Electronic – Calculate total harmonic distortion

A temperature of -50C is below -40C. Likewise for dB; -50dB is below, or less than, -40dB.

For power, -40dB is a ratio of \$\frac{1}{10,000}\$ while -50dB is a ratio of \$\frac{1}{100,000}\$.

Because the voltages are in rms. Remember that the rms voltage is the equivalent DC voltage source that delivers the same amount of power to a resistive load. The rms is given by

$$v_\mathrm{rms} = \sqrt{\frac{1}{T} \int_{0}^{T} v^2 \ \mathrm{d}t}.$$

In our case \$v\$ is a sum of pure sine waves. Hence it is easier to work in the frequency domain, if you do not have a burning desire of evaluating integrals containing products of sine waves.

The rms in the frequency domain is given by

$$V_\mathrm{rms} = \sqrt{\frac{1}{N^2} \sum_{i = 1}^N |V_i |^2},$$

where \$V_i\$ are the frequency components and \$N\$ is the number of samples. Now remember that a sine wave satisfies

$$|V| = \frac{A}{2} (\delta(f - f_0) + \delta(f + f_0)).$$

Hence a single harmonic has two components with amplitude \$A / 2\$. Thus

$$V_\mathrm{rms}^2 = \frac{1}{N^2} \frac{A^2}{2}.$$

Consider what happens when we take the rms of a waveform comprising two distinct harmonics with amplitudes \$A_1\$ and \$A_2\$. Then you have

$$V^2_\mathrm{rms} = \frac{1}{N^2} \left( \frac{A_1^2}{2} + \frac{A_2^2}{2} \right).$$

Substituting the rms of the first harmonic \$V_1\$ and the rms of the second harmonic \$V_2\$ yields

$$V_\mathrm{rms} = \sqrt{V_1^2 + V_2^2}.$$