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This can be solved in a far simpler (and practical EE) way: -

To get the RMS value of the composite waveform, you: -

• Square the individual parts (triangle and DC) to get the respective powers into a 1 Ω resistor
• Weight them individually with their duty cycle
• Add the two weighted powers together together and finally,
• Take the square root to get back to RMS voltage and lose the 1 Ω dependency.

1. For the triangle section, it's weighted power is \$\frac{4}{3}\times 7\div 15\$
2. For the DC part it's just \$4\times 8\div 15\$
3. Add them to get 2.755555
4. Take the square root to get 1.65998661307

Proof of triangle waveform RMS: -

This can be solved in a far simpler (and practical EE) way: -

To get the RMS value of the composite waveform, you: -

• Square the individual parts (triangle and DC) to get the respective powers into a 1 Ω resistor
• Weight them individually with their duty cycle
• Add the two weighted powers together and finally,
• Take the square root to get back to RMS voltage and lose the 1 Ω dependency.

1. For the triangle section, it's weighted power is \$\frac{4}{3}\times 7\div 15\$
2. For the DC part it's just \$4\times 8\div 15\$
3. Add them to get 2.755555
4. Take the square root to get 1.65998661307

Proof of triangle waveform RMS: -

This can be solved in a far simpler way: -

So, to get the RMS value of the waveform, you square the individual parts (triangle and DC) to get power into a 1 Ω resistor, weight them with their duty cycle, add them together and finally, take the square root to get back to RMS voltage and lose the 1 Ω dependency.

• For the triangle section, it's weighted power is \$\dfrac{4}{3}\times 7\div 15\$
• For the DC part it's just \$4\times 8\div 15\$
• Add them to get 2.755555
• Take the square root to get 1.65998661307

Proof of triangle waveform RMS: -

This can be solved in a far simpler (and practical EE) way: -

To get the RMS value of the composite waveform, you: -

• Square the individual parts (triangle and DC) to get the respective powers into a 1 Ω resistor
• Weight them individually with their duty cycle
• Add the two weighted powers together together and finally,
• Take the square root to get back to RMS voltage and lose the 1 Ω dependency.

1. For the triangle section, it's weighted power is \$\frac{4}{3}\times 7\div 15\$
2. For the DC part it's just \$4\times 8\div 15\$
3. Add them to get 2.755555
4. Take the square root to get 1.65998661307

Proof of triangle waveform RMS: -

This can be solved in a far simpler way: -

So, to get the RMS value of the waveform, you square the individual parts (triangle and DC) to get power into a 1 Ω resistor, weight them with their duty cycle, add them together and finally, take the square root.

• For the triangle section, it's weighted power is \$\dfrac{4}{3}\times 7\div 15\$
• For the DC part it's just \$4\times 8\div 15\$
• Add them to get 2.755555
• Take the square root to get 1.65998661307

Proof of triangle waveform RMS: -

This can be solved in a far simpler way: -

So, to get the RMS value of the waveform, you square the individual parts (triangle and DC) to get power into a 1 Ω resistor, weight them with their duty cycle, add them together and finally, take the square root to get back to RMS voltage and lose the 1 Ω dependency.

• For the triangle section, it's weighted power is \$\dfrac{4}{3}\times 7\div 15\$
• For the DC part it's just \$4\times 8\div 15\$
• Add them to get 2.755555
• Take the square root to get 1.65998661307

Proof of triangle waveform RMS: -