text:
## Concerning an Heuristic Point of View Toward the Emission and Transformation of Light.

by (Albert) (Einstein), 1905

(\Huge{A}) profound formal distinction exists between the theoretical concepts which physicists have formed regarding gases and other ponderable bodies and the Maxwellian theory of electromagnetic processes in so--called empty space. While we consider the state of a body to be completely determined by the positions and velocities of a very large, yet finite, number of atoms and electrons, we make use of continuous spatial functions to describe the electromagnetic state of a given volume, and a finite number of parameters cannot be regarded as sufficient for the complete determination of such a state. According to the Maxwellian theory, energy is to be considered a continuous spatial function in the case of all purely electromagnetic phenomena including light, while the energy of a ponderable object should, according to the present conceptions of physicists, be represented as a sum carried over the atoms and electrons. The energy of a ponderable body cannot be subdivided into arbitrarily many or arbitrarily small parts, while the energy of a beam of light from a point source $($according to the Maxwellian theory of light or, more generally, according to any wave theory$)$ is continuously spread an ever increasing volume.

The wave theory of light, which operates with continuous spatial functions, has worked well in the representation of purely optical phenomena and will probably never be replaced by another theory. It should be kept in mind, however, that the optical observations refer to time averages rather than instantaneous values. In spite of the complete experimental confirmation of the theory as applied to diffraction, reflection, refraction, dispersion, etc., it is still conceivable that the theory of light which operates with continuous spatial functions may lead to contradictions with experience when it is applied to the phenomena of emission and transformation of light.

It seems to me that the observations associated with blackbody radiation, fluorescence, the production of cathode rays by ultraviolet light, and other related phenomena connected with the emission or transformation of light are more readily understood if one assumes that the energy of light is discontinuously distributed in space. In accordance with the assumption to be considered here, the energy of a light ray spreading out from a point source is not continuously distributed over an increasing space but consists of a finite number of energy quanta which are localized at points in space, which move without dividing, and which can only be produced and absorbed as complete units.

In the following I wish to present the line of thought and the facts which have led me to this point of view, hoping that this approach may be useful to some investigators in their research.

### 1. Concerning a Difficulty with Regard to the Theory of Blackbody Radiation

We start first with the point of view taken in the Maxwellian and the electron theories and consider the following case. In a space enclosed by completely reflecting walls, let there be a number of gas molecules and electrons which are free to move and which exert conservative forces on each other on close approach: i.e. they can collide with each other like molecules in the kinetic theory of gases1. Furthermore, let there be a number of electrons which are bound to widely separated points by forces proportional to their distances from these points. The bound electrons are also to participate in conservative interactions with the free molecules and electrons when the latter come very close. We call the bound electrons "oscillators'': they emit and absorb electromagnetic waves of definite periods.

According to the present view regarding the origin of light, the radiation in the space we are considering $($radiation which is found for the case of dynamic equilibrium in accordance with the Maxwellian theory$)$ must be identical with the blackbody radiation --- at least if oscillators of all the relevant frequencies are considered to be present.

For the time being, we disregard the radiation emitted and absorbed by the oscillators and inquire into the condition of dynamical equilibrium associated with the interaction $($or collision$)$ of molecules and electrons. The kinetic theory of gases asserts that the average kinetic energy of an oscillator electron must be equal to the average kinetic energy of a translating gas molecule. If we separate the motion of an oscillator electron into three components at angles to each other, we find for the average energy $\overline{E}$ of one of these linear components the expression

\begin{equation*} \overline{E} = (R/N) ~ T, \end{equation*}

where $R$ denotes the universal gas constant, $N$ denotes the number of "real molecules'' in a gram equivalent, and $T$ the absolute temperature. The energy $\overline{E}$ is equal to two-thirds the kinetic energy of a free monatomic gas particle because of the equality the time average values of the kinetic and potential energies of the oscillator. If through any cause --in our case through radiation processes-- it should occur that the energy of an oscillator takes on a time-average value greater or less than $\overline{E}$, then the collisions with the free electrons and molecules would lead to a gain or loss of energy by the gas, different on the average from zero. Therefore, in the case we are considering, dynamic equilibrium is possible only when each oscillator has the average energy $\overline{E}$.

We shall now proceed to present a similar argument regarding the interaction between the oscillators and the radiation present in the cavity. Herr Planck has derived 2 the condition for the dynamics equilibrium in this case under the supposition that the radiation can be considered a completely random process 3. He found

\begin{equation*} (\overline{E}_{\nu}) = (L^3/ 8 \pi \nu^2) \rho_{\nu}, \end{equation*}

where $(\overline{E}_{\nu})$ is the average energy $($per degree of freedom$)$ of an oscillator with eigenfrequency $\nu$, $L$ the velocity of light, $\nu$ the frequency, and $\rho_{\nu} d \nu$ the energy per unit volume of that portion of the radiation with frequency between $\nu$ and $\nu + d\nu$.

If the radiation energy of frequency $\nu$ is not continually increasing or decreasing, the following relations must obtain:

\begin{equation*} (R/N)~ T = \overline{E} = \overline{E}{\nu} = (L^3/8 \pi \nu^2) \rho_{\nu}, \end{equation*} \begin{equation*} \rho_{\nu} = (R/N) (8\pi \nu^2/L^3) ~ T. \end{equation*}

These relations, found to be the conditions of dynamic equilibrium, not only fail to coincide with experiment, but also state that in our model there can be not talk of a definite energy distribution between ether and matter. The wider the range of wave numbers of the oscillators, the greater will be the radiation energy of the space, and in the limit we obtain

\begin{equation*} \int \limits^{\infty}_0~ \rho_{\nu}~ d \nu = \frac{R}{N} \cdot \frac{8 \pi}{L^3} \cdot T~ \int \limits^{\infty}_0 ~ \nu^2 ~d \nu = \infty. \end{equation*}

### 2. Concerning Planck's Determination of the Fundamental Constants

We wish to show in the following that Herr Planck's determination of the fundamental constants is, to a certain extent, independent of his theory of blackbody radiation.

Planck's formula, 4 which has proved adequate up to this point, gives for $\rho_{\nu}$

\begin{equation*} \rho_{\nu} = \frac{\alpha \nu^3}{e^{\beta \nu /T} - 1}, \end{equation*} \begin{equation*} \alpha = 6.10 \times 10^{-56}, \end{equation*} \begin{equation*} \beta = 4.866 \times 10^{-11}. \end{equation*} For large values of $T/\nu$; i.e. for large wavelengths and radiation densities, this equation takes the form

\begin{equation*} \rho_{\nu} = (\alpha/\beta) ~\nu^2 T. \end{equation*}

It is evident that this equation is identical with the one obtained in Sec. 1 from the Maxwellian and electron theories. By equating the coefficients of both formulas one obtains

\begin{equation*} (R/N) (8 \pi/L^3) = (\alpha/\beta) \end{equation*}

or

\begin{equation*} N = (\beta/\alpha) (8 \pi R/L^3) = 6.17 \times 10^{23}. \end{equation*}

i.e., an atom of hydrogen weighs $1/N$ grams $= 1.62 \times 10^{-24}$ g. This is exactly the value found by Herr Planck, which in turn agrees with values found by other methods.

We therefore arrive at the conclusion: the greater the energy density and the wavelength of a radiation, the more useful do the theoretical principles we have employed turn out to be: for small wavelengths and small radiation densities, however, these principles fail us completely.

In the following we shall consider the experimental facts concerning blackbody radiation without invoking a model for the emission and propagation of the radiation itself.

### 3. Concerning the Entropy of Radiation

The following treatment is to be found in a famous work by Herr W. Wien and is introduced here only for the sake of completeness.

