AC+DC signal equation=> Then Filter AC signal= output only DC signal visible

In summary: I hope that helps.In summary, the conversation involves discussing the process of removing an AC signal from a combination of AC and DC signals. The best solution would be to use a large capacitor to ground after a 10 K resistor, but it may not completely remove the AC component. The conversation also includes a discussion about posting a new thread with attachments and calculating the impedance of a parallel RC network, including the reactance of a capacitor.

What could be the best code in C to see the DC out put?

  • Lowpass filter

    Votes: 1 100.0%
  • FFT

    Votes: 0 0.0%

  • Total voters
    1
  • #1
sandhi
9
0
Dear friends,

I would like to know how can I remove an AC signal from a signal which is combination of AC+DC. AC signal 60V/50Hz DC=50V.

Please suggest me a C code or any algoritham or a filter which can remove AC signal completely and I should be able to see only DC signal in the out put.


with best regards,
Sandhi :(
 

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  • #2
You would not be able to filter out the 60 Hz component.

The AC input of 60 volts would overdrive the amplifier and the gain of 3.9 would make it worse if that was possible. The output would just be a 50 Hz square wave

If you had an input of 0.6 volts, it may be possible to reduce the AC component in the output, but not remove it completely. You could put a large capacitor to ground after the 10 K resistor.
Try 25 uF.
 
  • #3
I am sorry to tell that I have mentioned wrong value. It is 5V/1Hz Ac signal and 12 V DC signal.
 
  • #4
The 1 Hz signal would make things difficult, but there is a bigger problem.

12 volts DC into an amplifier with a gain of -3.9 will give an output of -46.8 volts.

Now, unless you have a supply of +/- 50 volts this is going to overload your amplifier.

Apart from that, filters for 1 Hz are not really practical. You can do calculations and design a filter, but the component values will become very large and expensive.
 
  • #5
I would like to know how can I post a new thread with attachment? I don't see option where I can send a new post with attachment.

with best regards
prashant
 
  • #6
Choose "new topic" at the top of the main page.

Fill in the template, if it is on the homework page.

When that opens go to the paperclip thing at the top of the edit screen.
Browse and find the attachment on your hard drive. Upload it.

Then put your cursor where you want the attachment to be. Go to the paperclip again and click on the name of the file you uploaded.
It will put a link to the attachment and you can see it if you go to "preview post" at the bottom of the edit screen.

This way, you can have various pictures or diagrams in the right place in your text.

Click on "submit new thread".
 
  • #7
Thank you, I have found it. I have posted my new thread about origin of Capacitance formula.
 
  • #8
Dear Sir,

I would like to know if you have received my mail few hours ago

Prashant Kumar
sandhi
 
  • #9
Yes, I received your mail.

At 0.5 Hz, the capacitor has a reactance of 106103 ohms. The net impedance of the capacitor and resistor combination is 9.956 Kohms. So, really the capacitor has no effect.

In series with 1.01 Mohm, the output is 9.56 / 1010 or 0.00946 times the input voltage or 0.047 volts. Without the capacitor, it would be 0.0495 volts.

0.5 Hz is 1 cycle every 2 seconds. This is a very low frequency. What are you actually doing?
 
  • #10
I am trying to calculate Rf by measuring the signal at Rm which I call it as Um. When there is no capacitance it is very simple to calculate Rf but with varying capacitance Ce and Rf I need to calculate du/dt to the signal measured at Rm i.e to Um volts. Um is almost like a squarewave.

Dear Sir, I would like to know the exact equation for Rf when I include capacitance Ce.

with regards,
Sandhi
 
  • #11
This is the homework section, so the rules apply. This type of problem is solved by a process of steps, not really one formula.

You might like to have a look at this:
https://www.physicsforums.com/showthread.php?t=362836

It should tell you how to work out the impedance of a parallel RC network. You can add currents as vectors if they are flowing in the same wire, but you can't add impedances if they are in parallel.

You will also need to work out the reactance of a capacitor.
This is given by:
d1093caebdb7b6d3f6adfd94d00d03ec.png


where F is in Hz and C is in Farads.
 

1. What is an AC+DC signal equation?

An AC+DC signal equation is a mathematical representation of a signal that contains both alternating current (AC) and direct current (DC) components. It is typically written in the form of V(t) = VDC + VACsin(ωt + φ), where V(t) is the total signal, VDC is the DC component, VAC is the AC component, ω is the angular frequency, and φ is the phase shift.

2. How does filtering an AC signal result in only the DC signal being visible?

Filtering an AC signal involves using a circuit or device to remove the AC component from the signal, leaving only the DC component. This can be achieved by using a high-pass filter, which blocks low frequencies (i.e. the AC component) and allows high frequencies (i.e. the DC component) to pass through.

3. What is the purpose of isolating the DC signal from an AC+DC signal?

The purpose of isolating the DC signal from an AC+DC signal is to extract useful information from the signal and eliminate any unwanted noise or interference. This can be important in various applications, such as in electronic circuits or signal processing.

4. Can you give an example of a device that uses the AC+DC signal equation and filtering to produce a DC output?

An example of a device that uses the AC+DC signal equation and filtering to produce a DC output is a rectifier. A rectifier is an electrical circuit that converts AC power to DC power, using a combination of diodes and capacitors to filter out the AC component and produce a smooth DC output.

5. How does the frequency of the AC signal affect the resulting DC output after filtering?

The frequency of the AC signal does not affect the DC output after filtering, as long as the filtering is done properly. This is because the filtering process removes the AC component, regardless of its frequency, leaving only the DC component. However, the amplitude and phase of the AC component may affect the overall amplitude and phase of the resulting DC output.

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