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How to tell if the circuit is a High or low pass filter?

  1. May 3, 2015 #1
    1. The problem statement, all variables and given/known data
    Draw the circuit diagram of a second-order series High pass filter RLC filter.
    Given:
    R = 50 ohm
    Q_s = .5
    f_0 = 30 MHz
    determine L and C

    2. Relevant equations
    f_0 = 1/(2*pi*sqrt(LC))
    Q_s = 2*pi*f_0*L/R = 1/(2*pi*f_0*R)


    3. The attempt at a solution
    I know how to find L and C, given two variables and the two equations listed above, its simply a plug and chug. But I am confused on how to draw the circuit diagram. Also, more generally, how do I determine if a circuit diagram is high or low pass? or first or second order? If I was given just the diagram, with no values, how would I be able to tell?


    As an example, looking at this diagram :

    ?temp_hash=0997bccec61c92aa606d93eb1f13750c.png
    I am told that it is a first order High Pass filter, but if I was not told, how would I be able to tell?
     

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  3. May 3, 2015 #2
    Look at the circuit in this way: the output voltage is given by the voltage partition between the resistor and the capacitor. The impedance of the resistor in constant, while that of the capacitor is [tex] Z= \frac{1}{i\omega} [/tex].
    What does this mean? If frequency increases, the impedance of the capacitor becomes lower and lower, while the resistor's one is unchanged; as a consequence, the voltage drop on the capacitor will decrease and a larger amount of the input voltage will be at the output node.
    In the extreme case, if [tex] \omega -> +\infty [/tex] the capacitor will act as a short circuit, and [tex] V_{o}= V_{in} [/tex]

    If you write a few equations you can find analitically what I told you :)
     
  4. May 3, 2015 #3
    I understand what you are saying, when frequency is high, the impedance of the capacitor is negligible. The output voltage is equal to the voltage across the resistor, which is equal to the input voltage. But how does that tell me if it is a high or low pass filter?

    All the equations have told me is that the capacitor becomes less and less effective as the frequency increases. How does this help me distinguish between high and low pass filters?
     
  5. May 3, 2015 #4

    Zondrina

    User Avatar
    Homework Helper

    Please note, the impedance of a capacitor is actually:

    $$Z_C = \frac{1}{j \omega C}$$

    Notice as ##\omega \rightarrow \infty, Z_C \rightarrow 0##. So the capacitor behaves like a short circuit for really high frequencies.

    We call this a "high pass filter" because it lets the high frequency signals through.

    What about low frequencies you say?

    As ##\omega \rightarrow 0, Z_C \rightarrow \infty##. So the capacitor will behave like an open circuit. This means zero current will flow through the circuit, and hence the voltage across the resistor is zero. So the resistor also behaves like a short circuit. Notice no low signal frequencies can get through.

    This is why we call it a high pass filter. It lets the high frequency signals pass, and low ones get blocked.

    If you reverse the order of the resistor and capacitor in the circuit, you obtain a low pass filter. I will leave the analysis of this circuit to you as I think it would be informative.

    A useful way to easily remember which filter is which, is to remember these:

    CR - High pass filter
    RC - Low pass filter

    Where the order of the letters actually tells you the order of the components in the circuit.
     
  6. May 4, 2015 #5
    Maybe the point is another: can you tell me what does "low/high pass" mean?
    I think your problem is to understand what we are refferring to when we use these expressions, otherwise you would already answered to yourself with your last post.
     
  7. May 4, 2015 #6
    ah thank you for the explanation! I understand now, was confused about which filters allowed what frequencies through. Now taking the same thought a step further, a second order filter is one that involves a capacitor, inductor, as well as a resistor. If we look at a first order high pass filter, then stick an inductor between the capacitor and resistor (so now its a CLR circuit) how does that affect it?

    I know that the impedance of an inductor is jwL so as the frequency goes to infinity, so does the impedance, meaning it becomes an open circuit. So what kind of filter is it now? I dont understand how to measure the V_out at high frequencies now that the capacitor is a short circuit and the inductor is an open circuit.
     
