Jul 19, 2020 15:07
Thank you for all the help
Jul 19, 2020 15:06
So yeah, the best way to do this is by calibrating at the reference plane of the antenna, but if I couldn't do it, just wanted to know how the probe's RL affects the overall measurement
Jul 19, 2020 15:05
But the uncertainty of the cable, since it has its own return loss, is what makes this hard to quantify
Jul 19, 2020 15:04
In theory, if the cable is 50 Ohm, all I care about is how much power is reflected back from the antenna. It's going to have a phase shift, but if I get 10dB for RL, I know the antenna is taking in 90% of the power.
Jul 19, 2020 15:02
I'll look in those things. And yes, that has been my question all along. I am interested in the return loss of the antenna (or magnitude of S11).
Jul 19, 2020 14:03
For on thing, I know things that work in theory don't happen just like that in practice but my thinking was: "Alright, I have this cable here which essentially plugs into the antenna. If the cable is truly 50 Ohms, the reflection coefficient should still have the same magnitude"
Jul 19, 2020 13:56
That's why i asked how to eliminate it if I could not calibrate at the end of the cable
Jul 19, 2020 13:56
Thanks for all the help and sorry this turned into a long discussion. It does shed some light, though. Yeah, the cable is uncertainty that I have in my measurement and what I wanted to get rid of.
Jul 19, 2020 13:52
In summary, the PCB connector is on board and literally right the to the antenna and its matching parts. So, when I plugged in the cable, if the cable was accounted for, all I should is the antenna on the PCB, nothing else
Jul 19, 2020 13:44
This is the flow of power from the SoC to the Antenna: Driver output-->Balun-->RF Connector-->Pi Network (matching for ant)--Ant
Jul 19, 2020 13:41
All of the components before the antenna, and all the way to the SoC output, are 50 Ohm (or at least they're specified to be). The only unknown is the antenna
Jul 19, 2020 13:40
The reasoning is the following. When the cable is plugged in into the on board connector, it isolates the antenna and is matching components from the SoC driver. If I can make the antenna look 50 Ohms, then that's one thing less to worry about.
Jul 19, 2020 13:37
When I plug in the cable, then it breaks that path and allows me to measure things like power coming out of the SoC. Depending on the orientation of the connector, it would also allow me to test the antenna. The antenna is right next to this on board connector
Jul 19, 2020 13:37
You are correct, the SoC and the antenna are connected via a PCB trace (50 Ohms). Now, in that path, there is a receptacle connector (murata MM8030-2610RJ3), which allows me to plug in the cable I've talked about. During normal operations, when the cable is not plugged in, the path between the SoC and the antenna is uninterrupted.
Jul 19, 2020 13:24
No, the cable is not part of the system, I used to measure the antenna's S11. The antenna is a PCB antenna that will be driven by a SoC on board. On the PCB I have a connector that allows me to test for power, etc, but during the normal operation, the cable is not part of the circuit
Jul 19, 2020 13:24
Thanks. I guess the problem with this approach is that the standards are SMA connectors to the VNA. In order to calibrate with the cable, and to the PCB at the location of the antenna, I would need to create a 'proper' short, open, and a load on the PCB itself at the point where the cable plugs in. Not the easiest thing to do since it's hard to re-create true shorts or opens at those frequencies, but I will give that a try.
Jul 19, 2020 13:24
Thanks for the answer. I do have the standards and calibrated the VNA with those. Now, to connect to the antenna on the PCB, I used the cable. So I did not calibrate at the end of the cable if that is what you mean, I actually calibrated at the VNA ports with the short, load, and open standards. I can do port extensions at the ports but do you mean I need to calibrate with the cable included?
 
