Properties of Positive Real Functions
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I am trying to understand the properties of positive real (PR) and strictly positive real (SPR) transfer functions. If given a transfer function I know how to determine whether or not the function is PR/SPR, however I'm not sure how to approach the following example.
Say we are given two positive real transfer functions:
$G_{1}(s)$ and $G_{2}(s)$
How can I tell if the sum $$ G_{1}(s) + G_{2}(s)$$
or the product $$ G_{1}(s) * G_{2}(s)$$
is PR/SPR?
Intuitively, I think the sum is SPR, but as for the product I am uncertain. Either way, is there a framework I can use to approach this or does anyone know of a proof I could look at?
functions linear-transformations control-theory stability-theory
asked Dec 8 '18 at 4:29
Chemical EngineerChemical Engineer
936
$endgroup$
add a comment |
$begingroup$
I am trying to understand the properties of positive real (PR) and strictly positive real (SPR) transfer functions. If given a transfer function I know how to determine whether or not the function is PR/SPR, however I'm not sure how to approach the following example.
Say we are given two positive real transfer functions:
$G_{1}(s)$ and $G_{2}(s)$
How can I tell if the sum $$ G_{1}(s) + G_{2}(s)$$
or the product $$ G_{1}(s) * G_{2}(s)$$
is PR/SPR?
Intuitively, I think the sum is SPR, but as for the product I am uncertain. Either way, is there a framework I can use to approach this or does anyone know of a proof I could look at?
functions linear-transformations control-theory stability-theory
asked Dec 8 '18 at 4:29
Chemical EngineerChemical Engineer
936
$endgroup$
$begingroup$
This is true for the sum but not for the product. Consider for example $G_1(s)=G_2(s)=frac{1}{s+1}$. Both $G_1,G_2$ are SPR but their product is not as $Re[G_1^2(jomega)]<0$ for $omega>1$.
$endgroup$
– RTJ
Dec 13 '18 at 17:30
add a comment |
$begingroup$
I am trying to understand the properties of positive real (PR) and strictly positive real (SPR) transfer functions. If given a transfer function I know how to determine whether or not the function is PR/SPR, however I'm not sure how to approach the following example.
Say we are given two positive real transfer functions:
$G_{1}(s)$ and $G_{2}(s)$
How can I tell if the sum $$ G_{1}(s) + G_{2}(s)$$
or the product $$ G_{1}(s) * G_{2}(s)$$
is PR/SPR?
Intuitively, I think the sum is SPR, but as for the product I am uncertain. Either way, is there a framework I can use to approach this or does anyone know of a proof I could look at?
functions linear-transformations control-theory stability-theory
asked Dec 8 '18 at 4:29
Chemical EngineerChemical Engineer
936
$endgroup$
I am trying to understand the properties of positive real (PR) and strictly positive real (SPR) transfer functions. If given a transfer function I know how to determine whether or not the function is PR/SPR, however I'm not sure how to approach the following example.
Say we are given two positive real transfer functions:
$G_{1}(s)$ and $G_{2}(s)$
How can I tell if the sum $$ G_{1}(s) + G_{2}(s)$$
or the product $$ G_{1}(s) * G_{2}(s)$$
is PR/SPR?
Intuitively, I think the sum is SPR, but as for the product I am uncertain. Either way, is there a framework I can use to approach this or does anyone know of a proof I could look at?
functions linear-transformations control-theory stability-theory
functions linear-transformations control-theory stability-theory
asked Dec 8 '18 at 4:29
Chemical EngineerChemical Engineer
936
asked Dec 8 '18 at 4:29
Chemical EngineerChemical Engineer
936
edited Dec 8 '18 at 6:33
Chemical Engineer
asked Dec 8 '18 at 4:29
Chemical EngineerChemical Engineer
936
asked Dec 8 '18 at 4:29
Chemical EngineerChemical Engineer
936
asked Dec 8 '18 at 4:29
Chemical EngineerChemical Engineer
936
936
$begingroup$
This is true for the sum but not for the product. Consider for example $G_1(s)=G_2(s)=frac{1}{s+1}$. Both $G_1,G_2$ are SPR but their product is not as $Re[G_1^2(jomega)]<0$ for $omega>1$.
