I am working on series resonant frequency and have been given an equation to compare with my results and have no idea how to use it, I haven't found it anywhere.

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1 + jQ [f/f°-f°/f]

I believe I can compare this to the results of Vout / Vin?

I think this is a transfer function (H) and should be equal?
it's the j that has me confused.

 

The lower case 'j' is used as the imaginary unit. It is nominally defined as the square root of negative 1. Alternatively it is the solution to:

\(x^2 + 1 = 0\)

 

I see, thanks for your speedy response.

Now I am to compare Vout/Vin with the answer I get from that equation.

In my case,

Vin = 100Vrms
Vout = 1.425 Vpeak (or divide by the square root of 2 to get Vrms)
Vout = 1.00Vrms

The frequency was 80Hz (although I tested many)
and the f0 (series resonant frequency) is 1591.55Hz

I feel like I have everything to plug into the formula and just can't do it!

 

You need to determine the Q of your circuit. Q is loosely defined as quiality of the tank (old term). If you were given the equation you posted, you should have been given the equation to calculate Q as well. Look in your book or notes and you will find it. From there you will have the information you need to plug into the equation. You also need to understand how the j modifies the phase of the vector of voltage or current.

 

Oh yes I forgot

The Q factor (which I believe is voltage magnification?) = 5

You are spot on though, I do need some help to understand how the j modifies the phase of the vector of voltage or current.

Vout / Vin = is giving me 0.01 which is a ratio of some kind?

I measured Peak voltage from an oscilloscope, does it have to be converted into RMS for the above formula?

If someone can walk me through the formula I don't understand once, then I will be able to repeat it for all my other results.

I'm new to the forum as you can see and i'm really impressed at how helpful people have been already! Thanks

 

As was stated j is the square root of -1; this at first makes no sense as to why it should be in an equation of electronics. However, devices as caps, or inductors provide a delayed effect on voltages and currents. In DC the voltages and currents are in line with each other, in AC as current travels through the devices the voltages and currents become out of sync, it is still there just that voltages can lead or lag the current by an angle (phasor). So if you set up a scope to look at the voltage at a certain point and the current at that same point, their peak value (instantaneous values) do not occur on the same vertical line on the scope.
At resonance the value associated with the j approaches 0 and our voltages and currents line up.
Walking you through the equation will not help you understand the point of the exercise. Look at en.wikipedia.org/wiki/Phasor for some insight into the shift the non - linear energy propogation through the circuit produces.

 

Thanks for your help, i'll let you know how I get on