Communication Systems/Wired Transmission

This page will discuss the topic of signal propagation through physical mediums, such as wires.

Transmission Line Equation

Many kinds of communication systems require signals at some point to be conveyed over copper wires.

The following analysis requires two assumptions:

• A transmission line can be decomposed into small, distributed passive electrical elements

• These elements are independent of frequency (i.e. although reactance is a function of frequency, resistance, capacitance and inductance are not)

These two assumptions limit the following analysis to frequencies up to the low MHz region. The second assumption is particularly difficult to defend since it is well known that the resistance of a wire increases with frequency because the conduction cross-section decreases. This phenomenon is known as the skin effect and is not easy to evaluate.

The purpose behind the following mathematical manipulation is to obtain an expression that defines the voltage (or current) at any time (t) along any portion (x) of the transmission line. Later, this analysis will be extended to include the frequency domain.

Recall the characteristic equations for inductors and capacitors:

$v=L\frac{\partial i}{\partial t}$ and $i=C\frac{\partial v}{\partial t}$

Kirchoff's Voltage Law

Kirchoff's voltage law (KVL) simply states that the sum of all voltage potentials around a closed loop equal zero. Or in other words, if you walked up a hill and back down, the net altitude change would be zero.

Applying KVL in the above circuit, we obtain:

$v\left(x,t\right)=R\mathrm{\Delta }xi\left(x,t\right)+L\mathrm{\Delta }x\frac{\partial i}{\partial t}\left(x,t\right)+v\left(x+\mathrm{\Delta }x,t\right)$

Rearranging:

$v\left(x,t\right)-v\left(x+\mathrm{\Delta }x,t\right)=R\mathrm{\Delta }xi\left(x,t\right)+L\mathrm{\Delta }x\frac{\partial i}{\partial t}\left(x,t\right)$

But the LHS (left hand side) of the above equation, represents the voltage drop across the cable element $\mathrm{\Delta }v$, therefor:

$\mathrm{\Delta }v=R\mathrm{\Delta }xi\left(x,t\right)+L\mathrm{\Delta }x\frac{\partial i}{\partial t}\left(x,t\right)$

Dividing through by $\mathrm{\Delta }x$, we obtain:

$\frac{\mathrm{\Delta }v}{\mathrm{\Delta }x}=Ri\left(x,t\right)+L\frac{\partial i}{\partial t}\left(x,t\right)$

The LHS is easily recognized as a derivative. Simplifying the notation:

$\frac{\partial v}{\partial x}=Ri+L\frac{\partial i}{\partial t}$

This expression has both current and voltage in it. It would be convenient to write the equation in terms of current or voltage as a function of distance or time.

Simplifying the Equation (trust me)

The first step in separating voltage and current is to take the derivative with respect to the position x (Equation 1):

$\frac{{\partial }^{2}v}{\partial x^2}=R\frac{\partial i}{\partial x}+L\frac{{\partial }^{2}i}{\partial x\partial t}$

The next step is to eliminate the current terms, leaving an expression with voltage only. The change in current along the line is equal to the current being shunted across the line through the capacitance C and conductance G. By applying KCL in the circuit, we obtain the necessary information (Equation 2):

$\frac{\partial i}{\partial x}=Gv+C\frac{\partial v}{\partial t}$

Taking the derivative with respect to time, we obtain (Equation 3):

$\frac{{\partial }^{2}i}{\partial x\partial t}=G\frac{\partial v}{\partial t}+C\frac{{\partial }^{2}v}{\partial t^2}$

Substituting (Equation 2) and (Equation 3) into (Equation 1), we obtain the desired simplification:

$\frac{{\partial }^{2}v}{\partial x^2}=R\left[Gv+C\frac{\partial v}{\partial t}\right]+L\left[G\frac{\partial v}{\partial t}+C\frac{{\partial }^{2}v}{\partial t^2}\right]$

Collecting the terms, we obtain:

The Transmission Line Equation for Voltage

$\frac{{\partial }^{2}v}{\partial x^2}=RGv+\left(RC+LG\right)\frac{\partial v}{\partial t}+LC\frac{{\partial }^{2}v}{\partial t^2}$

This equation is known as the transmission line equation. Note that it has voltage at any particular location x as a function of time t.

