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

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40 CHAPTER 2<br />

Aliasing phase error for a square wave using 4-tap DFT-detection.<br />

The deduction <strong>of</strong> the aliasing-caused phase error, which occurs if we use a square<br />

wave as modulation signal, detected <strong>with</strong> the 4-tap approach, works in the same<br />

way as the previous deduction. The only difference is that the square wave has an<br />

infinite frequency spectrum, which makes the notation more complex. Therefore,<br />

we will simplify wherever possible. A square wave ssq(t) can be spilt into a sum <strong>of</strong><br />

odd harmonics <strong>of</strong> the base frequency <strong>with</strong> decreasing amplitude for increasing<br />

frequency:<br />

ssq(<br />

t)<br />

= cos<br />

1<br />

− ⋅cos<br />

7<br />

1<br />

3<br />

( ω t − ϕ ) − ⋅ cos(<br />

3ω<br />

t − 3ϕ<br />

) + ⋅cos(<br />

5ω<br />

t − 5ϕ<br />

)<br />

0<br />

1<br />

9<br />

1<br />

5<br />

( 7ω<br />

t − 7ϕ<br />

) + ⋅cos(<br />

9ω<br />

t − 9ϕ<br />

) −...<br />

0<br />

0<br />

0<br />

In the Fourier domain this corresponds to the spectrum Ssq(f):<br />

Ssq(<br />

f)<br />

=<br />

−<br />

δ<br />

0<br />

0<br />

0<br />

j7ϕ<br />

ϕ<br />

( ω + 7ω<br />

) ⋅ e 0<br />

j5<br />

δ(<br />

ω + 5ω<br />

) ⋅ e 0 δ(<br />

ω + 3ω<br />

)<br />

+<br />

δ<br />

jϕ<br />

( ω + ω ) ⋅ e 0 δ(<br />

ω − ω )<br />

−j3ϕ<br />

− ϕ<br />

( ω − 3ω<br />

) ⋅ e 0<br />

j5<br />

δ(<br />

ω − 5ω<br />

) ⋅ e 0 δ(<br />

ω − 7ω<br />

)<br />

0<br />

3<br />

+<br />

0<br />

1<br />

0<br />

5<br />

+<br />

0<br />

−<br />

0<br />

−jϕ<br />

0 ⋅ e<br />

1<br />

0<br />

−j7ϕ<br />

0 ⋅ e<br />

7<br />

Equation 2.29<br />

⎪⎧<br />

j3ϕ<br />

1 δ<br />

⋅ e<br />

⎨...<br />

− 0 + 0 − 0<br />

2 ⎪⎩<br />

7<br />

5<br />

3<br />

0<br />

0<br />

0<br />

⎪⎫<br />

+ ... ⎬<br />

⎪⎭<br />

Equation 2.30<br />

Again, considering the integration <strong>of</strong> the sampling points, we multiply Ssq(f) <strong>with</strong><br />

sinc(π⋅f⋅T/2). This time multiplication <strong>with</strong> the sinc function eliminates all negative<br />

signs:

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