Chirp Signal – FFT & PSD in Matlab & Python

Key focus: Know how to generate a Chirp signal, compute its Fourier Transform using FFT and power spectral density (PSD) in Matlab & Python.

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Introduction

All the signals discussed so far do not change in frequency over time. Obtaining a signal with time-varying frequency is of main focus here. A signal that varies in frequency over time is called “chirp”. The frequency of the chirp signal can vary from low to high frequency (up-chirp) or from high to low frequency (low-chirp).

Observation

Chirp signals/signatures are encountered in many applications ranging from radar, sonar, spread spectrum, optical communication, image processing, doppler effect, motion of a pendulum, as gravitation waves, manifestation as Frequency Modulation (FM), echo location [1] etc.

Mathematical Description:

A linear chirp signal sweeps the frequency from low to high frequency (or vice-versa) linearly. One approach to generate a chirp signal is to concatenate a series of segments of sine waves each with increasing(or decreasing) frequency in order. This method introduces discontinuities in the chirp signal due to the mismatch in the phases of each such segments. Modifying the equation of a sinusoid to generate a chirp signal is a better approach.

The equation for generating a sinusoidal (cosine here) signal with amplitude A, angular frequency $latex \omega_0$ and initial phase $latex \phi$ is

$latex x(t)=A cos(\omega_0 t + \phi ) \quad\quad (1) &s=1$

This can be written as a function of instantaneous phase

$latex x(t)= A cos(\theta (t)) = A cos ( \omega_0 t+ \phi ) = A cos ( 2 \pi f_0 t+ \phi ) \quad (2) &s=1$

where $latex \theta (t) = \omega_0 t+ \phi$ is the instantaneous phase of the sinusoid and it is linear in time. The time derivative of instantaneous phase $latex \theta(t)$ is equal to the angular frequency $latex \omega$ of the sinusoid – which in case is a constant in the above equation.

$latex \omega_0 = \frac{d}{dt} \theta (t) \quad\quad (3)&s=1$

Instead of having the phase linear in time, let’s change the phase to quadratic form and thus non-linear.

$latex \theta(t) = 2 \pi \alpha t^2 + 2 \pi f_0 t + \phi \quad\quad (4) &s=1$

for some constant $latex \alpha$.

Therefore, the equation for chirp signal takes the following form,

$latex x(t) = A cos(\theta(t)) = A cos( 2 \pi \alpha t^2 + 2 \pi f_0 t + \phi ) \quad (5)&s=1$

The first derivative of the phase, which is the instantaneous angular frequency becomes a function of time, which is given by

$latex \omega_i(t)=\frac{d}{dt} \theta (t) = 4 \pi \alpha t + 2 \pi f_0 \quad\quad (6)&s=1$

The time-varying frequency in Hertz is given by

$latex f_i (t) = 2 \alpha t + f_0 \quad\quad (7)&s=1$

In the above equation, the frequency is no longer a constant, rather it is of time-varying nature with initial frequency given by $latex f_0$. Thus, from the above equation, given a time duration T, the rate of change of frequency is given by

$latex k = 2 \alpha = \frac{f_1-f_0}{T} \quad\quad (8)&s=1$

where, $latex f_0$ is the starting frequency of the sweep, $latex f_1$ is the frequency at the end of the duration T.

Substituting (7) & (8) in (6)

$latex \omega_i(t)=\frac{d}{dt} \theta (t) = 2 \pi (kt+f_0) \quad\quad (9) &s=1$

From (6) and (8)

$latex \begin{aligned} \theta (t) &= \int \omega_i(t) dt\\ &= 2 \pi \int (kt+f_0) dt = 2 \pi \left(k\frac{t^2}{2}+f_0 t \right) + \phi_0\\ &= 2 \pi \left(k\frac{t^2}{2}+f_0 t \right) + \phi_0 = 2 \pi \left(\frac{k}{2} t +f_0 \right) t + \phi_0\end{aligned}\quad\quad (10) &s=1$

where $latex \phi_0$ is a constant which will act as the initial phase of the sweep.

Thus the modified equation for generating a chirp signal (from equations (5) and (10)) is given by

$latex x(t) = A cos(\theta(t)) = A cos (2 \pi f(t) t + \phi_0) \quad \quad (11) &s=1$

where the time-varying frequency function is given by

$latex f(t)= \frac{k}{2} t +f_0 \quad \quad\quad \quad (12)&s=1$

Generation of Chirp signal, computing its Fourier Transform using FFT and power spectral density (PSD) in Matlab is shown as example, for Python code, please refer the book Digital Modulations using Python.

Generating a chirp signal without using in-built “chirp” Function in Matlab:

Implement a function that describes the chirp using equation (11) and (12). The starting frequency of the sweep is $latex f_0$ and the frequency at time $latex t_1$ is $latex f_1$. The initial phase $latex \phi_0$ forms the final part of the argument in the following function

function x=mychirp(t,f0,t1,f1,phase)
%Y = mychirp(t,f0,t1,f1) generates samples of a linear swept-frequency
%   signal at the time instances defined in timebase array t.  The instantaneous
%   frequency at time 0 is f0 Hertz.  The instantaneous frequency f1
%   is achieved at time t1.
%   The argument 'phase' is optional. It defines the initial phase of the
%   signal degined in radians. By default phase=0 radian
    
if nargin==4
    phase=0;
end
    t0=t(1);
    T=t1-t0;
    k=(f1-f0)/T;
    x=cos(2*pi*(k/2*t+f0).*t+phase);
end

