I have a question concern "linear sine sweep".

Here is the equation of linear sine sweep(from Wikipedia Chirp).

$$ x(t) = sin\biggl[ 2\pi( f_0 +\frac{f_1-f_0}{2T}t)t\biggr] $$


$f_0$ is start frequency

$f_1$ is end frequency

$T$ is total length of excitation time.

In my case, I would like to calculate the excited frequency at certain time frame.

Let me calculate it through an example.

$f_0 = 20 \text{ Hz}$ , $f_1=200 \text{ Hz}$, $T = 60\text{ s}$.

In this case, above equation can be written as below,

$$ x(t) = sin\biggl[ 2\pi( 20 +1.5t)t\biggr]. $$

If we estimate the frequency at time 30 s, I think we could calculate as 65 Hz.

For checking this is right or not, I checked the power spectrum by selecting time through 29.5 s - 30.5 s for frequency resolution 1 Hz.

However, from the power spectrum, the frequency was 112 Hz.

There is difference between simulation value and calculated value almost twice.

Also, I just checked 1 second(29.5 s - 30.5 s) Fourier transform absolute value, it also displayed the peak frequency as 112 Hz.

What is the correct value?

If the calculated value, 65 Hz, is wrong, is there any comment for correcting this?

Thank you in advance.

If there is any comment or question, please let me know.


Frequency is the first derivative of the phase so in this case that's f0+(f0-f1)*t/T. If you plug in the numbers you get 110Hz, which is what your measurement shows. Not sure how you got 65Hz for that ??

It's intuitively clear as well. The sweep starts at 20 Hz and ends at 200 Hz after 60s. After 30s you should be at the half way point between 20 and 200, which is 110 Hz.

$$\varphi (t) = 2 \pi \cdot (f_{0}+\frac{f_{1}-f_{0}}{2T}\cdot t)\cdot t = 2 \pi \cdot (f_{0}\cdot t+\frac{f_{1} -f_{0}}{2T}\cdot t^{2})$$

$$f(t)=\frac{1}{2 \pi}\cdot \frac{\partial \phi}{\partial t}=f_{0}+\frac{f_{1}-f_{0}}{T}t$$

  • 1
    $\begingroup$ Thank you for your answer. This is really helpful for me. My opinion, I thought the frequency which is variable of time is (20 + 1.5t) at my case. So I just multiply 30 into t, I got 65 Hz. Thank you. $\endgroup$ – actlee Jan 3 '14 at 16:51

See the recent topic (Generating a sound where end frequency is 1/2 of the start frequency) on the opposite case of constructing such a signal.

As used in the wikipedia article, but not explained, the local or instantaneous frequency is the derivative of the phase function.

In the interval $[t-Δt, t+Δt]$ the signal moves through $\frac{\phi( t+Δt)-\phi(t-Δt)}{2\pi}\approx\frac{\phi'(t)}{2\pi}\cdot 2Δt$ full periods of the sine function. So for small intervals, but large enough to contain still a few oscillations, $\frac{\phi'(t)}{2\pi}$ is a good approximation for the local frequency.


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