$endgroup$
– RTJ
Dec 13 '18 at 17:30
add a comment |
$begingroup$
This is true for the sum but not for the product. Consider for example $G_1(s)=G_2(s)=frac{1}{s+1}$. Both $G_1,G_2$ are SPR but their product is not as $Re[G_1^2(jomega)]<0$ for $omega>1$.
$endgroup$
– RTJ
Dec 13 '18 at 17:30
$begingroup$
This is true for the sum but not for the product. Consider for example $G_1(s)=G_2(s)=frac{1}{s+1}$. Both $G_1,G_2$ are SPR but their product is not as $Re[G_1^2(jomega)]<0$ for $omega>1$.
$endgroup$
– RTJ
Dec 13 '18 at 17:30
$begingroup$
This is true for the sum but not for the product. Consider for example $G_1(s)=G_2(s)=frac{1}{s+1}$. Both $G_1,G_2$ are SPR but their product is not as $Re[G_1^2(jomega)]<0$ for $omega>1$.
$endgroup$
– RTJ
Dec 13 '18 at 17:30
add a comment |
1 Answer
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$begingroup$
At any given $s$ the two transfer functions give you two complex numbers $rho_1+i,sigma_1$ and $rho_2+i,sigma_2$, with the constraints that $rho_1,rho_2geq0$.
For the summation you only have that it is strictly positive real iff $rho_1$ and $rho_2$ are not both simultaneously zero. For example $G_1(s)=alpha,G_2(s)$ with $alpha>0$ would not satisfy this, or the transfer functions already need to be strictly positive real.
For the multiplication the real part would become $rho_1,rho_2 - sigma_1,sigma_2$. Therefore, the fact that $G_1(s)$ and $G_2(s)$ are positive real does not give you enough information about their imaginary parts to tell if the result is even positive real.
answered Dec 8 '18 at 9:12
Kwin van der VeenKwin van der Veen
5,3552826
$endgroup$
$begingroup$
This makes sense for proving that the real parts are either positive, but don't we require other conditions to hold? Such as negative real poles and positive residues?
$endgroup$
– Chemical Engineer
Dec 8 '18 at 18:33
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
oldest
votes
$begingroup$
At any given $s$ the two transfer functions give you two complex numbers $rho_1+i,sigma_1$ and $rho_2+i,sigma_2$, with the constraints that $rho_1,rho_2geq0$.
For the summation you only have that it is strictly positive real iff $rho_1$ and $rho_2$ are not both simultaneously zero. For example $G_1(s)=alpha,G_2(s)$ with $alpha>0$ would not satisfy this, or the transfer functions already need to be strictly positive real.
For the multiplication the real part would become $rho_1,rho_2 - sigma_1,sigma_2$. Therefore, the fact that $G_1(s)$ and $G_2(s)$ are positive real does not give you enough information about their imaginary parts to tell if the result is even positive real.
answered Dec 8 '18 at 9:12
Kwin van der VeenKwin van der Veen
5,3552826
$endgroup$
$begingroup$
This makes sense for proving that the real parts are either positive, but don't we require other conditions to hold? Such as negative real poles and positive residues?
$endgroup$
– Chemical Engineer
Dec 8 '18 at 18:33
add a comment |
$begingroup$
At any given $s$ the two transfer functions give you two complex numbers $rho_1+i,sigma_1$ and $rho_2+i,sigma_2$, with the constraints that $rho_1,rho_2geq0$.
