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Analysis

I'm trying to find the resonant frequency for this circuit

schematic

simulate this circuit – Schematic created using CircuitLab

Writing up the node voltage equation for \$V_o \$ $$\frac{V_o-V_{in}}{Z_L}+\frac{V_o}{Z_C} + \frac{V_o}{R}=0 $$ And using that \$\frac{V_o-V_{in}}{Z_L}= \frac{1}{L} \cdot \displaystyle\int (V_o-V_{in} )\: dt \$ and \$\frac{V_o}{Z_C}=C \cdot \frac{dV_o}{dt} \$ brings us $$C \cdot \frac{dV_o}{dt} + \frac{1}{L} \cdot \displaystyle\int (V_o-V_{in} )\: dt + \frac{V_o}{R}=0$$ Dividing through with \$C \$, differentiating every term and moving \$V_{in} \$ to the right hand side gives me $$\frac{d^2V_o}{dt}+ \frac{1}{RC} \frac{dV_o}{dt} + \frac{1}{LC}V_o = \frac{1}{LC} V_{in}$$

Calculations

According to "Eletrical Engineering principles and applications by Hambley", the square root of the term before \$V_o \$ is called the undamped resonant frequency \$\omega_0 \$.

In this case the resonant frequency is $$\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{62 \text{uH} \cdot 63 \text{nF}}} = 0.5059 \: \text{MHz}$$ Also according to Hambley, at the resonance frequency the equivalent circuit impedance is purely resistive, so \$\Im{(Z_{eq})} = 0 \$.

The equivalent impedance of this circuit is $$Z_{eq} = Z_L + \frac{R \cdot Z_C}{R + Z_C} = sL + \frac{R}{sC(R+ \frac{1}{sC})}$$ Plugging \$s= j\omega_0 \$ and plugging in component value into the above equation gives me $$Z_{eq} = 15.14 + j11.57 \Omega$$

Question

Which clearly shows that the impedance isn't purely resistive. So my question is, why not? Is my equivalent impedance wrong, or perhaps my resonance frequency?

Edit

Source about resonant frequency

enter image description here

Figure 6.23

enter image description here

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  • \$\begingroup\$ Did not get into the details of your derivation. However, 1/SQRT(LC) is correct for series RLC or parallel RLC. Not sure if you can still use this formula as your circuit is a combination of both R||C in series with L. \$\endgroup\$
    – Wintermute
    Commented Feb 22, 2021 at 17:25
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    \$\begingroup\$ The circuit on the page is different from the circuit you posted. I guess this has something to do with the discrepancies. \$\endgroup\$ Commented Feb 22, 2021 at 18:32
  • \$\begingroup\$ @SredniVashtar Yeah you are probably right. But the way he wrote it just confuses me. "The resonant frequency is defined to be the frequency at which the impedance is purely resistive". So, is it only defined for this RLC circuit, or for every RLC circuit? \$\endgroup\$
    – Carl
    Commented Feb 22, 2021 at 18:38
  • \$\begingroup\$ The problem with how many textbooks treat resonance is that they usually consider only the two simple situations of series RLC and parallel RLC. Yours is neither. And as you can see, the frequency at which the impedance has an extremum, the frequency at which the impedance is real, and the frequency at which XL = XC are all different. In a series RLC circuit (the one on the page) the last two freqs are the same and the first tend to them for R->0. In your circuit R->0 will leave you with an inductor alone. O.t.o.h, R->infinity will make all frequencies converge and leave an ideal series LC. \$\endgroup\$ Commented Feb 22, 2021 at 23:28

1 Answer 1

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I'm trying to find the resonant frequency for this circuit

Try this calculator. I spent a lot of time getting it right LOL: -

enter image description here

The natural resonant frequency you calculated is in radians per second by the way. In hertz it is 80.52932 kHz.

Also according to Hambley, at the resonance frequency the equivalent circuit impedance is purely resistive

That isn't true from what I can tell.... If you look at this impedance matching calculator on the same basic website it shows at what frequency the input will be purely resistive: -

enter image description here

I've had to frig around to make the numbers match about right with the first calculator but, the upshot of what it is telling you is that the frequency where the input impedance is purely resistive is 50.63 kHz. And, at that frequency, the input resistance is 24.79 Ω.

There are full derivations on that page.

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  • \$\begingroup\$ Wow, that's actually a really useful you are linking to, I'll bookmark it so I don't forget it. Yeah you are right \$f_n = \frac{\omega_0}{2\pi}=\frac{0.5059 \text{MHz}}{2\pi}=80.529 \text{kHz} \$ so it seems my calculation of that is correct. However, can you explain why the equivalent impedance is not purely resistive at this frequency? \$\endgroup\$
    – Carl
    Commented Feb 22, 2021 at 18:18
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    \$\begingroup\$ @Carl that's the bit I'm trying to figure out. I know it isn't so give me a little time on that bit. \$\endgroup\$
    – Andy aka
    Commented Feb 22, 2021 at 18:19
  • \$\begingroup\$ Alright, thanks for clearing up, Andy - this has really helped me. I now realize that I misused the information from Hambley, I won't do that again. One last question though. Is the general way of finding the resonance frequency setting up the differential equation as I did in my question, and then looking at the term in front of \$V_o \$ or is there an alternative (aside from that handy calculator you linked to)? \$\endgroup\$
    – Carl
    Commented Feb 22, 2021 at 19:06
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    \$\begingroup\$ @Carl I'd solve it directly by using Laplace terms then manipulate the transfer function like on the website I linked. I don't bother starting with a differential equation. \$\endgroup\$
    – Andy aka
    Commented Feb 22, 2021 at 19:30

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