Symbol || in Math

shteii01

Joined Feb 19, 2010
4,644
You have an RC circuit, it has a resistor and capacitor. In order to analyze such circuit it is very convenient to change it into purely "resistive" circuit where all elements are resistors. The "resistance" of the capacitor is represented by impedance, Zc=1/(jwC). So we replace the capacitor with its impedance and now instead of resistor and capacitor we have resistor and "resistor". We know how to work with resistors so now we have an easier time analyzing the circuit.

In this case you have two "resistors" in parallel. R is in parallel with Zc, R || Zc or R || \(\frac{1}{j{\varpi}C}\)
 
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Thread Starter

Mechatronical

Joined Feb 15, 2014
30
You have an RC circuit, it has a resistor and capacitor. In order to analyze such circuit it is very convenient to change it into purely "resistive" circuit where all elements are resistors. The "resistance" of the capacitor is represented by impedance, Xc=1/(jwC). So we replace the capacitor with its impedance and now instead of resistor and capacitor we have resistor and "resistor". We know how to work with resistors so now we have an easier time analyzing the circuit.

In this case you have two "resistors" in parallel. R is in parallel with Xc, R || Xc or R || \(\frac{1}{j{\varpi}C}\)
Thank you once again. And please do keep in touch. Your help is appreciated. I've added a link to the equation for you to check with your statement:

So R || 1/jωC can be written as the following?

Output impedance = [R * 1/jωC] / [R + 1/jωC]
 

Thread Starter

Mechatronical

Joined Feb 15, 2014
30
You have an RC circuit, it has a resistor and capacitor. In order to analyze such circuit it is very convenient to change it into purely "resistive" circuit where all elements are resistors. The "resistance" of the capacitor is represented by impedance, Xc=1/(jwC). So we replace the capacitor with its impedance and now instead of resistor and capacitor we have resistor and "resistor". We know how to work with resistors so now we have an easier time analyzing the circuit.

In this case you have two "resistors" in parallel. R is in parallel with Xc, R || Xc or R || \(\frac{1}{j{\varpi}C}\)
Here's the link:

http://s27.postimg.org/q29byyccz/image.jpg
 

shteii01

Joined Feb 19, 2010
4,644
Thank you once again. And please do keep in touch. Your help is appreciated. I've added a link to the equation for you to check with your statement:

So R || 1/jωC can be written as the following?

Output impedance = [R * 1/jωC] / [R + 1/jωC]
Yes, and that is what Mr. Chips also showed you, minus the explanation.

As far as the picture. That one deals with the Output Resistance. The way I understand it is that Output Resistance is what the load sees when the load is attached to the circuit.

It is like finding Thevenin Equivalent Resistance. You take a pair of terminals, in our case terminals where capacitor is attached, and you check the circuit to see what sources you have. In our case we assume that we have independent voltage source Vi, such sources are replaced by short circuit. So now we look at the circuit and what do we see? We see that resistor R is in parallel with capacitor C. Vo is not a voltage source, it is just the voltage across the circuit element that we are interested in.

We replace the capacitor C with its equivalent impedance, Zc, so that it behave as another resistor. Now we have two resistors in parallel, R and Zc. And we know how to find equivalent resistance of two resistors in parallel, it is one of the first things taught in circuit analysis, Requivalent=R1*R2/(R1+R2), same formula that Mr. Chips showed to you. In our case it is Requivalent=R*Zc/(R+Zc) where Zc=1/(jwC).
 
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WBahn

Joined Mar 31, 2012
29,976
What does the symbol || equal to?
Is the following equation;

R || 1/jωC

equal to sqrt(R^2 + (1/jωC)^2)
These aren't quite equal.

The first is a notation for a resistive impedance in parallel with a capacitive impedance. The result is an impedance that is a complex number having both a magnitude and an angle.

The second equation gives only the magnitude of that complex number.
 

Thread Starter

Mechatronical

Joined Feb 15, 2014
30
These aren't quite equal.

The first is a notation for a resistive impedance in parallel with a capacitive impedance. The result is an impedance that is a complex number having both a magnitude and an angle.

The second equation gives only the magnitude of that complex number.
Thank you for your time.
 

Thread Starter

Mechatronical

Joined Feb 15, 2014
30
Yes, and that is what Mr. Chips also showed you, minus the explanation.

As far as the picture. That one deals with the Output Resistance. The way I understand it is that Output Resistance is what the load sees when the load is attached to the circuit.

It is like finding Thevenin Equivalent Resistance. You take a pair of terminals, in our case terminals where capacitor is attached, and you check the circuit to see what sources you have. In our case we assume that we have independent voltage source Vi, such sources are replaced by short circuit. So now we look at the circuit and what do we see? We see that resistor R is in parallel with capacitor C. Vo is not a voltage source, it is just the voltage across the circuit element that we are interested in.

