# Transient response second order cct

Discussion in 'Homework Help' started by Jess_88, May 4, 2011.

1. ### Jess_88 Thread Starter Member

Apr 29, 2011
174
1
Hey guys.

I'm trying to use transient anaysis to determine
1)Rise time
2)delay
3)setteling time
4)ripple factor

where R = 0.5K ohm, L = 100mH and C = 0.1uF

The only problem, is that I havent got the text book and the have been no lectures on this.
I'v been fineding it realy difficult to get any informasion on this cct trype of cct.

this is what I understand so far
I think its a band pass filter
the transfer function in H(jw) = v2/v1

I realy need some help getting started on this one.

I'v seen some formula on risetime ect but I dont understand them.

eg) rise time = 2.2 / w3dm( where w3db looks like a subscript... weired)

and what the heck is the magnitude responce???

it would be so great if I could get the lowdown on this cct.

thanks so much guys

Last edited: May 4, 2011
2. ### Jess_88 Thread Starter Member

Apr 29, 2011
174
1
ok so I have this so far

centre frequency (fo) = 1/(2pi(sqrt(LC))
=1591Hz

B/w found by
(voltage at fo)/sqrt2 = (lets call it "G")

20log(G/Vfo) = -3db

B/W = f2 - f1
= f2 - 1591H

... so how do I fined f2, V at fo

3. ### Jess_88 Thread Starter Member

Apr 29, 2011
174
1
ok, sorry for the freakout.
I think I understand a bit more now.
so this is what I have.

a = R/2L = 2500
wo = 10 000

V = L(di/dt) +Ir +Vc
io = C(dv/dt)

by second order derivative
V = CL(d2V/dt2) + CR(dv/dt) +Vc(di/dt) = C(d2V/dt2)

(d2V/dt2) + R/L(dv/dt) + Vc/LC = V/LC

0 = Vf + e^(-at)[A1coswd + A2sinwd]
wd = 9682.45t

now I know I can fined the values A1 and A2 relative to each other using differentiation but from what I understand, this "filter cct" will have a maximum voltage.

How would I go about finding the frequency at which the output voltage will be max?
From what I understand, I also want to determine frequencies at -3dB (20dB Vi/Vo) decay and -12dB decay which i'm also not sure about.

what is the significance of determining frequencies and voltages at -3dB and -12dB (what dose this indicate?)?

and what is the significance of dampening coefficient ( = (R/2)*(sqrt(C/L))).

I know its allot of questions (sorry), but it would be great if someone could explain some of this stuff.

thanks guys

4. ### Jess_88 Thread Starter Member

Apr 29, 2011
174
1
I'm having some trouble determining w1 and 12.

so far I have.

wo = 1/(sqr(LC)) = 10 000

fc = (1/(2pi*(sqr(LC)))) = 1591.54

Wpk = wo*(sqr(7/8))

Fpk = 1488Hz *(Wpk/(2*pi))

now this is my confusion.
From-
w1 = -(R/2L) + sqr[((R/2L)^2) + 1/LC] = 7.807kHz
w2 = (R/2L) + sqr[((R/2L)^2) + 1/LC] = 12.8kHz

but from my simulations.

Where did I go wrong?

cheers guys

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Your original post suggests the question relates to transient response parameters - rise time, settling time & so forth. Your last post appears to be looking at frequency response (Bode) plots. These are related but different. Are you wanting to look at frequency response or are you actually interested in transient behavior per your original post?

6. ### Jess_88 Thread Starter Member

Apr 29, 2011
174
1
I'm actually trying to do both for the same circuit. Sorry for not stating this originally.
As you can see, my understanding of both topics of minimal... I can't fined much information with regard to 2ed order frequency and time domain analysis so I'm finding it very difficult so analyse my graphs ext.

7. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
For the frequency response I would expect the peak value to occur when

$\omega=sqrt{\frac{1}{LC}-\frac{R^2}{2L^2}}$

Which gives ω=9354.14 rad/sec or 1488.76Hz

You have a peak on your frequency plot close to that frequency.

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8. ### Jess_88 Thread Starter Member

Apr 29, 2011
174
1
ok cool I can see that.
how would I go about determining w1 and w2 for the bandwidth?
This can been seen/determined from the graph right?
My calculated values for w1 = 7.807kHz and w2 = 12.8kHz look way out

9. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Firstly, the unit for ω is rads/sec not kHz. You must divide ω by 2π to convert.