Suppose we have radiation occupying a volume $v$. We assume that the observable properties of the radiation are completely determined when the radiation density $\rho(\nu)$ is given for all frequencies 5. Since radiation of different frequencies are to be considered independent of each other when there is no transfer of heat or work, the entropy of the radiation can be represented by

\begin{equation*} S = v \int \limits^{\infty}_0~ \varphi (\rho, \nu)~ d \nu, \end{equation*}

where $\varphi$ is a function of the variables $\rho$ and $\nu$.

$\varphi$ can be reduced to a function of a single variable through formulation of the condition that the entropy of the radiation is unaltered during adiabatic compression between reflecting walls. We shall not enter into this problem, however, but shall directly investigate the derivation of the function $\varphi$ from the blackbody radiation law.

In the case of blackbody radiation, $\rho$ is such a function of $\nu$ that the entropy is maximum for a fixed value of energy; i.e.,

\begin{equation*} \delta~ \int \limits^{\infty}_0~ \varphi~(\rho, \nu)~ d \nu = 0, \end{equation*}

providing

\begin{equation*} \delta~ \int \limits^{\infty}_0~ \rho d \nu = 0. \end{equation*}

From this it follows that for every choice of $\delta \rho$ as a function of $\nu$

\begin{equation*} \int \limits^{\infty}_0~ \left( \frac{\partial \varphi}{\partial \rho} - \lambda \right) \delta \rho d \nu = 0, \end{equation*}

where $\lambda$ is independent of $\nu$. In the case of blackbody radiation, therefore, $\partial \varphi/ \partial \rho$ is independent of $\nu$.

The following equation applies when the temperature of a unit volume of blackbody radiation increases by $dT$

\begin{equation*} dS = \int \limits^{\nu = \infty}_{\nu = 0}~ \left( \frac{\partial \varphi} {\partial \rho} \right)~d \rho d \nu, \end{equation*}

or, since $\partial \varphi / \partial \rho$ is independent of $\nu$.

\begin{equation*} dS = (\partial \varphi/ \partial \rho) ~ dE. \end{equation*}

Since $dE$ is equal to the heat added and since the process is reversible, the following statement also applies

\begin{equation*} dS = (1/T) ~ dE. \end{equation*}

By comparison one obtains

\begin{equation*} \partial \varphi/ \partial \rho = 1/T. \end{equation*}

This is the law of blackbody radiation. Therefore one can derive the law of blackbody radiation from the function $\varphi$, and, inversely, one can derive the function $\varphi$ by integration, keeping in mind the fact that $\varphi$ vanishes when $\rho = 0$.

### 4. Asymptotic form for the Entropy of Monochromatic Radiation at Low Radiation Density

From existing observations of the blackbody radiation, it is clear that the law originally postulated by Herr W. Wien,

\begin{equation*} \rho = \alpha \nu^3 e^{- \beta \nu/T}, \end{equation*}

is not exactly valid. It is, however, well confirmed experimentally for large values of $\nu/T$. We shall base our analysis on this formula, keeping in mind that our results are only valid within certain limits.

This formula gives immediately

\begin{equation*} (1/T) = - (1/\beta \nu) ~ \mbox{ln}~ (\rho/\alpha \nu^3) \end{equation*}

and then, by using the relation obtained in the preceding section,

\begin{equation*} \varphi(\rho, \nu) = - \frac{\rho}{\beta \nu} \left[ \mbox{ln} ~ \left( \frac{\rho} {\alpha \nu^3} \right) - 1 \right]. \end{equation*}

Suppose that we have radiation of energy $E$, with frequency between $\nu$ and $\nu + d \nu$, enclosed in volume $v$. The entropy of this radiation is:

\begin{equation*} S = v \varphi (\rho, \nu) d\nu = - \frac{E}{\beta \nu} \left[\mbox{ln}~ \left( \frac{E}{v \alpha \nu^3 d \nu} \right) - 1 \right]. \end{equation*}

If we confine ourselves to investigating the dependence of the entropy on the volume occupied by the radiation, and if we denote by $S_0$ the entropy of the radiation at volume $v_0$, we obtain

\begin{equation*} S - S_0 = (E/\beta \nu) ~ \mbox{ln}~ (v/v_0). \end{equation*}

This equation shows that the entropy of a monochromatic radiation of sufficiently low density varies with the volume in the same manner as the entropy of an ideal gas or a dilute solution. In the following, this equation will be interpreted in accordance with the principle introduced into physics by Herr Boltzmann, namely that the entropy of a system is a function of the probability its state.

### 5. Molecular--Theoretic Investigation of the Dependence of the Entropy of Gases and Dilute solutions on the volume

In the calculation of entropy by molecular--theoretic methods we frequently use the word "probability" in a sense differing from that employed in the calculus of probabilities. In particular "gases of equal probability" have frequently been hypothetically established when one theoretical models being utilized are definite enough to permit a deduction rather than a conjecture. I will show in a separate paper that the so-called "statistical probability" is fully adequate for the treatment of thermal phenomena, and I hope that by doing so I will eliminate a logical difficulty that obstructs the application of Boltzmann's Principle. here, however, only a general formulation and application to very special cases will be given.

If it is reasonable to speak of the probability of the state of a system, and furthermore if every entropy increase can be understood as a transition to a state of higher probability, then the entropy $S_1$ of a system is a function of $W_1$, the probability of its instantaneous state. If we have two noninteracting systems $S_1$ and $S_2$, we can write

\begin{equation*} S_1 = \varphi_1 (W_1), \end{equation*} \begin{equation*} S_2 = \varphi_2 (W_2). \end{equation*}

If one considers these two systems as a single system of entropy $S$ and probability $W$, it follows that

\begin{equation*} S = S_1 + S_2 = \varphi (W) \end{equation*}

and \begin{equation*} W = W_1 \cdot W_2. \end{equation*}

The last equation says that the states of the two systems are independent of each other.

From these equation it follows that

\begin{equation*} \varphi (W_1 \cdot W_2) = \varphi_1 (W_1) + \varphi_2 (W_2) \end{equation*}

and finally

\begin{equation*} \varphi_1(W_1) = C ~\mbox{ln} (W_1) + \mbox{const}, \end{equation*} \begin{equation*} \varphi_2(W_2) = C ~ \mbox{ln} (W_2) + \mbox{const}, \end{equation*} \begin{equation*} \varphi (W) = C ~ \mbox{ln} (W) + \mbox{const}. \end{equation*}

The quantity $C$ is therefore a universal constant; the kinetic theory of gases shows its value to be $R/N$, where the constants $R$ and $N$ have been defined above. If $S_0$ denotes the entropy of a system in some initial state and $W$ denotes the relative probability of a state of entropy $S$, we obtain in general

\begin{equation*} S - S_0 = (R/N) ~ \mbox{ln}~W. \end{equation*}

First we treat the following special case. We consider a number $(n)$ of movable points $($e.g., molecules$)$ confined in a volume $v_0$. Besides these points, there can be in the space any number of other movable points of any kind. We shall not assume anything concerning the law in accordance with which the points move in this space except that with regard to this motion, no part of the space $($and no direction within it$)$ can be distinguished from any other. Further, we take the number of these movable points to be so small that we can disregard interactions between them.

