Oct 02

# Theory behind Doppler frequency shift

Doppler frequency shift plays a major part in wireless communications, radar and many other systems. Sometimes Doppler shift is useful, like in radar detection systems, but sometimes it’s harmful, like in mobile wireless communications systems. Calculating the Doppler shift is very trivial.

$f_{Doppler} = \left(\frac{v}{c}\right) f_{T}$,

where v is the relative speed between transmitter and receiver, and the c is the speed of light, which is the speed the electromagnetic waves travel.

But why exactly do we see this frequency shift? What is the theory behind it? This is something we often take for granted and not pay much attention to. Actually, there isn’t much in to it. In reality there is actually no frequency shift. The frequency is the same. So what is it then? It’s all in the definition of frequency. Frequency is defined as:

$f = \frac{1}{2\pi} \frac{d\phi}{d t}$,

where $phi$ is the phase change. Therefore, frequency is the rate of change of phase. Then we divide by a scalar value to normalize for the angular phase. So how can we derive Doppler frequency from this? Take look at the picture below.

Here’s the situation. A person is shouting with a constant frequency of $f_THz$. This is heard by another person at some distant. Since both of them are stationary, the sound wave travels at a speed of c m/s, the speed of light. Now consider the distance this wave traveled in t seconds past the listener:

$R = ct$

Now we want to find out the frequency at which this sound wave was received at the receiver. So to find the rate of change of phase, we need to find how much phase changed during this t seconds it passed the listener. In a sinusoidal signal, each period is a phase change of $2pi$ (well the result is a phase difference of 0). So if we could find how many periods of the wave passed the listener in that time duration we could find the frequency. Since each period is the length of a wavelength, $lambda$, we can easily find the frequency as follows:

Number of periods passed: $P =\frac{R}{\lambda} =\frac{ct}{\lambda}$

Total phase change during this time t: $\phi = 2\pi\frac{ct}{\lambda}$

Frequency:  $f_R = \frac{1}{2\pi} \frac{d\phi}{d t} = \frac{c}{\lambda} = f_T Hz$

Final equality comes from $c=f\lambda$.

Now consider the second scenario. That person is now travelling in car while shouting at the same frequency $f_THz$. The car is travelling at a speed of v m/s. Now the sound wave travels at a speed of (c+v) m/s, it’s natural speed plus the speed of the car from which it is transmitted. Now let’s do the same calculation as earlier. Let’s calculate how many periods of this wave passed the (stationary) listener this time. Remember now the wave is travelling faster than earlier, so it must travel further than R meters in t seconds.

$R' = (c+v)t$

Number of periods passed: $P' =\frac{R'}{\lambda} =frac{(c+v)t}{\lambda}$

Total phase change during this time t: $\phi' = 2\pi\frac{(c+v)t}{\lambda}$

Frequency:  $f_{R'} = \frac{1}{2\pi} \frac{d\phi'}{d t} = \frac{(c+v)}{\lambda}$

$f_{R'} = \frac{c}{\lambda} +\frac{v}{\lambda} = f_T + v\frac{f_T}{c} = \left(1 + \frac{v}{c}\right)f_T$

This gives a Doppler frequency shift of $f_{Doppler} = \left(\frac{v}{c}\right) f_{T}$.

Remember that the sign of v changes with the relative velocity between the transmitter and the receiver, so that Doppler shift might be positive or negative. So that’s the simple theory behind Doppler frequency shift.