3D Time-of-flight distance measurement with custom - Universität ...

OPTICAL TOF RANGE MEASUREMENT 37
and algorithms in order to understand clearly and easily what happens to the phase
spectrum. In a next step we will transfer these results to a square wave. And lastly
we will simulate aliasing effects on several other waveforms, presuming DFT
detection with four sampling points.
An important boundary condition is the chosen integration interval for the sampling
point acquisition. As already discussed this is chosen to be half the modulation
period ∆t=T/2 in the following calculations. From Figure 2.8 we can conclude that,
for ∆t=T/2, all even harmonics of the basic frequency, namely 2f, 4f, 6f, … will be
suppressed and do not contribute to the demodulation result. This is because these
harmonics are eliminated by a zero crossing of the sinc-function. This simple
example shows the importance of the choice of ∆t for aliasing effects.
Aliasing phase error for 3f-harmonic using 4-tap DFT-detection.
In order to simplify explanations we use the expression “tap” for “sampling point”
from now on. We consider the input signal s(t) having a harmonic of triple base
frequency and amplitude “1”.
( ω t − ϕ ) + cos(
3ω
t − 3 )
s( t)
ϕ
= cos 0 0
0 0
Equation 2.22
The phase delay ϕ0 is caused by the distance dependent time delay in our TOF
application. This time delay causes a phase delay of 3ϕ0 in the 3f-harmonic
component.
Transformed into the Fourier domain we obtain the spectrum S(f):
S(
f)
=
1
2
{
δ
+ δ
−jϕ
( ω − ω ) 0
0 ⋅ e
jϕ
+ δ(
ω + ω ) 0
0 ⋅ e
( ω − 3ω
−j3ϕ
) ⋅ e 0 + δ(
ω + 3ω
j3ϕ
) ⋅ e 0 }
0
0
Equation 2.23
Now we consider the integration of the taps, i.e. convolution of s(t) with rect(t/∆t)
and multiplication of S(f) with sinc(2π⋅f⋅∆t/2)=sinc(π⋅f⋅T/2) resulting in:
Sint(
f)
=
1
2
2
with a =
π
{
a ⋅
+ b ⋅
and
− jϕ
( δ(
ω − ω ) e 0
0 ⋅
jϕ
+ δ(
ω + ω ) e 0
0 ⋅ )
− j3ϕ
( δ(
ω − 3ω
) e 0
0 ⋅
j3ϕ
+ δ(
ω + 3ω
) e 0
0 ⋅ )}
2 a
b = − = −
3π
3
Equation 2.24
The coefficients a and b are obtained by evaluation of sinc(π⋅f⋅T/2) at f=1/T and
f=3/T, respectively. By substituting b with a and considering -1 = e jπ = e -jπ we only