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Next: Electromagnetic Waves Up: Traveling Waves Previous: Transmission Lines


Normal Reflection and Transmission at Interfaces

Consider two uniform semi-infinite strings that run along the $ x$ -axis, and are tied together at $ x=0$ . Let the first string be of density per unit length $ \rho_1$ , and occupy the region $ x<0$ , and let the second string be of density per unit length $ \rho_2$ , and occupy the region $ x>0$ . The tensions in the two strings must be equal, otherwise the string interface would not be in force balance in the $ x$ -direction. Let $ T$ be the common tension. Suppose that a transverse wave of angular frequency $ \omega $ is launched from a wave source at $ x=-\infty$ , and propagates towards the interface. Assuming that $ \rho_1\neq \rho_2$ , we would expect the wave incident on the interface to be partially reflected, and partially transmitted. The frequencies of the incident, reflected, and transmitted waves are all the same, because this property of the waves is ultimately determined by the oscillation frequency of the wave source. Hence, in the region $ x<0$ , the wave displacement takes the form

$\displaystyle y(x,t)= A_i\,\cos(\omega\,t-k_1\,x-\phi_i) + A_r\,\cos(\omega\,t+k_1\,x-\phi_r).$ (438)

In other words, the displacement is a linear superposition of an incident wave and a reflected wave. The incident wave propagates in the positive $ x$ -direction, and is of amplitude $ A_i$ , wavenumber $ k_1=\omega/v_1$ , and phase angle $ \phi_i$ . The reflected wave propagates in the negative $ x$ -direction, and is of amplitude $ A_r$ , wavenumber $ k_1=\omega/v_1$ , and phase angle $ \phi_r$ . Here, $ v_1=\sqrt{T/\rho_1}$ is the phase velocity of traveling waves on the first string. In the region $ x>0$ , the wave displacement takes the form

$\displaystyle y(x,t)= A_t\,\cos(\omega\,t-k_2\,x-\phi_t).$ (439)

In other words, the displacement is solely due to a transmitted wave that propagates in the positive $ x$ -direction, and is of amplitude $ A_t$ , wavenumber $ k_2=\omega/v_2$ , and phase angle $ \phi_t$ . Here, $ v_2=\sqrt{T/\rho_2}$ is the phase velocity of traveling waves on the second string.

Let us consider the matching conditions at the interface between the two strings: that is, at $ x=0$ . First, because the two strings are tied together at $ x=0$ , their transverse displacements at this point must equal one another. In other words,

$\displaystyle y(0_-,t) = y(0_+,t),$ (440)

or

$\displaystyle A_i\,\cos(\omega\,t-\phi_i) +A_r\,\cos(\omega\,t-\phi_r)=A_t\,\cos(\omega\,t-\phi_t).$ (441)

The only way that the previous equation can be satisfied for all values of $ t$ is if $ \phi_i=\phi_r=\phi_t$ . This being the case, the common $ \cos(\omega\,t-\phi_i)$ factor cancels out, and we are left with

$\displaystyle A_i + A_r = A_t.$ (442)

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,

$\displaystyle \frac{1}{2} \omega^2 Z_1 (A_i^{ 2}-A_r^{ 2}) = \frac{1}{2} \omega^{ 2} Z_2 A_t^{ 2},$ (443)

where $ Z_1=\sqrt{\rho_1 T}$ and $ Z_2=\sqrt{\rho_2 T}$ are the impedances of the first and second strings, respectively. The previous expression reduces to

$\displaystyle Z_1\,(A_i+A_r)\,(A_i-A_r) = Z_2\,A_t^{\,2},$ (444)

which, when combined with Equation (442), yields

$\displaystyle Z_1\,(A_i-A_r) = Z_2\,A_t.$ (445)

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

$\displaystyle A_r$ $\displaystyle =\left(\frac{Z_1-Z_2}{Z_1+Z_2}\right) A_i,$ (446)
$\displaystyle A_t$ $\displaystyle = \left(\frac{2 Z_1}{Z_1+Z_2}\right) A_i.$ (447)

The coefficient of reflection, $ R$ , is defined as the ratio of the reflected to the incident energy flux: that is,

$\displaystyle R = \left(\frac{A_r}{A_i}\right)^2 = \left(\frac{Z_1-Z_2}{Z_1+Z_2}\right)^2.$ (448)

The coefficient of transmission, $ T$ , is defined as the ratio of the transmitted to the incident energy flux: that is,

$\displaystyle T = \frac{Z_2}{Z_1}\left(\frac{A_t}{A_i}\right)^2= \frac{4 Z_1 Z_2}{(Z_1+Z_2)^2}.$ (449)

It can be seen that

$\displaystyle R+T=1.$ (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, $ \rho_2$ , tends to infinity, so that $ Z_2=\sqrt{\rho_2 T}\rightarrow\infty$ . It follows from Equations (446) and (447) that $ A_r=-A_i$ and $ A_t=0$ . Likewise, Equations (448) and (449) yield $ R=1$ and $ T=0$ . Hence, the interface between the two strings is stationary (because it oscillates with amplitude $ A_t$ ), 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 $ \pi$ radians. In other words, $ A_r=-A_i$ . Thus, the resultant wave displacement on the string becomes

