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Periodendauer periodischer Signale

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$x(t) = A_1\cdot cos\Big(2\pi f_1\cdot t- \frac{2\pi}{360}\cdot \phi_1\Big)+A_2\cdot cos\Big(2\pi f_2\cdot t- \frac{2\pi}{360}\cdot \phi_2\Big)$

A1=0.50V
f1=1.00kHz
ϕ1=0.00Grad
A2=0.50V
f2=2.00kHz
ϕ2=90.00Grad
t=0.00s
1234567891011120.511.52−0.5−1−1.5−2
o+
tT
x(t)
x(t)= 0.5 x(t+T0)= 0.276 x(t+2T0)= 0.276
xmax=0.88 T0= 1