Re: [Scilab-users] Problems with Scilab routine "conv"

[Stéphane Mottelet]

Hello,

I have given a a more detailed answer on scilab discourse, with maths
stuff correctly typeset and plots comparing approximations:

https://scilab.discourse.group/t/approximate-a-convolution-product/

S.

Le 03/02/2023 à 11:52, Heinz Nabielek a écrit :

This is my latest code version: the 'convoluted secondary failure fraction' is 
nicely below primary failure,  but seems too low.
Heinz


m=2; // Weibull modulus in mechanism  #1
k=1E-7;  // corrosion rate(s-1) in mechanism  #1
n=500;   // number of time steps in hourly intervals
t=(1:n)';// timesteps in hours
ts= 3600*t;  // timesteps in seconds
PHI= 1 - exp(-((k*ts)^m)); // failure fraction in mechanism #1
deriv=m*k*((k*ts)^(m-1)) .* exp(-((k*ts)^m)); //derivative of PHI
phi=3600*deriv;  //derivative of PHI integrated over an hour
f= 1 - exp(-ts/3E5); // "pure" failure fraction in mechanism #2
F=(convol(phi,f)/n)(1:n);   //"convoluted" failure fraction mechanism #2
plot("nl",t, [PHI phi f F'],'-');xgrid;
legend('PHI primary failure','phi primary failure rate','pure secondary failure 
fraction','convoluted secondary failure fraction',4);
xlabel('time (hours)');
ylabel('failure fraction/ failure rate');




On 03.02.2023, at 11:24, Heinz Nabielek  wrote:

On 03.02.2023, at 11:13, Stéphane Mottelet  wrote:

Thanks for the code.

Just a remark on the notations, you should write :

F(T)  = Int_{0}^{T}  PHI(t) . f(T-t) . dt

i.e. not F(t) since t is mute.

However, you should pay attention to the delay notion associated with 
convolution and the relationships between discrete convolution and continuous 
convolution. I am not sure that the output of conv matches with a given 
discretization of the integral above. Maybe rectangle method, but I am not sure 
at all. Anyway, you should have F(0)=0 which does not seem to be the case in 
your graph.

F(t), of course.

But the formula is fundamentally wrong and one should have seen it already from 
a dimensional consideration.

Correct should be F(t)  = Int_{0}^{T}  (d PHI(t)/dt) . f(T-t) . dt

Hope you can help between conv and convol and possible something else.
Heinz







Dear SciLab Friends:

I have an object consisting of many (~10,000) small components that can fail in 
a statistical way during long-term operation at extreme conditions.

My primary failure model is described by PHI(t) going monotonically from zero 
to one at times from t=0 to T. In the computer, this is realized in n timesteps.

Mechanism 2: There is a secondary failure mechanism that starts only when a 
component has previously failed by mechanism 1 and occurs after some delay. The 
delay function is f(t) and final expression for the evolution of the mechanism 
2 failure is given by:

F(t)  = Int_{0}^{T}  PHI(t) . f(T-t) . dt

a classical convolution Faltungsintegral.

In Scilab, I write

F= conv(PHI, f, "same")/ n;

and the resulting function F looks reasonable for most of the time. Under some 
conditions, however, I can have

F>PHI

at early stages, but this is physically impossible. Because it comes later, F 
must always be below PHI.
What do I do wrong?
Heinz

_
PS: massively simplified code below..

k=1E-7;  // corrosion rate(s-1) in mechanism #1
n=100;   // number of timesteps
t=(1:n)'; // time in days
ts= 3600*t;  // time in seconds
PHI= 1 - exp(-((k*ts)^2));   // failure fraction in mechanism #1
plot("nl", t, PHI, 'r--');
f= 1 - exp(-ts/3E5); // "pure" failure fraction in mechanism #2
F=conv(PHI,f,"same")/length(t); //"convoluted" failure fraction mechanism #2
plot(t,F,'b--');
legend('failure fraction #1','subsequent failure fraction #2',4);
xlabel('time (hours)');
ylabel('failure fraction');
title('Component failure evolution #1 is followed by #2');


--
Stéphane Mottelet
Ingénieur de recherche
EA 4297 Transformations Intégrées de la Matière Renouvelable
Département Génie des Procédés Industriels
Sorbonne Universités - Université de Technologie de Compiègne
CS 60319, 60203 Compiègne cedex
Tel : +33(0)344234688
http://www.utc.fr/~mottelet

___
users mailing list - users@lists.scilab.org
Click here to unsubscribe: 
https://lists.scilab.org/mailman/listinfo/users
This email and any attachments are intended solely for the use of the 
individual or entity to whom it is addressed and may be confidential and/or 
privileged.