$\displaystyle y(x,t)$ $\displaystyle = A_i\,\cos(\omega\,t-k_1\,x-\phi_i) -A_i\,\cos(\omega\,t+k_1\,x+\phi_i)$    
  $\displaystyle =2 A_i \sin(k_1 x) \sin(\omega t-\phi_i),$ (451)

where use has been made of the trigonometric identity $ \cos a-\cos b \equiv 2 \sin[(a+b)/2] \sin[(b-a)/2]$ . (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, $ \rho_2$ , tends to zero, so that $ Z_2=\sqrt{\rho_2 T}\rightarrow 0$ . It follows from Equations (446) and (447) that $ A_r=A_i$ and $ A_t=2\,A_i$ . Likewise, Equations (448) and (449) yield $ R=1$ and $ T=0$ . 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, $ A_r=A_i$ . Thus, the resultant wave displacement on the string becomes

$\displaystyle y(x,t)$ $\displaystyle = A_i\,\cos(\omega\,t-k_1\,x-\phi_i) +A_i\,\cos(\omega\,t+k_1\,x+\phi_i)$    
  $\displaystyle =2\,A_i\,\cos(k_1\,x)\,\cos(\omega\,t-\phi_i),$ (452)

where use has been made of the trigonometric identity $ \cos a + \cos b\equiv 2\,\cos[(a+b)/2]\,\cos[(a-b)/2]$ . (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 $ \rho_1$ and $ \rho_2$ are separated by a short section of string of mass per unit length $ \rho_3$ . Let all three strings have the common tension $ T$ . Suppose that the first and second strings occupy the regions $ x<0$ and $ x>a$ , respectively. Thus, the middle string occupies the region $ 0\leq x\leq a$ . Moreover, the interface between the first and middle strings is at $ x=0$ , and the interface between the middle and second strings is at $ x=a$ . Suppose that a wave of angular frequency $ \omega $ is launched from a wave source at $ x=-\infty$ , and propagates towards the two interfaces. We would expect this wave to be partially reflected and partially transmitted at the first interface ($ x=0$ ), and the resulting transmitted wave to then be partially reflected and partially transmitted at the second interface ($ x=a$ ). Thus, we can write the wave displacement in the region $ x<0$ as

$\displaystyle y(x,t) = A_i\,\cos(\omega\,t-k_1\,x) + A_r\,\cos(\omega\,t+k_1\,x),$ (453)

where $ A_i$ is the amplitude of the incident wave, $ A_r$ is the amplitude of the reflected wave, and $ k_1=\omega/(T/\rho_1)^{1/2}$ . Here, the phase angles of the two waves have been chosen so as to facilitate the matching process at $ x=0$ . The wave displacement in the region $ x>a$ takes the form

$\displaystyle y(x,t) = A_t\,\cos(\omega\,t-k_2\,x-\phi_t),$ (454)

where $ A_t$ is the amplitude of the final transmitted wave, and $ k_2=\omega/(T/\rho_2)^{1/2}$ . Finally, the wave displacement in the region $ 0\leq x\leq a$ is written

$\displaystyle y(x,t) = A_+\,\cos(\omega\,t-k_3\,x) + A_-\,\cos(\omega\,t+k_3\,x),$ (455)

where $ A_+$ and $ A_-$ are the amplitudes of the forward- and backward-propagating waves on the middle string, respectively, and $ k_3=\omega/(T/\rho_3)^{1/2}$ . Continuity of the transverse displacement at $ x=0$ yields

$\displaystyle A_i + A_r=A_+ + A_-,$ (456)

where a common factor $ \cos(\omega\,t)$ has cancelled out. Continuity of the energy flux at $ x=0$ gives

$\displaystyle Z_1\,(A_i^{\,2}-A_r^{\,2}) = Z_3\,(A_+^{\,2}-A_-^{\,2}),$ (457)

so the previous two expressions can be combined to produce

$\displaystyle Z_1\,(A_i-A_r)= Z_3\,(A_+-A_-).$ (458)

Continuity of the transverse displacement at $ y=a$ yields

$\displaystyle A_+\,\cos(\omega\,t-k_3\,a)+ A_-\,\cos(\omega\,t+k_3\,a)= A_t\,\cos(\omega\,t-k_2\,a-\phi_t).$ (459)

Suppose that the length of the middle string is one quarter of a wavelength: that is, $ k_3\,a=\pi/2$ . Furthermore, let $ \phi_t=k_3\,a-k_2\,a$ . It follows that $ \cos(\omega t-k_3 a)= \sin(\omega t)$ , $ \cos(\omega t+k_3 a)=-\sin(\omega t)$ , and $ \cos(\omega t-k_2 a-\phi_t)=\sin(\omega t)$ . Thus, canceling out a common factor $ \sin(\omega t)$ , the previous expression yields

