2
$\begingroup$

I would like to understand what's happening to a signal emitted from a moving source and arriving to a moving receptor. But, when I am looking on internet about Doppler effect, I can only find equations linking received frequency to emitted frequency. But what I want to do here is to simulate the Doppler effect on a random signal in Matlab.

Introduction

My idea was to see the Doppler effect as a consequence of the movement between source and receptor, I started by writing something like this :

$$S_r(\phi(t)) = S_e(t)$$

Where $S_e$ is the emitted signal, $S_r$ the received signal and $\phi(t)$ a function giving the arrival time of the signal emitted by the source at time $t$, I think the quantity $\phi(t) - t$ is called TDOA sometimes. Since I'm using classic physics here, I have $\phi(t) = t + \frac{d_t(t)}{c}$ where $d_t(t)$ is the absolute distance travelled by the signal emitted at time $t$ between the source and the receiver.

Application to a simple problem

Let's now consider a moving emitter and a immobile receiver. To simplify things, both start at the same point with $d_e(0) = d_r(0) = 0$. Taking an immobile receiver will simplify the formula of $d_t(t)$, because in this case $d_t(t) = v_et$. Well I just have to apply my formula now, and I obtain

$$S_r\left(t + \frac{v_et}{c}\right) = S_e(t) \implies S_r(t) = S_e\left(t - \frac{v_et}{c}\right)$$

Application to periodic wave

Well, I tried to apply this simple approach to a periodic wave of frequency $f_e$ in order to try to find the equation (which is a standard équation about Doppler effect):

$$f_{r}={\frac {c}{c-v_{e}}}\cdot f_{e}$$

So, I just took $S_e(t) = \cos\left(2\pi f_e t \right)$. And then :

$$S_r(t) = \cos\left(2\pi f_e \left(t - \frac{v_et}{c}\right) \right) = \cos\left(2\pi f_e \frac{c - v_e}{c} t \right)$$

And I'm finally finding... $f_{r}=\frac {c-v_{e}}{c} \cdot f_{e}$.

So... the exact opposite of what I was supposed to find. And I don't understand why... (Same thing happens when I consider a moving receptor). So my first question would be to know where my mistake is... Because when I am simulating this approach with Matlab, I find the correct answer when using a periodic wave. So for me, this method seems to work...

Questions

  • Where is my mistake when applying my approach to periodic signals ?
  • Is my approach good enough to modelize Doppler effect on any kind of wave ? Can I generalize it even more ?
  • How can I generalize to introduce Special relativity in my equation (in order to work with fast objects like satellites) ?

Matlab source code

    %% Configuration
    vE = 80; % Source speed (m/s)
    c = 122; % Celerity (m/s)

    d0 = 0; % Initial distance between source and receiver (m)

    nT = 1500; % Number of visible periods

    Fc = 20; % Carrier frequency
    Tc = 1/Fc; % Carrier period

    Fs = 1000; % Sampling frequency

    %% Script

    % Create signal
    At_t = 0:1/Fs:nT*Tc;
    At = cos(2*pi*Fc*At_t);

    % Apply Doppler

    dp = abs(d0 - vE .* At_t);
    dt = dp ./ c;

    % Interpolation/Resampling
    do_At_t_temp = At_t + dt;

    do_At_t = min(do_At_t_temp):1/Fs:max(do_At_t_temp);
    do_At = interp1(do_At_t_temp, At, do_At_t);

    % Plot
    figure;
    plot(At_t, At); hold on;
    plot(do_At_t, do_At);
    grid;
    legend('Without doppler', 'With Doppler');

    figure;
    [pxx,f] = pwelch(At,[],[],[],Fs);
    plot(f, pxx); hold on
    [pxx,f] = pwelch(do_At,[],[],[],Fs);
    plot(f, pxx);
    legend('Without doppler', 'With Doppler');
    xlim([0, 3*Fc])
    grid;

    fprintf('Theorical values : %d Hz and %d Hz\n', round(Fc * c / (c - vE), 2), round(Fc * c / (c + vE), 2));

