Directional Coupler Numerical Problems
Directional Coupler Numerical Problems: 2079 Bhadra (BEI)
A \(20\text{ dB}\) directional coupler has a directivity of \(30\text{ dB}\). If the input power to the main arm is \(100\text{ mW}\), determine the power delivered to the coupled port and the isolated port. Assume all ports are matched.
Given Data
The coupling factor is
\[ C=20\text{ dB} \]The directivity is
\[ D=30\text{ dB} \]The input power is
\[ P_1=100\text{ mW} \]We need to determine the power delivered to the coupled port, \(P_c\), and the power delivered to the isolated port, \(P_i\).
Calculation of Coupled Port Power
The coupling factor of a directional coupler is defined as
\[ C=10\log_{10}\left(\frac{P_1}{P_c}\right) \]Substituting the given values,
\[ 20=10\log_{10}\left(\frac{100}{P_c}\right) \]Dividing both sides by \(10\),
\[ 2=\log_{10}\left(\frac{100}{P_c}\right) \]Taking the antilogarithm,
\[ 10^2=\frac{100}{P_c} \]Therefore,
\[ 100=\frac{100}{P_c} \]Hence,
\[ P_c=1\text{ mW} \]Thus, the power delivered to the coupled port is
\[ \boxed{P_c=1\text{ mW}} \]Calculation of Isolated Port Power
The directivity of a directional coupler is defined as the ratio of the power at the coupled port to the power at the isolated port. Therefore,
\[ D=10\log_{10}\left(\frac{P_c}{P_i}\right) \]Given
\[ D=30\text{ dB} \]Since \(P_c=1\text{ mW}\),
\[ 30=10\log_{10}\left(\frac{1}{P_i}\right) \]Dividing both sides by \(10\),
\[ 3=\log_{10}\left(\frac{1}{P_i}\right) \]Taking the antilogarithm,
\[ 10^3=\frac{1}{P_i} \]Therefore,
\[ P_i=\frac{1}{1000}\text{ mW} \] \[ P_i=0.001\text{ mW} \]Since \(1\text{ mW}=1000\ \mu\text{W}\),
\[ P_i=0.001\text{ mW}=1\ \mu\text{W} \]Hence, the power delivered to the isolated port is
\[ \boxed{P_i=0.001\text{ mW}} \] or\[ \boxed{P_i=1\ \mu\text{W}} \]
Final Answer
The power delivered to the coupled port is
\[ \boxed{P_c=1\text{ mW}} \]The power delivered to the isolated port is
\[ \boxed{P_i=0.001\text{ mW}=1\ \mu\text{W}} \]Therefore, the \(20\text{ dB}\) coupling factor causes \(1\text{ mW}\) of the \(100\text{ mW}\) input power to appear at the coupled port. Due to the finite \(30\text{ dB}\) directivity, only \(1\ \mu\text{W}\) appears at the isolated port.
Directional Coupler Numerical Problem: 2075 Bhadra (BEX)
A directional coupler has a coupling factor of \(10\text{ dB}\) and directivity of \(40\text{ dB}\). If the incident power is \(10\text{ W}\), calculate the power available at the coupled port, isolated port, and transmitted port.
Given Data
The coupling factor is
\[ C=10\text{ dB} \]The directivity is
\[ D=40\text{ dB} \]The incident power is
\[ P_1=10\text{ W} \]We need to determine the coupled port power \(P_c\), isolated port power \(P_i\), and transmitted port power \(P_t\).
