Mixing
Finding transfer function with Fast Fourier Fransform.
$begingroup$
I have two signals with input = a(t) and output = b(t) that have been sampled every 0.01s and as such the fast Fourier transform has been used on both and utilised to produce a transfer function.
The only problem is that currently I'm left with a massive list of complex values. How would one find a closed form type expression for such a system? I imagine I need to find the poles and zeros?
Edit: For clarity I would like to say that I used numpy.FFT (python language) and it produces an array of complex values. This was applied to both a(t) and b(t). The FFT of b was divided by the FFT of a. This leaves me with an array of complex numbers.
fourier-analysis signal-processing
asked Mar 19 '14 at 20:12
user3394391user3394391
1135
$endgroup$
add a comment |
$begingroup$
I have two signals with input = a(t) and output = b(t) that have been sampled every 0.01s and as such the fast Fourier transform has been used on both and utilised to produce a transfer function.
The only problem is that currently I'm left with a massive list of complex values. How would one find a closed form type expression for such a system? I imagine I need to find the poles and zeros?
Edit: For clarity I would like to say that I used numpy.FFT (python language) and it produces an array of complex values. This was applied to both a(t) and b(t). The FFT of b was divided by the FFT of a. This leaves me with an array of complex numbers.
fourier-analysis signal-processing
asked Mar 19 '14 at 20:12
user3394391user3394391
1135
$endgroup$
add a comment |
$begingroup$
I have two signals with input = a(t) and output = b(t) that have been sampled every 0.01s and as such the fast Fourier transform has been used on both and utilised to produce a transfer function.
The only problem is that currently I'm left with a massive list of complex values. How would one find a closed form type expression for such a system? I imagine I need to find the poles and zeros?
Edit: For clarity I would like to say that I used numpy.FFT (python language) and it produces an array of complex values. This was applied to both a(t) and b(t). The FFT of b was divided by the FFT of a. This leaves me with an array of complex numbers.
fourier-analysis signal-processing
asked Mar 19 '14 at 20:12
user3394391user3394391
1135
$endgroup$
I have two signals with input = a(t) and output = b(t) that have been sampled every 0.01s and as such the fast Fourier transform has been used on both and utilised to produce a transfer function.
The only problem is that currently I'm left with a massive list of complex values. How would one find a closed form type expression for such a system? I imagine I need to find the poles and zeros?
Edit: For clarity I would like to say that I used numpy.FFT (python language) and it produces an array of complex values. This was applied to both a(t) and b(t). The FFT of b was divided by the FFT of a. This leaves me with an array of complex numbers.
fourier-analysis signal-processing
fourier-analysis signal-processing
asked Mar 19 '14 at 20:12
user3394391user3394391
1135
asked Mar 19 '14 at 20:12
user3394391user3394391
1135
asked Mar 19 '14 at 20:12
user3394391user3394391
1135
asked Mar 19 '14 at 20:12
user3394391user3394391
1135
asked Mar 19 '14 at 20:12
user3394391user3394391
1135
1135
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let A(f) and B(f) be the FFT's from a(t) respectively b(t).
Then you're supposed to take the inverse FFT from B(f)/A(f).
It will result in an array of real values that identifies the transfer function with respect to time.
answered Mar 19 '14 at 20:48
Klaas van AarsenKlaas van Aarsen
4,3421822
$endgroup$
$begingroup$
Thank you for your reply @i-like-selenaa ... I've often seen transfer functions expressed as polynomials with s's in the numerator and denominator for the case of laplace transforms. Won't your suggestion provide me with an impulse response?
$endgroup$
– user3394391
Mar 19 '14 at 21:03
$begingroup$
I just ran the inverse on B/A and still have complex values. EDIT: Just noticed that they are of tiny order though like e^-14.
$endgroup$
– user3394391
Mar 19 '14 at 21:14
$begingroup$
Close enough. The transfer function convolves with the input function to yield the output function. If your input function would be the impulse function, the output function happens to be the same as the transfer function, which is why the impulse response is so important. Note that Laplace transforms are identical to Fourier transforms with $s=iomega$.
