Consider a sinusoidal singal with power of -45 dBm. What is the peak-to-peak voltage of this signal? Assume a load impedance of 50 Ω. Handwritten Calculation required.

V_p-p = 10^((-45-30)/20) * sqrt(50)

V_p-p = 0.0316 V

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The peak-to-peak voltage of a sinusoidal signal can be determined using the power and load impedance. In this case, the power of the signal is given as -45 dBm. To find the peak-to-peak voltage, we can convert the power from decibel-milliwatt (dBm) to watts using the formula P(W) = 10^(P(dBm)/10)/1000. Substituting the given power value (-45 dBm) into the formula, we get P(W) = 10^(-45/10)/1000 = 0.000000316 W. Now, using the formula P(W) = (V(pp)^2)/(4*R), where V(pp) is the peak-to-peak voltage and R is the load impedance, we can solve for V(pp). Rearranging the equation, we have V(pp)^2 = P(W) * 4 * R. Substituting the known values, V(pp)^2 = 0.000000316 * 4 * 50 = 0.0000632. Taking the square root of both sides, V(pp) = √(0.0000632) = 0.00795 volts. Hence, the peak-to-peak voltage of this sinusoidal signal, considering a load impedance of 50 Ω, is approximately 0.00795 volts.