Similarly for current, we obtain:

The Transmission Line Equation for Current

$\frac{{\partial }^{2}i}{\partial x^2}=RGi+\left(RC+LG\right)\frac{\partial i}{\partial t}+LC\frac{{\partial }^{2}i}{\partial t^2}$

But we're not quite done yet.

Solving the Transmission Line Equation

Historically, a mathematician would solve the transmission line equation for v by assuming a solution for v, substituting it into the equation, and observing whether the result made any sense. An engineer would follow a similar procedure by making an “educated guess” based on some laboratory experiments, as to what the solution might be. Today there are more sophisticated techniques used to find solutions. In this respect, the engineer may lag behind the mathematician by several centuries in finding applications for mathematical tools.

To solve the transmission line equation, we shall guess that the solution for the voltage function is of the form:

$v\left(t\right)=e^{j\omega t}{e}^{-\gamma x}$

The first term represents a unity vector rotating at an angular velocity of $\omega$ radians per second, in other words, a sine wave of some frequency. The second term denotes the sinusoid being modified by the transmission line, namely its amplitude decaying exponentially with distance. If we let $\gamma$ be a complex quantity, we can also include any phase changes which occur as the signal travels down the line.

The sine wave is used as a signal source because it is easy to generate, and manipulate mathematically. Euler’s Identity shows the relationship between exponential notation and trigonometric functions:

Euler's Identity

$e^{j\omega t}=\cos \left(\omega t\right)+j\sin \left(\omega t\right)$

Going back to our educated guess, we will let $\gamma =\alpha +j\beta$, therefore:

$v\left(t\right)=e^{j\omega t}{e}^{-\left(\alpha +j\beta \right)x}=e^{-\alpha x}{e}^{\left(\omega t-\beta x\right)j}$

The term $e^{-\alpha x}$ represents the exponential amplitude decay as this signal travels down the line. $\alpha$ is known as the attenuation coefficient and is expressed in Nepers per meter.

The term $e^{\left(\omega t-\beta x\right)j}$ represents the frequency of the signal at any point along the line. $\beta$ component is known as the phase shift coefficient, and is expressed in radians per meter.

Substituting our educated guess

$v\left(t\right)=e^{j\omega t}{e}^{-\left(\alpha +j\beta \right)x}$

into the transmission line equation for voltage, we obtain:

$\frac{{\partial }^2}{\partial x^2}\left[e^{j\omega t}{e}^{-\left(\alpha +j\beta \right)x}\right]=RG\left[e^{j\omega t}{e}^{-\left(\alpha +j\beta \right)x}\right]+\left(RC+LG\right)\frac{\partial }{\partial t}\left[e^{j\omega t}{e}^{-\left(\alpha +j\beta \right)x}\right]+LC\frac{{\partial }^2}{\partial t^2}\left[e^{j\omega t}{e}^{-\left(\alpha +j\beta \right)x}\right]$

This looks pretty intimidating, but if you can do basic differentials and algebra, you can do this!

Simplifying the Equation (trust me)

The idea now is to work through the math to see if we come up with a reasonable solution. If we arrive at a contradiction or an unreasonable result, it means that our educated guess was wrong and we have to do more experimenting and come up with a better guess as to how voltage and current travel down a transmission line.

Let's look at this equation one term at a time:

LHS = RHS Term 1 + RHS Term 2 + RHS Term 3

Starting with the left hand side (LHS) we get the following simplification:

$\frac{{\partial }^2}{\partial x^2}\left[e^{j\omega t}{e}^{-\left(\alpha +j\beta \right)x}\right]=\frac{\partial }{\partial x}\left[-\left(\alpha +j\beta \right)e^{j\omega t}{e}^{-\left(\alpha +j\beta \right)x}\right]={\left(\alpha +j\beta \right)}^2{e}^{j\omega t}{e}^{-\left(\alpha +j\beta \right)x}$

Believe it or not, the RHS Term 1 does not need simplifying.

