I went and finished other problems, I still can't finish this one, and it's killing me:(
  • #1
aNxello
13
0
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)
 
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  • #2
Can you show some of your work , to see what goes wrong ?
 
  • #3
View attachment 694
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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  • #4
I certainly cannot read that , can you try to write \(\displaystyle A_n , B_n \) to solve them together ?
 
  • #5
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.
 
  • #6
Let me try this, I will update with my results, thank you!
Also the range was already from 0 to Pi/Omega!
 
  • #7
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
 
  • #8
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
 
  • #9
aNxello said:
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)

aNxello said:
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:

FAQ: Current Through A half-wave rectifier

What is a half-wave rectifier?

A half-wave rectifier is a type of circuit that converts alternating current (AC) to direct current (DC). It utilizes a diode to only allow the positive half of the AC wave to pass through, resulting in a pulsating DC output.

How does a half-wave rectifier work?

A half-wave rectifier works by using a diode to block the negative half of the AC wave, resulting in a pulsating DC output. The diode acts as a one-way valve, allowing current to flow in only one direction.

What is the advantage of using a half-wave rectifier?

The main advantage of a half-wave rectifier is its simplicity. It requires only one diode and can be easily constructed using basic electronic components. It is also more efficient than using a full-wave rectifier in certain applications.

What is the disadvantage of using a half-wave rectifier?

The main disadvantage of a half-wave rectifier is its low efficiency. Since it only utilizes half of the AC wave, the resulting DC output is not as smooth as a full-wave rectifier. This can cause issues in sensitive electronic devices.

How is a half-wave rectifier used in practical applications?

A half-wave rectifier is commonly used in low-power applications such as battery chargers, small power supplies, and signal detectors. It can also be found in various household appliances that require low voltage DC power, such as LED lights and small motors.

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