1. **Problem statement:** Convert an absolute power ratio of 200 W to power gain in dB, and convert a power gain of 23.01 dB to an absolute power ratio. 2. **Formulas used:** - Power gain in dB: $$A_p(dB) = 10 \log_{10} \left( \frac{P_{out}}{P_{in}} \right)$$ - Absolute power ratio from dB: $$\frac{P_{out}}{P_{in}} = 10^{\frac{A_p(dB)}{10}}$$ 3. **Part i: Convert 200 W absolute power ratio to dB** - Given absolute power ratio: $$\frac{P_{out}}{P_{in}} = 200$$ - Apply formula: $$A_p(dB) = 10 \log_{10}(200)$$ - Calculate logarithm: $$\log_{10}(200) = \log_{10}(2 \times 10^2) = \log_{10}(2) + \log_{10}(10^2) = 0.3010 + 2 = 2.3010$$ - Multiply by 10: $$A_p(dB) = 10 \times 2.3010 = 23.01\, dB$$ 4. **Part ii: Convert 23.01 dB power gain to absolute power ratio** - Given power gain: $$A_p(dB) = 23.01$$ - Apply formula: $$\frac{P_{out}}{P_{in}} = 10^{\frac{23.01}{10}} = 10^{2.301}$$ - Calculate power ratio: $$10^{2.301} = 200$$ **Final answers:** - i. Power gain in dB: $$23.01\, dB$$ - ii. Absolute power ratio: $$200$$