If you are not one of the named recipients or have received this email in error,

(i) you should not read, disclose, or copy it,

(ii) please notify sender of your receipt by reply email and delete this email 
and all attachments,

(iii) Dassault Systèmes does not accept or assume any liability or 
responsibility for any use of or reliance on this email.


Please be informed that your personal data are processed according to our data 
privacy policy as described on our website. Should you have any questions 
related to 

Re: [Scilab-users] Problems with Scilab routine "conv"

[Samuel Gougeon]

Le 03/02/2023 à 21:25, Samuel Gougeon a écrit :
Le 03/02/2023 à 20:57, Samuel Gougeon a écrit :
.../...
Here is a draft proposal:
1) build the (let's say row) vector A = (dPHI/dt) of sampled data at
sampled values t
2) build the row vector B = f(-t) of sample data at t values
3) build the matrix C of (padded) A and the matrix D of (shiffted and
padded) D = B(T-t), with T as multiples of the dt step
   Each row of C and D corresponds to a T value.
4) compute E = (C .* D)*dt  when dt is the time step.
5) Set to zero all elements on (#) and above the diagonal of E.
   That's to cancel elements with t>T, to code the upper bound T of
the integral.
5) sum E along rows  (the actual integration. You can refine with a
trapeze method).

and that's it: the resulting column vector is your F(T).


As you know the formal expressions of f() and of phi(t), the other way is to 
use intg() in a for loop over T,
leading to something like this:

m = 2; // Weibull modulus in mechanism  #1
k = 1E-7;  // corrosion rate(s-1) in mechanism  #1
function  r = pseudoconv(t, T)
   r = 3600*m*k*((k*t)^(m-1)) * exp(-((k*t)^m)) * (1 - exp((T+t)/3E5));
endfunction
n = 500;
dT = 5;
F = zeros(1, n);
T = (0:(n-1))*dT;
for i = 1:n
   F(i) = intg(0, T(i), pseudoconv);

Sorry:

   F(i) = intg(0, T(i), list(pseudoconv, T(i)));



instead. That's the point preventing to use conv()!



end


This email and any attachments are intended solely for the use of the 
individual or entity to whom it is addressed and may be confidential and/or 
privileged.

If you are not one of the named recipients or have received this email in error,

(i) you should not read, disclose, or copy it,

(ii) please notify sender of your receipt by reply email and delete this email 
and all attachments,

(iii) Dassault Systèmes does not accept or assume any liability or 
responsibility for any use of or reliance on this email.


Please be informed that your personal data are processed according to our data 
privacy policy as described on our website. Should you have any questions 
related to personal data protection, please contact 3DS Data Protection Officer 
https://www.3ds.com/privacy-policy/contact/

___
users mailing list - users@lists.scilab.org
Click here to unsubscribe: 
https://lists.scilab.org/mailman/listinfo/users

Re: [Scilab-users] Problems with Scilab routine "conv"

[Samuel Gougeon]

Le 03/02/2023 à 20:57, Samuel Gougeon a écrit :
.../...
Here is a draft proposal:
1) build the (let's say row) vector A = (dPHI/dt) of sampled data at
sampled values t
2) build the row vector B = f(-t) of sample data at t values
3) build the matrix C of (padded) A and the matrix D of (shiffted and
padded) D = B(T-t), with T as multiples of the dt step
   Each row of C and D corresponds to a T value.
4) compute E = (C .* D)*dt  when dt is the time step.
5) Set to zero all elements on (#) and above the diagonal of E.
   That's to cancel elements with t>T, to code the upper bound T of
the integral.
5) sum E along rows  (the actual integration. You can refine with a
trapeze method).

and that's it: the resulting column vector is your F(T).


As you know the formal expressions of f() and of phi(t), the other way is to 
use intg() in a for loop over T,
leading to something like this:

m = 2; // Weibull modulus in mechanism  #1
k = 1E-7;  // corrosion rate(s-1) in mechanism  #1
function  r = pseudoconv(t, T)
   r = 3600*m*k*((k*t)^(m-1)) * exp(-((k*t)^m)) * (1 - exp((T+t)/3E5));
endfunction
n = 500;
dT = 5;
F = zeros(1, n);
T = (0:(n-1))*dT;
for i = 1:n
   F(i) = intg(0, T(i), pseudoconv);
end


Samuel





This email and any attachments are intended solely for the use of the 
individual or entity to whom it is addressed and may be confidential and/or 
privileged.

