On 11 Jan 2007 05:41:59 -0800, "Noway2" <no_spam_me2@hotmail.com>
wrote:
>
>Fitlike Min wrote:
>> "Noway2" <no_spam_me2@hotmail.com> wrote in message
>> news:1168454258.926519.320090@k58g2000hse.googlegroups.com...
>> >
>> > naebad wrote:
>> > > pruthvisri02 wrote:
>> > > > Hallo, I am relatively new to RF electronics,I need some support for
>> > > > modelling my blocks in PLL.The block we have problem is third order
>> > > > LoopFilter (L.F). Here we go --->
>> > > > The Transfer Function (T.F) for L.F -> (ds+1)/s(as**2+bs+c)
>> > > > we deduced the above T.F -> k1/s + k2/(s-r1)+k3(s-r2)
>> > > > where r1 and r2 are the roots of the quadratic equation in T.F &
>> a,b,c,d
>> > > > are constants.
>> > > > And I reduced it to time domain as
>> k1+k2*e(pow(-r1*t))+k3*e(pow(-r2*t)).
>> > > > and finally Vo across the output cap is ->
>> > > > Vo = [k1+k2*e(pow(-r1*t))+k3*e(pow(-r2*t))]* Current_In
>> > > > Current_In is the current flowing into the loop filter.
>> > > > with this method , i am not getting the required response.
>> > > > Could some one help me out , what's the mistake in this style ??
>> > > > Is there any other better method to solve the same ??
>> > > > Regards
>> > >
>> > > It's not a great idea to have to wrok out time-responses. Essentially
>> > > that's why Bode-Plots were invented and in particular PHASE-MARGIN! You
>> > > need only look at the pahse margin and this information will tell you
>> > > what the damping is in closed loop. A good phase margin may well be
>> > > about 60 degrees (maybe a bit optimistic at times). Time-domain stuff
>> > > is for amateurs and students - start working in the frequency domain.
>> > > Plot the open loop frequency response.
>> > >
>> > >
>> > > Naebad
>> >
>> > The statement that "Time-domain stuff is for amateurs and students" is
>> > a poor one that speaks volumes about your understanding.
>> >
>> Are you a student?
>>
>> F.
>>
>> Posted via a free Usenet account from http://www.teranews.com
>
>No, I am not a student, at least in the official sense. I have been a
>practicing engineer for about 10 years since completing undergrad
>school.
>
>There was a time when I would have probably agreed, at least in part,
>with Naebad's stament. About the time I graduated, my thinking would
>have been along the lines of "Why the hell would anyone ever want to
>bother with time domain." Since then, experience has changed my
>thinking.
And I'd back up that sentiment. Frequency-domain analysis is useful,
but there are things that can be learned from time-domain analysis
that can be crucial. "Time-domain stuff is for amateurs and students"
is not at all accurate in my experience.
Eric Jacobsen
Minister of Algorithms, Intel Corp.
My opinions may not be Intel's opinions.
http://www.ericjacobsen.org
Reply by Noway2●January 11, 20072007-01-11
Fitlike Min wrote:
> "Noway2" <no_spam_me2@hotmail.com> wrote in message
> news:1168454258.926519.320090@k58g2000hse.googlegroups.com...
> >
> > naebad wrote:
> > > pruthvisri02 wrote:
> > > > Hallo, I am relatively new to RF electronics,I need some support for
> > > > modelling my blocks in PLL.The block we have problem is third order
> > > > LoopFilter (L.F). Here we go --->
> > > > The Transfer Function (T.F) for L.F -> (ds+1)/s(as**2+bs+c)
> > > > we deduced the above T.F -> k1/s + k2/(s-r1)+k3(s-r2)
> > > > where r1 and r2 are the roots of the quadratic equation in T.F &
> a,b,c,d
> > > > are constants.
> > > > And I reduced it to time domain as
> k1+k2*e(pow(-r1*t))+k3*e(pow(-r2*t)).
> > > > and finally Vo across the output cap is ->
> > > > Vo = [k1+k2*e(pow(-r1*t))+k3*e(pow(-r2*t))]* Current_In
> > > > Current_In is the current flowing into the loop filter.
> > > > with this method , i am not getting the required response.
> > > > Could some one help me out , what's the mistake in this style ??
> > > > Is there any other better method to solve the same ??