This system, which, for example, can be an ideal gas or a dilute solution, possesses an entropy $S_0$. Let us imagine transferring all $n$ movable points into a volume $v$ $($part of the volume $v_0$$) without anything else being changed in the system. This state obviously possesses a different entropy (S), and now wish to evaluate the entropy difference with the help of the Boltzmann's Principle. We inquire: How large is the probability of the latter state relative to the original one? Or: How large is the probability that at a randomly chosen instant of time all n movable points in the given volume v_0 will be found by chance in the volume v? For this probability, which is a "statistical probability", one obviously obtains: \begin{equation*} W = (v/v_0)^n; \end{equation*} By applying the Boltzmann's Principle, one then obtains \begin{equation*} S - S_0 = R ~ (n/N) ~ \mbox{ln}~ (v/v_0). \end{equation*} It is noteworthy that in the derivation of this equation, from which one can easily obtain the law of Boyle and Gay--Lussac as well as the analogous law of osmotic pressure thermodynamically,6 no assumption had to be made as to a law of motion of the molecules. ### 6. Interpretation of the Expression for the volume Dependence of the entropy of Monochromatic Radiation in Accordance with Boltzmann's Principle In Sec. 4, we found the following expression for the dependence of the entropy of monochromatic radiation on the volume \begin{equation*} S - S_0 = (E/\beta \nu) ~ \mbox{ln}~ (v/v_0). \end{equation*} If one writes this in the from \begin{equation*} S - S_0 = (R/N) ~ \mbox{ln} \left[ (v/v_0)^{(N/R) (E/\beta \nu)} \right]. \end{equation*} and if one compares this with the general formula for the Boltzmann's principle \begin{equation*} S - S_0 = (R/N) ~ \mbox{ln} W, \end{equation*} one arrives at the following conclusion: If monochromatic radiation of frequency \nu and energy E is enclosed by reflecting walls in a volume v_0, the probability that the total radiation energy will be found in a volume v (part of the volume v_0$$)$ at any randomly chosen instant is

\begin{equation*} W = (v/v_0)^{(N/R) (E/\beta \nu)}. \end{equation*}

From this we further conclude that: Monochromatic radiation of low density $($within the range of validity of Wien's radiation formula$)$ behaves thermodynamically as though it consisted of a number of independent energy quanta of magnitude $R \beta \nu/N$.

We still wish to compare the average magnitude of the energy quanta of the blackbody radiation with the average translational kinetic energy of a molecule at the same temperature. The latter is $^3/_2 (R/N) T$, while, according to the Wien formula, one obtains for the average magnitude of an energy quantum

\begin{equation*} \int \limits^{\infty}_0~ \alpha \nu^3 e^{- \beta \nu/T} d\nu\bigg/ \int \limits^{\infty}_0~ \frac{N}{R \beta \nu} ~ \alpha \nu^3 e^{- \beta \nu/ T} d \nu = 3 (RT/N). \end{equation*}

If the entropy of monochromatic radiation depends on volume as though the radiation were a discontinuous medium consisting of energy quanta of magnitude $R \beta \nu/N$, the next obvious step is to investigate whether the laws of emission and transformation of light are also of such a nature that they can be interpreted or explained by considering light to consist of such energy quanta. We shall examine this question in the following.

### 7. Concerning Stokes's Rule

According to the result just obtained, let us assume that, when monochromatic light is transformed through photoluminescence into light of a different frequency, both the incident and emitted light consist of energy quanta of magnitude $R \beta \nu/N$, where $\nu$ denotes the relevant frequency. The transformation process is to be interpreted in the following manner. Each incident energy quantum of frequency $\nu_1$ is absorbed and generates by itself--at least at sufficiently low densities of incident energy quanta -- a light quantum of frequency $\nu_2$; it is possible that the absorption of the incident light quanta can give rise to the simultaneous emission of light quanta of frequencies $\nu_3, \nu_4$ etc., as well as to energy of other kinds, e.g., heat. It does not matter what intermediate processes give rise to this final result. If the fluorescent substance is not a perpetual source of energy, the principle of conservation of energy requires that the energy of an emitted energy quantum cannot be greater than that of the incident light quantum; it follows that

\begin{equation*} R~ \beta \nu_2/N \leq R~ \beta \nu_1/N \end{equation*}

or

\begin{equation*} \nu_2 \leq \nu_1. \end{equation*}

This is the well--known Stokes's Rule.

It should be strongly emphasized that according to our conception the quantity of light emitted under conditions of low illumination $($other conditions remaining constant$)$ must be proportional to the strength of the incident light, since each incident energy quantum will cause an elementary process of the postulated kind, independently of the action of other incident energy quanta. In particular, there will be no lower limit for the intensity of incident light necessary to excite the fluorescent effect.

According to the conception set forth above, deviations from Stokes's Rule are conceivable in the following cases:

1. when the number of simultaneously interacting energy quanta per unit volume is so large that an energy quantum of emitted light can receive its energy from several incident energy quanta;

2. when the incident $($or emitted$)$ light is not of such a composition that it corresponds to blackbody radiation within the range of validity of Wien's Law, that is to say, for example, when the incident light is produced by a body of such high temperature that for the wavelengths under consideration Wien's Law is no longer valid.

The last-mentioned possibility commands especial interest. According to the conception we have outlined, the possibility is not excluded that a "non-Wien radiation" of very low density can exhibit an energy behavior different from that of a blackbody radiation within the range of validity of Wien's Law.

### 8. Concerning the Emission of Cathode Rays Through Illumination of Solid Bodies

The usual conception that the energy of light is continuously distributed over the space through which it propagates, encounters very serious difficulties when one attempts to explain the photoelectric phenomena, as has been pointed out in Herr Lenard's pioneering paper 7.

According to the concept that the incident light consists of energy quanta of magnitude $R \beta \nu/N$, however, one can conceive of the ejection of electrons by light in the following way. Energy quanta penetrate into the surface layer of the body, and their energy is transformed, at least in part, into kinetic energy of electrons. The simplest way to imagine this is that a light quantum delivers its entire energy to a single electron: we shall assume that this is what happens. The possibility should not be excluded, however, that electrons might receive their energy only in part from the light quantum.

An electron to which kinetic energy has been imparted in the interior of the body will have lost some of this energy by the time it reaches the surface. Furthermore, we shall assume that in leaving the body each electron must perform an amount of work $P$ characteristic of the substance. The ejected electrons leaving the body with the largest normal velocity will be those that were directly at the surface. The kinetic energy of such electrons is given by

\begin{equation*} R~ \beta \nu/N - P. \end{equation*}

If the body is charged to a positive potential $\Pi$ and is surrounded by conductors at zero potential, and if $\Pi$ is just large enough to prevent loss of electricity by the body, if follows that:

\begin{equation*} \Pi \epsilon = R \beta \nu/N - P \end{equation*}

where $\epsilon$ denotes the electronic charge, or

\begin{equation*} \Pi E = R \beta \nu - P' \end{equation*}

where $E$ is the charge of a gram equivalent of a monovalent ion and $P'$ is the potential of this quantity of negative electricity relative to the body. 8

If one takes $E = 9.6 \times 10^3$, then $\Pi \cdot 10^{-8}$ is the potential in volts which the body assumes when irradiated in a vacuum.

In order to see whether the derived relation yields an order of magnitude consistent with experience, we take $P' = 0$, $\nu = 1.03 \times 10^{15}$ $($corresponding to the limit of the solar spectrum toward the ultraviolet$)$ and $\beta = 4.866 \times 10^{-11}$. We obtain $\Pi \cdot 10^7 = 4.3$ volts, a result agreeing in order magnitude with those of Herr Lenard.9

If the derived formula is correct, then $\Pi$, when represented in Cartesian coordinates as a function of the frequency of the incident light, must be a straight line whose slope is independent of the nature of the emitting substance.

As far as I can see, there is no contradiction between these conceptions and the properties of the photoelectric observed by Herr Lenard. If each energy quantum of the incident light, independently of everything else, delivers its energy of electrons, then the velocity distribution of the ejected electrons will be independent of the intensity of the incident light; on the other hand the number of electrons leaving the body will, if other conditions are kept constant, be proportional to the intensity of the incident light.10

Remarks similar to those made concerning hypothetical deviations from Stokes's Rule can be made with regard to hypothetical boundaries of validity of the law set forth above.