If you are not one of the named recipients or have received this email in error,

(i) you should not read, disclose, or copy it,

(ii) please notify sender of your receipt by reply email and delete this email 
and all attachments,

(iii) Dassault Systèmes does not accept or assume any liability or 
responsibility for any use of or reliance on this email.


Please be informed that your personal data are processed according to our data 
privacy policy as described on our website. Should you have any questions 
related to personal data protection, please contact 3DS Data Protection Officer 
https://www.3ds.com/privacy-policy/contact/

___
users mailing list - users@lists.scilab.org
Click here to unsubscribe: 
https://lists.scilab.org/mailman/listinfo/users

Re: [Scilab-users] Problems with Scilab routine "conv"

[Samuel Gougeon]

Le 03/02/2023 à 11:24, Heinz Nabielek a écrit :

On 03.02.2023, at 11:13, Stéphane Mottelet  wrote:

Thanks for the code.

Just a remark on the notations, you should write :

F(T)  = Int_{0}^{T}  PHI(t) . f(T-t) . dt

i.e. not F(t) since t is mute.

However, you should pay attention to the delay notion associated with 
convolution and the relationships between discrete convolution and continuous 
convolution. I am not sure that the output of conv matches with a given 
discretization of the integral above. Maybe rectangle method, but I am not sure 
at all. Anyway, you should have F(0)=0 which does not seem to be the case in 
your graph.


That's expected, since for a regular convolution as performed with
conv() or convol(), the integral bounds do not depend on the delay. They
depend only on the widths of both convoluted functions.

Neither F(t) = Int_{0}^{T} PHI(t) . f(T-t) . dt
nor F(t) = Int_{0}^{T} (d PHI(t)/dt) . f(T-t) . dt  are convolutions.
They are "elastically-limited" ones. That is to say: not convolutions at
all ;-)


F(t), of course.



no no, F(T), as Stéphane wrote it. After computing the integral over t,
t is used and over, and disappears from the result. Only the delay T
remains as parameter in the integrand, and as the variable in the result
F(T).


So now, your question might be likely: how is it possible to actually
compute
F(t) = Int_{0}^{T} (d PHI(t)/dt) . f(T-t) . dt
?

Here is a draft proposal:
1) build the (let's say row) vector A = (dPHI/dt) of sampled data at
sampled values t
2) build the row vector B = f(-t) of sample data at t values
3) build the matrix C of (padded) A and the matrix D of (shiffted and
padded) D = B(T-t), with T as multiples of the dt step
Each row of C and D corresponds to a T value.
4) compute E = (C .* D)*dt  when dt is the time step.
5) Set to zero all elements on (#) and above the diagonal of E.
That's to cancel elements with t>T, to code the upper bound T of
the integral.
5) sum E along rows  (the actual integration. You can refine with a
trapeze method).

and that's it: the resulting column vector is your F(T).

Best regards
Samuel

PS: that's an interesting problem
PS2: (#) otherwise F(0) won't be 0
___
users mailing list - users@lists.scilab.org
Click here to unsubscribe: 
https://lists.scilab.org/mailman/listinfo/users
This email and any attachments are intended solely for the use of the 
individual or entity to whom it is addressed and may be confidential and/or 
privileged.

If you are not one of the named recipients or have received this email in error,

(i) you should not read, disclose, or copy it,

(ii) please notify sender of your receipt by reply email and delete this email 
and all attachments,

(iii) Dassault Systèmes does not accept or assume any liability or 
responsibility for any use of or reliance on this email.


Please be informed that your personal data are processed according to our data 
privacy policy as described on our website. Should you have any questions 
related to personal data protection, please contact 3DS Data Protection Officer 
https://www.3ds.com/privacy-policy/contact/

Re: [Scilab-users] Problems with Scilab routine "conv"

[Stéphane Mottelet]

I can't figure out weither the problem resides in your model or in the 
computations, but when I had (many years ago) to use discrete convolution to 
approximate continuous convolution, I noticed that the best way to obtain 
coherent results is to use the composed midpoint rule to approximate the 
integral

h(t)=\int_0^t f(\tau) g(t-\tau) d\tau.

