Transfer Function - RLC Circuit

Hi,

I am stuck on this question...



The current source is throwing me off this, and I do not know exactly how to deal with it.

My answer seems very wrong -> V(0)/i(g) = Zr+Zl, so a hand would be greatly appreciated, as I am really stuck.

Thanks.

You are on the right track but have missed out the parallel combination with Z_c

Also note that V_o = V_g and from KCL, I_g = I_c + (I_RL)

I knew my answer is incorrect, as it does not make any sense.

If it was an ordinary voltage source instead of current source, you would have:

V(o)/V(i)= (Zl+Zr)/( (Zl+Zr) || Zc )

wouldn't you?

In this case, you must apply Thevenin/Norton theorems to convert the current source to a voltage source.

This means replacing the parallel impedance Z with an equivalent Thevenin series impedance Z_th and changing the current source to a voltage source.

V_o/I_g = Z then becomes equivalent to the series impedance Z_th for V_o/V_in.

For the circuit given, this is very simply achieved since the output voltage is across the whole of the load. So simply put the series/parallel combination in series with the source and replace I_g with V_g. Note that Z = Z_th in this problem.

Now work the transfer function as you would for H_(omega) = V_o/V_in.

I kind of understand what you are saying (I think); I will list my steps:

As you said, I did the source conversion, source I(g) converts into a Voltage source V(g) with series (source) impedance of Z(th), whereby Z(th):

Z(th) = Z(c) || ( Z(r) + Z(l) ).

So after this conversion, doing the transfer function as normal:

V(o)/V(g) = Z(s)/( Z(s) + Z(th) ) ---> where Z(s) is the total impedance of the original circuit, which is also equal to Z(th)...

So the right hand side would simply cancel down to 1/2.

Correct up to here? I hope its not completely incorrect...

Actually, I think it's me that led you down a wrong path. Sorry!!

Thinking about it, the answer is staring us in the face!

V_(o)/I_(g) = Z where Z is the total impedance of the LCR network.

Z = Z_(c) || ( R + Z_(L) ).

Z = Z_(c)(R + Z_(L)) / R + Z_(c) + Z_(L)

Z = [RZ_(c) + Z_(c)Z_(L)] / [R + Z_(c) + Z_(L)]

Where: Z_(c) = 1/jωC; Z_(L) = jωL

Finally substitute values for C = 0.1F; L = 2H and R=4 ohms and reduce.

Haha, that's alright mate, thanks for your help nonetheless, you are always the first (and only) person to reply to my threads anyway really!! Much appreciated!

If you look at my last post, and follow on from there, it leads to the same answer you got. So your answer is probably right - but it just doesn't seem right? (gut feeling)

How certain are you that this is correct?

I'm certain this is correct but it does not lead to the answer you gave. Your reduction to a scalar is incorrect because the expression is an impedance and therefore has magnitude and phase frequency dependence.

Don't forget that the RLC circuit has reactive components. The impedance comprises real and imaginary parts: (R_eq + jX_eq) and |Z|ej^Φ

I will have a look at deriving the full answer but it may not be for a few hours when I get some time later today.

Thank you so much for your help mate.

Your answer is definitely correct. Why?

i(g) = i(1) + i(2)

(1) i(1) = V(o)/Z(c)
(2) i(2) = V(o)/(Z(r)+Z(l))

And substitute 1 & 2 into the first equation. That leads to your answer... I am not sure if there is any other method