What is the equivalent circuit for pulse generator pulse transmission? - Page 1
Forum section: electronics / testgear

author: Rene
author_rank: Regular Contributor
timestamp: January 29, 2016, 05:23:39 pm
content: Hello.I am sure the title of my post is more confusing than helpful so let me explain. I own a Siglent SDG1025 function generator, one of the options in that function generator is to generate pulses. This is great, but I am wondering how are the pulses outputted.For example, take a look at attached picture “Model1.png”. That model would suggest that when a pulse is generated, the generator would close both switches at the exact same time, leave them pressed for the defined amount of time and then release both of them at the exact same time. When this happens, I would end up with two waves travelling down (one positive and one negative).Now take a look at attached picture “Model2.png”. That model would suggest that when a pulse is generated, the generator will only close the positive terminal for the defined amount of time and then release the switch. This would mean that there would only be one wave traveling down (the positive) because the negative wave would have already traveled all the way down the line.I really have no idea if this matters or not, I am currently trying to get a little more intuition about how transmission lines work so I started thinking about the effect of transmission lines and pulse generator equipment.Just curious.Thanks.

author: Wuerstchenhund
author_rank: Super Contributor
timestamp: January 29, 2016, 05:55:47 pm
content: Quote from: Rene on January 29, 2016, 05:23:39 pmThat isn't how signal generators like your Siglent work. On these cheap DDS generators pulses are usually created the same as square waves, with (compared with a real pulse generator) pretty slow edges, i.e. the transition from 0V to Vs is relatively slow. This creates a lot of frequency components outside the harmonic pattern that would make up a 'true' pulse.I know the comparison of a pulse with a simple switch is tempting, but if you want to talk about transmission lines you have to talk about frequency components. All non-sinusoid waveforms (i.e. square waves, triangle waves, pulses, whatever) are made up of a combination of sine waves at different frequencies. For transmission lines, it's these frequency components which are relevant, in addition to its DC properties.Even the simple "pulse generator" in your drawings is affected. You'd be tempted to think that the rise time when closing the switch would be zero, i.e. an instant jump from 0V to Vs. In reality, even with a perfect switch, the voltage level is somewhat slowly (not as slow as on you Siglent, though) increasing, caused by the various inductive anc capacitive properties of the transmission line. It's still very fast, but not an instant.QuoteIf that matters depends on what's on the other end of your transmission line, i.e. what is the voltage potential. If there's no voltage difference between both ends of a wire then no current is flowing, thus there won't be any EM emissions on that line, which means if the open end is really open then in principle(!) there is no voltage difference between the "generator" GND and the open end of the GND wire, thus it doesn't matter if there are one or two switches.

author: hibone
author_rank: Regular Contributor
timestamp: January 29, 2016, 11:58:56 pm
content: I am probably looking at the easy side of things but, if I understand correctly the meaning of the question, I guess the correct answer is number 2.Since the braid is always connected to ground and acts as a voltage reference, the pulse travels on the central conductor of the coax. That is true if the one side of the coax not connected to the pulse generator remains unconnected. Clearly if you put a resistor across the tip and the ring, the pulse will travel along the central conductor and then along the braid.Of course its amplitude will decrease depending on its spectral content.

author: uncle_bob
author_rank: Unknown
timestamp: January 30, 2016, 03:02:14 am
content: HiAs mentioned above, signal generators do not run switches in their outputs. The proper model is a controlled voltage source with a resistor in series with it. The resistor models the "output impedance" of the generator. The voltage steps to form the output waveform. In the most simple case (no DC offset) the voltage source is zero volts except when a pulse is going out.In the sense of your drawings, number 2 is the better representation. One side of the BNC connector is always grounded to the case of the instrument. The only thing that is "fiddled with" is the center conductor. The gotcha with your drawing is that the generator *drives* out zero volts just as much as it drives the "pulse high" voltage. If a switch was used, the voltage on the output would simply float between pulse high outputs.Now, transmission line wise, you have a wave that moves down a line on the rising edge. There is a second wave that travels down the line on the falling edge. The current associated with these edges travels in the upper line in each of your drawings. An equal and opposite current travels in the lower line in each of your drawings. If the line is not terminated in it's characteristic impedance, the wave will bounce when it hits the far end. If when it bounces back to the source, that is not the characteristic impedance, the wave will bounce yet again. (That's the short version ... the full long version takes a few weeks ...).Not sure if that helps or just confuses things ....Bob