For two causal signals t->f(t)], t-> g(t) with respective support [0,N*δ] and  
[0,M*δ], sampled in the middle of each [i*δ,(i+1)*δ] interval, define the discrete 
sequences

f_i = f((i-1/2)*δ), i=1...N
g_i = g((i-1/2)*δ), i=1..M

then define h_i = h((i*δ) for i=0...N+M by

h_0 = 0
h_i  = δ*\sum_{k=1}^i \tilde f_{i-k+1}*g_k, i = 1...N+M-1
h_{N+M} = 0

h_i for i = 1...N+M-1 is the approximation of the integral by the composed 
midpoint rule and is directly given in Scilab by δ*conv(f,g) where f and g are 
defined above. Here is an example:

function y=f(t)
   y=sin(2*%pi*t);
endfunction

function y=g(t)
   y=sin(%pi*t);
endfunction

δ=0.01;
N=50;  // t->f(t) has support [0,0.5]
M=100; // t->g(t) has support [0,1]

h=[0 δ*conv(f((1/2:N)*δ), g((1/2:M)*δ)) 0];

th=(0:N+M)*δ; // t->h(t) has support [0,1.5]
tf=(0:N)*δ;
tg=(0:M)*δ;

clf
plot(tf,f(tf),tg,g(tg),th,h)
legend f g f*g


S.

Le 03/02/2023 à 11:52, Heinz Nabielek a écrit :

This is my latest code version: the 'convoluted secondary failure fraction' is 
nicely below primary failure,  but seems too low.
Heinz


m=2; // Weibull modulus in mechanism  #1
k=1E-7;  // corrosion rate(s-1) in mechanism  #1
n=500;   // number of time steps in hourly intervals
t=(1:n)';// timesteps in hours
ts= 3600*t;  // timesteps in seconds
PHI= 1 - exp(-((k*ts)^m)); // failure fraction in mechanism #1
deriv=m*k*((k*ts)^(m-1)) .* exp(-((k*ts)^m)); //derivative of PHI
phi=3600*deriv;  //derivative of PHI integrated over an hour
f= 1 - exp(-ts/3E5); // "pure" failure fraction in mechanism #2
F=(convol(phi,f)/n)(1:n);   //"convoluted" failure fraction mechanism #2
plot("nl",t, [PHI phi f F'],'-');xgrid;
legend('PHI primary failure','phi primary failure rate','pure secondary failure 
fraction','convoluted secondary failure fraction',4);
xlabel('time (hours)');
ylabel('failure fraction/ failure rate');





On 03.02.2023, at 11:24, Heinz Nabielek 
 wrote:

On 03.02.2023, at 11:13, Stéphane Mottelet 
 wrote:


Thanks for the code.

Just a remark on the notations, you should write :

F(T)  = Int_{0}^{T}  PHI(t) . f(T-t) . dt

i.e. not F(t) since t is mute.

However, you should pay attention to the delay notion associated with 
convolution and the relationships between discrete convolution and continuous 
convolution. I am not sure that the output of conv matches with a given 
discretization of the integral above. Maybe rectangle method, but I am not sure 
at all. Anyway, you should have F(0)=0 which does not seem to be the case in 
your graph.


F(t), of course.

But the formula is fundamentally wrong and one should have seen it already from 
a dimensional consideration.

Correct should be F(t)  = Int_{0}^{T}  (d PHI(t)/dt) . f(T-t) . dt

Hope you can help between conv and convol and possible something else.
Heinz








Dear SciLab Friends:

I have an object consisting of many (~10,000) small components that can fail in 
a statistical way during long-term operation at extreme conditions.

My primary failure model is described by PHI(t) going monotonically from zero 
to one at times from t=0 to T. In the computer, this is realized in n timesteps.

Mechanism 2: There is a secondary failure mechanism that starts only when a 
component has previously failed by mechanism 1 and occurs after some delay. The 
delay function is f(t) and final expression for the evolution of the mechanism 
2 failure is given by:

F(t)  = Int_{0}^{T}  PHI(t) . f(T-t) . dt

a classical convolution Faltungsintegral.

In Scilab, I write

F= conv(PHI, f, "same")/ n;

and the resulting function F looks reasonable for most of the time. Under some 
conditions, however, I can have

F>PHI

at early stages, but this is physically impossible. Because it comes later, F 
must always be below PHI.
What do I do wrong?
Heinz

_
PS: massively simplified code below..

k=1E-7;  // corrosion rate(s-1) in mechanism #1
n=100;   // number of timesteps
t=(1:n)'; // time in days
ts= 3600*t;  // time in seconds
PHI= 1 - exp(-((k*ts)^2));   // failure fraction in mechanism #1
plot("nl", t, PHI, 'r--');
f= 1 - exp(-ts/3E5); // "pure" failure fraction in mechanism #2
F=conv(PHI,f,"same")/length(t); //"convoluted" failure fraction mechanism #2
plot(t,F,'b--');
legend('failure fraction #1','subsequent failure fraction #2',4);
xlabel('time (hours)');
ylabel('failure fraction');
title('Component failure evolution #1 is followed by #2');


