Quick Electrical Engineering Question

[Chriscross3234]

So I'm studying for my Intro to Electrical Engineering class and is there a way to visually identify between a first order lowpass filter and a first order highpass filter on a circuit? Is it just the way the inductor/capacitor is arranged on the circuit?

From what I gather, for an RC circuit, if the resistor comes first and the capacitor follows, then it is lowpass. If the capacitor comes first and the resistor follows, then it is highpass. Same follows for an RL circuit however it's the other way around.

Does this seem right?

 

[Analog]

Originally posted by: Chriscross3234
So I'm studying for my Intro to Electrical Engineering class and is there a way to visually identify between a first order lowpass filter and a first order highpass filter on a circuit? Is it just the way the inductor/capacitor is arranged on the circuit?

From what I gather, for an RC circuit, if the resistor comes first and the capacitor follows, then it is lowpass. If the capacitor comes first and the resistor follows, then it is highpass. Same follows for an RL circuit however it's the other way around.

Does this seem right?

You are essentially correct.

Always treat a cap as an open at low freq. and a short at hi freq. A LPF will have a cap to ground. A hpf will have a resistor to gnd.

 

[Drako]

Originally posted by: Analog

Originally posted by: Chriscross3234
So I'm studying for my Intro to Electrical Engineering class and is there a way to visually identify between a first order lowpass filter and a first order highpass filter on a circuit? Is it just the way the inductor/capacitor is arranged on the circuit?

From what I gather, for an RC circuit, if the resistor comes first and the capacitor follows, then it is lowpass. If the capacitor comes first and the resistor follows, then it is highpass. Same follows for an RL circuit however it's the other way around.

Does this seem right?

You are essentially correct.

Always treat a cap as an open at low freq. and a short at hi freq. A LPF will have a cap to ground. A hpf will have a resistor to gnd.

This.

Check this out, it's got some very useful Op amp circuits

 

[Chriscross3234]

Yeah, looking at it now, what Analog said makes perfect sense, Thanks!

 

[JDub02]

hook up a speaker. lots of bass = LPF .. no bass = HPF

😉

 

[frostedflakes]

Yup, just remember that capacitors pass high frequencies and block low, whereas inductors block high frequencies and pass low.

 

[TecHNooB]

I don't know if you've reached phasors yet, but resistors, capacitors, and inductors can be seen as impedances where..

R = R
C = 1/(jwC)
L = jwL
j = imaginary number (same as i in the math world 😛)

If you can accept the above for now, the explanation is simple 🙂

Measuring V_out from between R and C in both cases..

Config 1:
AC_Vin ---- R ---- C ----- Ground

Config 2:
AC_Vin ---- C ---- R ----- Ground

Notice that the impedance for a capacitor is 1/(jwC) where w = 2*pi*f.

As frequency -> infinity, the impedance approaches zero.
As frequency -> 0, the impedance approaches infinity.

In Config 1:
When the frequency is low, the impedance for C is high, so C looks like an open circuit. So Vout is approximately Vin.

When the frequency is high, C looks like a short circuit to ground, so your voltage across C is zero.

In Config 2:
When the frequency is low, C looks like an open circuit, so you are measuring from R to ground with no voltage source tied to it (because of the open circuit behavior of C at low frequencies). Therefore, Vin is zero.

When the frequency is high, C looks like an short circuit, so all the voltage falls across R. Therefore, Vout = Vin.

 

[Chriscross3234]

Yeah TecH, we went over phasors and are now learning about frequency response and bode plots, hence LPF and HPF. A lot of what you said clarified what I am learning 😉.

Also one more question. For an RC circuit, is the break frequency (aka half power freq) the same for both HPF and LPF, ie, f = 1/(2(pi)RC) ? And, I am assuming that the break frequency for an RL circuit is f = R/(2(pi)L).

I know this is a very generic question that skims the surface of what is actually going on in order to determine the actual break frequency, but my professor very quickly went over how to obtain these equations and I don't think we will have to show how we got them on the exam 😛

 

[oiprocs]

Engineers don't make crazy money. If you want crazy money, be prepared to disregard everything in you learn in your EE courses.

 

[Chriscross3234]

Originally posted by: scorpious
Engineers don't make crazy money. If you want crazy money, be prepared to disregard everything in you learn in your EE courses.

Unfortunately, my intro EE courses are required before I get to my actual degree courses, which is biomedical engineering, and I'm sure the concepts learned in EE will most likely show up in my field of engineering. BTW I'm not in it purely for the money. If I was I would change my major to business 😀

 

[Gibson486]

Originally posted by: Chriscross3234

Originally posted by: scorpious
Engineers don't make crazy money. If you want crazy money, be prepared to disregard everything in you learn in your EE courses.

