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You missing one, very important step. In order to get the impulse response of your filter while using the sweep sine, you must get the output of your system $h[n]$ to excitation with sweep-sine signal $x[n]$:

$y[n]=h[n]\star x[n]$

In your case that's the filtering with filter function. Now in order to get the impulse response you must convolve your signal with the inverse filter $f[n]$. Generally that's the time-inverted sweep signal (although special kind of pre-conditioning, amplitude scaling, etc. might be done as well):

$h[n]=y[n]\star f[n]$

Now you have your impulse response that you can use for further analysis. Please take a look at this answeranswer tackling very similar problem.

For more theory please see:

A. Farina - Advancements in Impulse Response Measurements by Sine Sweeps

A. Farina - Simultaneous Measurement of Impulse Response and Distortion with a Swept-Sine Technique

You missing one, very important step. In order to get the impulse response of your filter while using the sweep sine, you must get the output of your system $h[n]$ to excitation with sweep-sine signal $x[n]$:

$y[n]=h[n]\star x[n]$

In your case that's the filtering with filter function. Now in order to get the impulse response you must convolve your signal with the inverse filter $f[n]$. Generally that's the time-inverted sweep signal (although special kind of pre-conditioning, amplitude scaling, etc. might be done as well):

$h[n]=y[n]\star f[n]$

Now you have your impulse response that you can use for further analysis. Please take a look at this answer tackling very similar problem.

For more theory please see:

A. Farina - Advancements in Impulse Response Measurements by Sine Sweeps

A. Farina - Simultaneous Measurement of Impulse Response and Distortion with a Swept-Sine Technique

You missing one, very important step. In order to get the impulse response of your filter while using the sweep sine, you must get the output of your system $h[n]$ to excitation with sweep-sine signal $x[n]$:

$y[n]=h[n]\star x[n]$

In your case that's the filtering with filter function. Now in order to get the impulse response you must convolve your signal with the inverse filter $f[n]$. Generally that's the time-inverted sweep signal (although special kind of pre-conditioning, amplitude scaling, etc. might be done as well):

$h[n]=y[n]\star f[n]$

Now you have your impulse response that you can use for further analysis. Please take a look at this answer tackling very similar problem.

For more theory please see:

A. Farina - Advancements in Impulse Response Measurements by Sine Sweeps

A. Farina - Simultaneous Measurement of Impulse Response and Distortion with a Swept-Sine Technique

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jojeck
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You missing one, very important step. In order to get the impulse response of your filter while using the sweep sine, you must get the output of your system $h[n]$ to excitation with sweep-sine signal $x[n]$:

$y[n]=h[n]\star x[n]$

In your case that's the filtering with filter function. Now in order to get the impulse response you must convolve your signal with the inverse filter $f[n]$. Generally that's the time-inverted sweep signal (although special kind of pre-conditioning, amplitude scaling, etc. might be done as well):

$h[n]=y[n]\star f[n]$

Now you have your impulse response that you can use for further analysis. Please take a look at this answer tackling very similar problem.

For more theory please see:

A. Farina - Advancements in Impulse Response Measurements by Sine Sweeps

A. Farina - Simultaneous Measurement of Impulse Response and Distortion with a Swept-Sine Technique