--
Stéphane Mottelet
Ingénieur de recherche
EA 4297 Transformations Intégrées de la Matière 

Re: [Scilab-users] Problems with Scilab routine "conv"

[Stéphane Mottelet]

I can't figure out weither the problem resides in your model or in the 
computations, but when I had (many years ago) to use discrete convolution to 
approximate continuous convolution, I noticed that the best way to obtain 
coherent results is to use the composed midpoint rule to approximate the 
integral

h(t)=\int_0^t f(\tau) g(t-\tau) d\tau.

For two causal signals t->f(t)], t-> g(t) with respective support [0,N*δ] and  
[0,M*δ], sampled in the middle of each [i*δ,(i+1)*δ] interval, define the discrete 
sequences

f_i = f((i-1/2)*δ), i=1...N
g_i = g((i-1/2)*δ), i=1..M

then define h_i = h((i*δ) for i=0...N+M by

h_0 = 0
h_i  = δ*\sum_{k=1}^i \tilde f_{i-k+1}*g_k, i = 1...N+M-1
h_{N+M} = 0

h_i for i = 1...N+M-1 is the approximation of the integral by the composed 
midpoint rule and is directly given in Scilab by δ*conv(f,g) where f and g are 
defined above. Here is an example:

function y=f(t)
   y=sin(2*%pi*t);
endfunction

function y=g(t)
   y=sin(%pi*t);
endfunction

δ=0.01;
N=50;  // t->f(t) has support [0,0.5]
M=100; // t->g(t) has support [0,1]

h=[0 δ*conv(f((1/2:N)*δ), g((1/2:M)*δ)) 0];

th=(0:N+M)*δ; // t->h(t) has support [0,1.5]
tf=(0:N)*δ;
tg=(0:M)*δ;

clf
plot(tf,f(tf),tg,g(tg),th,h)
legend f g f*g

[cid:part1.Q0mLKf6X.Y7yxiCKD@utc.fr]


S.


Le 03/02/2023 à 11:52, Heinz Nabielek a écrit :

This is my latest code version: the 'convoluted secondary failure fraction' is 
nicely below primary failure,  but seems too low.
Heinz


m=2; // Weibull modulus in mechanism  #1
k=1E-7;  // corrosion rate(s-1) in mechanism  #1
n=500;   // number of time steps in hourly intervals
t=(1:n)';// timesteps in hours
ts= 3600*t;  // timesteps in seconds
PHI= 1 - exp(-((k*ts)^m)); // failure fraction in mechanism #1
deriv=m*k*((k*ts)^(m-1)) .* exp(-((k*ts)^m)); //derivative of PHI
phi=3600*deriv;  //derivative of PHI integrated over an hour
f= 1 - exp(-ts/3E5); // "pure" failure fraction in mechanism #2
F=(convol(phi,f)/n)(1:n);   //"convoluted" failure fraction mechanism #2
plot("nl",t, [PHI phi f F'],'-');xgrid;
legend('PHI primary failure','phi primary failure rate','pure secondary failure 
fraction','convoluted secondary failure fraction',4);
xlabel('time (hours)');
ylabel('failure fraction/ failure rate');





On 03.02.2023, at 11:24, Heinz Nabielek 
 wrote:

On 03.02.2023, at 11:13, Stéphane Mottelet 
 wrote:



Thanks for the code.

Just a remark on the notations, you should write :

F(T)  = Int_{0}^{T}  PHI(t) . f(T-t) . dt

i.e. not F(t) since t is mute.

However, you should pay attention to the delay notion associated with 
convolution and the relationships between discrete convolution and continuous 
convolution. I am not sure that the output of conv matches with a given 
discretization of the integral above. Maybe rectangle method, but I am not sure 
at all. Anyway, you should have F(0)=0 which does not seem to be the case in 
your graph.



F(t), of course.

But the formula is fundamentally wrong and one should have seen it already from 
a dimensional consideration.

Correct should be F(t)  = Int_{0}^{T}  (d PHI(t)/dt) . f(T-t) . dt

Hope you can help between conv and convol and possible something else.
Heinz








Dear SciLab Friends:

I have an object consisting of many (~10,000) small components that can fail in 
a statistical way during long-term operation at extreme conditions.

