Can you check my transfer functions?

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fouadalnoor

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Hello guys, I just tried to answer the questions attached...

I have also attached the solutions for part a) but I got a negative answer for Wo^2, am I doing anything wrong?

Thanks for your input

Fouad.
 

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I have done the H(jw) only.

It should give you a starting point.
 

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ljcox said:
I have done the H(jw) only.

It should give you a starting point.
Click to expand...

Thanks for the reply!

Umm I did get that result. If you take a look at the Solutions PDF you'll see that when I try to re-arrange it for wo then I get stuck because wo^2 = -4.142x10^6.

In order to get w0 we do: |H(jwo)| = 1/sqrt(2) *|H(jw)|max

when w = 0 then |H(jw)|max = 1

Thus: |H(jwo)| = 1/sqrt(2)

and now solving for wo (as shown on your equation):

1/(1-Wo^2LC) = 1/sqrt(2)

which means

Wo^2 = -4.142x10^6.

And thus my problem...

Hope you can help!
 
emm you sure?

If we do this:

H(jw) = Zc/Zc+ZL

H(jw) = 1/1+(ZL/Zc)

And since ZL = jwL and Zc = 1/jwC (same as -j/wC)

then ZL/Zc = -w^2LC

So H(jw) = 1/1-w^2LC

same as before? or did I go wrong somewhere with my signs?
 
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Yes, you're right.

I re-checked it this morning & found it is a negative.

BTW jwL & -j/wC are reactances, not impedances.

I'll now re-read your .pdf.

I did a simulation see attachments.
 

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I put a 100 Ohm resistor in series with the inductor and the simulation is attached.

I have now done the maths for this case. The phase angle agrees with the simulation.

But I have only done it at ω = 0, ω = 1/LC (reasonance) and as ω > ∞.

It is a long time since I did jω calculations, so I'm a bit rusty.

I hope this helps. You can set r = 0 for your situation.
 

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Hi there,


Here's what i got:

H(jw)=1/(j*w*C*R-w^2*C*L+1)

Amplitude:
Ampl=1/sqrt(w^2*C^2*R^2+(1-w^2*C*L)^2)

Phase Angle:
ph1=taninv(0/1)=0
ph2=taninv((w*C*R)/(1-w^2*C*L))
ph=ph1-ph2=-taninv((w*C*R)/(1-w^2*C*L))
This equation for phase however requires watching out for what quadrant the argument for taninv is in, and so it really requires the two argument inverse tangent function:
ph=-taninv(w*C*R, 1-w^2*C*L) [note two arguments for this inverse tangent function taninv(y,x)]
 
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Hello,```

It's been ages I didn't work with those ..

If there's any incorrect technical terms or something, feel free to correct me.

The attenuation is 20Log(YouKnowTheDrill)..

EDIT: To be read in the order 1, 3, 2, 4, 5, 6

All my best,
 

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Here is my version and I have inserted some comments & questions on your 4.jpg.
 

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Hello back,

The answer is really simple.

It is 1/(1-10^2*10^-9*10^6) , right ? .. This becomes 1/(1-10^-1), right ?


What happens if you multiply the Num and Denum by 10 ?

It becomes 10/(10-1)= 10/9. Voilà.


You can do it anyway, really.. Your version is obviously correct..

I like putting results as nice clean fractions, and only calculate their value at the moment where I need the "value".

This way, anything that can be simplified, like 9/2 * 2*220 will be simplified so you don't use your calculator unless you "have to"..calculation time is diminshed.. ERRORS are diminished both in involontary mistakes AND in error propagation.

I didn't get back to you promptly, I am studying hardcore right now for the final exam (I missed the hole year, and only have 3 weeks to go through the hole program of 6 courses.. I'm doing 12 to 13 hours a day PURE studying now No resting inbetween + No eating or drinking.. Ooohaa.

So any further questions, I'll check my mail in a week or so..

Good luck,
 

Yes that is the way I do it also.

Often, you don't need a calculator.
 
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