Simplifying the RHS Term 2, we obtain:

$\left(RC+LG\right)\frac{\partial }{\partial t}\left[e^{j\omega t}{e}^{-\left(\alpha +j\beta \right)x}\right]=\left(RC+LG\right)j\omega \left(e^{j\omega t}{e}^{-\left(\alpha +j\beta \right)x}\right)$

Simplifying the RHS Term 3, we obtain:

$LC\frac{{\partial }^2}{\partial t^2}\left[e^{j\omega t}{e}^{-\left(\alpha +j\beta \right)x}\right]=LC\frac{\partial }{\partial t}\left[j\omega e^{j\omega t}{e}^{-\left(\alpha +j\beta \right)x}\right]=-LC{\omega }^2{e}^{j\omega t}{e}^{-\left(\alpha +j\beta \right)x}$

Let's put it all back together:

${\left(\alpha +j\beta \right)}^2{e}^{j\omega t}{e}^{-\left(\alpha +j\beta \right)x}=RG\left(e^{j\omega t}{e}^{-\left(\alpha +j\beta \right)x}\right)+\left(RC+LG\right)j\omega \left(e^{j\omega t}{e}^{-\left(\alpha +j\beta \right)x}\right)-LC{\omega }^2\left(e^{j\omega t}{e}^{-\left(\alpha +j\beta \right)x}\right)$

Note that each of the four terms contain the expression $e^{j\omega t}{e}^{-\left(\alpha +j\beta \right)x}$.

Therefore we end up with:

${\left(\alpha +j\beta \right)}^2=RG+\left(RC+LG\right)j\omega -LC{\omega }^2$

And this can be further simplified to:

Attenuation and Phase Shift Coefficients

$\alpha +j\beta =\gamma =\sqrt{\left(R+j\omega L\right)\left(G+j\omega C\right)}$

This result is not self contradictory or unreasonable. Therefore we conclude that our educated guess was right and we have successfully found an expression for attenuation and phase shift on a transmission line as a function of its distributed electrical components and frequency.

Lossless Transmission Line

Signal loss occurs by two basic mechanisms: signal power can be dissipated in a resistor [or conductance] or signal currents may be shunted to an AC ground via a reactance. In transmission line theory, a lossless transmission line does not dissipate power. Signals, will still gradually diminish however, as shunt reactances return the current to the source via the ground path. For the power loss to equal zero, R = G = 0. This condition occurs when the transmission line is very short. An oscilloscope probe is an example of a very short transmission line. The transmission line equation reduces to the voltage equation:

$\frac{{\partial }^{2}v}{\partial x^2}=LC\frac{{\partial }^{2}v}{\partial t^2}$

and the current equation:

$\frac{{\partial }^{2}i}{\partial x^2}=LC\frac{{\partial }^{2}i}{\partial t^2}$

To determine how sinusoidal signals are affected by this type of line, we simply substitute a sinusoidal voltage or current into the above expressions and solve as before, or we could take a much simpler approach. We could start with the solution for the general case:

$\alpha +j\beta =\gamma =\sqrt{\left(R+j\omega L\right)\left(G+j\omega C\right)}$

Let R = G = 0, and simplify:

$\alpha +j\beta =\sqrt{\left(j\omega L\right)\left(j\omega C\right)}=\omega {\left(LC\right)}^{2}j$

Equating the real and imaginary parts:

$\alpha =0$

$\beta =\omega \sqrt{LC}$

This expression tells us that a signal travelling down a lossless transmission line, experiences a phase shift directly proportional to its frequency.

Phase Velocity

A new parameter, known as phase velocity, can be extracted from these variables:

$V_p=\frac{1}{\sqrt{LC}}=\frac{\omega }{\beta }$ meters per second

Phase velocity is the speed at which a fixed point on a wavefront, appears to move. In the case of wire transmission lines, it is also the velocity of propagation., typically: 0.24c < Vp < 0.9c .

The distance between two identical points on a wavefront is its wavelength ($\lambda$) and since one cycle is defined as 2$\pi$ radians:

$\lambda =\frac{2\pi }{\beta }$ and $\omega =2\pi f$

therefore:

$V_p=\lambda f$

In free space, the phase velocity is 3 x 10⁸ meters/sec, the speed of light. In a cable, the phase velocity is somewhat lower because the signal is carried by electrons. In a waveguide transmission line, the phase velocity exceeds the speed of light.

