[SOLVED] center frequency of RLC band pass fillter

[greenjuice]

There are series and parallel RLC band pass fillter, as you know.

In series RLC band pass fillter, as frequency gets bigger, size of inductor's and

capacitor's impedance are same but their polarity is different at special point.

That special point is center frequency

because sum of inductor's and capacitor's impedance is zero. so

resistance have all input voltage.

But I don't know why parallel RLC band pass fillter has center frequency at point

at which size of inductor's and capacitor's impedance are same.

Please explain why this happen conceptually.

Thank you

 

[doraemon]

Hello!

First the intuitive part of things:

If you consider the parallel LC assocuation I guess you can agree that at DC, the
L part will be a short circuit (if we assume its resistance to be 9).
-> At that point, the output is 0
I think you can also agree that at high frequencies, C will act as a short circuit,
-> At that point also the output is 0.
In between, there will be a point where the response is maximal.
Where is it? You may remember that the impedance of the coil is L omega or 2 pi f l.
You may also remember that the impedance of the capacitor is 1/(C omega) = 1/2pifc.
Now I also suppose you know that the parallel association of 2 impedances z1 and z2
gives you an equivalent impedance z such as 1/z = 1/z1 + 1/z2.

If you replace this by L and C impedances, you will find the equivalent impedance
as a function of omega.

Now you the response of your filter will be maximal where this impedance is maximal,
therefore where 1/z is minimal.

The method is then to calculate the derivative of 1/z and try to find where this derivative
goes to 0. At this point, if the derivative is 0, mathematically it means that you have either
a max or a min or your function. The slope is 0, it can therefore be at the top or at the bottom
of a shell-like curve.

Now you have the method, it's your turn to do your homework.

Dora.

 

[jiripolivka]

The simplest way to get the BPF center frequency is to tune each of the two (or more) LC resonators to the desired frequency, then adjust their coupling. For under-critical coupling the center frequency stays as for the single LC resonator, for over-critical coupling the frequency response splits.

 

[KlausST]

Hi,

Another point of view...

See the bottom picture. All three devices are in series all three devices see the same current.
With low frequencies the impedance of the capacitor is high, thus low current flow.
With increasing frequencies the impedance of the capacitor decreases and thus the current increases
Now see it from the opposite side of frequency
With high frequencies the impedance if the inductance is high, thus the current is low.
With decreasing frequency the impedance of the inductance decreases and thus the current increases.

If you draw a chart with frequency on the x axis and current on the y axis, then you see on the right and left side low current. Both increasing to the center.
So it is obvious that there must be a maximum current inbetween.
Now the voltage across the capacitor is 90° lagging with respect to the current.
But the voltage across the inductance is 90° leading.
So in total they are 180° to each other. In other words they compensate each other.
To fully compensate, the amplitude of both voltages need to be the same.
With the same current it means the impedance of both impedances must be the same size.

Now you can use mathematics to solve the formula where you set both impedances to be equal.
You find resonance frequency. It is the frequency, where you have least overall impedance, maximum current.

Maybe Xc = XL = 1000 ohms, R = 100 ohms. Uin = 1 V rms.
Xc and XL compensate each other resulting in 0 Ohms in combination. Remaining the 100 Ohms.
Current = 1V/100 Ohms = 10mA.
But the same current is flowing through C and L, and causes a voltage of 1000 Ohms x 10mA = 10 V.
A typical resonance: With only 1V input you will see 10V at both L and C.

Klaus

 

[greenjuice]

To doraemon,

Um. At parallel BPF,

I know parallel association of impedances(inductor, capacitor). And when association is maximal,

response of BPF is maximal.

I want to know what happened at that maximal point such as interaction of inductor and capacitor.

When we find center frequency at BPF, first we find transfer function H(s). And we find Hmax(H(s)=1)

last H(jw)=1 <- w is center frequency. so we take it.

