rc-implementation of Laplace transfer function?
Coolens Hugo
Is it possible to implement the Laplace transfer function:
vo(s)/vi(s)=(1+5s)/(1+5s+s*s) using only r and c components?
please e-mail your response also to coo...@kiho.be
regards
Hugo
Andy Borsa
Coolens Hugo <coo...@kiho.be> writes:
> Is it possible to implement the Laplace transfer function:
> vo(s)/vi(s)=(1+5s)/(1+5s+s*s) using only r and c components?
>
should know how to do the math.
--
Andy Borsa -- !!!The Universe is discreetly analog!!!
Gastone Demalde'
an...@moose.mv.com (Andy Borsa) wrote:
>Coolens Hugo <coo...@kiho.be> writes:
>> Is it possible to implement the Laplace transfer function:
>> vo(s)/vi(s)=(1+5s)/(1+5s+s*s) using only r and c components?
>A little ultra-simple math will answer this for you. As a student, you
>should know how to do the math.
I don't find this a simple problem to resolve.
Could you explain me the layout of the RC circuit that implements this function
of transfer and like do you have done to draw such circuit?
Saluti da Parma (Italy)
William E. Sabin
This function does not quite satisfy the Hurwitz criteria for a
realizable network. If the 1 in the denominator is changed to 0.96, then
it can be done because of cancellation of terms in the numerator and
denominator. The result is .2 Farads in parallel with 1.04 Ohms.
Bill W0IYH
Gastone Demalde'
"William E. Sabin" <sab...@crpl.cedar-rapids.lib.ia.us> wrote:
>This function does not quite satisfy the Hurwitz criteria for a
>realizable network.
I am sorry, I don't know the Hurwitz criteria :-(
Could you explain me what does it tell?
> If the 1 in the denominator is changed to 0.96, then
>it can be done because of cancellation of terms in the numerator and
>denominator.
It is always a good idea simplify the problems ;-)
Saluti da Parma (Italy)
Hugo Coolens
Coolens Hugo <coo...@kiho.be> writes:
Is it possible to implement the Laplace transfer function:
vo(s)/vi(s)=(1+5s)/(1+5s+s*s) using only r and c components?
Andy Borsa (an...@moose.mv.com answers:
>
> >A little ultra-simple math will answer this for you. As a student, you
> >should know how to do the math.
>
This kind of answer doesnt impress me at all.
If it is that simple for you, you should at least answer the question with
yes or no and draw the schematic if the answer is yes, do you dare to
take the challenge?
William E Sabin DID try to answer it by simplifying the problem, he
changed the nominator of the transfer function to s*s+5*s+0.98, this way
the transfer function became a first order one, however this is not what
I want because the original transfer function has a zero and Williams
function does not.
Later on, William stated that it was not possible to implement because the
function did not obey the Hurwitz criteria for physical networks. I have
looked in a textbook for the Hurwitz criteria and the only thing I found about them was
that they are a means to determine if a transfer function may be stable,
which is not the same as whether it can be implemented with r and c
components.
By the way, this tranfer function does describe a stable system and in
my opinion it obeys the Hurwitz criteria (and nominator coefficients are real
and positive)
Gaston Demalde' wrote
> I don't find this a simple problem to resolve
I still agree with him
Hugo Coolens
coo...@kiho.be
William E. Sabin
Hugo Coolens wrote:
>
>
> Later on, William stated that it was not possible to implement because the
> function did not obey the Hurwitz criteria for physical networks. I have
> looked in a textbook for the Hurwitz criteria and the only thing I found about them was
> that they are a means to determine if a transfer function may be stable,
> which is not the same as whether it can be implemented with r and c
> components.
> By the way, this tranfer function does describe a stable system and in
> my opinion it obeys the Hurwitz criteria (and nominator coefficients are real
>
The Hurwitz criteria are mainly not about stability, they are about
*realizability*, which is a different matter. When you try to synthesize
a netwok from your original function you come up with one negative
resistor and one negative capacitor, in addition to one real resistor and
one real capacitor. The negative components cannot be realized with
*PASSIVE* components. Suggest you study the Hurwitz criteria more
carefully because you have not properly understood them. If you look at
your function you will find that the exponents in the numerator and the
denominator are not in the right order for a realizable network. This is
partly what Hurwitz is about. If the 1 in the denominator is changed to
0.96 the problem goes away.
Bill W0IYH
Andy Borsa
Hugo Coolens <coo...@allserv.kiho.be> writes:
>
>
>
> Coolens Hugo <coo...@kiho.be> writes:
> Is it possible to implement the Laplace transfer function:
> vo(s)/vi(s)=(1+5s)/(1+5s+s*s) using only r and c components?
