Consider an ideal full-bridge single-phase DC-AC inverter with a DC bus voltage magnitude of 1000 V. The inverter output voltage v(t) shown below, is obtained when diagonal switches of the inverter are switched with 50% duty cycle. The inverter feeds a load with a sinusoidal current given by, i(t)=10sin(ωt-π/3),  where ω=2π/t. The active power, in watts, delivered to the load is _____. (round off to nearest integer)

The correct answer is 3183.

1. Analysis of the Output Voltage Waveform:
From the given waveform of the output voltage $v\left(t\right)$, we observe that it is a symmetric square wave with a period of $T$ and a peak magnitude equal to the DC bus voltage $V_{dc}=1000\text{V}$.

The voltage waveform can be mathematically represented over one cycle as:
$v\left(t\right)=1000\text{V}\text{for}0<t<0.5T$
$v\left(t\right)=-1000\text{V}\text{for}0.5T<t<T$

2. Fourier Series Expansion of the Voltage:
For a square wave with odd symmetry, the Fourier series expansion consists only of odd harmonic sine terms:
$v\left(t\right)=\sum _{n=1,3,5,...}^{\infty }\frac{4V_{dc}}{n\pi }sin\left(n\omega t\right)$

The fundamental frequency component ($n=1$) of the output voltage is:
$v_1\left(t\right)=V_{1m}sin\left(\omega t\right)=\frac{4V_{dc}}{\pi }sin\left(\omega t\right)$

Substituting $V_{dc}=1000\text{V}$:
$v_1\left(t\right)=\frac{4000}{\pi }sin\left(\omega t\right)\text{V}$

3. Load Current and Active Power Calculation:
The current delivered to the load is purely sinusoidal at the fundamental frequency:
$i\left(t\right)=10sin(\omega t-\frac{\pi }{3})\text{A}$

Since the load current has no higher-order harmonics, active power is only delivered by the fundamental component of the voltage. The active power $P$ is given by:
$P=V_{1,rm}{I}_{rms}cos\left(\varphi \right)$

where:
- The RMS value of the fundamental voltage is $V_{1,rms}=\frac{V_{1m}}{\sqrt{2}}=\frac{4000}{\pi \sqrt{2}}\text{V}$
- The RMS value of the load current is $I_{rms}=\frac{I_m}{\sqrt{2}}=\frac{10}{\sqrt{2}}\text{A}$
- The phase angle difference is $\varphi =\frac{\pi }{3}$

Substitute these values into the power equation:
$P=\left(\frac{4000}{\pi \sqrt{2}}\right)\times \left(\frac{10}{\sqrt{2}}\right)\times cos\left(\frac{\pi }{3}\right)$

Since $cos\left(\frac{\pi }{3}\right)=0.5$:
$P=\frac{40000}{2\pi }\times 0.5=\frac{10000}{\pi }\text{W}$

Evaluating this numerically:
$P\approx 3183.09886\text{W}$

Rounding to the nearest integer, we get:
$P=3183\text{W}$