Normal Reflection and Transmission at Interfaces

(438) |

In other words, the displacement is a linear superposition of an

(439) |

In other words, the displacement is solely due to a

Let us consider the matching conditions at the interface between the two strings: that is, at . First, because the two strings are tied together at , their transverse displacements at this point must equal one another. In other words,

(440) |

or

(441) |

The only way that the previous equation can be satisfied for all values of is if . This being the case, the common factor cancels out, and we are left with

Second, because the two strings lack an energy dissipation mechanism, the energy flux into the interface must match that out of the interface. In other words,

(443) |

where and are the impedances of the first and second strings, respectively. The previous expression reduces to

(444) |

which, when combined with Equation (442), yields

Equations (442) and (445) can be solved to give

The

The

It can be seen that

(450) |

In other word, any incident wave energy that is not reflected is transmitted.

Suppose that the density per unit length of the second string, , tends to infinity, so that . It follows from Equations (446) and (447) that and . Likewise, Equations (448) and (449) yield and . Hence, the interface between the two strings is stationary (because it oscillates with amplitude ), and there is no transmitted energy. In other words, the second string acts exactly like a fixed boundary. It follows that when a transverse wave on a string is incident at a fixed boundary then it is perfectly reflected with a phase shift of radians. In other words, . Thus, the resultant wave displacement on the string becomes

(451) |

where use has been made of the trigonometric identity . (See Appendix B.) We conclude that the incident and reflected waves interfere in such a manner as to produce a standing wave with a node at the fixed boundary.

Suppose that the density per unit length of the second string, , tends to zero, so that . It follows from Equations (446) and (447) that and . Likewise, Equations (448) and (449) yield and . Hence, the interface between the two strings oscillates at twice the amplitude of the incident wave (that is, the interface is a point of maximal amplitude oscillation), and there is no transmitted energy. In other words, the second string acts exactly like a free boundary. It follows that when a transverse wave on a string is incident at a free boundary then it is perfectly reflected with zero phase shift. In other words, . Thus, the resultant wave displacement on the string becomes

(452) |

where use has been made of the trigonometric identity . (See Appendix B.) We conclude that the incident and reflected waves interfere in such a manner as to produce a standing wave with an anti-node at the free boundary.

Suppose that two strings of mass per unit length and are separated by a short section of string of mass per unit length . Let all three strings have the common tension . Suppose that the first and second strings occupy the regions and , respectively. Thus, the middle string occupies the region . Moreover, the interface between the first and middle strings is at , and the interface between the middle and second strings is at . Suppose that a wave of angular frequency is launched from a wave source at , and propagates towards the two interfaces. We would expect this wave to be partially reflected and partially transmitted at the first interface ( ), and the resulting transmitted wave to then be partially reflected and partially transmitted at the second interface ( ). Thus, we can write the wave displacement in the region as

(453) |

where is the amplitude of the incident wave, is the amplitude of the reflected wave, and . Here, the phase angles of the two waves have been chosen so as to facilitate the matching process at . The wave displacement in the region takes the form

(454) |

where is the amplitude of the final transmitted wave, and . Finally, the wave displacement in the region is written

(455) |

where and are the amplitudes of the forward- and backward-propagating waves on the middle string, respectively, and . Continuity of the transverse displacement at yields

where a common factor has cancelled out. Continuity of the energy flux at gives

(457) |

so the previous two expressions can be combined to produce

Continuity of the transverse displacement at yields

(459) |

Suppose that the length of the middle string is one quarter of a wavelength: that is, . Furthermore, let . It follows that , , and . Thus, canceling out a common factor , the previous expression yields

Continuity of the energy flux at gives

(461) |

so the previous two equations can be combined to generate

Equations (456) and (462) yield

(463) |

whereas Equations (458) and (460) give

(464) |

Hence, combining the previous two expression, we obtain

(465) | ||

(466) |

Finally, the overall coefficient of reflection is

(467) |

whereas the overall coefficient of transmission becomes

(468) |

Suppose that the impedance of the middle string is the geometric mean of the impedances of the two outer strings: that is, . In this case, it follows, from the previous two equations, that and . In other words, there is no reflection of the incident wave, and all of the incident energy ends up being transmitted across the middle string from the leftmost to the rightmost string. Thus, if we want to transmit transverse wave energy from a string of impedance to a string of impedance (where ) in the most efficient manner possible--that is, with no reflection of the incident energy flux--then we can achieve this by connecting the two strings via a short section of string whose length is one quarter of a wavelength, and whose impedance is . This procedure is known as

The previous analysis of the reflection and transmission of transverse waves at an interface between two strings is also applicable to the reflection and transmission of other types of wave incident on an interface between two media of differing impedances. For example, consider a transmission line, such as a co-axial cable. Suppose that the line occupies the region , and is terminated (at ) by a load resistor of resistance . Such a resistor might represent a radio antenna [which acts just like a resistor in an electrical circuit, except that the dissipated energy is radiated, rather than being converted into heat energy (Fitzpatrick 2088)]. Suppose that a signal of angular frequency is sent down the line from a wave source at . The current and voltage on the line can be written

(469) | ||

(470) |

where is the amplitude of the incident signal, the amplitude of the signal reflected by the load, the characteristic impedance of the line, and . Here, is the characteristic phase velocity at which signals propagate down the line. (See Section 7.5.) The resistor obeys Ohm's law, which yields

(471) |

It follows, from the previous three equations, that

(472) |

Hence, the coefficient of reflection, which is the ratio of the power reflected by the load to the power sent down the line, is

(473) |

Furthermore, the coefficient of transmission, which is the ratio of the power absorbed by the load to the power sent down the line, takes the form

(474) |

It can be seen, by comparison with Equations (448) and (449), that the load terminating the line acts just like another transmission line of impedance . It follows that power can only be efficiently sent down a transmission line, and transferred to a terminating load, when the impedance of the line matches the effective impedance of the load (which, in this case, is the same as the resistance of the load). In other words, when there is no reflection of the signal sent down the line (i.e., ), and all of the signal energy is therefore absorbed by the load (i.e., ). As an example, a