EECS 216 LECTURE NOTES
CORRELATION BETWEEN SINUSOIDAL SIGNALS
Given: x(t) = Acos(ωt +θx) and y(t) = Bcos(ωt +θy),−∞ < t < ∞.
Goal: To determine the phase difference |θx−θy|from data x(t) and y(t).
Soln: Compute the CORRELATION
COEFFICIENT =ρxy =CN(x, y) = C(x, y)/pE(x)E(y)
where: C(x, y) = RT
0x(t)y(t)dt and E(x) = C(x, x) and T=2π
ω=period.
Then: ρ(x, y) = CN(x, y) = cos(θx−θy)→ |θx−θy|.
Proof: E(x) = RT
0A2cos2(ωt +θx) = RT
0
A2
2(1 + cos(2ωt + 2θx)) = TA2
2.
using: cos2x=1
2(1 + cos(2x)). Compare to rms derivation.
C(x, y) = RT
0AB cos(ωt +θx) cos(ωt +θy)dt =AB
2RT
0cos(2ωt +θx+θy)dt
+AB
2RT
0cos(θx−θy)dt =TAB
2cos(θx−θy)
using: cos(x) cos(y) = 1
2cos(x+y) + 1
2cos(x−y) and M(sinusoid)=0.
Then: ρxy =CN(x, y) = C(x,y)
√E(x)E(y)=(T AB/2) cos(θx
−θy)
√(T A2/2)(T B2/2) = cos(θx−θy).
Note: (1) θx=θy⇔ρxy = 1; (2) |θx−θy|=π
2⇔ρxy = 0:
The sin and cos functions are orthogonal over one period.
Given: x(t) = Acos(2πfxt) and y(t) = Bcos(2πfyt),−∞ < t < ∞.
Goal: To determine the frequency difference |fx−fy|from data x(t) and y(t).
Soln: CORRELATION
COEFFICIENT =ρxy =CN(x, y) = C(x,y)
√E(x)E(y)=sinc((fx−fy)T)
where: sinc(x) = sin(πx)
πx has peak at x= 0; smaller at half-integer x.
Proof: E(x) = TA2
2and E(y) = TB2
2as above. But now we have:
C(x, y) = AB RT/2
−T/2cos(2πfxt) cos(2πfyt)dt =AB
2RT/2
−T/2cos(2π(fx+fy)t)dt
+AB
2RT/2
−T/2cos(2π(fx−fy)t)dt =TAB
2[sinc((fx+fy)T) + sinc((fx−fy)T)]
using: 1
TRT/2
−T/2cos(ωt)dt =sin(ωt)
ωT |T /2
−T/2=sin(ωT /2)
ωT /2=sinc(f T ) (ω= 2πf ).
Now: (fx+fy)>> (fx−fy)→sinc((fx+fy)T)<< sinc((fx−fy)T).
Then: ρxy =CN(x, y) = C(x,y)
√E(x)E(y)=(T AB/2)sinc(fx
−fy)T
√(T A2/2)(T B2/2) =sinc(fx−fy)T.
Note: (1) fx=fy⇔ρxy = 1; (2) Including phase in this →mess.