In the foregoing it has been assumed that the energy of at least some of the quanta of the incident light is delivered completely to individual electrons. If one does not make this obvious assumption, one obtains, in place of the last equation:

\begin{equation*} \Pi E + P' \leq R \beta \nu. \end{equation*}

For fluorescence induced by cathode rays, which is the inverse process to the one discussed above, one obtains by analogous considerations:

\begin{equation*} \Pi E + P' \geq R \beta \nu. \end{equation*}

In the case, of the substances investigated by Herr Lenard, $PE$ 11 is always significantly greater than $R \beta \nu$, since the potential difference, which the cathode rays must traverse in order to produce visible light, amounts in some cases to hundreds and in others to thousands of volts 12. It is therefore to be assumed that the kinetic energy of an electron goes into the production of many light energy quanta.

### 9. Concerning the Ionization of Gases by Ultraviolet Light Solid Bodies

We shall have to assume that, the ionization of a gas by ultraviolet light, an individual light energy quantum is used for the ionization of an individual gas molecule. From this follows immediately that the work of ionization $($i.e., the work theoretically needed for ionization$)$ of a molecule cannot be greater than the energy of an absorbed light quantum capable of producing this effect. If one denotes by $J$ the $($theoretical$)$ work of ionization per gram equivalent, then it follows that:

\begin{equation*} R ~\beta \nu \geq J. \end{equation*}

According to Lenard's measurements, however, the largest effective wavelength for air is approximately $1.9 \times 10^{-5}$ cm: therefore:

\begin{equation*} R ~\beta \nu = 6.4 \cdot 10^{12}~ \mbox{erg} \geq J. \end{equation*}

An upper limit for the work of ionization can also be obtained from the ionization potentials of rarefied gases. According to J. Stark 13 the smallest observed ionization potentials for air $($at platinum anodes$)$ is about 10 V 14. One therefore obtains $9.6 \times 10^{12}$ as an upper limit for $J$, which is nearly equal to the value found above.

There is another consequence the experimental testing of which seems to me to be of great importance. If every absorbed light energy quantum ionizes a molecule, the following relation must obtain between the quantity of absorbed light $L$ and the number of gram molecules of ionized gas $j$:

\begin{equation*} j = L/R \beta \nu. \end{equation*}

If our conception is correct, this relationship must be valid for all gases which $($at the relevant frequency$)$ show no appreciable absorption without ionization.

#### FOOTNOTES

1. This assumption is equivalent to the supposition that the average kinetic energies of gas molecules and electrons are equal to each other at thermal equilibrium. It is well known that, with the help of this assumption, Herr Drude derived a theoretical expression for the ratio of thermal and electrical conductivities of metals.

2. M. Planck, Ann. Phys. 1, 99 $($1900$)$.

3. This problem can be formulated in the following manner. We expand the $Z$ component of the electrical force $(Z)$ at an arbitrary point during the time interval between $t = 0$ and $t = T$ in a Fourier series in which $A_{\nu} \geq 0$ and $0 \leq \alpha_{\nu} \leq 2 \pi$: the time $T$ is taken to be very large relative to all the periods of oscillation that are present:

\begin{equation*} Z = \sum^{\nu = \infty}_{\nu = 1} A_{\nu} \sin \left( 2 \pi \nu \frac{t}{T} + \alpha_{\nu} \right), \end{equation*}

If one imagines making this expansion arbitrary often at a given point in space at randomly chosen instants of time, one will obtain various sets of values of $A_{\nu}$ and $\alpha_{\nu}$. There then exist for the frequency of occurrence of different sets of values of $A_{\nu}$ and $\alpha_{\nu}$ $($statistical$)$ probabilities $dW$ of the form:

\begin{equation*} dW = f (a_1, A_2, \ldots, \alpha_1, \alpha_2, \ldots) dA_1 dA_2 \ldots d\alpha_1 d \alpha_2 \ldots, \end{equation*}

The radiation is then as disordered as conceivable if

\begin{equation*} f (A_1, A_2, \ldots \alpha_1, \alpha_2, \ldots) = F_1 (A_1) F_2 (A_2) \ldots f_1 (\alpha_1) f_2 (\alpha_2) \ldots, \end{equation*}

i.e., if the probability of a particular value of $A$ or $\alpha$ is independent of other values of $A$ or $\alpha$. The more closely this condition is fulfilled $($namely, that the individual pairs of values of $A_{\nu}$ and $\alpha_{\nu}$ are dependent upon the emission and absorption processes of specific groups of oscillators$)$ the more closely will radiation in the case being considered approximate a perfectly random state.

4. M. Planck, Ann. Phys. 4, 561 $($1901$)$.

5. This assumption is an arbitrary one. One will naturally cling to this simplest assumption as long as it is not controverted experiment.

6. If $E$ is the energy of the system, one obtains:

\begin{equation*} - d \cdot (E - TS) = pdv = TdS = RT \cdot (n/N) \cdot (dv/v); \end{equation*}

therefore

\begin{equation*} pv = R \cdot (n/N) \cdot T. \end{equation*}

7. P. Lenard, Ann. Phys., 8, 169, 170 $($1902$)$.

8. If one assumes that the individual electron is detached from a neutral molecule by light with the performance of a certain amount of work, nothing in the relation derived above need be changed; one can simply consider $P'$ as the sum of two terms.

9. P.Lenard, Ann. Phys. 8, pp. 163, 185, and Table I, Fig. 2 $($1902$)$.

10. P. Lenard, Ref. 9, p. 150 and p. 166--168.

11. Should be $\Pi E$ $($translator's note$)$.

12. P. Lenard, Ann. Phys., 12, 469 $($1903$)$.

13. J. Stark, Die Electrizitet in Gasen $($Leipzig, 1902, p. 57$)$.

14. In the interior of gases the ionization potential for negative ions is, however, five times greater.

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# Interpretation

Purpose

To show how DC-IP data are processed and inverted to reveal meaningful information about the earth structure (conductivity and chargeability).

## Linearization

For the inversion of IP data, often linearzed approach is used [Sei59] [OL94a]. The measured voltage, $$V$$ will be a function of conductivity. We consider two end members of conductivity at inifinte and zero frequency: $$\sigma_{\infty}$$ and $$\sigma_0$$. Here conductivity at zero frequency can be expressed as

$\sigma_0 = \sigma_{\infty} (1-\eta)$

Considering measured voltage at zero frequency or late on-time, the voltage will be a function of $$\sigma_0 \ (x, y, z)$$ :

$V_0 = V[\sigma_0] = V[\sigma_{\infty}(1-\eta)]$

By expanding this in terms of $$\sigma_{\infty}$$, we obtain

$V_0 = V[\sigma_{\infty}(1-\eta)] = V[\sigma_{\infty}] - \frac{\partial V}{\partial \sigma_{\infty}} \sigma_{\infty} \eta + O \Big( (\sigma_{\infty} \eta)^2 \Big)$

IP datum is considered as a small perturbation ($$\eta \ll 1$$), and can be written as

$d^{IP} = V_0 - V_{\infty} = - \frac{\partial V}{\partial \sigma_{\infty}} \sigma_{\infty} \eta + O \Big( (\sigma_{\infty} \eta)^2 \Big)$

Ignoring second order term yields

$\begin{split}d^{IP} = - \frac{\partial V}{\partial \sigma_{\infty}} \sigma_{\infty} \eta = - \frac{\partial V}{\partial \ log (\sigma_{\infty})} \eta \\ = \frac{\partial V}{\partial \ log (\rho_{\infty})} \eta\end{split}$

Considering apparent chargeability, $$\eta_a$$, we obtain

(494)$d^{IP} = \eta_a = \frac{V_0 - V_{\infty} }{V_0} = \frac{\partial log \ (V)}{\partial \ log (\rho_{\infty})} \eta$

Finally, IP data can be expressed as a linear equation

(495)$\mathbf{d}^{IP} = \mathbf{J} \boldsymbol{\eta}$

where $$\mathbf{J}$$ is a sensitivity matrix.

