A string of impedance Z1 2kgs is joined to a second string

A string of impedance Z_1 = 2kg/s is joined to a second string of impedance Z_2. The strings are along the z-direction and an incident wave y_1 = (0.1m) cos((2 pi (100)/s)t - k_1 z) is launched in string one moving towards positive z and towards the junction of the two strings. Find Z_2 if 20% of the incident power is reflected and the reflected wave has the same orientation as the incident wave. Find Z_2 if 20% of the incident power is reflected and the reflected wave is inverted relative to the incident wave.

Solution

A) POWER OF REFLECTED WAVE = (Z1 - Z2 ) ² /  (Z1 +Z2 ) ² * POWER OF INCIDENT WAVE

(Z1 - Z2 ) ² /  (Z1 +Z2 ) ^{2 =} POWER OF REFLECTED WAVE / POWER OF INCIDENT WAVE

=>   (2 - Z2 ) ² /  (2 +Z2 ) ² = 0.2

=> (2 - Z2 ) / (2 +Z2 ) = 0.4472

=> Z2 = 0.7640

B)

POWER OF REFLECTED WAVE = (Z2 - Z1 ) ² /  (Z2 +Z1 ) ² * POWER OF INCIDENT WAVE

(Z2 - Z1 ) ² /  (Z1 +Z2 ) ^{2 =} POWER OF REFLECTED WAVE / POWER OF INCIDENT WAVE

=>   ( Z2-2 ) ² /  (2 +Z2 ) ² = 0.2

=> (Z2 - 2 ) / (2 +Z2 ) = 0.4472

=> Z2 = 5.23