> Andy Borsa (an...@moose.mv.com answers:
>>
>> >A little ultra-simple math will answer this for you. As a student, you
>> >should know how to do the math.
>>
> This kind of answer doesnt impress me at all.
> If it is that simple for you, you should at least answer the question with
> yes or no and draw the schematic if the answer is yes, do you dare to
> take the challenge?
>
OK. William Sabin was correct. You can't realize this function with
just RC's. Here's 3 rules for realizibility of an RC impedance or
admittance function -
1. Poles and zeros must be simple, ie, lie on the negative-real
s-plane axis.
2. Pole and zeros must alternate.
3. Z(infinity) >= 0, or Y(0) >= 0.
The function in question satisfies 1, but misses only slightly on 2.
zero: s = -0.2
poles: s = -0.208712, -4.79129
Best theory on this subject I know of -
Synthesis of Passive Networks
by Ernst A. Guillemin
1957 and later prinbtings by John Wiley and Sons, no ISBN on my
copy.
William E. Sabin
Andy Borsa wrote:
>
> Hugo Coolens <coo...@allserv.kiho.be> writes:
> >
> >
> >
> > Coolens Hugo <coo...@kiho.be> writes:
> > Is it possible to implement the Laplace transfer function:
> > vo(s)/vi(s)=(1+5s)/(1+5s+s*s) using only r and c components?
> > Andy Borsa (an...@moose.mv.com answers:
>
> just RC's. Here's 3 rules for realizibility of an RC impedance or
> admittance function -
>
> 1. Poles and zeros must be simple, ie, lie on the negative-real
> s-plane axis.
>
> 2. Pole and zeros must alternate.
>
> 3. Z(infinity) >= 0, or Y(0) >= 0.
>
> The function in question satisfies 1, but misses only slightly on 2.
>
> zero: s = -0.2
> poles: s = -0.208712, -4.79129
than 1.0 for omega from 0 to 1. A passive RC network cannot do this. The
reason for the greater than 1 is that negative R and C are involved.
Bill
Hugo Coolens
On Fri, 16 Aug 1996, William E. Sabin wrote:
> Andy Borsa wrote:
> >
> > Hugo Coolens <coo...@allserv.kiho.be> writes:
> > >
> > >
> > >
> > > Coolens Hugo <coo...@kiho.be> writes:
> > > Is it possible to implement the Laplace transfer function:
> > > vo(s)/vi(s)=(1+5s)/(1+5s+s*s) using only r and c components?
> > > Andy Borsa (an...@moose.mv.com answers:
> >
> > OK. William Sabin was correct. You can't realize this function with
> > just RC's. Here's 3 rules for realizibility of an RC impedance or
> > admittance function -
> > 2. Pole and zeros must alternate.
Can you explain me what you mean bij alternate poles and zeroes
> > The function in question satisfies 1, but misses only slightly on 2.
> >
> > zero: s = -0.2
> > poles: s = -0.208712, -4.79129
> > Also, the function has a frequency response magnitude slightly greater
> than 1.0 for omega from 0 to 1. A passive RC network cannot do this. The
> reason for the greater than 1 is that negative R and C are involved.
>
> Bill
>
>
This is certainly not true Bill, I know of a passive RC-network which
amplifies. Its transfer function is:
Vo(s)/Vi(s)=(1+3s)/(1+3s+s*s)
Can you synthesize it and show me your method is successfull in this case
too?
Finally I want to notice also that I am in fact looking for voltage transfer
functions and in fact not for impedance functions.
Thanks for the help so far,
regards,
Hugo
Richard Karlquist
In article <321430...@crpl.cedar-rapids.lib.ia.us>,
William E. Sabin <sab...@crpl.cedar-rapids.lib.ia.us> wrote:
>Andy Borsa wrote:
>>
>> Hugo Coolens <coo...@allserv.kiho.be> writes:
>> >
>> >
>> >
>> > Coolens Hugo <coo...@kiho.be> writes:
>> > Is it possible to implement the Laplace transfer function:
>> > vo(s)/vi(s)=(1+5s)/(1+5s+s*s) using only r and c components?
>> > Andy Borsa (an...@moose.mv.com answers:
>>
>> OK. William Sabin was correct. You can't realize this function with
>> just RC's. Here's 3 rules for realizibility of an RC impedance or
>> admittance function -
The original problem asked for a *transfer* function to be
realized. You are giving rules for a *driving point* function,
which are more restrictive than with a transfer function.
>>
>> 1. Poles and zeros must be simple, ie, lie on the negative-real
>> s-plane axis.
For example the twin-T notch filter has transmission zeroes on
the jw axis.
>>
>> 2. Pole and zeros must alternate.
