Series RLC filter (trap filter)

In summary, the book says that at resonance the voltage on the resistor (the Vout) = the whole signal, but the question is asking if the voltage on the resistor (the Vout) = the whole signal when the circuit is at resonance. So the answer is no, the voltage on the resistor (the Vout) does not equal the whole signal at resonance.
  • #1
Lisa...
189
0
Hey! I need to solve the following problem:

a) Show the trap filter below acts to reject signals at a frequency
[tex] \omega = \frac{1}{\sqrt{LC}} [/tex]

http://img418.imageshack.us/img418/8168/rlc2wf.gif [Broken]

b) How does the width of the frequency band rejected depend on the resistance R?


For a) I thought that at resonance (when

[tex] \omega = \frac{1}{\sqrt{LC}} [/tex] )

the reactance of L & C = 0 and therefore Z= R. In a series circuit the current is the same everywhere, but the voltage divides itself according to V= IR. Therefore the voltage on L & C is 0 (R=0) and the voltage on R= IR. But is that correct? I think Vout is measured on R and NOT L & C, but how would I know? Btw what is the function of that grounded symbol, what does it mean and is it relevant for this question?

As for b) I know that
[tex] Q= \frac{\omega_0}{\Delta \omega} [/tex]

so

[tex] \Delta \omega = \frac{\omega_0}{Q} [/tex]

with

[tex] Q= \frac{\omega_0 L}{R} [/tex]

Substitution gives:

[tex] \Delta \omega = \frac{\omega_0}{\frac{\omega_0 L}{R}} [/tex]

[tex] =\frac{R}{L}[/tex]


[tex] \Delta f= \frac{\Delta \omega}{2 \pi}[/tex]

[tex] = \frac{\frac{R}{L}}{2 \pi} [/tex]

[tex]= \frac{R}{2 \pi L}[/tex]


Though my textbook says that


[tex] \Delta \omega = \frac{R}{2L} [/tex]


Could anybody please tell me what I'm doing wrong?!
 
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  • #2
How are they defining delta-omega? Are they maybe defining it as one side of the width of the stop band, and you are defining it as the full width?

BTW, in your first part, since the reactive impedance of the series LC goes toward zero at resonance, think of the R as the top part of a voltage divider and the LC as the bottom part of the voltage divider. What do you get for the transfer function of a voltage divider as the bottom impedance goes towards zero?
 
  • #3
berkeman said:
How are they defining delta-omega? Are they maybe defining it as one side of the width of the stop band, and you are defining it as the full width?

In my book you have this graph that shows the average power supplied by the generator to the series combination as a function of generator frequency. The resonance width (delta omega) is the frequency difference between the two points on the curve where the power is half its maximum value... so I guess they are defining it as the full width?

So... where did I go wrong in part b?

berkeman said:
BTW, in your first part, since the reactive impedance of the series LC goes toward zero at resonance, think of the R as the top part of a voltage divider and the LC as the bottom part of the voltage divider. What do you get for the transfer function of a voltage divider as the bottom impedance goes towards zero?

Do you mean this equation:
http://img286.imageshack.us/img286/3289/naamloos6tv.gif [Broken]

Well that one will equal 1, so Vout= Vin :grumpy: ?
So the voltage on the resistor (the Vout) = the whole signal, so the whole signal gets transferred at resonance? But that doesn't match the question, I mean it is given that at resonance the whole signal is blocked...
 
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  • #4
The impedance of a parallel LC goes to infinity at resonance, and the impedance of a series LC goes to zero at resonance. So | Vo/Vi | goes to zero at the resonance of the circuit you show. The equation that you wrote is for when the LC is at the Vin spot, and the R is from Vout to ground.
 
  • #5
Ah I see, thanks!
But what about this delta omega? Is my thinking correct and the answer in my book wrong or not?
 
  • #6
Well, to check the delta-omega, you can write the equation for the input impedance at Vi, and solve for the two omega values where the input impedance goes to SQRT(2) of the minimum impedance (which is R). Subtract the two omegas to see if you get your previous answer or the book's answer.
 
  • #7
This is what I've done. Please tell me if I've done it okay, cause I need to deliver it in 2 hours...:

V= IR, so Vin= IZ = I [tex] \sqrt{R^2 + (wl - \frac{1}{wc})^2} [/tex]

Vin= IR [tex] \sqrt{1 + (\frac{wl - 1/wc}{R})^2} [/tex] so

[tex]\frac{(wl - 1/wc)}{R}[/tex] = + or - 1 to give sqrt(2)

[tex]\frac{wl}{R} - \frac{1}{Rwc}[/tex] = + or - 1

[tex]\frac{w^2 l C}{RwC} - \frac{1}{Rwc} + or - \frac{RwC}{RwC} =0 [/tex]

so

[tex]w^2 l C + or - RwC - 1 = 0 [/tex]

using the abc formula
with
a= LC
b= + or - RC
c= -1

gives

[tex] w = \frac{+ or - RC + or - \sqrt{R^2 C^2 + 4 LC}}{2LC} [/tex]

Substracting

[tex] w = \frac{-RC + or - \sqrt{R^2 C^2 + 4 LC}}{2LC} [/tex]

from

[tex] w = \frac{+ RC + or - \sqrt{R^2 C^2 + 4 LC}}{2LC} [/tex]

gives

[tex] w= \frac{2 RC}{2 LC} [/tex]

which is:

[tex] w= \frac{R}{L} [/tex]


So may I conclude that the book is wrong? Or did I make a mistake somewhere?!
 
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  • #8

1. What is a Series RLC filter?

A Series RLC filter, also known as a trap filter, is a type of electronic filter that is used to block certain frequencies while allowing others to pass through. It is made up of a resistor (R), inductor (L), and capacitor (C) connected in series.

2. How does a Series RLC filter work?

A Series RLC filter works by taking advantage of the impedance properties of the components. The inductor has a high impedance to high frequencies, while the capacitor has a low impedance to high frequencies. By carefully choosing the values of R, L, and C, the filter can be designed to block or attenuate specific frequencies.

3. What are the applications of a Series RLC filter?

A Series RLC filter has a wide range of applications, including in audio equipment, power supplies, and communication systems. It is commonly used to remove unwanted noise or interference from electronic signals and to shape the frequency response of a circuit.

4. How do I calculate the values for a Series RLC filter?

The values for a Series RLC filter can be calculated using the following equations:
- Resonant frequency (f0) = 1 / (2π√(LC))
- Quality factor (Q) = √(L / RC)
- Bandwidth (BW) = f0 / Q
Once the desired values for f0 and Q are determined, the values for R, L, and C can be calculated using these equations.

5. What are the advantages of using a Series RLC filter?

There are several advantages to using a Series RLC filter, including its simplicity and low cost. It also has a sharp roll-off rate, meaning it can effectively block or attenuate unwanted frequencies. Additionally, the filter can be easily tuned by adjusting the values of R, L, and C to suit different applications.

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