## Two stage method

For the inversion of time domain IP data, two-stage method is often used [OL94a]. The first step is DC inversion, and here we consider a voltage at at late on-time ($$V_0$$) as DC datum. By inverting DC data, we recover conductivity model, $$\sigma_{est}$$. Second step is IP inversion. Voltages at late off-times ($$V_s$$) are considered as IP data, and often voltages in specific time window is integrated. The recovered conductivity is used to form sensitivity matrix, and using the linear equation (Eq. (495)) we invert IP data, and recover chargeability.

Note

Note that in practice there are signficant electromagnetic (EM) induction effects right after current is turend on, hence obtaining $$V_{\infty}$$ is challenging. Similarly, right after the current is turned off, significant EM induction effects exists, and this is the reason why we only consider late off-time voltages as IP data.

## Example

We illustrate a synthetic two stage inversion (DC-IP inversion) example. Voltages at zero and infinite freqencies are computed: $$V_0$$ and $$V_{\infty}$$ using $$\sigma_0$$ and $$\sigma_{\infty}$$, respectively. DC datum is considered as $$V_0$$, and apparent chargeabiility is calculated using Eq. (494). Conductivity and chargeability models are presented in Fig. 174 (a) and (c), respectively.

### DC-IP data

A gradient array survey is used to obtain both DC and IP data. A plan view of the resistivity model and electrode geometry is shown in Fig. 161 (a). The survey consists of a grid of 13 x 13 receivers located between a 450 meter dipole current source. Each receiver is a 20 meter dipole. The corresponding apparent resistivity map is shown in Fig. 161 (b). Due to the multiple prisms having different resistivity values from the background, the obtained apparent resistivity map is fairly complicated. In constrast, only the center prism is chargeable, hence a single apparent chargeability anomaly is obtained. From these apparent resistivity and chargeability maps, we can recognize conductive and chargeable unit, and its hoizontal location and extent, whereas depth information the unit is lacked.

A dipole-dipole survey is carried out along a line that passes directly above the conductive prism. Different from the gradient array case, we assumed all prisms are chargeable. Fig. 175 (a) and (c) show conductivity and chargeability sections, respectively. The resulting pseudosections are shown in Fig. 175 (b) and (d). Due to the near surface prisms both DC and IP pseudosections are complicated, and hard to be interpreted just by looking at them. For following DC-IP inversions, we use these DC and IP data obtained with the dipole-dipole array.

### 2D Inversion of DC-IP data

Following the two stage inversion method we first invert DC data, and obtain conductivity model. Then by using the recovered conductivity, we form a sensitivity matrix for an IP inversion, invert IP data, and recover chargeability model.

The DC and IP data are inverted using a standard Gauss-Newton framework. This is outlined in Inversion. The DC data are the measured voltages ($$V_0$$) and the goal is to find an electrical conductivity that satisfactorily reproduces these data and agrees with a priori geologic structure and petrophysical constraints. Although the geology is 3D, we invert the data using a 2D inversion algorithm. Parameters used for the inversion of the dipole-dipole data (Fig. 174 (b)) are provided in Table 4.

Number of sources 20 195 $$2\%|d| (percentage) + 2 \times 10^{-4} V$$ (floor) $$10 \times 10 \times 10$$ meters $$0.01$$ S/m $$0.01, 1, 1, 1$$

Fig. 176 (b) presents the recovered 2D conductivity model after convergence of the algorithm. Multiple conductors and a resistor are well imaged in 2D section.

For the IP inversion, the same inversion parameters are used except for uncertainty and reference model. Percentage and floor for the uncertainty are set to 0 $$\%$$ and 0.3 mV/V. The reference chargeability used here is zero. Fig. 176 (d) shows the reocovered 2D chargeability model, and all chargealbe prisms either on the surface and at depth are imaged well.

Note

Considering time domain IP data in practice, we cannot measure neither $$V_0$$ and $$V_{\infty}$$. Hence, often an IP datum is considered as an integration of late off-time voltages (named apparent chargeability): $$\eta_a = \frac{1}{V_0}\int V_s \ (t) \ dt$$. Therefore, a recovered chargeability model by inverting the apparent chargeability will not correspond to the chargeability, $$\eta=\frac{\rho_{0}-\rho_{\infty}}{\rho_0}$$.

Rather, the recovered charegeability model from the inversion should be considered as “pseudo-chargeability” reflecting some chargeability information captured in a chosen window.

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# D and JK flip flop

Discussion in 'Homework Help' started by bhuvanesh, Aug 27, 2014.

1. ### bhuvanesh Thread Starter Member

Aug 10, 2013
268
2
how is that said"D flip flop next value depends on both present state and input values ,but JK flip flop depends only on present input values"
please explain me or if u feel little bore give right link where i will be answered

2. ### Brownout Well-Known Member

Jan 10, 2012
2,375
998
It's actually the other way around. The "D" f/f depends only on the input value. The JK flip flop depends on the input value and present state under certain circumstances. For example, if j=1 and k=1, the next state will be ~(present state)

bhuvanesh likes this.
3. ### tshuck Well-Known Member

Oct 18, 2012
3,531
675
I think you've gotten those mixed up.

A D flip flop depends on D - that is, at the sensitive clock edge, the output gets the value of D, regardless of what the output was.
The truth take for a D flip flop:
Column 1 Column 2 Column 3
Clock D Qnext
Rising edge 0 0
Rising edge 1 1
Non-Rising X Q

This results in the characteristic equation for the D flip flop as: $Q_{next} = D$

The JK flip flop, on the other hand, does depend on the current output - just look at its characteristic equation: $Q_{next} = J\overline{Q}+\overline{K}Q$
As you can see, $Q_{next}$ depends on Q, meaning the next state is dependent on the current state.

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4. ### bhuvanesh Thread Starter Member

Aug 10, 2013
268
2
when designing is considered,for JK f/f we have another extra table called excitation table ,which is not while designing D f/f.could you explain what and why the excitaion table is used in JK f/f when designing

5. ### tshuck Well-Known Member

Oct 18, 2012
3,531
675
Okay, let's take a look at the relevant tables (I can't remember how to do row spanning...):

Column 1 Column 2 Column 3 Column 4 Column 5 Column 6 Column 7 Column 8 Column 9
Characteristic table Excitation table
J K Comment Qnext Q J K Comment Qnext
0 0 hold state Q 0 0 X No Change 0
0 1 reset 0 0 1 X Set 1
1 0 set 1 1 X 1 Reset 0
1 1 toggle Q 1 X 0 No Change 1

The excitation table tells you what Qnext is, given J, K, and Q.

Here is a post I made regarding using JK flip flops to design a synchronous counter - going through it may help you.

bhuvanesh likes this.
6. ### bhuvanesh Thread Starter Member

Aug 10, 2013
268
2
most answers are in the word itself,but i am always beating above the bush
1)characteristic table show character of that f/f
2)excitation table show next state output with not only J and K and but also with Q(present state)
right?

7. ### tshuck Well-Known Member

Oct 18, 2012
3,531
675
With respect to the table above, the characteristic table describes the output, given inputs (this is why Qnext is sometimes given as Q), where the excitation table shows the state of signals of concern that affect the output, Qnext (since the current state affects the next state, Q is listed along with J and K).

bhuvanesh likes this.