> > > > Regards
> > >
> > > It's not a great idea to have to wrok out time-responses. Essentially
> > > that's why Bode-Plots were invented and in particular PHASE-MARGIN! You
> > > need only look at the pahse margin and this information will tell you
> > > what the damping is in closed loop. A good phase margin may well be
> > > about 60 degrees (maybe a bit optimistic at times). Time-domain stuff
> > > is for amateurs and students - start working in the frequency domain.
> > > Plot the open loop frequency response.
> > >
> > >
> > > Naebad
> >
> > The statement that "Time-domain stuff is for amateurs and students" is
> > a poor one that speaks volumes about your understanding.
> >
> Are you a student?
>
> F.
>
> Posted via a free Usenet account from http://www.teranews.com
No, I am not a student, at least in the official sense. I have been a
practicing engineer for about 10 years since completing undergrad
school.
There was a time when I would have probably agreed, at least in part,
with Naebad's stament. About the time I graduated, my thinking would
have been along the lines of "Why the hell would anyone ever want to
bother with time domain." Since then, experience has changed my
thinking.
Reply by Fitlike Min●January 11, 20072007-01-11
"Noway2" <no_spam_me2@hotmail.com> wrote in message
news:1168454258.926519.320090@k58g2000hse.googlegroups.com...
>
> naebad wrote:
> > pruthvisri02 wrote:
> > > Hallo, I am relatively new to RF electronics,I need some support for
> > > modelling my blocks in PLL.The block we have problem is third order
> > > LoopFilter (L.F). Here we go --->
> > > The Transfer Function (T.F) for L.F -> (ds+1)/s(as**2+bs+c)
> > > we deduced the above T.F -> k1/s + k2/(s-r1)+k3(s-r2)
> > > where r1 and r2 are the roots of the quadratic equation in T.F &
a,b,c,d
> > > are constants.
> > > And I reduced it to time domain as
k1+k2*e(pow(-r1*t))+k3*e(pow(-r2*t)).
> > > and finally Vo across the output cap is ->
> > > Vo = [k1+k2*e(pow(-r1*t))+k3*e(pow(-r2*t))]* Current_In
> > > Current_In is the current flowing into the loop filter.
> > > with this method , i am not getting the required response.
> > > Could some one help me out , what's the mistake in this style ??
> > > Is there any other better method to solve the same ??
> > > Regards
> >
> > It's not a great idea to have to wrok out time-responses. Essentially
> > that's why Bode-Plots were invented and in particular PHASE-MARGIN! You
> > need only look at the pahse margin and this information will tell you
> > what the damping is in closed loop. A good phase margin may well be
> > about 60 degrees (maybe a bit optimistic at times). Time-domain stuff
> > is for amateurs and students - start working in the frequency domain.
> > Plot the open loop frequency response.
> >
> >
> > Naebad
>
> The statement that "Time-domain stuff is for amateurs and students" is
> a poor one that speaks volumes about your understanding.
>
> pruthvisri02 wrote:
> > Hallo, I am relatively new to RF electronics,I need some support for
> > modelling my blocks in PLL.The block we have problem is third order
> > LoopFilter (L.F). Here we go --->
> > The Transfer Function (T.F) for L.F -> (ds+1)/s(as**2+bs+c)
> > we deduced the above T.F -> k1/s + k2/(s-r1)+k3(s-r2)
> > where r1 and r2 are the roots of the quadratic equation in T.F & a,b,c,d
> > are constants.
> > And I reduced it to time domain as k1+k2*e(pow(-r1*t))+k3*e(pow(-r2*t)).
> > and finally Vo across the output cap is ->
> > Vo = [k1+k2*e(pow(-r1*t))+k3*e(pow(-r2*t))]* Current_In
> > Current_In is the current flowing into the loop filter.
> > with this method , i am not getting the required response.
> > Could some one help me out , what's the mistake in this style ??
> > Is there any other better method to solve the same ??
> > Regards
>
> It's not a great idea to have to wrok out time-responses. Essentially
> that's why Bode-Plots were invented and in particular PHASE-MARGIN! You
> need only look at the pahse margin and this information will tell you
> what the damping is in closed loop. A good phase margin may well be
> about 60 degrees (maybe a bit optimistic at times). Time-domain stuff
> is for amateurs and students - start working in the frequency domain.
