Transfer Function of LRC Filter

In summary: I am looking for the transfer function of thevenin equivalent. I am not really sure what you are trying to find. Generally, I think it is easiest to take the circuit from the time domain and redraw in the "s" frequency domain with Laplace transform. Then you solve for what ever with regular techniques, and take the inverse transform so you are back in the t domain.
  • #1
mrjeffy321
Science Advisor
877
1
A circuit was constructed with an inductor, capacitor, and resistor all wired in series. A sinusoidal input signal was put through the circuit and the output voltage was measured across the resistor as a function of the input angular frequency (ω).

My task is to find the transfer function of this circuit….to find the ratio of the output voltage to the input voltage.

The inductor and capacitor has frequency dependent impedances, but the resistor’s impedance is constant for all frequencies.
Since all the circuit components are wires in series, the total impedance of the circuit is simply the sum of each component’s impedance.

Impedance…
of resistor = R,
of capacitor = -j / ωC,
of inductor = jωL,

The transfer function cab be found as,
R / (R + jωL + -j / ωC)

By multiply the denominator by its complex conjugate in order to remove the imaginary parts from the bottom of the fraction, we get,
R [R - j(ωL – 1/(ωC))] / (R^2 + (ωL – 1/(ωC))^2
If we say that X = R and Y = (ωL – 1/(ωC)), then we can re-write this as,
(X^2 + jXY) / (X^2 + Y^2)
And we can divide everything by X to get,
(X + jY) / (X + Y^2 / X)
Then to get the magnitude of the output voltage, we need to find the magnitude of the numerator,
sqrt (X^2 + Y^2) / (X + Y^2 / X)

But when I graph this transfer function on the same graph with my actual data points (on a logarithmic scale), I do not get anything which comes even close.
My theoretical values from the above function give me a downward sloping line, but my data points (and knowledge of what should happen) show an increasing function with some peak value when then decreases back down towards zero again. Since the impedances of the capacitor and inductor are frequency dependent, there will be some frequency (the resonance frequency) where the two should cancel each other out and the only impedance at this frequency should be that of the resistor (yields maximum power). But this is not happening in my theoretical function.

Where did I go wrong?

I was told I am aiming to get a transfer function which looks like,
(X – jY) / (1 + Y^2)
For some values of X and Y.


EDIT:

If I keep playing with the equation, I can simplify it down to:
|H(ω)| = R / sqrt (R^2 + (ωL – 1/ωC)^2)

But this just gives me the same graph I had before, which looks nothing like my data points or how I know the graph should behave.
 
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  • #2
So you are looking for the transfer of the thevenin equivalent? I am not really sure what your method is, or what you are trying to find. Generally, I think it is easiest to take the circuit from the time domain and redraw in the "s" frequency domain with Laplace transform. Then you solve for what ever with regular techniques, and take the inverse transform so you are back in the t domain.
 
  • #3
I think you meant
(X^2 - jXY) / (X^2 + Y^2)
not
(X^2 + jXY) / (X^2 + Y^2)
but fixing that would only change the phase of your transfer function not the amplitude. Apart from that it looks right to me.

Two things to check:

1. Have you got consistent units for R L and C?

2. Check the resonant frequency when ωL = 1/ωC is what you expected it to be. (ω is in rad/sec, not Hz, of course).
 
  • #4
I am trying to find the transfer function in terms of the angular frequency of the input signal across the circuit.

How I found what I think to be the transfer function was to first find the voltage across the capacitor (this is the output voltage), then find the voltage over the entire circuit. The voltage across the resistor (say it has a resistance, R) is R * I, for some current I. The voltage over the entire circuit, for that same current I, is I * (R + jωL – j/ωC).
The transfer function is:
H(ω) = V_out / V_in = (I * R) / (I * (R + jωL – j/ωC))
The current, I, cancels out, leaving me with just a ratio of the impedances,
H(ω) = R / (R + jωL – j/ωC)

If I multiply the bottom by the complex conjugate, I get
H(ω) = R * [R - j(ωL – 1/(ωC))] / (R^2 + (ωL – 1/(ωC))^2

This would be the overall transfer function, taking into account both the Real and Imaginary parts. But now I want to graph this transfer function on the same axis I am graphing my actual data points collected as a function of frequency (Hz), so I will need to convert this into a magnitude.