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# Random Coin Toss in Haskell

Learn You a Haskell gives an exercise to implement the following function:

-- make a function that simulates tossing a function 3 times
import System.Random

tossCoin3 :: Int -> (Bool, Bool, Bool)
tossCoin3 x =
let (num1, gen1) = random (mkStdGen x)   :: (Int, StdGen)
(num2, gen2) = random (gen1)         :: (Int, StdGen)
(num3, _)    = random (gen2)         :: (Int, StdGen)

in (convertToBool num1, convertToBool num2, convertToBool num3)
where convertToBool y = if y mod 2 == 0 then True else False

I would've mapped a list of [num1, num2, num3] with the convertToBool, and then convert to a 3-tuple. But then I saw this post.

# Converting a Bool to a Bool with an if

if tests Bool values and returns results. You’re using it to return True if the input is True, and False if the input is False:

convertToBool y = if y mod 2 == 0 then True else False

But that’s just the identity function. You can remove the if:

convertToBool y = y mod 2 == 0

# Useless parentheses

random (gen1)
random (gen2)

# Generating random Ints rather than the type you wish to get directly

You’re using random to generate an Int. That’s fine if you want an Int—but you want a Bool. Sure, you can calculate a Bool out of one, but random can also give you a Bool directly. This would simplify your implementation.

I feel it’s necessary to point out the dangers of limiting random numbers to a range with modulo. In this case, it’s not actually a problem, but pretend you want to generate a number between 0 and 6, and the random number generator you have gives you numbers between 0 and 9. If you used modulo, you’d map:

0, 6 --> 0
1, 7 --> 1
2, 8 --> 2
3, 9 --> 3
4    --> 4
5    --> 5
6    --> 6

That’s not a very even distribution. The problem arises because 7 and 10 do not share a factor. It’s not relevant to you because you’re trying to generate something in the range 0 to 1 (two values) and two is a factor of the number of values in Int, but it’s something to be aware of nonetheless.

# Dependence on mkStdGen

Right now, you let the user pass in a seed that is then passed to mkStdGen. That works, I suppose, but it would be more flexible if you pushed that responsibility to the user, because then the user can create an StdGen however they desire (with, say, getStdGen). Your signature would then look like this:

coinToss3 :: StdGen -> (Bool, Bool, Bool)

You could even change your definition to allow any RandomGen instance:

coinToss3 :: (RandomGen g) => g -> (Bool, Bool, Bool)

# Potential cleanup with MonadRandom and Applicative

At your current point in Learn You A Haskell, you haven’t learned about Monad or Applicative yet, so you may want to ignore this for now and take a look at it later, but using MonadRandom and Applicative, your code can be considerably shortened:

import Control.Applicative
coinToss3 = evalRand $(,,) <$> random <*> random <*> random


text:
# Combinations problem

## Homework Statement

From 10 tickets of luck, only 3 are winning. If we buy 6 tickets, how many possibilities we have at least to have 1 winning ticket in our hand?

## Homework Equations

$$C_n^k=\frac{n!}{k!(n-k)!}$$

## The Attempt at a Solution

6 tickets from 10, we can choose on $C_1_0^6$. There are 4 tickets left. [itex]C_7^3[/tex] is the tickets which are not winning. Is the right answer:

$$C_1_0^6$$ - $$C_7^3$$

?


text:
# What is the inverse of f(x)=pi*arccos(x-1)+1?

$y = \cos \left(\frac{x - 1}{\pi}\right) + 1$

#### Explanation:

$y = \pi \arccos \left(x - 1\right) + 1$

To take an inverse, you switch the x and y in this equation and solve for y:

$x = \pi \arccos \left(y - 1\right) + 1$

$\frac{x - 1}{\pi} = \arccos \left(y - 1\right)$

$\cos \left(\frac{x - 1}{\pi}\right) = \cos \left(\arccos \left(y - 1\right)\right)$

$\cos \left(\frac{x - 1}{\pi}\right) = y - 1$

$y = \cos \left(\frac{x - 1}{\pi}\right) + 1$
Solved!


text:
# Uniform convergence of generalised Fourier series

Suppose $$u_n$$ is an orthonormal basis of smooth functions on $$S^1$$.

Does there exist a smooth function $$u$$ such that the generalised Fourier series $$u=\sum_{n\in\mathbb{N}} \langle u,u_n\rangle u_n$$ does not converge uniformly?

We of course have uniform convergence for the standard Fourier basis by integration by parts, but I cannot find a counterexample (or a proof of uniform convergence) for the general case.

• I'm not quite sure where the quantifiers go in your question; but, for example, if we fix some point $p \in S^1$, then the set of smooth functions vanishing at $p$ is dense in $L^2$, so we can find an orthonormal basis $u_n$ consisting entirely of such functions. If $u(p) \ne 0$ then the series will not even converge pointwise. May 16, 2019 at 3:40
• Thanks for your answer Nate. I tried to make a similar idea work, but ran into problems establishing the existence of a smooth orthonormal base of such functions (as opposed to merely $L^2$). Is this obvious? May 16, 2019 at 3:46
• Let $E$ be the set of such functions. Being a subset of the separable metric space $L^2$, $E$ is separable, so we can find a sequence $v_1, v_2, \dots$ which is dense in $E$ and therefore also dense in $L^2$. Now apply Gram-Schmidt to the sequence $v_n$. The resulting sequence $u_n$ will be orthonormal, and its linear span will be the same as that of the $v_n$, which is dense in $L^2$. May 16, 2019 at 3:48
• Ah of course, this was a silly question. Thank you. If you make that comment an answer I can accept it. May 16, 2019 at 3:52

For instance, fix $$x \in S^1$$ and consider the space $$C^\infty_x$$ of smooth functions that vanish at $$x$$. This is a dense subspace of $$L^2(S^1)$$, so by choosing a countable dense subset of $$C^\infty_x$$ and applying Gram-Schmidt, we can find a sequence $$u_1, u_2, \dots \in C^\infty_x$$ which is an orthonormal basis for $$L^2(S^1)$$. Then if $$u(x) \ne 0$$, the series evaluated at $$x$$ converges to 0, not to $$u(x)$$.


text:
The classical theory of Raman effect, also called the polarizability theory, was developed by G. Placzek in 1934. I shall discuss it briefly here. It is known from electrostatics that the electric field $E$ associated with the electromagnetic radiation induces a dipole moment $mu$ in the molecule, given by
$\mu = \alpha E$ …….(1)
where $\alpha$ is the polarizability of the molecule. The electric field vector $E$ itself is given by
$E = E_0 \sin \omega t = E_0 \sin 2\pi \nu t$ ……(2)
where $E_0$ is the amplitude of the vibrating electric field vector and $nu$ is the frequency of the incident light radiation.