Distortionless Transmission Line

A distortionless line does not distort the signal phase, but does introduce a signal loss. Since common transmission lines are not super conductors, the signal will decrease in amplitude but retain the same shape as the input. This characteristic is essential for long cable systems.

Phase distortion does not occur if the phase velocity V_p is constant at all frequencies.

By definition, a phase shift of 2$\pi$ radians occurs over one wavelength $\lambda$.

Since

$V_p=\lambda f\lambda =\frac{2\pi }{\beta }f=\frac{\omega }{2\pi }$

Then:

$V_p=\frac{2\pi }{\beta }\times \frac{\omega }{2\pi }=\frac{\omega }{\beta }$

This tells us that in order for phase velocity V_p to be constant, the phase shift coefficient $\beta$, must vary directly with frequency $\omega$.

Recall

$\gamma =\sqrt{\left(R+j\omega L\right)\left(G+j\omega C\right)}=\alpha +j\beta$

The problem now is to find $\beta$. This can be done as follows:

$\gamma =\sqrt{\left(\frac{R+j\omega L}{j\omega L}\right)\left(j\omega L\right)\left(\frac{G+j\omega C}{j\omega C}\right)\left(j\omega C\right)}=j\omega \sqrt{LC}\sqrt{1+\frac{R}{j\omega L}}\sqrt{1+\frac{G}{j\omega C}}$

It may seem that we have lost $\beta$, but do not give up. The 2nd and 3rd roots can be expanded by means of the Binomial Expansion.

Recall:

${\left(1+x\right)}^n=1+nx+\frac{n\left(n-1\right)}{2!}{x}^2+\frac{n\left(n-1\right)\left(n-2\right)}{3!}{x}^3+\cdots$

In this instance n = 1/2. Since the contribution of successive terms diminishes rapidly, $\gamma$ is expanded to only 3 terms:

$\gamma \approx j\omega \sqrt{LC}\left(1+\frac{1}{2}\frac{R}{j\omega L}-\frac{1}{8}{\left(\frac{R}{j\omega L}\right)}^2\right)\left(1+\frac{1}{2}\frac{G}{j\omega C}-\frac{1}{8}{\left(\frac{G}{j\omega C}\right)}^2\right)$

This may seem complex, but remember it is only algebra and it will reduce down to simple elegance. Expanding the terms we obtain:

$\gamma \approx j\omega \sqrt{LC}\left\{\begin{array}{c}1+\frac{1}{2}\frac{G}{j\omega C}-\frac{1}{8}{\left(\frac{G}{j\omega C}\right)}^2+\frac{1}{2}\frac{R}{j\omega L}-\frac{1}{4}\frac{RG}{{\omega }^{2}LC}\\ -\frac{1}{16}\frac{R}{j\omega L}{\left(\frac{G}{j\omega C}\right)}^2-\frac{1}{8}{\left(\frac{R}{j\omega L}\right)}^2\\ -\frac{1}{16}{\left(\frac{R}{j\omega L}\right)}^2\frac{G}{j\omega C}+\frac{1}{64}{\left(\frac{R}{j\omega L}\right)}^2{\left(\frac{G}{j\omega C}\right)}^2\end{array}\right\}$

Since $\gamma =\alpha +j\beta$, we merely have to equate the real and imaginary terms to find $\beta$.

$\beta \approx \omega \sqrt{LC}\left\{1+\underset{\mathrm{D}\mathrm{i}\mathrm{f}\mathrm{f}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{o}\mathrm{f}\mathrm{s}\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{s}}{\underbrace{\frac{1}{8}{\left(\frac{G}{\omega C}\right)}^2-\frac{1}{4}\frac{RG}{{\omega }^{2}LC}+\frac{1}{8}{\left(\frac{R}{\omega L}\right)}^2}}+\underset{\mathrm{V}\mathrm{e}\mathrm{r}\mathrm{y}\mathrm{s}\mathrm{m}\mathrm{a}\mathrm{l}\mathrm{l}}{\underbrace{\frac{1}{64}{\left(\frac{R}{\omega L}\right)}^2{\left(\frac{G}{\omega C}\right)}^2}}\right\}$

Or

$\beta \approx \omega \sqrt{LC}\left\{1+\frac{1}{8}{\left(\frac{R}{\omega L}-\frac{G}{\omega C}\right)}^2\right\}$

Note that if $\frac{R}{\omega L}=\frac{G}{\omega C}$ then $\beta \approx \omega \sqrt{LC}$

From this we observe that $\beta$ is directly proportional to $\omega$.