Unfortunately, my intro EE courses are required before I get to my actual degree courses, which is biomedical engineering, and I'm sure the concepts learned in EE will most likely show up in my field of engineering. BTW I'm not in it purely for the money. If I was I would change my major to business 😀

yes, they will show up...although it will not be as intense as your courses are. In EE, most of the work is debugging what you have.

As for manking money...yeah, engineers make alot out of college, but it's only because of supply and demand. For every engineer that graduates, there are like 5 business majors that join them. That said, you reach the ceiling rather quickly in engineering. To pass the ceiling, you have to essentially throw away your technical hat and put on your business cap and you can't look back. Once you make that transition, people start to question whether or not your "lost your engineering skills". With that said...you will not make the REAL buck wokring under someone else. You need to do things on your own regardless of what major you pick.

 

[frostedflakes]

It's actually not too difficult if you wanted to derive the break frequencies. First you'd need to find the magnitude response (not difficult for simple first order filters), then you can just substitute in 1/sqrt(2) (aka -3dB) for the magnitude and solve for f. The break frequencies you listed are correct.

As far as money goes, I think senior engineers and project managers can make decent money. I'd assume most people are not in engineering for the money, though.

 

[TecHNooB]

Okay, so you know your transfer function H(jw) or H(s) is equal to X_out(s)/X_in(s) where X can be any combination of voltage and current.

Since power has a square relation with I and R, that is, P = I^2 * R or V^2 / R, when you reach the break frequency, you have half the power, which implies 1/sqrt(2) times the voltage/current.

Think about it this way. Set R for both power equations equal to 1 (since the R and 1/R term only scales the power linearly). It's the square relationship we care about.

Then, V^2 = P or I^2 = P. At half power, your output voltage will be half of what it was before. So V1^2 = P is now V2^2 = 1/2*P where V1 must be equal to V2 = V1/sqrt(2).

Anyways, deriving the break frequencies is pretty easy. Basically, set your the magnitude of your transfer function equal to 1/sqrt(2) and solve for w or f.

From the break frequencies you have stated, I will assume your circuits are configured this way:

Circuit 1:
V_in ------ C ------ R ---- Ground

Circuit 2:
V_in ------ R ------- L ---- Ground

We will measure V_out from between C and R, and R and L.

** Note that it is important how these parts are arranged because it will effect your transfer function equation. Knowing where V_out is measured relative to V_in is important as well.

----------------------------------------------------------------------------------------------

Circuit 1 has this equation (it's simple voltage division):

Vout = Vin * (R/(R + Z_C)) = Vin * (R /(R + 1/jwC))

Vout/Vin = H(jw) = jwRC / (1 + jwRC) ~ Now find the magnitude.

|H(jw)| = (wRC)^2 / sqrt(1 + (wRC)^2) === 1/sqrt(2) at Break Freq.

If you set wRC = 1, |H(jw)| = 1/(sqrt(1+1) = 1/sqrt(2).

Therefore, when wRC = 1, we have our break frequency.

2*pi*f*R*C = 1 --> f = 1/(2*pi*R*C)

----------------------------------------------------------------------------------------------

Circuit 2 has this equation (again, simple voltage division):

Vout = Vin * (Z_L/(Z_L + R)) = Vin * jwL / (jwL + R)

Vout/Vin = H(jw) = jwL / (jwL + R) <------ Our transfer function

|H(jw)| = wL / sqrt((wL)^2 + R^2) = 1/sqrt(2) at Break Freq.

Notice if you set wL = R, you get |H(jw)| = R / sqrt(R^2 + R^2) = 1/sqrt(2)

wL = R --> 2*pi*f*L = R --> f = R/(2*pi*L)


Imaginary numbers (you can derive these easily by plotting the on the Real/Imaginary Axis):
|a + bj| = sqrt(a^2 + b^2)
|(a+bj)/(c+dj)| = |a+bj| / |c+dj|
|aj| = a

 

[krylon]

Originally posted by: Chriscross3234

Originally posted by: scorpious
Engineers don't make crazy money. If you want crazy money, be prepared to disregard everything in you learn in your EE courses.

Unfortunately, my intro EE courses are required before I get to my actual degree courses, which is biomedical engineering, and I'm sure the concepts learned in EE will most likely show up in my field of engineering. BTW I'm not in it purely for the money. If I was I would change my major to pre-law😀

fixed that for you