# Instananeous frequency of a chirp signal is halved?

• Luongo
In summary, the conversation discusses using MATLAB to research and analyze a chirp signal. The speaker takes the FFT of the chirp, sets the upper half of the FT to 0, and takes the IFFT to recover the original chirp. They then use a formula for calculating the instantaneous frequency, but notice that their results do not match what they expected. After discussing possible errors in their formula, they realize their mistake and correct it.
Luongo
I'm doing some research with MATLAB, where i have a simple chirp. y = sin(x.^2)
i take the FFT of the chirp, set the upper half of the FT of the signal to 0, and take the IFFT of this to recover the original chirp y with real and imaginary parts.
then i use the formula for calculation of the instantaneous frequency:

(s1.*ds2 - s2.*ds1)./(s1.^2 +s2.^2)
where s1 is the real part of y, s2 is imag of y
and ds1 and ds2 are the derivatives of s1 and s2 respectively.

It turns out i get my inst. freq to be y = x. but i expected y = 2x.
Is there any reason why my inst. frequency is missing the factor of 2? I have no idea what I'm doing wrong that results in the missing factor.
thanks!
any help is greatly appreciated.

here is my MATLAB code:
function []=chirp()

x = 0:0.01:5;
y = sin(x.^2);
y = sin(x.^2)-mean(y); %eliminate DC offset subtract mean
plot(x,y,'k')
hold on
y = fft(y);
y = 2*y; %double the amplitude
y(250:end) = 0; %cut off upper half the signal
y = ifft(y);
%double amplitude of the chirp signal
s1 = real(y);
plot(x,s1, 'p')
s2 = imag(y);
%plot(x,s2, 'r')
ds1 = diff(s1)./0.01;
ds1(501) = 0;
ds2 = diff(s2)./0.01;
ds2(501) = 0;
%derivative plots:
%plot(x,ds1,'y');
%plot(x,ds2,'k');
%obtain instantaneous frequency
n = (s1.*ds2 - s2.*ds1);
d = (s1.^2 +s2.^2+ (max(s1).^2));
q = n./d;
plot(x,q, 'g')

How did you arrive at your formula for the instantaneous frequency? I suspect that it might be off.

If I calculate the instantaneous frequency using a different approach, with the analytical signal 'y' using your code:

dy = diff(y)/0.01;
dy(501) = 0;
instfrq = imag(dy./y);

I get something that tracks 2*x instead. Here I used the imaginary part of the log-derivative of the analytic signal:

$\hat{y} = A e^{i\varphi(t)}$
$\frac{d\hat{y}}{dt}=i \frac{d\phi(t)}{dt} A e^{i\varphi(t)}$
$\rightarrow \frac{d\phi(t)}{dt} = \frac{\frac{d\hat{y}}{dt}}{i \hat{y}}$

it turns out i made a mistake with my epsilion correction factor, it works now though. thanks

## 1. What is the meaning of "instananeous frequency of a chirp signal is halved"?

The instantaneous frequency of a chirp signal refers to the frequency of the signal at a specific moment in time. When it is halved, it means that the frequency decreases by half compared to the previous moment.

## 2. Why is it important to measure the instantaneous frequency of a chirp signal?

The instantaneous frequency of a chirp signal is important because it gives information about the rate of change of the signal. This can be useful in various applications, such as radar systems, sonar, and medical imaging.

## 3. How is the instantaneous frequency of a chirp signal calculated?

The instantaneous frequency of a chirp signal is calculated by taking the derivative of the phase of the signal with respect to time. This can be done using mathematical formulas or through signal processing techniques.

## 4. Can the instantaneous frequency of a chirp signal be negative?

Yes, the instantaneous frequency of a chirp signal can be negative. This can occur when the frequency is decreasing over time, such as in a downward chirp signal.

## 5. How does halving the instantaneous frequency of a chirp signal affect the overall signal?

Halving the instantaneous frequency of a chirp signal will cause the signal to have a lower frequency overall. This can result in a longer wavelength, a slower rate of change, and potentially a longer duration of the signal.

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