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Circuits

06/24/2015 5:56 PM

Determine the transfer function in terms of R, and L (in addition to the frequency) and the type of the filter Find the limits of the magnitude of the transfer function for very low and very high frequencies.
Determine the cutoff frequency

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#1

Re: circuits

06/24/2015 5:58 PM

Find the information in your textbook, we don't do homework here.

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#3
In reply to #1

Re: circuits

06/24/2015 6:19 PM

Well i tried and yhis is where am lost ((1/3)+(jlw/3R))/(1+jlw2/(3R)) How to continue

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#5
In reply to #3

Re: circuits

06/24/2015 6:47 PM

Ok, so you've done some work, can you explain what that equation tells you?

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#13
In reply to #5

Re: circuits

06/26/2015 12:00 PM

well if i continue with my equation ill have a high pass filter because as w >0 hv> 1/3 and as w>infinity hv> max =1 but my doughts are about the equation im used to jww"/(1+jw/w") or k/(1+jw/w") how to transform my equation to one of the preceding form?

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#2

Re: Circuits

06/24/2015 6:07 PM
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#4

Re: Circuits

06/24/2015 6:36 PM
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#6

Re: Circuits

06/24/2015 6:51 PM

wheres the zener?

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#7

Re: Circuits

06/24/2015 7:18 PM

Try some internet homework help sites.

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#8

Re: Circuits

06/24/2015 8:09 PM

Only because you have already shown some effort here...

As ω→∞ what do you expect this network transfer to approach?

As ω→0 what do you expect this network transfer to approach?

I will give you one freebie here. There is only one frequency dependent component so at most this will be a single pole configuration but not necessarily only a high pass or low pass filter.

What parts of this circuit resemble a high or low pass circuit and if either what will be the corner frequency?

Once you've correctly answered these questions, you should be able to sketch a Bode transfer function.

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#9

Re: Circuits

06/25/2015 10:27 PM

Why don't YOU determine it?

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#10

Re: Circuits

06/26/2015 2:54 AM

Us helping you will mean probably that you really won't understand it. If you did not understand it in the classroom, with a tutor who should answer your questions (did you even ask?), you have less chance to understand it here....

Not fully understanding the problem, will leave a small but important "hole" in your overall knowledge.

But as you progress further, you will usually find that "hole" will affect more and more of your understanding.

It could at least cause your pass mark to be seriously reduced......or to even fail you....

In your position, I would approach your tutor/professor for extra coaching lessons, not go to CR4.....

You appear to have only joined CR4 for that one and only reason.......which I also find to be negative for both you and us......

Go and learn it properly......

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#11

Re: Circuits

06/26/2015 11:37 AM

Normally I try to break the terms down to (1+j*w*xyz) so you can easily pull out the corner frequencies (as w becomes very large the j term dominates, the corner is at 1/xyz, with w very small the 1 dominates). Note you may also have terms that are second order - but try and put it into standard format anyways.

Barring that - use brute force and select frequencies to sketch (Bode plot) your function with magnitude and phase shift. It is often beneficial to do the numerator and denominator separately so you can see the poles and zeroes.

I learnt from an old book - Lathi - Signals, Systems and Controls - the appendixes are excellent aid. There are many other good books.

You need to get into the habit of arranging your equations into standard format so you can see the standard terms or equations.

If you show the development of how you got your final equation it can also be beneficial to you - especially if you are being marked.

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#14
In reply to #11

Re: Circuits

06/26/2015 12:06 PM

well if i continue with my equation ill have a high pass filter because as w >0 hv> 1/3 and as w>infinity hv> max =1 but my doughts are about the equation im used to jww"/(1+jw/w") or k/(1+jw/w") how to transform my equation to one of the preceding form?

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Guru

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#15
In reply to #14

Re: Circuits

06/26/2015 12:19 PM

Your familiar forms don't fit.

put the numerator and denominator into a familiar form, instead of the whole equation.

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#12

Re: Circuits

06/26/2015 11:43 AM

Just another thought to help you visualize - you know at very low frequencies the L is essentially a short - so calculate the transfer with the L shorted.

At very high frequencies the L is open circuit - repeat first step.

So now all you need to figure is what happens through a very narrow range of frequencies between those two limits.

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#16

Re: Circuits

06/26/2015 1:28 PM

Try this:

Zs=R||(R+L*s)=R*(R+L*s)/(2*R+L*s)

(Vin-Vout)/R=Vout/Zs

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