Good question 

Consider an open TL, you can find the expression for the input impedance in any microwave book for example Pozar's.  What happens when the frequency is very small? 

It is related to frequency.  What you see as open in circuit theory is not really an open in high frequency.  It is a complicated network of R,  L and C

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I can place the question differently.... In terms of Energy, it is considered a lossless 50 ohm transmission line but if it has 50 ohm impedance then how one can get same signal at the other as supplied at one end?

Dear Mr. A. Sikdar,

First let me correct you: 

Any lossy transmission line is represented as RLGC, whereas lossless line is represented as L and C only. 

why we fail to measure Z0 with a multimeter

because multimeter checks V & I and finds that inner and outer connductor has high insulation in case of a coaxial cable. Also Z0 becomes dominant when circuit dimensions are comparable to wavelength. so u check the insulation rather than measuring Z0, as a matter of fact we made a RF multimeter which measures L and C of a coaxial line and if you divide and take sqrt , you will arrive at 50 ohms.

Real and complex Z0

As Z0 is the sqrt (L/C) , where L and C are just numbers, please remember they are not reactances , they are just some values which is dependent on geometry , therefore Z0 of coaxial line is resistive and frequency independent. But in real life some parasitics may come in to trouble you.

how one can get same signal at the other as supplied at one end?

Recall maximum power transfer theorem ,

mind u Maxwell is never proved to be wrong till today ...

if your coaxial cable were infinitely long and open at the end you would also measure 50 Ohm with a normal Ohmmeter...to mimic tis case you may terminate it with a real resistor (e.g. 50 Ohm) and you measure 50 ohm. But  when its open at the end and say 1 meter long you see at the normal ohmmeter the steady state situation for this case. If you use a fast Ohm meter with nanosecond resolution (i.e. TDR scope) then you really see in the first  few nanoseconds the char impedance (50 Ohm) and later to transition to infinity (if the cable is open)

The way I think of it is that if you send a brief pulse through the coax cable, and you terminate the cable with a resistor of the same value as the "characteristic impedance" of that coax cable, then you will see no reflections of the pulse at the opposite end of the cable. The energy of that pulse will be completely absorbed by the terminating resistor.

If you use a terminating resistor of a different value than the characteristic impedance, you will see reflections.

A low loss coax cable is one in which the center conductor and shield offer very little resistance. So that just about all of the energy of that transmitted pulse will be absorbed by the (correctly sized) terminating resistor. So for example, if you do get reflections in a low loss cable, these reflections will bounce back and forth between the ends of the cable for a longer period of time, before becoming fully attenuated.

If you apply an infinitely long "pulse," using a typical volt-ohmmeter, you will see only the steady state response. At steady state, sure enough, any reflections which might have been noticeable will have vanished. The DC source of the VOM will only measure the resistance of the center conductor and shield, if the end of the cable is shorted out, or the (hopefully close to) infinite resistance to DC of the dielectric used, to keep the center conductor from shorting out with the shield, throughout the length of the coax cable.

But as Fritz Caspers said, if the coax is infinitely long, and you apply a DC voltage to it, then the wavefront will never reach the end of the cable, there will be no reflections, and the measured impedance should be the characteristic impedance.

Transmission lines, like this coax cable, are modeled as a series of inductors along the full length of the center conductor, and a series of capacitors between center conductor and shield. (And series resistance too, along the center conductor, and resistance parallel to the capacitors between center conductor and shield, for real-world lossy cables.) So you can see what happens when DC is applied to one end. As the wavefront travels along the coax, it meets inductors which initially resist the current flow, and it meets discharged capacitors that look like short circuits, and want to be charged. Soon each inductor becomes fully conducting, and each capacitor becomes an open circuit. This continues to happen as long as it takes for that voltage wavefront to reach the end of the cable. The effect of the gradual current increase through the inductors along the path, and the gradual current decrease through the capacitors along the path, if the cable were infinitely long and very low loss, will look to the ohmmeter just like a resistance of the value of the characteristic impedance.

The multimeter of volt-ampere meter  is time invariant parameter measuring device, while cable impedance what you are talking is for time-variant signal or more precisely it is referred to High frequency parameter.  Do not confuse your self between two different aspects. The cable is specially built such that at high frequancy & over a wide band, it real part remains 50 ohm and imaginary part is zero. I hope this suffice to your doubt

I have a different explanation that could help you to understand the fact clearly, the infinity long transmission line or cable is ideally replaced with its characteristic impedance 50 ohm for calculative use for a single unit length; but when you put a ohmmeter for a short line then the expected value is nearly zero (I mean very small), in practice the value is too small to measure by your ohmmeter hence it hows undetermined that you have interpreted as infinity (i.e. it is infinity towards the zero or smallest value not towards the highest value).

The impedance definition is the relation between E and H fields for TEM or free space propagation (Maxwell equation). 50 ohms is the optimal impedance to have low losses and maximum avalaible power in coaxial cable. The global circuit analysis need to include propagation (transient and stationnary waves) and we generaly used S (scattering) parameters. For DC multimeter (frequency=0, ...) the circuit response is the stationnary circuit analysis. There is many exercice (3rd year of licence) or review to explain this.

The characteristic impedance definition is not the relation between voltage and current at the input of the line. This would be the input impedance of the cable for a given length and a given load at the cable output. Generally speaking this impedance doesn't match Z0.

The characteristic impedance of the cable, modeled as a transmission line, isn't the TEM mode impedance either. As suggested before, the TEM mode impedance is the intrinsic impedance of the dielectric.

The characteristic impedance (Z0) is a parameter belonging to the distributed parameter model called 'Transmission line', which is a circuit model for the TEM mode. As it is well known, one of the limiting factors of the circuit theory is that it only holds when the circuit is much smaller than the wavelength. Thus, it cannot handle propagation effects or delays.