$\displaystyle A_+-A_- = A_t.$ (460)

Continuity of the energy flux at $ x=a$ gives

$\displaystyle Z_3\,(A_+^{\,2}-A_-^{\,2}) = Z_2\,A_t^{\,2}.$ (461)

so the previous two equations can be combined to generate

$\displaystyle Z_3\,(A_+ + A_-)= Z_2\,A_t.$ (462)

Equations (456) and (462) yield

$\displaystyle A_i + A_r = \frac{Z_2}{Z_3}\,A_t,$ (463)

whereas Equations (458) and (460) give

$\displaystyle A_i-A_r=\frac{Z_3}{Z_1}\,A_t.$ (464)

Hence, combining the previous two expression, we obtain

$\displaystyle A_r$ $\displaystyle = \left(\frac{Z_1 Z_2-Z_3^{ 2}}{Z_1 Z_2+Z_3^{ 2}}\right)A_i,$ (465)
$\displaystyle A_t$ $\displaystyle =\left(\frac{2 Z_1 Z_3}{Z_1 Z_2+Z_3^{ 2}}\right) A_i.$ (466)

Finally, the overall coefficient of reflection is

$\displaystyle R = \left(\frac{A_r}{A_i}\right)^2 = \left(\frac{Z_1 Z_2-Z_3^{ 2}}{Z_1 Z_2+Z_3^{ 2}}\right)^2,$ (467)

whereas the overall coefficient of transmission becomes

$\displaystyle T = \frac{Z_2}{Z_1}\left(\frac{A_t}{A_i}\right)^2= \frac{4 Z_1 Z_2 Z_3^{ 2}}{(Z_1 Z_2+Z_3^{ 2})^2}=1-R.$ (468)

Suppose that the impedance of the middle string is the geometric mean of the impedances of the two outer strings: that is, $ Z_3=\sqrt{Z_1 Z_2}$ . In this case, it follows, from the previous two equations, that $ R=0$ and $ T=1$ . 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 $ Z_1$ to a string of impedance $ Z_2$ (where $ Z_2\neq Z_1$ ) 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 $ \sqrt{Z_1 Z_2}$ . This procedure is known as impedance matching.

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 $ x<0$ , and is terminated (at $ x=0$ ) by a load resistor of resistance $ R_L$ . 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 $ \omega $ is sent down the line from a wave source at $ x=-\infty$ . The current and voltage on the line can be written

$\displaystyle I(x,t)$ $\displaystyle = I_i\,\cos(\omega\,t-k\,x) + I_r\,\cos(\omega\,t+k\,x),$ (469)
$\displaystyle V(x,t)$ $\displaystyle = I_i\,Z\,\cos(\omega\,t-k\,x) - I_r\,Z\,\cos(\omega\,t+k\,x),$ (470)

where $ I_i$ is the amplitude of the incident signal, $ I_r$ the amplitude of the signal reflected by the load, $ Z$ the characteristic impedance of the line, and $ k=\omega/v$ . Here, $ v$ is the characteristic phase velocity at which signals propagate down the line. (See Section 7.5.) The resistor obeys Ohm's law, which yields

$\displaystyle V(0,t) = I(0,t)\,R_L.$ (471)

It follows, from the previous three equations, that

$\displaystyle I_r = \left(\frac{Z-R_L}{Z+R_L}\right) I_i.$ (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

$\displaystyle R = \left(\frac{I_r}{I_i}\right)^2=\left(\frac{Z-R_L}{Z+R_L}\right)^2.$ (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

$\displaystyle T = 1-R = \frac{4 Z R_L}{(Z+R_L)^2}.$ (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 $ R_L$ . 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 $ Z=R_L$ there is no reflection of the signal sent down the line (i.e., $ R=0$ ), and all of the signal energy is therefore absorbed by the load (i.e., $ T=1$ ). As an example, a half-wave antenna (i.e., an antenna whose length is half the wavelength of the emitted radiation) has a characteristic impedance of $ 73 \Omega$ (Held and Marion 1995). Hence, a transmission line used to feed energy into such an antenna should also have a characteristic impedance of $ 73 \Omega$ . Suppose, however, that we encounter a situation in which the impedance of a transmission line, $ Z_1$ , does not match that of its terminating load, $ Z_2$ . Can anything be done to avoid reflection of the signal sent down the line? It turns out, by analogy with the previous analysis, that if the line is connected to the load via a short section of transmission line whose length is one quarter of the wavelength of the signal, and whose characteristic impedance is $ Z_3=\sqrt{Z_1 Z_2}$ , then there is no reflection of the signal: that is, all of the signal power is absorbed by the load. A short section of transmission line used in this manner is known as a quarter-wave transformer.


next up previous
Next: Electromagnetic Waves Up: Traveling Waves Previous: Transmission Lines
Richard Fitzpatrick 2013-04-08