The following wrapper script utilizes the above function and generates a chirp with starting frequency $latex f_0=1Hz$ at the start of the time base and $latex f_1=25 Hz$ at $latex t_1=1s$ which is the end of the time base. From the PSD plot, it can be ascertained that the signal energy is concentrated only upto 25 Hz

fs=500; %sampling frequency
t=0:1/fs:1; %time base - upto 1 second

f0=1;% starting frequency of the chirp
f1=fs/20; %frequency of the chirp at t1=1 second
x = mychirp(t,f0,1,f1); 
subplot(2,2,1)
plot(t,x,'k');
title(['Chirp Signal']);
xlabel('Time(s)');
ylabel('Amplitude');

FFT and power spectral density

As with other signals, describes in the previous posts, let’s plot the FFT of the generated chirp signal and its power spectral density (PSD).

L=length(x);
NFFT = 1024;
X = fftshift(fft(x,NFFT));
Pxx=X.*conj(X)/(NFFT*NFFT); %computing power with proper scaling
f = fs*(-NFFT/2:NFFT/2-1)/NFFT; %Frequency Vector

subplot(2,2,2)
plot(f,abs(X)/(L),'r');
title('Magnitude of FFT');
xlabel('Frequency (Hz)')
ylabel('Magnitude |X(f)|');
xlim([-50 50])


Pxx=X.*conj(X)/(NFFT*NFFT); %computing power with proper scaling
subplot(2,2,3)
plot(f,10*log10(Pxx),'r');
title('Double Sided - Power Spectral Density');
xlabel('Frequency (Hz)')
ylabel('Power Spectral Density- P_{xx} dB/Hz');
xlim([-100 100])

X = fft(x,NFFT);
X = X(1:NFFT/2+1);%Throw the samples after NFFT/2 for single sided plot
Pxx=X.*conj(X)/(NFFT*NFFT);
f = fs*(0:NFFT/2)/NFFT; %Frequency Vector
subplot(2,2,4)
plot(f,10*log10(Pxx),'r');
title('Single Sided - Power Spectral Density');
xlabel('Frequency (Hz)')
ylabel('Power Spectral Density- P_{xx} dB/Hz');

For Python code, please refer the book Digital Modulations using Python
Chirp signal FFT and power spectral density in Matlab

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References:

[1] Patrick Flandrin,“Chirps everywhere”,CNRS — Ecole Normale Supérieure de Lyon

Topics in this chapter

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Books by the author

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24 thoughts on “Chirp Signal – FFT & PSD in Matlab & Python”

  1. Does anyone know any way to produce a chirp in Matlab that follows a custom polynomial rather than the pre-built in functions provided in the chirp() function?

    Does the chirp() function allow for more customization than I am aware of?

    I want to produce a chirp based on the following polynomial (determined using polyfit() on a set of data points)…

    y2 = p(1).*x.^10 + p(2).*x.^9 + p(3).*x.^8 + p(4).*x.^7 + p(5).*x.^6 + p(6).*x.^5 + p(7).*x.^4 + p(8).*x.^3 + p(9).*x.^2 + p(10).*x + p(11) ;

    Reply
  2. Hi Mathuranathan. Thanks for the detailed explanation. Is there a typo? Formula (2) correctly states “x(t) = Acos(ÆŸ(t))”. But immediately after that you state “where ÆŸ(t) = ω0 + Ф is the instantaneous phase and it is linear in time.” Looks like there’s a ‘t’ missing: I think it should read “where ÆŸ(t) = ω0t + Ф is the instantaneous phase and it is linear in time.”

    Reply
  3. Hey nice article. I have a question: why is the amplitude of the second plot (magnitude of fft) not one? Because all the frequencys in your chirp signal have the amplitude of one? Am I missing something

    Reply
  4. Thank you Mathuranathan

    My question could I multiply the the values of frequency (in second graph) by the ratio (1/0.1, which is the amplitude of of chirp signal on the amplitude of FFt), and then we have a frequency-acceleration graph?
    Regards

    Reply
  5. I want to add into my question that there seems to be a difference between Eq(6) and Eq(8). Can you explain this difference?

    Reply
  6. Hello Mathuranathan

    Great article with proper detailed explanations about chirps! Regarding PSD, may I know why you have scaled the magnitude^2 by NFFT twice (psd = X.conj(X)/(NFFTNFFT))?

    Reply
  7. Hello Mathuranathan

    plot(f,abs(X)/(L),’r’);

    Why abs(X) is divide by (L)? Can’t we just write plot(f,abs(X),’r’); only?

    Thank You.

    Reply
    • FFT is taken on a vector of size L. Hence the amplitude is normalized by dividing by L. If the shape of the amplitude spectrum is important than the actual amplitude values, then the scaling factor can be ignored.

      Reply
    • Sinusoidal signal is a smooth repetition of oscillation at a fixed frequency. Chirp signal is also a sinusoidal signal whose frequency continuously increases (or decreases).

      Reply
    • Magnitude response is the magnitude of the complex value of the frequency response vs. frequency.
      It is used to see how a system affects the amplitudes of frequencies in a signal.

      A variable’s amplitude is just a measure of change relative to its central position, whereas a variable’s magnitude is a measure of distance or quantity regardless of direction.

      Reply
    • The power spectral density tell us the distribution of power of the signal vs. frequency. The PSD plots indicate most of the power is concentrated from 1Hz to 25 Hz which conforms with the intended configuration of the chirp in the time domain.

      Reply

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