> Plot the open loop frequency response.
>
>
> Naebad
The statement that "Time-domain stuff is for amateurs and students" is
a poor one that speaks volumes about your understanding.
Reply by naebad●January 9, 20072007-01-09
pruthvisri02 wrote:
> Hallo, I am relatively new to RF electronics,I need some support for
> modelling my blocks in PLL.The block we have problem is third order
> LoopFilter (L.F). Here we go --->
> The Transfer Function (T.F) for L.F -> (ds+1)/s(as**2+bs+c)
> we deduced the above T.F -> k1/s + k2/(s-r1)+k3(s-r2)
> where r1 and r2 are the roots of the quadratic equation in T.F & a,b,c,d
> are constants.
> And I reduced it to time domain as k1+k2*e(pow(-r1*t))+k3*e(pow(-r2*t)).
> and finally Vo across the output cap is ->
> Vo = [k1+k2*e(pow(-r1*t))+k3*e(pow(-r2*t))]* Current_In
> Current_In is the current flowing into the loop filter.
> with this method , i am not getting the required response.
> Could some one help me out , what's the mistake in this style ??
> Is there any other better method to solve the same ??
> Regards
It's not a great idea to have to wrok out time-responses. Essentially
that's why Bode-Plots were invented and in particular PHASE-MARGIN! You
need only look at the pahse margin and this information will tell you
what the damping is in closed loop. A good phase margin may well be
about 60 degrees (maybe a bit optimistic at times). Time-domain stuff
is for amateurs and students - start working in the frequency domain.
Plot the open loop frequency response.
Naebad
Reply by Noway2●January 9, 20072007-01-09
pruthvisri02 wrote:
> Hallo, I am relatively new to RF electronics,I need some support for
> modelling my blocks in PLL.The block we have problem is third order
> LoopFilter (L.F). Here we go --->
> The Transfer Function (T.F) for L.F -> (ds+1)/s(as**2+bs+c)
> we deduced the above T.F -> k1/s + k2/(s-r1)+k3(s-r2)
> where r1 and r2 are the roots of the quadratic equation in T.F & a,b,c,d
> are constants.
> And I reduced it to time domain as k1+k2*e(pow(-r1*t))+k3*e(pow(-r2*t)).
> and finally Vo across the output cap is ->
> Vo = [k1+k2*e(pow(-r1*t))+k3*e(pow(-r2*t))]* Current_In
> Current_In is the current flowing into the loop filter.
> with this method , i am not getting the required response.
> Could some one help me out , what's the mistake in this style ??
> Is there any other better method to solve the same ??
> Regards
First, If I understand you correctly, you are tying to determine the
time domain output voltage, given the input current. You say you have
a transfer function. Are you sure that the transfer function relates
the output voltage to input current? Did you perform the circuit
analysis yourself or were you given the transfer function? If you
performed the analysis this may be a high probablility algebra error
point.
Second, Make sure you understand and apply the partial fraction
expansion correctly, which is what I am assuming you are doing based on
your statements. This is another place it is really easy to make an
algebra or procedureal error, especially if you have repeated or
complex roots.
If all else fails, see if you can find a math program that will perform
the algebra and give you a correct answer. This could help you
determine where your error lies. If you do get a program, don't fall
to the tempation of letting it do all your work for you or else you
will do very poorly come test time as you will still need to be able to
calculate these systems by hand.
Reply by pruthvisri02●January 9, 20072007-01-09
Hallo, I am relatively new to RF electronics,I need some support for
modelling my blocks in PLL.The block we have problem is third order
LoopFilter (L.F). Here we go --->
The Transfer Function (T.F) for L.F -> (ds+1)/s(as**2+bs+c)
we deduced the above T.F -> k1/s + k2/(s-r1)+k3(s-r2)
where r1 and r2 are the roots of the quadratic equation in T.F & a,b,c,d
are constants.
And I reduced it to time domain as k1+k2*e(pow(-r1*t))+k3*e(pow(-r2*t)).
and finally Vo across the output cap is ->
Vo = [k1+k2*e(pow(-r1*t))+k3*e(pow(-r2*t))]* Current_In
Current_In is the current flowing into the loop filter.
with this method , i am not getting the required response.
Could some one help me out , what's the mistake in this style ??
Is there any other better method to solve the same ??
Regards