Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

transforming phase slope into delay

yefj

Advanced Member level 4
Joined
Sep 12, 2019
Messages
1,267
Helped
1
Reputation
2
Reaction score
3
Trophy points
38
Activity points
7,679
Hello, a single LC cell was made to test delay.the phase increases linearly.how do i find the delay my square wave will get from this slope?
my slope is 1.8 degrees for each 1Mhz.
Thanks.

1715409471402.png
ran
 
This has a discussion of group delay in it, attached.


Regards, Dana.
 

Attachments

  • L2_2_f10n.pdf
    650.5 KB · Views: 53
Hello Dana ,by the definition of group delay
group-delay=1.8degree/1MHz=1.8Mega degrees*seconds

so if I put a pulse threw it, how do I know the time my pulse will be shifted , because the units are degrees*time
 
Last edited:
Yes in group delay each sine component of the pulse is delayed according
to its frequency, the delay in radians. So you know the fundamental freq,
its period, is 2 * pi radians, and its harmonics as well.

From the prior ref post :

1715425719782.png


Delta t can be calced from freg and phase shift in radians.

Regards, Dana.
 
Hello Dana , so my LC has 38Mhz bandwidth. Suppose i want to enter a pulse with rise/fall time of 100ns.
Which means the BW of the pulse is 3.5Mhz.
I have simulated the pulse below and i got a delay of 5ns.
Given this example
How do i calculate this 5ns delay from the group delay of group-delay=1.8degree/1MHz

1715427243084.png


1715428785144.png
 
This filter arrangement turns any repeating waveform into a sine (or sine-like). It presents a certain impedance for any given frequency. The resulting impedance is the sum of L impedance and C impedance. And somehow we can contribute the effect of R1 which has its ohm value.

The result across R1 is attenuated amplitude. The percentage of attenuation is related to the sine (or cosine) of incoming amplitude. Or it could be initial sine-shaped component of incoming waveform.
--- Updated ---

Hmmm... I'm finding the above is merely a rough idea of the theory.
Near resonance the delay is 90 degrees yet there's no attenuation .
At 3 or 4 times resonant freq the delay reaches maximum attenuation and 180 degree delay.
 
Last edited:
Your pulse fundamental is 3.5 Mhz or 286 nS.

dT = phase (radians) / [ 2 x pi x freq *(cyc/sec) ]

What is your phase measurement at 3.5 mhz ?

Note is your source generator Z 50 ohms as well ?


Regards, Dana.
 
1715447910589.png


You have 6 degrees of phase lag = ~ .1 radian

dT = phase (radians) / [ 2 x pi x freq *(cyc/sec) ]

dT = .1 / (2 x pi x 3.5 Mhz) = 4.5 nS

Which is in rough sim agreement with transient delay sim.


Regards, Dana.
 
Last edited:
Hello Dana , my pulse has rise time/fall time is 100ns
Ton=700ns and period is 2u.
In simple 50% duty cycle its f=1/period.But here is a more complex case.
How did you find its fundamental?
Thanks.
 
I used your 3.5 Mhz as pulse period. Which is fundamental sine, then there
are 3 x 5 x 7 x......harmonics in it....

I used your 100 nS as Tr and Tf.

Regards, Dana.
 
Last edited:

LaTeX Commands Quick-Menu:

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top