My primary failure model is described by PHI(t) going monotonically from zero 
to one at times from t=0 to T. In the computer, this is realized in n timesteps.

Mechanism 2: There is a secondary failure mechanism that starts only when a 
component has previously failed by mechanism 1 and occurs after some delay. The 
delay function is f(t) and final expression for the evolution of the mechanism 
2 failure is given by:

F(t)  = Int_{0}^{T}  PHI(t) . f(T-t) . dt

a classical convolution Faltungsintegral.

In Scilab, I write

F= conv(PHI, f, "same")/ n;

and the resulting function F looks reasonable for most of the time. Under some 
conditions, however, I can have

F>PHI

at early stages, but this is physically impossible. Because it comes later, F 
must always be below PHI.
What do I do wrong?
Heinz

_
PS: massively simplified code below..

k=1E-7;  // corrosion rate(s-1) in mechanism #1
n=100;   // number of timesteps
t=(1:n)'; // time in days
ts= 3600*t;  // time in seconds
PHI= 1 - exp(-((k*ts)^2));   // failure fraction in mechanism #1
plot("nl", t, PHI, 'r--');
f= 1 - exp(-ts/3E5); // "pure" failure fraction in mechanism #2
F=conv(PHI,f,"same")/length(t); //"convoluted" failure fraction mechanism #2
plot(t,F,'b--');
legend('failure fraction #1','subsequent failure fraction #2',4);
xlabel('time (hours)');
ylabel('failure fraction');
title('Component failure evolution #1 is followed by #2');





--
Stéphane Mottelet
Ingénieur de recherche
EA 

Re: [Scilab-users] Problems with Scilab routine "conv"

[Heinz Nabielek]

This is my latest code version: the 'convoluted secondary failure fraction' is 
nicely below primary failure,  but seems too low.
Heinz


m=2; // Weibull modulus in mechanism  #1
k=1E-7;  // corrosion rate(s-1) in mechanism  #1
n=500;   // number of time steps in hourly intervals
t=(1:n)';// timesteps in hours
ts= 3600*t;  // timesteps in seconds
PHI= 1 - exp(-((k*ts)^m)); // failure fraction in mechanism #1
deriv=m*k*((k*ts)^(m-1)) .* exp(-((k*ts)^m)); //derivative of PHI
phi=3600*deriv;  //derivative of PHI integrated over an hour
f= 1 - exp(-ts/3E5); // "pure" failure fraction in mechanism #2
F=(convol(phi,f)/n)(1:n);   //"convoluted" failure fraction mechanism #2
plot("nl",t, [PHI phi f F'],'-');xgrid;
legend('PHI primary failure','phi primary failure rate','pure secondary failure 
fraction','convoluted secondary failure fraction',4);
xlabel('time (hours)');
ylabel('failure fraction/ failure rate');



> On 03.02.2023, at 11:24, Heinz Nabielek  wrote:
>
> On 03.02.2023, at 11:13, Stéphane Mottelet  wrote:
>>
>> Thanks for the code.
>>
>> Just a remark on the notations, you should write :
>>
>> F(T)  = Int_{0}^{T}  PHI(t) . f(T-t) . dt
>>
>> i.e. not F(t) since t is mute.
>>
>> However, you should pay attention to the delay notion associated with 
>> convolution and the relationships between discrete convolution and 
>> continuous convolution. I am not sure that the output of conv matches with a 
>> given discretization of the integral above. Maybe rectangle method, but I am 
>> not sure at all. Anyway, you should have F(0)=0 which does not seem to be 
>> the case in your graph.
>
> F(t), of course.
>
> But the formula is fundamentally wrong and one should have seen it already 
> from a dimensional consideration.
>
> Correct should be F(t)  = Int_{0}^{T}  (d PHI(t)/dt) . f(T-t) . dt
>
> Hope you can help between conv and convol and possible something else.
> Heinz
>
>
>
>
>
>
>> Dear SciLab Friends:
>>
>> I have an object consisting of many (~10,000) small components that can fail 
>> in a statistical way during long-term operation at extreme conditions.
>>
>> My primary failure model is described by PHI(t) going monotonically from 
>> zero to one at times from t=0 to T. In the computer, this is realized in n 
>> timesteps.
>>
>> Mechanism 2: There is a secondary failure mechanism that starts only when a 
>> component has previously failed by mechanism 1 and occurs after some delay. 
>> The delay function is f(t) and final expression for the evolution of the 
>> mechanism 2 failure is given by:
>>
>> F(t)  = Int_{0}^{T}  PHI(t) . f(T-t) . dt
>>
>> a classical convolution Faltungsintegral.
>>
>> In Scilab, I write
>>
>> F= conv(PHI, f, "same")/ n;
>>
>> and the resulting function F looks reasonable for most of the time. Under 
>> some conditions, however, I can have
>>
>> F>PHI
>>
>> at early stages, but this is physically impossible. Because it comes later, 
>> F must always be below PHI.
>> What do I do wrong?
>> Heinz
>>
>> _
>> PS: massively simplified code below..
>>
>> k=1E-7;  // corrosion rate(s-1) in mechanism #1
>> n=100;   // number of timesteps
>> t=(1:n)'; // time in days
>> ts= 3600*t;  // time in seconds
>> PHI= 1 - exp(-((k*ts)^2));   // failure fraction in mechanism #1
>> plot("nl", t, PHI, 'r--');
>> f= 1 - exp(-ts/3E5); // "pure" failure fraction in mechanism #2
>> F=conv(PHI,f,"same")/length(t); //"convoluted" failure fraction mechanism #2
>> plot(t,F,'b--');
>> legend('failure fraction #1','subsequent failure fraction #2',4);
>> xlabel('time (hours)');
>> ylabel('failure fraction');
>> title('Component failure evolution #1 is followed by #2');