Calculation of Coupled Port Power
The coupling factor of a directional coupler is defined as
\[ C=10\log_{10}\left(\frac{P_1}{P_c}\right) \]Substituting the given values,
\[ 10=10\log_{10}\left(\frac{10}{P_c}\right) \]Dividing both sides by \(10\),
\[ 1=\log_{10}\left(\frac{10}{P_c}\right) \]Taking the antilogarithm,
\[ 10=\frac{10}{P_c} \]Therefore,
\[ P_c=1\text{ W} \]Hence, the power available at the coupled port is
\[ \boxed{P_c=1\text{ W}} \]Calculation of Isolated Port Power
The directivity of a directional coupler is defined as
\[ D=10\log_{10}\left(\frac{P_c}{P_i}\right) \]Given \(D=40\text{ dB}\) and \(P_c=1\text{ W}\),
\[ 40=10\log_{10}\left(\frac{1}{P_i}\right) \]Dividing both sides by \(10\),
\[ 4=\log_{10}\left(\frac{1}{P_i}\right) \]Taking the antilogarithm,
\[ 10^4=\frac{1}{P_i} \]Therefore,
\[ P_i=\frac{1}{10^4}\text{ W} \] \[ P_i=0.0001\text{ W} \]Converting watts into milliwatts,
\[ P_i=0.0001\times1000\text{ mW} \] \[ P_i=0.1\text{ mW} \]Hence, the power available at the isolated port is
\[ \boxed{P_i=0.1\text{ mW}} \]Calculation of Transmitted Port Power
For the ideal matched directional coupler assumed in this problem, the input power is divided between the transmitted port and the coupled port. Therefore, the transmitted power can be calculated by subtracting the coupled power from the incident power:
\[ P_1=P_t+P_c \]Rearranging for the transmitted power gives
\[ P_t=P_1-P_c \]Substituting the known values,
\[ P_t=10-1 \] \[ \boxed{P_t=9\text{ W}} \]The isolated port power is extremely small compared with the incident and coupled powers. Under the ideal matched-coupler assumption used for this numerical problem, it is neglected when calculating the transmitted power.
Final Answer
The power available at the coupled port is
\[ \boxed{P_c=1\text{ W}} \]The power available at the isolated port is
\[ \boxed{P_i=0.0001\text{ W}=0.1\text{ mW}} \]The transmitted power is
\[ \boxed{P_t=9\text{ W}} \]Therefore, the final results are
\[ \boxed{ P_c=1\text{ W},\qquad P_i=0.1\text{ mW},\qquad P_t=9\text{ W} } \]Directional Coupler Numerical Problem: 2073 Magh (BEX)
In a \(3\text{ dB}\) directional coupler, calculate the power output at port 2 and port 3 if \(20\text{ mW}\) power is applied at port 1. Also calculate the insertion loss.
Given Data
The input power is
\[ P_1=20\text{ mW} \]The coupling factor is
\[ C=3\text{ dB} \]Since this is a \(3\text{ dB}\) directional coupler, the input power is divided equally between the through port and the coupled port. Therefore,
\[ P_2=P_3 \]For the port numbering used in this question, port 2 is the transmitted port and port 3 is the coupled port.
Power at Port 2
For a \(3\text{ dB}\) directional coupler, half of the input power is transmitted to the through port. Therefore,
\[ P_2=\frac{P_1}{2} \]Substituting the input power,
\[ P_2=\frac{20}{2} \]Hence, the power output at port 2 is
\[ \boxed{P_2=10\text{ mW}} \]Power at Port 3
The remaining half of the input power is coupled to port 3. Therefore,
\[ P_3=\frac{P_1}{2} \]Substituting the given input power,
\[ P_3=\frac{20}{2} \]Hence, the power output at port 3 is
\[ \boxed{P_3=10\text{ mW}} \]Thus, the \(20\text{ mW}\) input power is equally divided between the transmitted and coupled ports:
\[ \boxed{P_2=P_3=10\text{ mW}} \]Calculation of Insertion Loss
The insertion loss from port 1 to port 2 is defined as
\[ IL=-10\log_{10}\left(\frac{P_2}{P_1}\right) \]Substituting the values of \(P_2\) and \(P_1\),
\[ IL=-10\log_{10}\left(\frac{10}{20}\right) \]Therefore,
\[ IL=-10\log_{10}\left(\frac{1}{2}\right) \]Since
\[ \log_{10}\left(\frac{1}{2}\right)=-0.3010 \]we obtain
\[ IL=-10(-0.3010) \] \[ IL=3.01\text{ dB} \]Hence, the insertion loss from port 1 to port 2 is
\[ \boxed{IL\approx3.01\text{ dB}} \]Final Answer
The output power at port 2 is
\[ \boxed{P_2=10\text{ mW}} \]The output power at port 3 is
\[ \boxed{P_3=10\text{ mW}} \]The insertion loss from port 1 to port 2 is
\[ \boxed{IL\approx3.01\text{ dB}} \]Therefore, for the ideal \(3\text{ dB}\) directional coupler, the \(20\text{ mW}\) input power is divided equally between the transmitted and coupled ports, giving \(10\text{ mW}\) at each port. The transmission from port 1 to port 2 corresponds to an insertion loss of approximately \(3.01\text{ dB}\).