  8. May 4, 2015 #7

    LvW

    User Avatar

    Do you know the simple voltage divider rule?
    In this case, you have a voltage divider consisting of a series impedance that resembles a series resonant circuit (short circuit with Z=0 at the resonant frequency) and an ohmic resistor.
    Hence, there is one single frequency for which the input voltage is identical to the output voltage. For all other frequencies, the output voltage across the resistor is smaller.
    What kind of filtering would you expect?
     
  9. May 4, 2015 #8
    The voltage drop across the inductor plus the voltage drop across the resistor would equal the voltage source (voltage divider)

    So at high frequencies, the CLR would behave like a LR circuit, which is a first order low pass filter. but since this only occurs at high frequencies, would this mean that it is neither a high nor low pass filter? Since the output voltage would not be equal to the input voltage at very high or very low frequencies, it dont think it can be classified as either?

    So applying the same theory to a RC circuit:
    - at high frequencies, the capacitor would act as a short circuit, meaning that the voltage drop across the resistor would have to equal the source. This means that it is not a high pass filter since the output voltage is 0. Does this imply that it is a low pass filter? More generally, can a filter be neither high nor low? According to my textbook, it states that a RC circuit is indeed a first order low pass filter, but what I am confused about is that the output voltage still does not quite equal the input voltage.

    - at low frequencies the capacitor would act as an open circuit. Using the voltage divider, the voltage drop across the resistor plus the voltage drop across the capacitor (that is now an open circuit) would have to equal the input voltage. The output voltage is equal to the voltage drop across the open circuit, however, this is not the same as the input voltage as it is only equal to the input voltage minus the voltage across the resistor.

    Also - can you tell if a circuit is high or low pass by looking at the transfer function H(f)?

    H(f) = output voltage / input voltage

    So what values of H(f) would classify a circuit to be high and what values of H(f) would classify a circuit to be low pass?


    --
    Sorry for the long post and thank you to all that have helped me so far!
     
  10. May 5, 2015 #9

    LvW

    User Avatar

    The voltage drop across the inductor plus the voltage drop across the resistor would equal the voltage source (voltage divider)

    Why did you ignore (forget) the drop across the capacitor?

    So at high frequencies, the CLR would behave like a LR circuit, which is a first order low pass filter. but since this only occurs at high frequencies, would this mean that it is neither a high nor low pass filter? Since the output voltage would not be equal to the input voltage at very high or very low frequencies, it dont think it can be classified as either?


    Correct - it is neither a lowpass nor a high-pass. But for very high frequencies it is very similar to a lowpass response.

    So applying the same theory to a RC circuit:
    - at high frequencies, the capacitor would act as a short circuit, meaning that the voltage drop across the resistor would have to equal the source. This means that it is not a high pass filter since the output voltage is 0. Does this imply that it is a low pass filter? More generally, can a filter be neither high nor low? According to my textbook, it states that a RC circuit is indeed a first order low pass filter, but what I am confused about is that the output voltage still does not quite equal the input voltage.

    A simple RC circuit is a classical first-order lowpass. The output voltage is identical to the input if there is no current through the capacitor. This happens only at DC (f=0).

    - at low frequencies the capacitor would act as an open circuit. Using the voltage divider, the voltage drop across the resistor plus the voltage drop across the capacitor (that is now an open circuit) would have to equal the input voltage. The output voltage is equal to the voltage drop across the open circuit, however, this is not the same as the input voltage as it is only equal to the input voltage minus the voltage across the resistor.


    No - the input voltage equals the sum of both voltages across both elements only if there is a current (and a corresponding voltage drop). However, for f=0 there is no steady-state current at all.

    Also - can you tell if a circuit is high or low pass by looking at the transfer function H(f)?


    Yes. You can see what will happen with the magnitude of the function for very low and/or very high frequencies.

    H(f) = output voltage / input voltage

    So what values of H(f) would classify a circuit to be high and what values of H(f) would classify a circuit to be low pass?

    There are no "values" classifying the function. You have to analyze the response for ALL frequencies between zero and infinite.
     
  11. May 5, 2015 #10

    psparky

    User Avatar
    Gold Member

    To OP:

    All the above that was told to you is true.

    Now it just takes time and experience to soak it all in. Be patient.

    Soon enough, you will be the one answering the questions from a newbie!!!
     
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