Network Theory Test 3

Concept:

For the sum of sinusoids with different frequencies as shown:

\(v\left( t \right) = {v_0} + {v_1}\sin {\omega _1}t + {v_2}\sin {\omega _2}t+ \ldots+ {v_n}\sin {\omega _n}t\)

The RMS value is calculated as:

\({v_{rms}} = \sqrt {v_0^2 + \frac{{v_1^2}}{2} + \frac{{v_2^2}}{2} + \ldots + \frac{{v_n^2}}{2}} \)

If the frequency of any two sinusoids is the same, we cannot directly apply the above formula.

We first combine them to a single sinusoid representation.

Application:

In v₁(t), the frequency two sinusoids have the same frequency, i.e. ω = 200.

∴ v₁(t) in the phasor domain can be written as:

V₁ = 2 + 3∠0° + 4∠- 90°

V₁ = 2 + (3 – 4j)

\(V_1 = 2 + 5\angle {\rm{ta}}{{\rm{n}}^{ - 1}}\left( { - \frac{4}{3}} \right)\)

V₁ = 2 – 5∠ -53.23°

v₁(t) = 2 – 5 cos (200t – 53.23°)

Now, we can use the general formula to calculate the RMS value as:

\({v_{1rms}} = \sqrt {{2^2} + \frac{{{5^2}}}{{2}}} \)

\({v_{1rms}} = \sqrt {4 + 12.5} \)

v_{1 rms} = 4.06 V

In signal v₂(t), the frequency of all the signals are different, therefore the RMS value will be:

\({v_{rms2}} = \sqrt {{2^2} + \frac{{{3^2}}}{2} + \frac{{{4^2}}}{2}} \)

v_{2 rms}  = 4.06 V

Now, the required ratio of the two RMS values is:

\(\frac{{{v_{1rms}}}}{{{v_{2rms}}}} = 1\)