For the summation you only have that it is strictly positive real iff $rho_1$ and $rho_2$ are not both simultaneously zero. For example $G_1(s)=alpha,G_2(s)$ with $alpha>0$ would not satisfy this, or the transfer functions already need to be strictly positive real.
For the multiplication the real part would become $rho_1,rho_2 - sigma_1,sigma_2$. Therefore, the fact that $G_1(s)$ and $G_2(s)$ are positive real does not give you enough information about their imaginary parts to tell if the result is even positive real.
answered Dec 8 '18 at 9:12
Kwin van der VeenKwin van der Veen
5,3552826
$endgroup$
$begingroup$
This makes sense for proving that the real parts are either positive, but don't we require other conditions to hold? Such as negative real poles and positive residues?
$endgroup$
– Chemical Engineer
Dec 8 '18 at 18:33
add a comment |
$begingroup$
At any given $s$ the two transfer functions give you two complex numbers $rho_1+i,sigma_1$ and $rho_2+i,sigma_2$, with the constraints that $rho_1,rho_2geq0$.
For the summation you only have that it is strictly positive real iff $rho_1$ and $rho_2$ are not both simultaneously zero. For example $G_1(s)=alpha,G_2(s)$ with $alpha>0$ would not satisfy this, or the transfer functions already need to be strictly positive real.
For the multiplication the real part would become $rho_1,rho_2 - sigma_1,sigma_2$. Therefore, the fact that $G_1(s)$ and $G_2(s)$ are positive real does not give you enough information about their imaginary parts to tell if the result is even positive real.
answered Dec 8 '18 at 9:12
Kwin van der VeenKwin van der Veen
5,3552826
$endgroup$
At any given $s$ the two transfer functions give you two complex numbers $rho_1+i,sigma_1$ and $rho_2+i,sigma_2$, with the constraints that $rho_1,rho_2geq0$.
For the summation you only have that it is strictly positive real iff $rho_1$ and $rho_2$ are not both simultaneously zero. For example $G_1(s)=alpha,G_2(s)$ with $alpha>0$ would not satisfy this, or the transfer functions already need to be strictly positive real.
For the multiplication the real part would become $rho_1,rho_2 - sigma_1,sigma_2$. Therefore, the fact that $G_1(s)$ and $G_2(s)$ are positive real does not give you enough information about their imaginary parts to tell if the result is even positive real.
answered Dec 8 '18 at 9:12
Kwin van der VeenKwin van der Veen
5,3552826
answered Dec 8 '18 at 9:12
Kwin van der VeenKwin van der Veen
5,3552826
answered Dec 8 '18 at 9:12
Kwin van der VeenKwin van der Veen
5,3552826
answered Dec 8 '18 at 9:12
Kwin van der VeenKwin van der Veen
5,3552826
5,3552826
$begingroup$
This makes sense for proving that the real parts are either positive, but don't we require other conditions to hold? Such as negative real poles and positive residues?
$endgroup$
– Chemical Engineer
Dec 8 '18 at 18:33
add a comment |
$begingroup$
This makes sense for proving that the real parts are either positive, but don't we require other conditions to hold? Such as negative real poles and positive residues?
$endgroup$
– Chemical Engineer
Dec 8 '18 at 18:33
$begingroup$
This makes sense for proving that the real parts are either positive, but don't we require other conditions to hold? Such as negative real poles and positive residues?
$endgroup$
– Chemical Engineer
Dec 8 '18 at 18:33
$begingroup$
This makes sense for proving that the real parts are either positive, but don't we require other conditions to hold? Such as negative real poles and positive residues?
$endgroup$
– Chemical Engineer
Dec 8 '18 at 18:33
add a comment |
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$begingroup$
This is true for the sum but not for the product. Consider for example $G_1(s)=G_2(s)=frac{1}{s+1}$. Both $G_1,G_2$ are SPR but their product is not as $Re[G_1^2(jomega)]<0$ for $omega>1$.
$endgroup$
– RTJ
Dec 13 '18 at 17:30