The transmission line model or distributed parameter circuit element is proposed to incorporate propagation effects in the circuit theory, so that it can be extended to the study of such effects when they occur as  single mode propagation in TEM transmission lines. In this case, when potentials for E and H can be uniquely defined - as V and I - the model 'Transmission line' is defined as an additional circuit element with its constitutive relations, which include propagation and reflection phenomena characterization.

The parameter Z0 is part of the circuit model, and is defined as the quotient between the complex amplitude of the incident voltage wave phasor and the complex amplitude of the incident current wave phasor in sinusoidal steady state - or under monochromatic excitation, or in the Fourier domain, as you wish. Also, it is usually defined as the ratio from voltage phasor to current phasor in a semi-infinite line. This last definition holds because in such a line there is no reflected wave and I dislike it because many times it leads to confusion. Mainly because people tend to misunderstood the implications of the semi-infiniteness condition.

In fact, if you load the line with a matched load - a 50 Ohms one or a 75 Ohms one for TV cables - so that there is no reflected wave in the line, and you make your measurement with the ohmmeter at DC, you will measure whatever Z0 corresponds. However, even in this case you will not be measuring the line's characteristic impedance, but the input impedance of the matched line. Though both values actually are the same. 

Of course, all the discussion holds in the case of the lossless line, so the 50 Ohm of Z0 have nothing to do with the resistors in the lumped parameter model for the short section of lossy line. Those 50 Ohm of Z0 don't mean the existence of Joules effect anywhere in the physical cable.

Dear Sikdar,

Prof de-Oliva-Rubio has given an in-depth, detailed and correct answer to your query. However, if you want to "visualize" those 50 ohms, the following experiment is proposed. Imagine you have a signal generator of those commonly available in teaching electronics labs. Imagine you apply a square wave to a long cable (say, 200 m long) terminated with an open circuit. The time spent by the electromagnetic disturbance to reach the end of the cable is t = d / c, where d is the length of the cable (200 m) and "c" is the "speed of light" in the dielectric material filling the space between conductors. A typical value for this time delay could be 0.8 microseconds. The period of the square wave applied to the input port of the cable should be significantly larger than 0.8 microseconds (let's say 5 microseconds). Since the output impedance of the signal generator is 50 ohms, the disturbance will not be reflected at the generator end of the cable, in such a way that after 2t = 1.6 microseconds the steady state will be reached.Imagine that the step voltage generated by the signal generator (in open circuit conditions) has an amplitude of 10 volts.  If you visualize the voltage at the input port of the cable using an oscilloscope, you would see a jump from zero to 5 volts and, after 1.6 microseconds you would get the second and final jump to about 10 volts (i.e., the steady state is reached after two steps, each of 5 volts). Why the first step is 5 volts? Well, the reason is that, for the time interval 0 < t < 1.6 microseconds the generator can not "know" what is the termination of the cable (an open circuit in this example). The response that you observe is exactly the same that you would have obtained by loading the generator with a 50 ohms resistive load. BUT this is true only during the first 1.6 microseconds. This "apparent" resistance is exactly the characteristic impedance of the cable. Obviously you can not extract this information from the steady state response.

By the way, eperiements of this kind are performed by my students in the lab to analyze the behavior of transmission lines using the time domain evolution of pulses along long transmission line sections (this makes possible to use relatively cheap electronic lab hardware).

Dear Medina,

Thanks for such a nice explanation. I have performed the similar experiment to measure the cable impedance in my lab. I have been successful to understand the behaviour for 50 ohm cable terminated by 50 ohm resistance. But, when I tried to understand the same behaviour for 75 ohm cable terminated by 75 ohm impedance, it did not respond similarly. I have a doubt that does the connector conspire in such measurement as all the connector I was using were 50 ohm connector? can you suggest some technique to measure the 75 ohm impedance of the cable by this technique?

Thanks............

Dear Sikdar,

If you excite your terminated transmission line (75 ohms characteristic impedance loaded with a 75 ohms resistor) the input impedance is 75 ohms. Since the output impedance of your signal generator is 50 ohms, the voltage level at the cable input port should be 75 / (50+ 75) times the open circuit voltage output of your source (i.e., the voltage measured WITHOUT the transmission line. By the way, if the cable length is long you should ground the two ends of the cable. Otherwise currents can flow over the external part of the external conductor in such a way that the measured input voltage cna be distorted with respect to the one computed with the standard transmission line theory.

Regards,

Francisco

I will look and let you know....Thanks @ Dr. Medina

It is the Characteristic impedance for RF waves to calculate the RF losses due to its inductance and capacitance nature.

Of course, with "speed of light" I mean the speed of the transverse electromagnetic waves in the dielectric medium (for the standard TV cables we use in the students lab this is 2.45 x 10^8 m/s more or less)

Keep in mind that V is not "E", unless the load is perfect resistance. V is an alternating waveform meeting a variety of "Z" dependant on frequency. E is a dc Coaxial cable characteristic impedance is an outcome of applied ac theory.

That one speaks of coaxial impedance rather than resistance distinguishes the difference.

The point of a typically 50 ohm impedance of the signal supply is well made. Maximum power transfer occurs when signal source impedance matches the black box load impedance. Coaxial cable is a 'special case' and designed for that...not by accident. A 50 ohm impedance termination to a 75 ohm characteristic coaxial presents an 'aberration'.

The impedance of cable depends on the following factors: losses and power handling capacity. Considering the power loss vs impedance, the loss is minimal for 77ohm.

Considering power handling capacity, maximum power handling is observed for 30ohm. Impedance of 50 ohm is in the middle of that. So 50ohm is considered as standard value of impedance.