Unfortunately even when converted to frequency as kHz you'll note these aren't -3dB points on the mid-band gain.

10. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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With an appreciable amount of algebraic manipulation, the upper -3dB frequency can be found from...

$\omega=sqrt{\frac{1}{LC}-\frac{R^2}{2L^2}+\frac{1}{2LC}sqrt{(8+\frac{R^4C^2}{L^2}-\frac{4R^2C}{L})}}$

Which gives an ω=14845.094 rad/sec or 2362.67 Hz

The -3dB point is off the mid-band gain of 1 [0dB], rather than the peak at 1488.76Hz. These values agree well with your graphical plot.

I can post the derivation for the equation above but I'm not sure it would be of much value.

11. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
A "purist" might argue that the -3dB point should be taken off the peak response value - not the mid-band gain of 0dB. There's some reasonable justification for this.

The peak response value at 1488.8Hz is +6.3dB. Can you prove this?
So if one searches for a solution for frequency where the gain is 3.3dB [i.e. 3dB down from 6.3dB] then one arrives at values of

f1=995Hz
f2=1855Hz

Again the solution requires working through some algebraic gymnastics.

So take your pick as to which answer better satisfies the original problem definition.

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12. ### Jess_88 Thread Starter Member

Apr 29, 2011
174
1
Do you think you could post derivation?
I'm still getting the wrong values...

****** Sorry, I just posted and I can see you have done more here... Just give me a min to work further

13. ### Jess_88 Thread Starter Member

Apr 29, 2011
174
1
ok so on my graph, f1 and f2 are taken from the dB response at 6.6dB and 3.3dB while
-3db is at 2362.67Hz right?
Also by looking at the graph there is only w1 and no w2 right?

Would I also be correct in assuming the +6.6dB is due to the given dampening ratio being 0.25?

I'm having trouble calculating these values... I keep getting values out by at least 500Hz. Could you post your calculations?

14. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Perhaps we should find some common ground ...

I have the complex frequency domain transfer function G(ω) as

$G(\omega)=\frac{1}{[(1-\omega^2 LC) +j\omega RC]}$

The transfer function magnitude value |G(ω)| is therefore

$|G(\omega)|=\frac{1}{sqrt{(1-\omega^2 LC)^2+\omega^2R^2C^2}}$

At peak response frequency f=1488.8 Hz or ω=9354.4 rads/sec

|G(ω)|=2.066 or 6.3dB

We want to find the -3dB points where G(ω)=3.29 dB (exact -3.01dB point)

A |G(ω)|dB gain of 3.29dB is equal to a |G(ω)|=1.461

We then have to solve for ω values giving

$|G(\omega)|=\frac{1}{sqrt{(1-\omega^2 LC)^2+\omega^2R^2C^2}}=1.461$

or

$(1-\omega^2 LC)^2+\omega^2R^2C^2=\frac{1}{(1.461)^2}$

15. ### Jess_88 Thread Starter Member

Apr 29, 2011
174
1
oh ok, I see now. Thanks .

Now I have one more thing to try and work out... The bode plot :S

From what I understand, it should be the same as a low pass filter.
Just a little bit confused as to the drawing of the response (as I cant create a simulation).

Is it a straight line from 0dB until the frequency cutoff, then a 6dB rise to peaking between f1 and f2, then a 60 degree fall?

16. ### t_n_k AAC Fanatic!

Mar 6, 2009
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The gain roll-off will approach 40dB per decade rather than 60dB per decade.

You've shown a partial Bode magnitude plot in your earlier post ...???

17. ### t_n_k AAC Fanatic!

Mar 6, 2009
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The Bode plot looks like this ....

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18. ### Jess_88 Thread Starter Member

Apr 29, 2011
174
1
Oh thanks thats great! .

19. ### Jess_88 Thread Starter Member

Apr 29, 2011
174
1
Sorry, where did I show a partial Bode magnitude plot???
Would this be from my image earlier posted in this thread?
Thanks for bringing this to my attention... I would assume this would be the dB response from the image right?

20. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
That's right - in post #4 - the dB plot.