We replace the capacitor C with its equivalent impedance, Xc, so that it behave as another resistor. Now we have two resistors in parallel, R and Xc. And we know how to find equivalent resistance of two resistors in parallel, it is one of the first things taught in circuit analysis, Requivalent=R1*R2/(R1+R2), same formula that Mr. Chips showed to you. In our case it is Requivalent=R*Xc/(R+Xc) where Xc=1/(jwC).

But what if the resistor values are different, R1 and R2?
Then, for a low pass and high pass filter, to which one of the equations to I apply the R1 and R2?

input = R+ 1/jwC
output = R||1/jw
 

WBahn

Joined Mar 31, 2012
29,976
But what if the resistor values are different, R1 and R2?
Then, for a low pass and high pass filter, to which one of the equations to I apply the R1 and R2?

input = R+ 1/jwC
output = R||1/jw
Huh?

Could you please explain exactly what you are trying to do?

At this point it sounds like you are just trying to have someone tell you some universal rule that will let you be an equation monkey churning a crank on a bunch of formulas without understanding any of them. I'm sure that's not what you want, but I get the feeling you are trying to get at something that isn't apparent to us based on the questions you are actually asking.
 

shteii01

Joined Feb 19, 2010
4,644
But what if the resistor values are different, R1 and R2?
Then, for a low pass and high pass filter, to which one of the equations to I apply the R1 and R2?

input = R+ 1/jwC
output = R||1/jw
When I wrote R1*R2/(R1+R2), R1 represent R in your picture, R2 represent impedance of the capacitor.
 

WBahn

Joined Mar 31, 2012
29,976
But what if the resistor values are different, R1 and R2?
Then, for a low pass and high pass filter, to which one of the equations to I apply the R1 and R2?

input = R+ 1/jwC
output = R||1/jw
Okay, so I was finally able to get your link to load (it's preferred that you upload images to the AAC server to that they are directly accessible and also archived along with the post) and have a better understanding of what you are trying to do.

They key is not knowing which equation to use, but in knowing what the concepts of input impedance and output impedance refer to.

For a filter, the input impedance is the impedance seen by the driving signal source. For a bare filter (not part of a circuit), the assumption is that the output is unloaded (i.e., open circuited). The output impedance is the impedance seen looking back into the output with the input signal source set to zero. For a voltage signal, that means replacing it with a short circuit (i.e., setting it's output to 0V).

With that in mind, to you see how for the circuit you have that input impedance is the two components in series while the output impedance is the two components in parallel? Also, do you see that this is not a universal statement? That if the filter topology is different that the equations will be different?
 

Thread Starter

Mechatronical

Joined Feb 15, 2014
30
Okay, so I was finally able to get your link to load (it's preferred that you upload images to the AAC server to that they are directly accessible and also archived along with the post) and have a better understanding of what you are trying to do.

They key is not knowing which equation to use, but in knowing what the concepts of input impedance and output impedance refer to.

For a filter, the input impedance is the impedance seen by the driving signal source. For a bare filter (not part of a circuit), the assumption is that the output is unloaded (i.e., open circuited). The output impedance is the impedance seen looking back into the output with the input signal source set to zero. For a voltage signal, that means replacing it with a short circuit (i.e., setting it's output to 0V).

With that in mind, to you see how for the circuit you have that input impedance is the two components in series while the output impedance is the two components in parallel? Also, do you see that this is not a universal statement? That if the filter topology is different that the equations will be different?
After some research, I believe that what I would have to relate to in regards to passive filters is impedance matching. However, for active filters, there is not output impedance. At least in regards to voltage buffers.

(That said, the given equations are satisfactory for passive filters.)

So I would like to ask you to kindly look at my latest thread with the following question:

I want to calculate the Q factor of a Band Pass Filter, which would subsequently give me the bandwidth. Poles determine the Q factor of the system. How do I use the poles to determine the Q factor?

What I'm asking for is a mathematical procedure or equation with a relationship between poles and Q factor.

I've added a link below of the transfer function of the Band Pass Filter with respect to time:

http://upload.wikimedia.org/math/1/f...e1ae214ad6.png
 

shteii01

Joined Feb 19, 2010
4,644
I've added a link below of the transfer function of the Band Pass Filter with respect to time:

http://upload.wikimedia.org/math/1/f...e1ae214ad6.png
Forgive me for being pedantic, but that is transfer function with respect to frequency.

The function is H(s). s is input, H(s) is output. You saying that s is time, that is wrong. s is jw, j is a constant and has no units, w is angular frequency and has units of radians per second.
 
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