Thus, from Eqs. (1) & (2),
$\mu= \alpha E_0 \sin 2\pi\nu t$ …..(3)
Such an oscillating dipole emits radiation of its own oscillation with a frequency $nu$ , giving the Rayleigh scattered beam. If, however, the polarizability varies slightly with molecular vibration, we can write
$\alpha =\alpha_0 + \frac {d\alpha} {dq} q$ …..(4)
where the coordinate q describes the molecular vibration. We can also write q as:
$q=q_0 \sin 2\pi \nu_m t$ …..(5)

Where $q_0$ is the amplitude of the molecular vibration and $\nu_m$ is its (molecular) frequency. From Eqs. 4 & 5, we have
$\alpha =\alpha_0 + \frac {d\alpha} {dq} q_0 \sin 2\pi \nu_m t$ …..(6)
Substituting for $alpha$   in (3), we have
$\mu= \alpha_0 E_0 \sin 2\pi \nu t + \frac {d\alpha}{dq} q_0 E_0 \sin 2\pi \nu t \sin 2\pi \nu_m t$ …….(7)
Making use of the trigonometric relation $\sin x \sin y = \frac{1}{2} [\cos (x-y) -\cos (x+y) ]$ this equation reduces to:
$\mu= \alpha_0 E_0 \sin 2\pi \nu t + \frac {1}{2} \frac {d\alpha}{dq} q_0 E_0 [\cos 2 \pi (\nu – \nu_m) t – \cos 2\pi (\nu+\nu_m) t]$ ……(8)
Thus, we find that the oscillating dipole has three distinct frequency components:

1• The exciting frequency $nu$ with amplitude $\alpha_0 E_0$ $2•$ \nu – \nu_m

3• $\nu + \nu_m$ (2 & 3 with very small amplitudes of $\frac {1}{2} \frac {d\alpha}{dq} q_0 E_0$ . Hence, the Raman spectrum $of a vibrating molecule consists of a relatively intense band at the incident frequency and two very weak bands at frequencies slightly above and below that of the intense band.$

If, however, the molecular vibration does not change the polarizability of the molecule then $(d\alpha / dq )=0$ so that the dipole oscillates only at the frequency of the incident (exciting) radiation. The same is true for the molecular rotation. We conclude that for a molecular vibration or rotation to be active in the Raman Spectrum, it must cause a change in the molecular polarizability, i.e., $d\alpha/dq \ne 0$ …….(9)
Homonuclear diatomic molecules such as $\mathbf {H_2 , N_2 , O_2}$ which do not show IR Spectra since they don’t possess a permanent dipole moment, do show Raman spectra since their vibration is accompanied by a change in polarizability of the molecule. As a consequence of the change in polarizability, there occurs a change in the induced dipole moment at the vibrational frequency.

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text:
# How can we difference that which is initial velocity and final velocity

Dear Student,
Initial velocity is the velocity an object begins with.
Final velocity is the velocity at which the object ends up in.
Say a car is travelling at 13 m/s and then crashes into a wall stopping it it 's initial velocity was 13 m/s and it 's final velocity is at 0 m/s.
Say a car starts off at 5 m/s but then accelerates from 5 m/s to 10 m/s. It 's initial velocity would be 5 m/s but it 's resulting or final velocity is 10 m/s.
Regards

• 0
What are you looking for?


text:
# Figure 3

 Pseudorapidity interval dependence of the second-order cumulants of net-pions (left) and net-kaons (right) normalized to the means (see text). The ALICE data are shown as solid black circles while the blue solid and dashed lines indicate the results from HIJING~ model calculations with and without resonance contributions, respectively. The error bars represent statistical uncertainties and the boxes around the data points represent the total systematic uncertainties.


text:
# Output of a discrete-time LTI system different form than input?

This question is related to this one. I'm going through old exams for a 2nd year systems and transforms course, and came across this question. I'm posting this question just in case my other question doesn't actually represent this issue properly...

The question states:

Suppose that we are given the following information about an LTI system: if the input to the system is $x_1[n]=(1/6)^n u[n]$, then the output is $y_1[n]=[a(1/2)^n +10(1/3)^n]u[n]$. If $x_2[n]=(-1)^n$, then the output is $y_2[n]=(7/4)(-1)^n$ where $a$ is a real number. Determine the transfer function of this system and the value of the number $a$.

The solution proceeds by determining the z-transform of $x_1$ and $y_1$ and stating the transfer function as $H(z)=Y_1(z)/X_1(z)$.

However, it seems to me that it's impossible for this system to generate an output of $(1/2)^n+(1/3)^n$ if the input is $(1/6)^n$,and hence the question is somewhat contrived if not outright bogus - see the discussion pertaining to my other question.

What am I missing here?

• Your confusion stems from your ignoring of the unit step. – Jazzmaniac Aug 6 '15 at 16:46
• Could you please elaborate? – Westerley Aug 6 '15 at 16:54
• @Jazzmaniac means the $u[n]$ part of the input signal. Your input is not $(1/6)^n$. It's that, multiplied by the unit step. – Peter K. Aug 6 '15 at 16:55
• Sorry - that part I got; what I meant was what effect that has on the question... But now I wonder if what the question is indicating is that $y_1[n]$ is the transient response of the system? – Westerley Aug 6 '15 at 16:56

If you compute the $\mathcal{Z}$-transforms of $x_1[n]$ and of $y_1[n]$ you get (as already shown in Brian's answer)

$$X_1(z)=\frac{1}{1-\frac16z^{-1}}\\ Y_1(z)=\frac{a}{1-\frac12z^{-1}}+\frac{10}{1-\frac13z^{-1}}=(a+10)\frac{1-\frac{a+15}{3(a+10)}z^{-1}}{(1-\frac12z^{-1})(1-\frac13z^{-1})}$$

The system's transfer function must be

$$H(z)=\frac{Y_1(z)}{X_1(z)}=(a+10)\frac{\left(1-\frac{a+15}{3(a+10)}z^{-1}\right)(1-\frac16z^{-1})}{(1-\frac12z^{-1})(1-\frac13z^{-1})}\tag{1}$$

From the relation between $x_2[n]$ and $y_2[n]$ we know that $H(-1)=7/4$, because for an input signal $b^n$, the output signal of a discrete-time LTI system with transfer function $H(z)$ is $H(b)\cdot b^n$ (see also this answer). If you plug that into $(1)$ you indeed get $a=-9$. With this value for $a$, the transfer function becomes

$$H(z)=\frac{(1-2z^{-1})(1-\frac16z^{-1})}{(1-\frac12z^{-1})(1-\frac13z^{-1})}\tag{2}$$

Note that the output signal $y_1[n]$ only reflects the system's poles, i.e. the transient response, because the steady state response is zero due to the system's zero at $z=1/6$. Note that the system's reaction to the input signal $x_3[n]=(\frac16)^n$ (without the step function!) is $y_3[n]=H(\frac16)\cdot(\frac16)^n=0$. Since your input signal is switched on at $n=0$ (due to the step function) you see a decaying transient response (because the system is stable), but no other contribution from the input signal.

It is important to realize that you can't determine a transfer function from the relation between $x_2[n]$ and $y_2[n]$ (this was the mistake in Brian's answer, see also my comment there). That relation only determines the transfer function for one value of $z$, namely $z=-1$. Note that the sequence $x_2[n]=(-1)^n$ has no $\mathcal{Z}$-transform. The function $1/(1+z^{-1})$ is the $\mathcal{Z}$-transform of $(-1)^n\cdot u[n]$!

1. The output of an LTI system just has the same form as the input for the eigenfunctions $a^n$ (without the step function!), not for other functions. As mentioned before, the corresponding output is $H(a)\cdot a^n$, where $H(z)$ is the system's transfer function.

2. A sinusoidal input signal only excites the system at one single frequency, so from the output signal you can only determine the system's response at that single frequency. The signal $(-1)^n$ is such a (discrete-time) sinusoidal signal. So from the corresponding output, only the value $H(-1)$ can be determined, nothing else.

3. If the input signal contains all frequencies, which is the case for a signal that is switched on at a certain time (i.e., multiplied by a step function), the system's transfer function can be determined from the output signal and the given input signal. This is the case for the input $x_1[n]$, where you can simply determine $H(z)$ from $X(z)$ and $Y(z)$. Since the step function contains all frequencies, all signals multiplied by it also contain all frequencies. If you excite a system with such a signal, you can determine the system's response at all frequencies. In general, a system's response can be determined only at those frequencies that are present in the input signal. If the input signal contains a single frequency (a sinusoid), you only get information about that single frequency. In order to determine the complete transfer function, you need an input signal containing all frequencies, such as an impulse or a step (or any signal multiplied by a step).