>>
>> 3. Z(infinity) >= 0, or Y(0) >= 0.
>>
>> The function in question satisfies 1, but misses only slightly on 2.
>>
>> zero: s = -0.2
>> poles: s = -0.208712, -4.79129
>> Also, the function has a frequency response magnitude slightly greater
>than 1.0 for omega from 0 to 1. A passive RC network cannot do this. The
>reason for the greater than 1 is that negative R and C are involved.
A passive RC network with a transfer impedance greater than unity
magnitude *is* possible. An example was published in
"Tele-Tech/Electronic Industries" around 1955. I have the article
somewhere in my files. The published circuit is a two terminal
pair network (i.e. the input and output grounds are not in common).
However, the problem as stated above does not require common grounds.
Rick Karlquist
rka...@scd.hp.com
>Bill
William E. Sabin
> >> >
> >> > Coolens Hugo <coo...@kiho.be> writes:
> >> > Is it possible to implement the Laplace transfer function:
> >> > vo(s)/vi(s)=(1+5s)/(1+5s+s*s) using only r and c components?
> A passive RC network with a transfer impedance greater than unity
> magnitude *is* possible. An example was published in
> "Tele-Tech/Electronic Industries" around 1955. I have the article
> somewhere in my files. The published circuit is a two terminal
> pair network (i.e. the input and output grounds are not in common).
> However, the problem as stated above does not require common grounds.
>
He wants a *voltage transfer function*, not an *impedance tranfer
function*. Not only that, Hugo in my opinion is a "pest" who is just
amusing himself.
Bill
Roy McCammon
rka...@scd.hp.com (Richard Karlquist) wrote:
>A passive RC network with a transfer impedance greater than unity
>magnitude *is* possible. An example was published in
>"Tele-Tech/Electronic Industries" around 1955. I have the article
>somewhere in my files. The published circuit is a two terminal
>pair network (i.e. the input and output grounds are not in common).
>However, the problem as stated above does not require common grounds.
>
>Rick Karlquist
>rka...@scd.hp.com
I haven't done the math, but I think here is an RC circuit with more than
unity voltage gain.
N1 -----R--+--R--+--R--+- N2 input between terminals N1,N0
| | | output between terminals N1,N2
C C C
| | | The voltage at N2-N0 at some frequency
N0 --------+-----+-----+ is 180 degree phased with respect to
N1-N0, so the voltage from N2 to N1 has
got to be greater than the voltage from
N1 to N0.
Then flip the circuit over and consider N1 as ground, N0 as input and
N2 as output and you have an RC circuit with a common ground and greater
than unity gain.
As I say, I didn't do the math, so if I'm wrong, I'll fess up up without
whining.
Opinions expressed herein are my own and may not represent those of my employer.
Hugo Coolens
You are quite right about this, the version of the filter you describe is
the high pass version, you get a low pass version by changing r's and c's.
For an explanation without maths 3 sections is easiest, however two sections
are already sufficient to obtain voltage gain. By the way, the voltage
transfer function a the two section low pass version is:
(1+3s)/(1+3s+s*s) which gives 0.65dB gain around 0.1Hz
It is indeed this filter which makes me wonder if there are other
rc-filters with voltage gain or formulated otherwise, can you realise
e.g. (1+5s)/(1+5s+s*s) voltage transfer function with only r's and c's?
regards,
Hugo
Hugo Coolens
On Thu, 29 Aug 1996, John Quinlan wrote:
> Hugo Coolens <coo...@allserv.kiho.be> wrote:
>
> <...>
> >It is indeed this filter which makes me wonder if there are other
> >rc-filters with voltage gain or formulated otherwise, can you realise
> >e.g. (1+5s)/(1+5s+s*s) voltage transfer function with only r's and c's?
> <...>
>
> I've seen at least five or six different r-c circuits with
> voltage gain. A four-component version is shown below:
>
>
> ----------R2----------
> | |
> | |
> ----+---R1-----+------C1---+--------
> |
> Vi C2 Vo
> |
> ---------------+--------------------
>
> If you add a little complexity to this circuit you can get gain
> without phase shift! I believe that several such circuits were
> patented some 40 or 45 years ago because you could build an
> oscillator using one less tube (valve) than it would otherwise
> take using conventional circuits.
>
> If you assign the component values below to the circuit above,
> you should achieve the desired transfer function (but I could
> have goofed up in the calculations):
>
> R1 = 3, R2 = 1, C1 = 1, C2 = 1/3
>
> Regards ...JPQ
>
>
>
Thanks for the reply John. However the filter you present here is just
the low pass two section filter earlier mentioned in this newsgroup
slightly differently drawn. However if you could present us the other 4
or 5 filters you know about I'd be very glad.
regards,
Hugo