$endgroup$
– Klaas van Aarsen
Mar 19 '14 at 21:14
$begingroup$
Yep. Those tiny order imaginary values would be rounding errors as you may have surmised.
$endgroup$
– Klaas van Aarsen
Mar 19 '14 at 21:19
$begingroup$
I just took a plot to see what it would look like (with just the real values) and it's a spikey mess! Do I want to try and fit a function to this?
$endgroup$
– user3394391
Mar 19 '14 at 21:22
|
show 4 more comments
$begingroup$
If you are starting from numerical data and want to estimate an analytical expression for it, then you will have to perform some sort of model fit to the data. For signal modeling, you can search for topics such as least squares, Pade approximations, and Prony's method to name a few.
Here are some slides I quickly found with google that wil provide a starting point for you.
answered Mar 21 '14 at 22:31
AnonSubmitter85AnonSubmitter85
2,85221420
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let A(f) and B(f) be the FFT's from a(t) respectively b(t).
Then you're supposed to take the inverse FFT from B(f)/A(f).
It will result in an array of real values that identifies the transfer function with respect to time.
answered Mar 19 '14 at 20:48
Klaas van AarsenKlaas van Aarsen
4,3421822
$endgroup$
$begingroup$
Thank you for your reply @i-like-selenaa ... I've often seen transfer functions expressed as polynomials with s's in the numerator and denominator for the case of laplace transforms. Won't your suggestion provide me with an impulse response?
$endgroup$
– user3394391
Mar 19 '14 at 21:03
$begingroup$
I just ran the inverse on B/A and still have complex values. EDIT: Just noticed that they are of tiny order though like e^-14.
$endgroup$
– user3394391
Mar 19 '14 at 21:14
$begingroup$
Close enough. The transfer function convolves with the input function to yield the output function. If your input function would be the impulse function, the output function happens to be the same as the transfer function, which is why the impulse response is so important. Note that Laplace transforms are identical to Fourier transforms with $s=iomega$.
$endgroup$
– Klaas van Aarsen
Mar 19 '14 at 21:14
$begingroup$
Yep. Those tiny order imaginary values would be rounding errors as you may have surmised.
$endgroup$
– Klaas van Aarsen
Mar 19 '14 at 21:19
$begingroup$
I just took a plot to see what it would look like (with just the real values) and it's a spikey mess! Do I want to try and fit a function to this?
$endgroup$
– user3394391
Mar 19 '14 at 21:22
|
show 4 more comments
$begingroup$
Let A(f) and B(f) be the FFT's from a(t) respectively b(t).
Then you're supposed to take the inverse FFT from B(f)/A(f).
It will result in an array of real values that identifies the transfer function with respect to time.
answered Mar 19 '14 at 20:48
Klaas van AarsenKlaas van Aarsen
4,3421822
$endgroup$
$begingroup$
Thank you for your reply @i-like-selenaa ... I've often seen transfer functions expressed as polynomials with s's in the numerator and denominator for the case of laplace transforms. Won't your suggestion provide me with an impulse response?
$endgroup$
– user3394391
Mar 19 '14 at 21:03
$begingroup$
I just ran the inverse on B/A and still have complex values. EDIT: Just noticed that they are of tiny order though like e^-14.
$endgroup$
– user3394391
Mar 19 '14 at 21:14
$begingroup$
Close enough. The transfer function convolves with the input function to yield the output function. If your input function would be the impulse function, the output function happens to be the same as the transfer function, which is why the impulse response is so important. Note that Laplace transforms are identical to Fourier transforms with $s=iomega$.
$endgroup$
– Klaas van Aarsen
Mar 19 '14 at 21:14
$begingroup$
Yep. Those tiny order imaginary values would be rounding errors as you may have surmised.
$endgroup$
– Klaas van Aarsen
Mar 19 '14 at 21:19
$begingroup$
I just took a plot to see what it would look like (with just the real values) and it's a spikey mess! Do I want to try and fit a function to this?