___
users mailing list - users@lists.scilab.org
Click here to unsubscribe: 
https://lists.scilab.org/mailman/listinfo/users
This email and any attachments are intended solely for the use of the 
individual or entity to whom it is addressed and may be confidential and/or 
privileged.

If you are not one of the named recipients or have received this email in error,

(i) you should not read, disclose, or copy it,

(ii) please notify sender of your receipt by reply email and delete this email 
and all attachments,

(iii) Dassault Systèmes does not accept or assume any liability or 
responsibility for any use of or reliance on this email.


Please be informed that your personal data are processed according to our data 
privacy policy as described on our website. Should you have any questions 
related to personal data protection, please contact 3DS Data Protection Officer 
https://www.3ds.com/privacy-policy/contact/

Re: [Scilab-users] Problems with Scilab routine "conv"

[Heinz Nabielek]

On 03.02.2023, at 11:13, Stéphane Mottelet  wrote:
>
> Thanks for the code.
>
> Just a remark on the notations, you should write :
>
> F(T)  = Int_{0}^{T}  PHI(t) . f(T-t) . dt
>
> i.e. not F(t) since t is mute.
>
> However, you should pay attention to the delay notion associated with 
> convolution and the relationships between discrete convolution and continuous 
> convolution. I am not sure that the output of conv matches with a given 
> discretization of the integral above. Maybe rectangle method, but I am not 
> sure at all. Anyway, you should have F(0)=0 which does not seem to be the 
> case in your graph.

F(t), of course.

But the formula is fundamentally wrong and one should have seen it already from 
a dimensional consideration.

Correct should be F(t)  = Int_{0}^{T}  (d PHI(t)/dt) . f(T-t) . dt

Hope you can help between conv and convol and possible something else.
Heinz






> Dear SciLab Friends:
>
> I have an object consisting of many (~10,000) small components that can fail 
> in a statistical way during long-term operation at extreme conditions.
>
> My primary failure model is described by PHI(t) going monotonically from zero 
> to one at times from t=0 to T. In the computer, this is realized in n 
> timesteps.
>
> Mechanism 2: There is a secondary failure mechanism that starts only when a 
> component has previously failed by mechanism 1 and occurs after some delay. 
> The delay function is f(t) and final expression for the evolution of the 
> mechanism 2 failure is given by:
>
> F(t)  = Int_{0}^{T}  PHI(t) . f(T-t) . dt
>
> a classical convolution Faltungsintegral.
>
> In Scilab, I write
>
> F= conv(PHI, f, "same")/ n;
>
> and the resulting function F looks reasonable for most of the time. Under 
> some conditions, however, I can have
>
> F>PHI
>
> at early stages, but this is physically impossible. Because it comes later, F 
> must always be below PHI.
> What do I do wrong?
> Heinz
>
> _
> PS: massively simplified code below..
>
> k=1E-7;  // corrosion rate(s-1) in mechanism #1
> n=100;   // number of timesteps
> t=(1:n)'; // time in days
> ts= 3600*t;  // time in seconds
> PHI= 1 - exp(-((k*ts)^2));   // failure fraction in mechanism #1
> plot("nl", t, PHI, 'r--');
> f= 1 - exp(-ts/3E5); // "pure" failure fraction in mechanism #2
> F=conv(PHI,f,"same")/length(t); //"convoluted" failure fraction mechanism #2
> plot(t,F,'b--');
> legend('failure fraction #1','subsequent failure fraction #2',4);
> xlabel('time (hours)');
> ylabel('failure fraction');
> title('Component failure evolution #1 is followed by #2');
___
users mailing list - users@lists.scilab.org
Click here to unsubscribe: 
https://lists.scilab.org/mailman/listinfo/users
This email and any attachments are intended solely for the use of the 
individual or entity to whom it is addressed and may be confidential and/or 
privileged.