This script seems to give the correct frequency shift (from 20Hz to 12.08 Hz)

enter image description here

$\endgroup$
2
  • 1
    $\begingroup$ The Cassini-Huygens Mission has a good example of applying the Doppler shift to both the carrier and the non-periodic data stream thespacereview.com/article/306/1 $\endgroup$
    – DJohnM
    May 23, 2020 at 21:08
  • 1
    $\begingroup$ The line $S_r(t+v_e/c)=S_e(t)\implies S_r(t)=S_e(t-v_e/c)$ is wrong. $\endgroup$
    – Felipe
    May 23, 2020 at 21:09

1 Answer 1

2
$\begingroup$

Let's consider a more general version of the problem first. Suppose the distance between the emitter and receiver is $d(t)$; we'll allow this to be an arbitrary function of time. We'll also suppose that the amplitude of the signal emitted as a function of time is $S(t)$, again allowing it to be an arbitrary function of time. Suppose the signal has a speed $c$, which is constant regardless of frequency. The signal emitted at time $t$ will be received at time $t+d(t)/c$, since the signal has to cross the distance $d(t)$ to get to the receiver. So we can write:

$$S(t)=S_r(t+d(t)/c)\equiv S_r(f(t))$$

where $f(t)=t+d(t)/c$. As long as $f(t)$ is invertible, we can then solve for the signal at the receiver by finding the inverse function for $f(t)$:

$$S_r(t)=S(f^{-1}(t))$$

So let's apply this to a stationary receiver at the origin, and an emitter moving with a constant velocity $v$ in a straight line directly toward or away from the receiver. Then $d(t)=|x_0+vt|$ for some initial position $x_0$, which means that

$$f(t)=t+\frac{1}{c}|x_0+vt|$$

This gives us two separate piecewise functions: one when $t>-\frac{x_0}{v}$ and one when $t<-\frac{x_0}{v}$. Let's label these

$$f_1(t)=t+\frac{x_0}{c}+\frac{v}{c}t=\frac{x_0}{c}+\left(1+\frac{v}{c}\right)t$$ and

$$f_2(t)=t-\frac{x_0}{c}-\frac{v}{c}t=-\frac{x_0}{c}+\left(1-\frac{v}{c}\right)t$$

Inverting each, we have:

$$f_1^{-1}(t)=\frac{t-x_0/c}{1+v/c}$$

and

$$f_2^{-1}(t)=\frac{t+x_0/c}{1-v/c}$$

which means that

$$S_r(t)=\begin{cases}S\left(\frac{t-x_0/c}{1+v/c}\right)&\text{for }t>-\frac{x_0}{v}\\S\left(\frac{t+x_0/c}{1-v/c}\right)&\text{for }t<-\frac{x_0}{v}\end{cases}$$

So this is the formula for a general, non-periodic signal $S$ emitted by an observer moving directly toward or away from you at speed $v$, starting from $x_0$. If we plug in a periodic function, say, $S(t)=A\cos(\omega t)$, then we have:

$$S_r(t)=\begin{cases}A\cos\left(\frac{\omega}{1+v/c}t-\frac{\omega x_0/c}{1+v/c}\right)&\text{for }t>-\frac{x_0}{v}\\A\cos\left(\frac{\omega}{1-v/c}t+\frac{\omega x_0/c}{1-v/c}\right)&\text{for }t<-\frac{x_0}{v}\end{cases}$$

When the emitter is receding from the observer, then we either have $x_0>0$ and $v>0$ or $x_0<0$ and $v<0$. This means $\frac{x_0}{v}$ is always positive, which in turn means that, for all positive $t$, we have that $t>0>-\frac{x_0}{v}$. So for a receding emitter, we use the top equation, meaning that the frequency heard from a receding emitter is

$$f_r=\frac{f}{1+v/c}$$

which is lower than the emitted frequency, as expected.

In turn, for an approaching emitter, we either have $x_0>0$ and $v<0$ or $x_0<0$ and $v>0$ (and it will only approach for a finite amount of time before passing the receiver and beginning to recede). This means that $\frac{x_0}{v}$ is negative, meaning that there is a certain time window when it is possible that $0<t<-\frac{x_0}{v}$. In that time window (i.e. the window of time when the emitter is approaching), the frequency heard at the receiver is, as you can see,

$$f_r=\frac{f}{1-v/c}$$

which is higher than the emitted frequency, again as expected.

$\endgroup$
1
  • $\begingroup$ This is a complete answer. Thanks a lot. So, using Special relativity, I presume I could find f(t) function using Lorentz transformations ? $\endgroup$
    – graille
    May 24, 2020 at 18:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.