strange output on CRO while measuring output of 45Mhz - Page 1
Forum section: electronics / beginners

author: Vindhyachal.takniki
author_rank: Frequent Contributor
timestamp: February 08, 2019, 03:17:56 pm
content: I am measuring output of 45Mhz of programmable oscillator of:http://www.cardinalxtal.com/products/programmable-oscillator-31. When I measuring it on gwinstel gds-11028 DSO,.2. Gnd loop is mimized, like this:https://goo.gl/images/QqHD2d3. Probes are conencted directly on oscillator output pins4. Vdd = 3.3V5. When on 1x probe: Signal is sine wave riding on 1.6Vdc. Peaks of sine wave dont touch the 3.3V & 0V limtis.6. When on 10x probe, signal is sine wave which seems to ride on dc also, but this time negative peak of sine goes below zero volt7. Why is that so?8. Also what is output of this oscillator, we have TTL version, is it sine wave?I thought it would be square wave.

author: Zero999
author_rank: Super Contributor
timestamp: February 08, 2019, 04:23:18 pm
content: I couldn't find the gwinstel gds-11028, using Google. Presumably you mean the GDS-1102B?https://www.testequity.com/documents/pdf/GDS-1000B-brochure.pdfIt's a 100MHz oscilloscope and you're measuring a 45MHz squarewave. An ideal square wave only contains odd harmonics, so you have 45MHz, plus the 3rd, 5th, 7th etc. harmonics 135MHz, 225MHz, 315MHz, but your oscilloscope will have a filter blocking everything below 100MHz, so you just see the fundamental of 45MHz.In other words, this is exactly what you should expect. To see anything resembling a square wave you need a much wider bandwidth oscilloscope.

author: alsetalokin4017
author_rank: Super Contributor
timestamp: February 08, 2019, 04:25:09 pm
content: The short answer is that you can't really measure a 45 MHz square wave with a 100 MHz oscilloscope.

author: Mechatrommer
author_rank: Super Contributor
timestamp: February 08, 2019, 05:05:08 pm
content: Quote from: Vindhyachal.takniki on February 08, 2019, 03:17:56 pmwith unterminated 1X coax cable, so you probably looking at one point in standing wave.with 10X probe, no reflection in probe cable, there maybe a little bit of loading in the circuit though. it goes -ve i guess due to overshoot in square wave signal or a little reflection due to mismatched cable. without properly measuring it with the right tool, we are actually measuring a funny business... attached is an example of 5V 125MHz oscillator clock in not so good terminated transmission line. if we let loose the termination it will become haywire. ymmv..

author: David Hess
author_rank: Super Contributor
timestamp: February 09, 2019, 03:54:22 am
content: Check to make sure the x10 probe is compensated correctly or it can increase the amplitude of high frequency signals.

author: Zero999
author_rank: Super Contributor
timestamp: February 09, 2019, 09:37:18 am
content: Another option is a low-Z probe, which can be made with a piece of 50Ω co-axial cable and some resistors. It will have an impedance of 500Ω over a wide range of frequencies. At low frequencies, the ×10 probe will have a higher impedance, but at high frequencies, above 20MHz or so, its impedance drops below that of the low-Z probe, which remains constant.The above is just for future reference. Your oscillator should be capable of driving a low-Z or ×10 probe, with no problems. As previously noted, it's the oscilloscope's limited bandwidth which is causing the square wave to appear sinusoidal, rather than the probe.

author: taydin
author_rank: Frequent Contributor
timestamp: February 09, 2019, 10:14:01 am
content: Quote from: Zero999 on February 08, 2019, 04:23:18 pmI haven't seen such a filter in a number of oscilloscopes I have used. But the analog front end will start attenuating signals beyond its stated bandwidth, which effectively behaves like a filter, but that would only be a very mild filter. When I apply a 0 dBm, 4 GHz sine wave to my 1 GHz scope, I am still comfortably seeing the signal with about 23 dB attenuation.Spectrum analyzers always have a very sharp cutoff filter at their rated bandwidth, because without that filter, frequencies above the bandwidth will raise the noise floor within the rated operating bandwidth.