I think,
'H(s)=1' is 'Vout/vin =1'. so Vout = Vin at center frequency and resistance has no voltage at that
frequency because parallel association of inductor and capacitor has all voltage, Vin.

Is it right? I am not sure my analysis.

I will waiting your reply. Thank you.

- - - Updated - - -
To KlausST,

Your view is very helpful! I can understand why series BPF has center frequency at which impedances of

inductor and capacitor are same.

And I think at parallel BPF, using your analysis, inductor and capacitor have same voltage.

So current across the inductor is 90 degree lagging with respect to the voltage
and current across the capacitor is 90 degree leading with respect to the voltage.

So current directions of inductor and capacitor are different each other.

It means if it is ideal case(no energy loss), current circulates infinitely in LC circuit.

is it right?

if then, what determines maximal voltage in LC circuit at center frequency.

Thank you for your advise and I'll wait your reply.

- - - Updated - - -
To jiripolivka,

yes. parallel BPF has LC resonator at center frequency.

But I can't understand adjusting their coupling and critical coupling.

Could you explain more detail?

 

[jiripolivka]

To doraemon,

Um. At parallel BPF,

I know parallel association of impedances(inductor, capacitor). And when association is maximal,

response of BPF is maximal.

I want to know what happened at that maximal point such as interaction of inductor and capacitor.

When we find center frequency at BPF, first we find transfer function H(s). And we find Hmax(H(s)=1)

last H(jw)=1 <- w is center frequency. so we take it.

I think,
'H(s)=1' is 'Vout/vin =1'. so Vout = Vin at center frequency and resistance has no voltage at that
frequency because parallel association of inductor and capacitor has all voltage, Vin.

Is it right? I am not sure my analysis.

I will waiting your reply. Thank you.

- - - Updated - - -
To KlausST,

Your view is very helpful! I can understand why series BPF has center frequency at which impedances of

inductor and capacitor are same.

And I think at parallel BPF, using your analysis, inductor and capacitor have same voltage.

So current across the inductor is 90 degree lagging with respect to the voltage
and current across the capacitor is 90 degree leading with respect to the voltage.

So current directions of inductor and capacitor are different each other.

It means if it is ideal case(no energy loss), current circulates infinitely in LC circuit.

is it right?

if then, what determines maximal voltage in LC circuit at center frequency.

Thank you for your advise and I'll wait your reply.

- - - Updated - - -
To jiripolivka,

yes. parallel BPF has LC resonator at center frequency.

But I can't understand adjusting their coupling and critical coupling.

Could you explain more detail?

Dear friend,

coupling is described in detail in good textbook on flters. There are many kinds of coupling circuits and many design methods. My best advice is to read the theory, then make experiments with real LC circuits. You need a sweep signal generator and a detector with an oscilloscope. Experiments are simple and the result is that you learn much more than by models and simulations.
As a test circuit you can use any old AM receiver IF band-pass filter, or an UHF TV tuner. You can find good instruments at a school laboratory.

 

[KlausST]

Hi

To KlausST,

Your view is very helpful! I can understand why series BPF has center frequency at which impedances of

inductor and capacitor are same.

And I think at parallel BPF, using your analysis, inductor and capacitor have same voltage.

So current across the inductor is 90 degree lagging with respect to the voltage
and current across the capacitor is 90 degree leading with respect to the voltage.

So current directions of inductor and capacitor are different each other.

It means if it is ideal case(no energy loss), current circulates infinitely in LC circuit.

is it right?

if then, what determines maximal voltage in LC circuit at center frequency.

Thank you for your advise and I'll wait your reply.

Consider both currents in C and L compensate each other to zero. Then there is no current flow. No current means no voltage drop across R. Then the output voltage is the same as input voltage.
In any other case there will be attenuation. Both frequency ends on a diagram will show zero output voltage.

***

Do you have a simulation program? I think in (free) LTspice it should be possible to simulate this.

Klaus