$endgroup$
– user3394391
Mar 19 '14 at 21:22
|
show 4 more comments
$begingroup$
Let A(f) and B(f) be the FFT's from a(t) respectively b(t).
Then you're supposed to take the inverse FFT from B(f)/A(f).
It will result in an array of real values that identifies the transfer function with respect to time.
answered Mar 19 '14 at 20:48
Klaas van AarsenKlaas van Aarsen
4,3421822
$endgroup$
Let A(f) and B(f) be the FFT's from a(t) respectively b(t).
Then you're supposed to take the inverse FFT from B(f)/A(f).
It will result in an array of real values that identifies the transfer function with respect to time.
answered Mar 19 '14 at 20:48
Klaas van AarsenKlaas van Aarsen
4,3421822
answered Mar 19 '14 at 20:48
Klaas van AarsenKlaas van Aarsen
4,3421822
answered Mar 19 '14 at 20:48
Klaas van AarsenKlaas van Aarsen
4,3421822
answered Mar 19 '14 at 20:48
Klaas van AarsenKlaas van Aarsen
4,3421822
4,3421822
$begingroup$
Thank you for your reply @i-like-selenaa ... I've often seen transfer functions expressed as polynomials with s's in the numerator and denominator for the case of laplace transforms. Won't your suggestion provide me with an impulse response?
$endgroup$
– user3394391
Mar 19 '14 at 21:03
$begingroup$
I just ran the inverse on B/A and still have complex values. EDIT: Just noticed that they are of tiny order though like e^-14.
$endgroup$
– user3394391
Mar 19 '14 at 21:14
$begingroup$
Close enough. The transfer function convolves with the input function to yield the output function. If your input function would be the impulse function, the output function happens to be the same as the transfer function, which is why the impulse response is so important. Note that Laplace transforms are identical to Fourier transforms with $s=iomega$.
$endgroup$
– Klaas van Aarsen
Mar 19 '14 at 21:14
$begingroup$
Yep. Those tiny order imaginary values would be rounding errors as you may have surmised.
$endgroup$
– Klaas van Aarsen
Mar 19 '14 at 21:19
$begingroup$
I just took a plot to see what it would look like (with just the real values) and it's a spikey mess! Do I want to try and fit a function to this?
$endgroup$
– user3394391
Mar 19 '14 at 21:22
|
show 4 more comments
$begingroup$
Thank you for your reply @i-like-selenaa ... I've often seen transfer functions expressed as polynomials with s's in the numerator and denominator for the case of laplace transforms. Won't your suggestion provide me with an impulse response?
$endgroup$
– user3394391
Mar 19 '14 at 21:03
$begingroup$
I just ran the inverse on B/A and still have complex values. EDIT: Just noticed that they are of tiny order though like e^-14.
$endgroup$
– user3394391
Mar 19 '14 at 21:14
$begingroup$
Close enough. The transfer function convolves with the input function to yield the output function. If your input function would be the impulse function, the output function happens to be the same as the transfer function, which is why the impulse response is so important. Note that Laplace transforms are identical to Fourier transforms with $s=iomega$.
$endgroup$
– Klaas van Aarsen
Mar 19 '14 at 21:14
$begingroup$
Yep. Those tiny order imaginary values would be rounding errors as you may have surmised.
$endgroup$
– Klaas van Aarsen
Mar 19 '14 at 21:19
$begingroup$
I just took a plot to see what it would look like (with just the real values) and it's a spikey mess! Do I want to try and fit a function to this?
$endgroup$
– user3394391
Mar 19 '14 at 21:22
$begingroup$
Thank you for your reply @i-like-selenaa ... I've often seen transfer functions expressed as polynomials with s's in the numerator and denominator for the case of laplace transforms. Won't your suggestion provide me with an impulse response?
$endgroup$
– user3394391
Mar 19 '14 at 21:03
$begingroup$
Thank you for your reply @i-like-selenaa ... I've often seen transfer functions expressed as polynomials with s's in the numerator and denominator for the case of laplace transforms. Won't your suggestion provide me with an impulse response?