Finding |H(ω)|, I get,
|H(ω)| = R / sqrt (R^2 + (ωL – 1/ωC)^2)

But when I graph this, I am getting unexpected results.
I am not sure if I am doing something wrong in the derivation or not.
Does my H(ω) look correct? I am inclined to think that it is not since in the lab it hints at finding it in the form of (X + jY) / (1 + Y^2) for X and Y as some constants and/or functions of known values of L, R, and/or C.

EDIT:
AlephZero said:
I think you meant
(X^2 - jXY) / (X^2 + Y^2)
not
(X^2 + jXY) / (X^2 + Y^2)
but fixing that would only change the phase of your transfer function not the amplitude. Apart from that it looks right to me.

Two things to check:

1. Have you got consistent units for R L and C?

2. Check the resonant frequency when ωL = 1/ωC is what you expected it to be. (ω is in rad/sec, not Hz, of course).
Yes, it looks like I accidentally let a + slip in where a - should be.

Unit wise, I think everything looks good. I have Ohms on top and Ohms - Ohms on bottom, making the overall thing unitless, as would be expected.

In my data, there is a pretty clear resonance frequency spike in the output voltage. In my equation,
|H(ω)| = R / sqrt (R^2 + (ωL – 1/ωC)^2)
there is also, theoretically, a maximum when ωL – 1/ωC since that would make |H(ω)| = 1 and at all other values for ω, |H(ω)| < 1.
 
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  • #5
I think I might have been right and not have known it.
I kept playing around with it and re-graphed it. Now I am getting something much more reasonable with a peak power in the right place and everything.

Now the only problem is that, although the general shape of the line corresponds to my data, the actual values are quite different and the experimental resonance frequency is several thousand Hertz away from the theoretical resonance frequency.
What might account for this discrepancy? It is not a small amount, almost 150 kHz difference?
Also, even at the experimental resonance frequency, my voltage output is only on the order of ½ its theoretical value; the experimental transfer function never gets to 1, it stops at about 0.53, what might be the cause of this. I am just trying to figure out how my measurements went so far off the mark…..these are no small errors here.
 
  • #6
Apart from just making a mistake with the measurements, one likely reason is the difference between real and ideal components.

Real inductors have resistance, and the inductance can vary with frequency and the current passing through the inductor.

Capacitors and resistors also have component tolerances of course, but apart from that they tend to behave closer to ideal components than inductors do.
 

1. What is a transfer function?

A transfer function is a mathematical representation of the relationship between the input and output of a system. In the context of an LRC filter, the transfer function describes how the input signal is filtered and transformed into the output signal.

2. What is an LRC filter?

An LRC filter is an electrical circuit comprised of inductors (L), resistors (R), and capacitors (C) that is used to filter out specific frequencies from an input signal. It is commonly used in electronic devices to remove unwanted noise or interference from a signal.

3. How is the transfer function of an LRC filter calculated?

The transfer function of an LRC filter can be calculated using the formula H(s) = Vout(s)/Vin(s), where H(s) is the transfer function, Vout(s) is the output signal, and Vin(s) is the input signal. This formula takes into account the values of the components in the circuit, as well as the frequency of the input signal.

4. What is the significance of the transfer function in an LRC filter?

The transfer function of an LRC filter allows us to analyze and understand the behavior of the filter. It provides information about the filter's frequency response, gain, and phase shift. This information is crucial in designing and optimizing the performance of electronic circuits.

5. How does the transfer function affect the performance of an LRC filter?

The transfer function determines the frequency response of an LRC filter, which affects its ability to filter out unwanted frequencies. A higher gain at the desired frequency indicates a more effective filter. Additionally, the phase shift caused by the transfer function can impact the accuracy and timing of the output signal.

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