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Simple question about frequency

  1. Feb 1, 2015 #1

    Im quite new to frequency and how it effects a circuit with both capacitors and inductor. I have created a simple circuit in Multisim and run an AC-analysis.
    Pictures of both circuit and analysis is attachments.

    What is happening here.

    And arent the circuit also depending on the time (When the capacitors is getting fully charged)?

    What I think is happening;

    1. With lov frequency both the capacitor will have very high resistance, and the inductor will have very lov. Because of JWL, and 1/Jwl W= rad/s
    And all the voltage will be the same, because now current will run tru the circuit.

    2. Then with higher frequency the capacitors will start charging, and therefore it will take som voltage.

    But is it not time depending, only frequency depending? bilde77.png
  2. jcsd
  3. Feb 2, 2015 #2


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    I am not quite sure what your problem is - however, you must strictly discriminate between analyses in the TIME and FREQUENCY domain.
    You have performed an AC analysis (frequency domain) and, therefore, you cannot expect to see any time dependence in the simulation result.
    Your output voltage is measured across a series LC resonant block which has an impedance minimum (zero) at the resonant frequency.
    As a consequence, the output voltage goes down to zero at this frequency.
    Very far below and above this frequency either the capacitor or the inductor, respectively, provide a very large impedance - hence, the complex impedance divider between R and the LC block exhibits (nearly) no damping and the output approaches "1".
    Please note that it is common practice to display such results using logarithmic units (dB).
    In case you want to study the time behaviour of the circuit you have two choices (TRAN analysis):
    * Response to a voltage step at the input,
    * Response to a sinusoidal signal (fixed frequency) .
  4. Feb 2, 2015 #3
    Simulations won't help you if you don't understand anything about theory behind them.
    Theory in this case is the application of the phasors in a steady state AC circuits analysis.
    How quickly will you learn this depends on the math background you have.
    http://www.ece.msstate.edu/~donohoe/ece3183ac_analysis.pdf [Broken]
    Last edited by a moderator: May 7, 2017
  5. Feb 3, 2015 #4
    Thanks a lot for the response. I now understand much better. Do you know why its common pratice to display results using logarithmic units (db9?
  6. Feb 3, 2015 #5
    Thanks. I will read the PDF. I have learn some stedy state before, but its getting more complex now, with different frequency. Im sure its a good ide to read it over again then.
    Last edited by a moderator: May 7, 2017
  7. Feb 3, 2015 #6


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    Your question concerns the so called BODE diagram. Here, the magnitude is displayed in dB and the phase in deg - both over a log. spaced frequency axis.
    This has many advantages. One adavantage is that it is very easy to verify the relation between amplitude (magnitude) and phase. I recommend to read some book chapters or other contributions using the keyword "Bode diagram".
  8. Feb 3, 2015 #7


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    The main reason for using a dB scale is that you can cover a huge range of values. This is along the same lines as the use of 'Standard Form' - e.g. 2.34 X1012, which avoids writing masses and masses of zeros (2340000000000). The dB scale is similarly based on the base ten logarithm of a number. Confusing until you get so used to it that you can't do without it.
  9. Feb 5, 2015 #8


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    the graph you have of Vout is the steady state magnitude and phase of the signal Vout in your circuit vs frequency. if you did a time analysis at various frequencies, you would find that there would be some delay before the outut mag and phase settled at their steady state value

    that means that with a 1k frequency input, your output will be one value after some time. If you were to use a signal of a different frequency this value will be different.

    Lets look at the circuit. at DC, current will not move through a capacitor, so it makes sense that Vout will equal the voltage input (zero current through the resistor, so zero voltage drop)
    at daylight ("inf" frequency) current will not move through an inductor, so it makes sense that Vout will equal the voltage input (zero current through the resistor, so zero voltage drop)
  10. Feb 6, 2015 #9
    Thanks for a very good explanation. That makes so much sense now
    Thanks for a good, simple explanation. That makes so more sense now.
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