Therefore the requirement for distortionless transmission is:

RC = LG

This is one of the essential design characteristics for a broadband coax cable network.

If we equate the real terms, we obtain:

$\alpha \approx \sqrt{RG}$

So there is a reason to study algebra after all!

The Frequency Domain

Signal analysis is often performed in the frequency domain. This tells us how the transmission line affects the spectral content of the signals they are carrying.

To determine this, it is necessary to find the Fourier Transform of the transmission line equation. Recall:

$\frac{{\partial }^{2}v}{\partial x^2}=RGv+\left(RC+LG\right)\frac{\partial v}{\partial t}+LC\frac{{\partial }^{2}v}{\partial t^2}$

and recall (hopefully) the Fourier Transform (which converts the time domain to the frequency domain):

$𝔽\left\{f\left(t\right)\right\}=F\left(\omega \right)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-j\omega t}f\left(t\right)dt$

To prevent this analysis from ‘blowing up’, we must put a stipulation on the voltage function namely, that it vanishes to zero at an infinite distance down the line. This comprises a basic boundary condition.

$\text{let}v\to 0\text{as}x\to \mathrm{\infty }$

This stipulation is in agreement with actual laboratory experiments. It is well known that the signal magnitude diminishes as the path lengthens.

Likewise, a time boundary condition, that the signal was zero at some time in the distant past and will be zero at some time in the distant future, must be imposed.

$\text{let}v\to 0\text{as}t\to \mathrm{\infty }$

Although engineers have no difficulty imposing these restrictions, mathematical purists, are somewhat offended. For this and other reasons, other less restrictive transforms have been developed. The most notable in this context, is the Laplace transform, which does not have the same boundary conditions.

Having made the necessary concessions in order to continue our analysis, we must find the Fourier Transform corresponding to the following terms:

$𝔽\left\{v\right\}𝔽\left\{\frac{\partial v}{\partial t}\right\}𝔽\left\{\frac{{\partial }^{2}v}{\partial t^2}\right\}$

$\text{Let:}𝔽\left\{v\right\}=V$

Then applying the transform on the derivative, we obtain:

$𝔽\left\{\frac{\partial v}{\partial t}\right\}=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-j\omega t}\frac{\partial v}{\partial t}dt$

This equation can be solved by using integration by parts:

$\int udv=uv-\int vdu$

$\text{let}u=e^{-j\omega t}\therefore du=-j\omega e^{-j\omega t}$

$\text{and}dv=\frac{\partial v}{\partial t}\therefore v=v$

$\therefore 𝔽\left\{\frac{\partial v}{\partial t}\right\}=e^{-j\omega t}{v|}_{-\mathrm{\infty }}^{\mathrm{\infty }}-\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}v\left(-j\omega e^{-j\omega t}\right)dt$

Applying the boundary conditions when t goes to infinity makes the 1st term disappear.

$\therefore 𝔽\left\{\frac{\partial v}{\partial t}\right\}=j\omega \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-j\omega t}vdt$

Note that the resulting integral is simply the Fourier Transform. In other words:

$𝔽\left\{\frac{\partial v}{\partial t}\right\}=j\omega 𝔽\left\{v\right\}=j\omega V$

similarly:

$𝔽\left\{\frac{{\partial }^{2}v}{\partial t^2}\right\}={\left(j\omega \right)}^2𝔽\left\{v\right\}={\left(j\omega \right)}^{2}V$

We can now write the transmission line equation in the frequency domain:

$\frac{{\partial }^{2}V}{\partial x^2}=RGV+\left(RC+LG\right)j\omega V+LC{\left(j\omega \right)}^{2}V$

where:

$V=V\left(\omega \right)=𝔽\left\{v\left(t\right)\right\}$

Rearranging the terms, we obtain:

$\frac{{\partial }^{2}V}{\partial x^2}=\left[RG+\left(RC+LG\right)j\omega +\left(j\omega L\right)\left(j\omega C\right)\right]V$

or

$\frac{{\partial }^{2}V}{\partial x^2}=\left[\left(R+j\omega L\right)\left(G+j\omega C\right)\right]V$

since:

$\sqrt{\left(R+j\omega L\right)\left(G+j\omega C\right)}=\alpha +j\beta =\gamma$

then

$\frac{{\partial }^{2}V}{\partial x^2}={\gamma }^{2}V$

or

$\frac{{\partial }^{2}V}{\partial x^2}-{\gamma }^{2}V=0$

This represents the most general form of the transmission line equation in the frequency domain. This equation must now be solved for V to observe how voltage (or current) varies with distance and frequency. This can be done by assuming a solution of the form:

$V=\underset{\text{forward}\text{wave}}{\underbrace{A{e}^{-\gamma x}}}+\underset{\text{reverse}\text{wave}}{\underbrace{B{e}^{\gamma x}}}$

These terms represent an exponential decay as the signal travels down the transmission line. If we ignore any reflections, assuming that the cable is infinitely long or properly terminated, this simplifies to:

$V=V_0{e}^{-\gamma x}$

To verify whether this assumption is correct, substitute it into the equation, and see if a contradiction occurs. If there is no contradiction, then our assumption constitutes a valid solution.

$\frac{{\partial }^2}{\partial x^2}{V}_0{e}^{-\gamma x}-{\gamma }^2{V}_0{e}^{-\gamma x}=0$

$\frac{\partial }{\partial x}\left(-{\gamma }^2{V}_0{e}^{-\gamma x}\right)-{\gamma }^2{V}_0{e}^{-\gamma x}=0$

${\gamma }^2{V}_0{e}^{-\gamma x}-{\gamma }^2{V}_0{e}^{-\gamma x}=0$

$0=0$

Thus we validate the assumed solution. This tells us that in the frequency domain, the voltage or current on a transmission line decays exponentially:

$V=V_0{e}^{-\gamma x}$

where:

$\gamma =\sqrt{\left(R+j\omega \right)\left(G+j\omega \right)}=\left|\gamma \right|\mathrm{\angle }\phi =\alpha +j\beta$

$\gamma =\text{ propagation}\text{constant}$

$\alpha =\text{ attenuation}\text{coeficient}$

$\beta =\text{phase}\text{coefficient}$

In exponential notation, a sinusoid may be represented by a rotating unity vector, of some frequency:

$e^{j\omega t}=\cos \omega t+j\sin \omega t$

Note that the magnitude of this function is 1, but the phase angle is changing as a function of t.

If we let: $V_0=e^{j\omega t}$

Then: $V_0=e^{j\omega t}{e}^{-\gamma x}=e^{j\omega t}{e}^{-\left(\alpha +j\beta \right)x}=\underset{\text{attenuation}vs.x}{\underbrace{e^{-\alpha x}}}\stackrel{\text{phase}vs\text{.}t\text{and}x}{\overbrace{e^{j\left(\omega t-\beta x\right)}}}$

This result is quite interesting because it is the same solution for the transmission line equation in the time domain. The term $e^{-\alpha x}$ represents an exponential decay. The signal is attenuated as length x increases. The amount of attenuation is defined as:

Attenuation in Nepers: $N=\left|\ln e^{-\alpha x}\right|=\alpha x$

Attenuation in dB: $=20\log e^{-\alpha x}\approx 8.68589\alpha x$

This allows us to determine the attenuation at any frequency at any point in a transmission line, if we are given the basic line parameters of R, L, G, & C.

The term $e^{j\left(\omega t-\beta x\right)}$ represents a rotating unity vector since:

$e^{j\left(\omega t-\beta x\right)}=\cos \left(\omega t-\beta x\right)+j\sin \left(\omega t-\beta x\right)$

The phase angle of this vector is $\beta x$ radians.

Characteristic Impedance

The characteristic impedance of a transmission line is also known as its surge impedance, and should not be confused with its resistance. If a line is infinitely long, electrical signals will still propagate down it, even though the resistance approaches infinity. The characteristic impedance is determined from its AC attributes, not its DC ones.

Recall from our earlier analysis: