I need to find the transfer function of this RL circuit. I've tried to figure it out but I think something went wrong. Any help would be apprecciated!

 

So far everything looks good. What do you expect to get?

 

The problem is that when I plot the transfer function I get the following Bode diagrams
so the circuit is a high-pass filter but the position of the R and L components is different from the classic RL high-pass filter.

Am I wrong?

So far everything looks good. What do you expect to get?

 

hi vmm,
Have you considered that the R2 value maybe the internal resistance of the Inductor.?

What do you consider is wrong with the Bode Plot.?

E

 

Thanks for the tip! I didn't think about that! So can I assume that this is an ordinary RL high-pass filter and calculate the cutoff frequency, right?

hi vmm,
Have you considered that the R2 value maybe the internal resistance of the Inductor.?

What do consider is wrong with the Bode Plot.?

E

 

hi,
If the Inductor has a known resistance, then the impedance of the total R & L should be included, as the Vout signal is across this total impedance, not just the inductor impedance.

E

What are the given values of R1 , R2 and L.??

 

hi,
Compare these two plots, one with Inductor resistance and one with no inductor resistance.

The Transfer functions must be different for the two examples.???
E

 

Yes, the transfer function is different (as matter of fact I've got a different function) but both are high pass filter because both pass all frequencies above the cuttoff frequency. Am I right?

hi,
Compare these two plots, one with Inductor resistance and one with no inductor resistance.

The Transfer functions must be different for the two examples.???
EView attachment 242021

 

Just by looking at your schematic and transfer function, we can see that DC gain is:

K_dc = R2/(R1 + R2)

And we will have one Zero and one Pole. And from what I see you decided to use R1= R2 = 10 Ω and L = 1H and this is why you have a DC gain at -6dB (0.5 V/V) and Fc =√(Fz*Fp) = 2.25Hz (due to the fact that the zero and pole are no well spread apart from each other).

 

Just a quick note....

For a single RL or RC series circuit possibly also with resistive load...

When the inductor is in series with the output the response will be low pass.
When the inductor is in parallel with the output the response will be high pass.
Since the capacitor is the dual of the inductor, the opposite is true:
When the cap is in series with the output the response is high pass, and
when the cap is in parallel with the output the response will be low pass.
The main reason for this is because of the way the impedance changes for each element as the frequency changes. You should probably graph that also so you get a feel for what is happening.

You can use these ideas to get a sanity check of your computed response. it's more qualitative than quantitative but helps as a quick check.