To calculate the resistances for a Tee Attenuator:
\[ R1 = Z0 \cdot \frac{(10^{L/20} - 1)}{(10^{L/20} + 1)} \]
\[ R2 = Z0 \cdot \frac{2 \cdot 10^{L/20}}{(10^{L/20})^2 - 1} \]
\[ R3 = Z0 \]
Where:
A Tee Attenuator is a type of passive electronic device that reduces the power of a signal without distorting its waveform. It is designed in the shape of the letter "T" and consists of three resistors. The main purpose of a Tee Attenuator is to lower the amplitude of an electronic signal, either for safety reasons or to match the input impedance of the devices to which the signal is sent. It is commonly used in radio, television, and audio systems.
Let's assume the following values:
Using the formula:
\[ R1 = 50 \cdot \frac{(10^{10/20} - 1)}{(10^{10/20} + 1)} = 50 \cdot \frac{(3.1623 - 1)}{(3.1623 + 1)} = 50 \cdot \frac{2.1623}{4.1623} = 25.98 \text{ Ohms} \]
\[ R2 = 50 \cdot \frac{2 \cdot 10^{10/20}}{(10^{10/20})^2 - 1} = 50 \cdot \frac{2 \cdot 3.1623}{(3.1623)^2 - 1} = 50 \cdot \frac{6.3246}{9.9999 - 1} = 35.25 \text{ Ohms} \]
\[ R3 = 50 \text{ Ohms} \]
The resistances are approximately \(R1 = 25.98\) Ohms, \(R2 = 35.25\) Ohms, and \(R3 = 50\) Ohms.
Let's assume the following values:
Using the formula:
\[ R1 = 75 \cdot \frac{(10^{20/20} - 1)}{(10^{20/20} + 1)} = 75 \cdot \frac{(10 - 1)}{(10 + 1)} = 75 \cdot \frac{9}{11} = 61.36 \text{ Ohms} \]
\[ R2 = 75 \cdot \frac{2 \cdot 10^{20/20}}{(10^{20/20})^2 - 1} = 75 \cdot \frac{2 \cdot 10}{(10)^2 - 1} = 75 \cdot \frac{20}{99} = 15.15 \text{ Ohms} \]
\[ R3 = 75 \text{ Ohms} \]
The resistances are approximately \(R1 = 61.36\) Ohms, \(R2 = 15.15\) Ohms, and \(R3 = 75\) Ohms.