Question

sign a Losstess matcking ne 7wor eslgn 7 752

Answer 1

We see that the source impedance is higher than the load impedance hence we use the LC circuit given in the figure below:

RLARE L RSMALLER

At resonance frequency the Q-factor of R_larger and C branch should be equal to Q-factor of L and R_smaller. Thus

$Q_s = \frac{X_L}{R_{smaller}} = \sqrt{\frac{R_{larger}}{R_{smaller}}-1}$

Given R_larger = 75 ohms and R_smaller = 20 ohms hence

75 Rs 1-V2.751.658 20

Thus

1.658 Rsmaller 1.658 x 20-33.16 S2

Since

$X_L = \omega L = 33.16$

33.16 33.16 2 × π × 107

Similarly

$Q_p = \frac{R_{larger}}{X_C} = \sqrt{\frac{R_{larger}}{R_{smaller}}-1}$

.. _ = 1.658

hence

75 1.658 c =-= 45.23Ω

Since

45.2362 ,

$C =\frac{1}{45.23 \omega} = \frac{1}{45.23\times 2\pi\times 10^7} \approx 0.352\; nF$