Lumped-element delay line formed with an array of equal-value capacitors and inductors.

I was told that the above circuit is used to delay signal. The time constant per element is t = sqrt(LC). The expression was provided by the book I'm reading, but it did not include how the time constant was derived. Can someone please provide an explanation of how the circuit work and how to derive said expression?

Note: I took a course in electronics 3 years ago, so the more details you can provide, the better.

  • \$\begingroup\$ Welcome to EE.SE. Hint: t is in seconds, L is in Henries, C is in Farads. Obviously fractional values are normally used. Do the math and sum them together to get total 't' (delay time). It is also a LPF so it is 1/t to get the maximum pass frequency. \$\endgroup\$
    – user105652
    May 30, 2018 at 3:39

1 Answer 1


A lumped element delay line is mimicing real life and real life is a transmission line (t-line). Examples are (but not limited to): -

  • Coaxial cable
  • Twisted pair
  • A waveguide
  • A PCB track and ground plane

So, in real life a t-line has the following relationships: -

enter image description here

Picture part of this slide show.

All the above can be reasonably easily proven using transmission-line theory and the telegrapher's equations.

The expression was provided by the book I'm reading, but it did not include how the time constant was derived

The expression in the book is based entirely on the distributed inductance and capacitance of a real t-line and so it is accurate, but only to a certain degree. As frequency rises (or rise/fall times shorten) the lumped-element model becomes less realistic.

For example if I made a single element LC and chose the ratio of L to C to be 2500, the characteristic impedance would be \$\sqrt{2500}\$ = 50 ohm and so I feed the circuit from a 50 ohm source: -

enter image description here

I can choose R = 50, L = 250 nH and C = 100 pF. I've chosen L and C to be these values because they are broadly what a 1 metre length of 50 ohm coax will be. I get this result: -

enter image description here

Tool source.

In the lower part of the image is the step response and I've set the cursor to be about 50% of the 1 volt step applied. The time to reach this point is about 6.5 ns. If I did the calculation: -

$$t_D = \sqrt{LC}$$

I get a value of 5 ns. Not a million miles off and hopefully you can see that lumped element lines are not quite what the theory states because they are trying to mimic a real t-line.

  • 1
    \$\begingroup\$ I am actually asking about the analysis of the circuit. Like base on the description of the circuit, it sounds like if I input a V(t) signal on the circuit, then the output should be V(t - tau), where tau = sqrt(LC) for this case. So I was wondering how to arrive at that answer. \$\endgroup\$ Jul 7, 2018 at 22:25
  • \$\begingroup\$ Why not get a free simulator and try it out or ask a more explicit question that I can focus on. \$\endgroup\$
    – Andy aka
    Jul 8, 2018 at 0:11
  • \$\begingroup\$ What should I do when someone answers my question? \$\endgroup\$
    – Andy aka
    Dec 13, 2022 at 12:46
  • \$\begingroup\$ Hello, there is a formula for a delay with LC ,what if i cascade severl LC , what is the total delay then, and what is the theory behind it? Thanks. \$\endgroup\$
    – rocko445
    Mar 26 at 21:24
  • \$\begingroup\$ @rocko445 Each LC low-pass stage delays by \$\sqrt{LC}\$ seconds. Regarding the theory I have an answer here \$\endgroup\$
    – Andy aka
    Mar 27 at 10:25

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