• I generally follow what you've written; how you (and @brian) arrived at the transfer function makes sense. However, I'm still not seeing how to 'reconcile' the explanation here with the notion discussed in my other question of the output being of the same form as the input. I suspect the 'notion' is too simplistic... – Westerley Aug 7 '15 at 6:18
• Could you kindly elaborate on why you can't determine transfer function from the relation between $x_2[n]$ and $y_2[n]$? The qualitative difference betwen $x_1[n] -> y_1[n]$ and $x_2[n] -> y_2[n]$ seems to be the step function... – Westerley Aug 7 '15 at 6:20
• On that note, why does the step function's presence/absence in the input make such a big difference? – Westerley Aug 7 '15 at 6:21

NOTE: The answer is not complete; however, it seems the problem cannot be solved.

We have the following z-transforms of the input signals: $$x_1[n]=(1/6)^n\cdot u[n]\quad\longleftrightarrow\quad X_1(z)=\frac{1}{1-(1/6)\cdot z^{-1}}$$ $$x_2[n] = (-1)^n\quad\longleftrightarrow\quad X_2(z)=\frac{1}{1+z^{-1}}$$

For the output signals we have the following z-transforms: $$y_1[n] = [a(1/2)^2 + 10(1/3)^n]u[n] \quad\longleftrightarrow\quad Y_1(z) = a\frac{1}{1-(1/2)\cdot z^{-1}} + 10\frac{1}{1-(1/3)\cdot z^{-1}}$$ $$y_2[n] = (7/4)(-1)^n\quad\longleftrightarrow\quad Y_2(z) = (7/4)\frac{1}{1+z^{-1}}$$

For both pairs of signals we know that the transfer function equals $H(z)=\frac{Y(z)}{X(z)}$. From $\{X_2(z),Y_2(z)\}$ we know that $H(z) = 7/4$ (Which is trivial due to linearity, i.e., it follows also directly from $\{x_2[n],y_2[n]\}$).

What remains is to solve $Y_1(z) = H(z)\cdot X_1(z)$ for $a$; however, I don't get a proper result.

EDIT: Another way to appraoch this last step is to solve the following: $$a\frac{2}{2-z^{-1}} + b\frac{3}{3-z^{-1}} = c\frac{6}{6-z^{-1}}$$ where you may choose $b=10$ if desired.

In matrix representation: $$\mathbf{A}\left[\begin{array}{1}a\\b\\c\end{array}\right] = \mathbf{b}$$ the matrix $\mathbf{A}$ is a full rank size 3x3 matrix, which shows that no scalar solution for $H(z)=c$ with $b=10$ exists.

EDIT2: The solution seems to be that $H(-1)=7/4$ and $a=-9$. We already found that $H(z)=7/4$ for any $z$, based on the signals $\{X_2(z), Y_2(z)\}$. When evaluating $\{X_1(z),Y_1(z)\}$ for $z=-1$, the solution $a=-9$ follows easily. However, for another value of $z$, say $z=-2$ we obtain $a\approx -9.375$.

Conclusion, the decomposition only holds for $z=-1$.

• I get exactly the same thing... I think something got incorrectly copied over somewhere. – CMDoolittle Aug 6 '15 at 18:12
• I've edited to state exactly what the question says; the changes (insignificant, in my mind) are in italics... – Westerley Aug 6 '15 at 18:24
• What I still don't get is how multiplying by $u[n]$ allows this system to generate an output involving $(1/2)^n + (1/3)^n$ with an input of $(1/6)^n$. Why does the discussion from the related question not apply? – Westerley Aug 6 '15 at 18:25
• To determine the value of $a$, the solution I have sets $(7/4)=H(-1)$ and proceeds from there, resulting in a value of $a=-9$. But again, I'm trying to understand the underlying concepts... – Westerley Aug 6 '15 at 18:28
• The signals $a^n$ and $a^n u[n]$ are different. You can use signals of the form $a^n$ to decompose a signal $x[n]$, leading to the representation $X(a)=\sum_n x[n]a^n$ ; however, you cannot do this when a step function is involved. One way go get this is that the signal $x[n]$ may be non-zero for negative time-indices, say for $n=-1$. With the signal $a^n$ you can represent such signals; however, with $a^nu[n]$ you cannot since for any $a$ the signal $a^{-1}u[-1]$ is zero. – Brian Aug 6 '15 at 18:32


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### 功能性离子液体萃取水溶液中Cu2+:实验与理论

1. a 华东理工大学结构可控先进功能材料及其制备教育部重点实验室和化学系 上海 200237;
b 华东理工大学国家环境保护化工过程环境风险评价与控制重点实验室 上海 200237
• 收稿日期:2014-11-12 出版日期:2015-02-14 发布日期:2015-01-29
• 通讯作者: 彭昌军, cjpeng@ecust.edu.cn; Tel &Fax: 021-64252218;杨强, qyang@ecust.edu.cn, Tel &Fax: 021-64252748. E-mail:cjpeng@ecust.edu.cn;qyang@ecust.edu.cn
• 基金资助:

项目受国家自然科学基金(Nos. 21103047, 21136004)和中央高校基本科研业务费资助(No. 222201313001).

### Extraction of Copper from Aqueous Solution with Functional Ionic Liquids: Experiment and Theoretical Calculation

Liu Mengyinga, Che Jianinga, Wu Weihonga, Lu Yunxianga, Peng Changjuna, Liu Honglaia, Lu Haob, Yang Qiangb, Wang Hualinb

1. a Key Laboratory for Advanced Materials and Department of Chemistry, East China University of Science and Technology, Shanghai 200237, China;
b Environmental Protection Key Laboratory of Environmental Risk Assessment and Control on Chemical Process, East China University of Science and Technology, Shanghai 200237, China
• Received:2014-11-12 Online:2015-02-14 Published:2015-01-29
• Supported by:

Project supported by the National Natural Science Foundation of China (Nos. 21103047, 21136004) and the Fundamental Research Funds for the Central Universities of China (No. 222201313001).

This work studied the extraction of copper (II) ions from aqueous solution with thiourea-appended imidazolium hydrophobic ionic liquids and the extraction mechanism by experiment and theory. The influence of parameters affecting the extraction of copper ion, such as the metal ion concentrations, volume ratio between aqueous solution and ionic liquid, contact time, sodium chloride and pH, as well as alkyl chain length was analyzed. In the case of room temperature, the volume ratio between aqueous/IL phases was 50, the extraction efficiencies >95% could be obtained for copper ion with all the ionic liquids [CnMPSM][PF6] (n=4, 6, 8). The results also suggest that n=4 did the same work on the extraction efficiency with n=6 but higher than n=8, while pH and the salt in the solution had little effect on the extraction efficiency for the extraction with [HMPSM][PF6]. The grafted functional group significantly enhanced the extraction efficiency for Cu2+ from 20% to over 99% compared with traditional ILs. To the IL [HMPSM][PF6], the content of imidazolium cation in the aqueous solution before and after extraction with functional IL reduces from 1.24% to 0.85% which means the coordination effect between functional group and the metal ion restrains the release of [HMPSM]+ cation from functional IL to the aqueous solution, while it increases from 0.63% to 0.87% for traditional IL, which is consistent with the results caused by the cation exchange mechanism. In the theory part, the lanl2dz-ECP basis set was employed for transition metal Cu, whereas for the remaining atoms 6-31G (d,p) was applied, cation-anion interaction energies of the ILs and the binding energies between Cu (II) ion and the ILs were calculated, also the surface properties most-negative-surface electrostatic potential (Vs,min) and the lowest surface average local ionization energy (īs,min), were determined by the Multiwfn 2.4 program. All the calculation results show that the sulfur atom from cation is easier to attract the metal ion electrostatically and covalently, thus leads to the high efficiency of extraction.