If you are not one of the named recipients or have received this email in error,

(i) you should not read, disclose, or copy it,

(ii) please notify sender of your receipt by reply email and delete this email 
and all attachments,

(iii) Dassault Systèmes does not accept or assume any liability or 
responsibility for any use of or reliance on this email.


Please be informed that your personal data are processed according to our data 
privacy policy as described on our website. Should you have any questions 
related to personal data protection, please contact 3DS Data Protection Officer 
https://www.3ds.com/privacy-policy/contact/

Re: [Scilab-users] Problems with Scilab routine "conv"

[Stéphane Mottelet]

Thanks for the code.

Just a remark on the notations, you should write :

F(T)  = Int_{0}^{T}  PHI(t) . f(T-t) . dt

i.e. not F(t) since t is mute.

However, you should pay attention to the delay notion associated with
convolution and the relationships between discrete convolution and
continuous convolution. I am not sure that the output of conv matches
with a given discretization of the integral above. Maybe rectangle
method, but I am not sure at all. Anyway, you should have F(0)=0 which
does not seem to be the case in your graph.

S.

Le 03/02/2023 à 04:53, Heinz Nabielek a écrit :

Dear SciLab Friends:

I have an object consisting of many (~10,000) small components that can fail in 
a statistical way during long-term operation at extreme conditions.

My primary failure model is described by PHI(t) going monotonically from zero 
to one at times from t=0 to T. In the computer, this is realized in n timesteps.

Mechanism 2: There is a secondary failure mechanism that starts only when a 
component has previously failed by mechanism 1 and occurs after some delay. The 
delay function is f(t) and final expression for the evolution of the mechanism 
2 failure is given by:

F(t)  = Int_{0}^{T}  PHI(t) . f(T-t) . dt

a classical convolution Faltungsintegral.

In Scilab, I write

F= conv(PHI, f, "same")/ n;

and the resulting function F looks reasonable for most of the time. Under some 
conditions, however, I can have

F>PHI

at early stages, but this is physically impossible. Because it comes later, F 
must always be below PHI.
What do I do wrong?
Heinz

_
PS: massively simplified code below..

k=1E-7;  // corrosion rate(s-1) in mechanism #1
n=100;   // number of timesteps
t=(1:n)'; // time in days
ts= 3600*t;  // time in seconds
PHI= 1 - exp(-((k*ts)^2));   // failure fraction in mechanism #1
plot("nl", t, PHI, 'r--');
f= 1 - exp(-ts/3E5); // "pure" failure fraction in mechanism #2
F=conv(PHI,f,"same")/length(t); //"convoluted" failure fraction mechanism #2
plot(t,F,'b--');
legend('failure fraction #1','subsequent failure fraction #2',4);
xlabel('time (hours)');
ylabel('failure fraction');
title('Component failure evolution #1 is followed by #2');


--
Stéphane Mottelet
Ingénieur de recherche
EA 4297 Transformations Intégrées de la Matière Renouvelable
Département Génie des Procédés Industriels
Sorbonne Universités - Université de Technologie de Compiègne
CS 60319, 60203 Compiègne cedex
Tel : +33(0)344234688
http://www.utc.fr/~mottelet

___
users mailing list - users@lists.scilab.org
Click here to unsubscribe: 
https://lists.scilab.org/mailman/listinfo/users
This email and any attachments are intended solely for the use of the 
individual or entity to whom it is addressed and may be confidential and/or 
privileged.

If you are not one of the named recipients or have received this email in error,

(i) you should not read, disclose, or copy it,

(ii) please notify sender of your receipt by reply email and delete this email 
and all attachments,

(iii) Dassault Systèmes does not accept or assume any liability or 
responsibility for any use of or reliance on this email.


Please be informed that your personal data are processed according to our data 
privacy policy as described on our website. Should you have any questions 
related to personal data protection, please contact 3DS Data Protection Officer 
https://www.3ds.com/privacy-policy/contact/