Jun 10, 2020 22:26
One problem with a differentiator circuit is that high frequency noise gets amplified. Say you have a low frequency signal, \$\sin(2\pi f_o t)\$, and some high frequency noise riding on it, \$\sin(2\pi f_n t)\$. The differentiation will produce \$2\pi f_o\cos(2\pi f_o t)+2\pi f_n\cos(2\pi f_n t)\$. Look at the last term. Unlike an integrator where you'd have a \$\dfrac{1}{f_n}\$ factor.
 

 Calculating SNR for a chart, manually

Calculating SNR for a chart, manually or using MATLAB
Mar 20, 2017 02:23
I hope it helps
Mar 20, 2017 02:21
t is time
Mar 20, 2017 02:21
But like I said, that's how I would do that if I am given a plot like yours and I have no way to extract the points with matlab
Mar 20, 2017 02:20
That's just an example, with 6 points, you can use as many as you need
Mar 20, 2017 02:19
You may need to rename, snr to maybe SNR, since there is a function called snr in matlab
Mar 20, 2017 02:18
This is what I would do.
t = [t1,t2,t3,t4,t5,t6];
signal = [s1,s2,s3,s4,s5,s6];
noise = [n1,n2,n3,n4,n5,n6];

snr = 20*log10(signal./noise);
plot(t,snr)
Mar 20, 2017 02:14
It's math. Take the x,y coordinates on plots b and c
Mar 20, 2017 02:13
That's what I am explaining to you
Mar 20, 2017 02:13
What is that you need to plot?
Mar 20, 2017 02:12
I don't know what you need then
Mar 20, 2017 02:11
And find SNR = 20 log(signal./noise)
Mar 20, 2017 02:10
The signal vector comes from plot b. Choose the amplitudes corresponding to the times in t. Same for plot c, that is the noise vector
Mar 20, 2017 02:09
Also create a vector signal and a vector noise. Those need to have the same number of points as t
Mar 20, 2017 02:09
You don't need all the points, choose a number of them, I don't know what else to tell you, that's what I would do
Mar 20, 2017 02:08
Well, you don't have a choice if those are the graphs you are given
Mar 20, 2017 02:08
So you can create a vector, say t = [t1, t2, t3,...,tn], t1, t2, t3 are just the times you see in both plot b and c
Mar 20, 2017 02:06
From the plots, you can see the amplitudes at specific times
Mar 20, 2017 02:04
Are you looking to plot the SNR of the plots you showed in your question?
Mar 20, 2017 02:02
Ok
 
Jun 23, 2016 22:23
Yes, you bet. I will edit the answer. Thank you!
Jun 23, 2016 21:29
I know what you mean. I will edit it
Jun 23, 2016 21:28
tell me what you think
Jun 23, 2016 21:28
Take a look at this:
https://www.precisionmicrodrives.com/application-notes/ab-022-pwm-frequency-for-linear-motion-control
Jun 23, 2016 21:26
Then you don't have to go that low in frequency*
Jun 23, 2016 21:23
For that then, you can restrict the range of PWM values, say in %, so that you are linear in terms of current drawn by the motor. Say you want to be linear when the duty cycle is between 45% to 80%, then you have to go that low in frequency. But that's a different story. I do get your point.
Jun 23, 2016 21:21
All that I am saying is that you can make a compromise here if you want to be linear in terms of current also
Jun 23, 2016 21:20
I get your point, I really do.
Jun 23, 2016 21:19
For linearity in terms of current, you need a lower pwm frequency. You may find an optimum point where you are somehwhat linear in terms of current and speed
Jun 23, 2016 21:17
I wanted to answer the question but referring to linearity in term of the current and the duty cycle
Jun 23, 2016 21:16
Well the question asked about linearity. You easily have linearity in terms of speed in relation to the duty cycle if the pwm frequency is high enough. Since the motor acts as a low pass filter, then you essentially get only the DC component (the average of your pwm signal) because the high freq components of the pwm are filtered out.
Jun 23, 2016 21:10
That's why you want to use a high pwm freq, so that the motor's speed doesn't swing a broad range of values.
Jun 23, 2016 21:09
Yes?