$endgroup$
– user3394391
Mar 19 '14 at 21:03
$begingroup$
I just ran the inverse on B/A and still have complex values. EDIT: Just noticed that they are of tiny order though like e^-14.
$endgroup$
– user3394391
Mar 19 '14 at 21:14
$begingroup$
I just ran the inverse on B/A and still have complex values. EDIT: Just noticed that they are of tiny order though like e^-14.
$endgroup$
– user3394391
Mar 19 '14 at 21:14
$begingroup$
Close enough. The transfer function convolves with the input function to yield the output function. If your input function would be the impulse function, the output function happens to be the same as the transfer function, which is why the impulse response is so important. Note that Laplace transforms are identical to Fourier transforms with $s=iomega$.
$endgroup$
– Klaas van Aarsen
Mar 19 '14 at 21:14
$begingroup$
Close enough. The transfer function convolves with the input function to yield the output function. If your input function would be the impulse function, the output function happens to be the same as the transfer function, which is why the impulse response is so important. Note that Laplace transforms are identical to Fourier transforms with $s=iomega$.
$endgroup$
– Klaas van Aarsen
Mar 19 '14 at 21:14
$begingroup$
Yep. Those tiny order imaginary values would be rounding errors as you may have surmised.
$endgroup$
– Klaas van Aarsen
Mar 19 '14 at 21:19
$begingroup$
Yep. Those tiny order imaginary values would be rounding errors as you may have surmised.
$endgroup$
– Klaas van Aarsen
Mar 19 '14 at 21:19
$begingroup$
I just took a plot to see what it would look like (with just the real values) and it's a spikey mess! Do I want to try and fit a function to this?
$endgroup$
– user3394391
Mar 19 '14 at 21:22
$begingroup$
I just took a plot to see what it would look like (with just the real values) and it's a spikey mess! Do I want to try and fit a function to this?
$endgroup$
– user3394391
Mar 19 '14 at 21:22
|
show 4 more comments
$begingroup$
If you are starting from numerical data and want to estimate an analytical expression for it, then you will have to perform some sort of model fit to the data. For signal modeling, you can search for topics such as least squares, Pade approximations, and Prony's method to name a few.
Here are some slides I quickly found with google that wil provide a starting point for you.
answered Mar 21 '14 at 22:31
AnonSubmitter85AnonSubmitter85
2,85221420
$endgroup$
add a comment |
$begingroup$
If you are starting from numerical data and want to estimate an analytical expression for it, then you will have to perform some sort of model fit to the data. For signal modeling, you can search for topics such as least squares, Pade approximations, and Prony's method to name a few.
Here are some slides I quickly found with google that wil provide a starting point for you.
answered Mar 21 '14 at 22:31
AnonSubmitter85AnonSubmitter85
2,85221420
$endgroup$
add a comment |
$begingroup$
If you are starting from numerical data and want to estimate an analytical expression for it, then you will have to perform some sort of model fit to the data. For signal modeling, you can search for topics such as least squares, Pade approximations, and Prony's method to name a few.
Here are some slides I quickly found with google that wil provide a starting point for you.
answered Mar 21 '14 at 22:31
AnonSubmitter85AnonSubmitter85
2,85221420
$endgroup$
If you are starting from numerical data and want to estimate an analytical expression for it, then you will have to perform some sort of model fit to the data. For signal modeling, you can search for topics such as least squares, Pade approximations, and Prony's method to name a few.
Here are some slides I quickly found with google that wil provide a starting point for you.
answered Mar 21 '14 at 22:31
AnonSubmitter85AnonSubmitter85
2,85221420
answered Mar 21 '14 at 22:31
AnonSubmitter85AnonSubmitter85
2,85221420
answered Mar 21 '14 at 22:31
AnonSubmitter85AnonSubmitter85
2,85221420
answered Mar 21 '14 at 22:31
AnonSubmitter85AnonSubmitter85
2,85221420
2,85221420
add a comment |
add a comment |
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