Hi Thomas You can certainly expect some insertion loss from the filter.In the attached sweep of mine the input signal from the tracking generator is at -20dBm so the insertion loss within the passband is -3.65dB or thereabouts. Also the passband of mine is something like 3.6KHz wide.
What you need to check is that the loss is uniform through the passband - so in my image you are trying to ascertain how flat the top of the curve is - we are aiming for something as flat as possible with steep sides.
How did you match your crystals? Also thinking about your measurements; if you are using an AD9851 eBay module then I am pretty sure they have an output impedance of 200R which won't be a good match into the 50R filter input unless followed by an appropriate buffer or other transformation. You should also consider terminating the scope in 50R on the filter output as here:
http://www.qrp.pops.net/rf-workbench-1.aspThe above is good article on basic RF measurement; it also covers calculating dB from 'scope peak to peak readings.
Mark G0MGX On 21/01/2014 10:22, Thomas Sarlandie wrote:
Hi,I (think I) understand the maths. It's more an experimental problem. I thought I was supposed to measure the ratio of output to input. From your email I understand that I should just do output to max output and that will work too.I will do that and send more results. Thanks for the quick reply! ThomasOn Tuesday, January 21, 2014, Ashhar Farhan <farhanbox@xxxxxxxxx <mailto:farhanbox@xxxxxxxxx>> wrote:the scope is giving you a reading in voltage, so, the first step is to convert it into a power reading. For instance, if the peak of the filter is 100mv, and then it dips down to 50 mv, then we have a 2:1 ratio in voltage. If you square the ratio of the voltage, that gives you the power ration, In this case, it would be 4:1. Now, 4 is about 6 db. so, we can say that there is a ripple of 6 db in your filter (which is not very good!). You can find a number of calculators to convert voltage into db values on the net. I just do the sums in my head. - f On Tue, Jan 21, 2014 at 2:53 PM, Thomas Sarlandie <thomas@xxxxxxxxxxxxx <javascript:_e({}, 'cvml', 'thomas@xxxxxxxxxxxxx');>> wrote: Hi everyone, I built the crystal filter tonight on a piece of pcb and experimented with it by injecting a signal from an arduino controlled ad9850. Following the input and output on my oscilloscope, I can tell that the filter is indeed very efficient at eliminating anything not in the 20mhz range (it's my first time building such a filter, the experience is impressive). When I zoom in around 20mhz, I see that I get the biggest output value for 20.000.080, 20.002.600 and 20.004.400. Between those values the output is a bit smaller but still much bigger than a few hundreds hertz above or below. I cannot really convert this in dB because in that range, the measured input voltage (on the scope) is slightly larger than the input. Clearly, the power is going through. How would you terminate the input and output to be able to measure the filter bandwidth properly? Is the center frequency going to change when I plug the filter in the rest of the circuit? (the input and output impedance are clearly not controlled in my current setup). How could I adjust it (except getting another bunch of crystals and trying again?) I see Farhan listed a section of the ARRL Handbook about QER filters. I will read up on that. Sorry for the very basic questions. This is my first attempt at a full radio project. Very challenging - loving it! 72, thomas - kk6aht
Attachment:
Crystal Filter Sweep.png
Description: PNG image
