How to measure phase difference of 2 signals using IQ demodulator

Click For Summary
To measure the phase difference between two 40MHz signals using an AD8333 IQ demodulator, one signal should be applied to the local oscillator (LO) input while the other is connected to the RF input. The I and Q outputs can be displayed on an oscilloscope to visualize the phase relationship, or digitized for further analysis. The phase can be calculated using the arctan function on the I and Q outputs. However, the AD8333 divides the LO input by 4, necessitating a PLL VCO running at four times the input frequency for accurate measurements. This setup will allow for effective phase difference determination between the two signals.
mtxx
Messages
1
Reaction score
0
for one section of my project, i need to know the phase difference between two signal.
they are both 40MHz, but out of phase by a certain angle.

i have a AD8333 board from Analog Devices,
it is a dual I/Q demodulator consist of double-balanced gilbert cell mixers.

The board has 2 RF inputs, a LO input, and 4 output (I1, I2, Q1, Q2).

I have an oscilloscope to display the results.

according to wikipedia http://en.wikipedia.org/wiki/Phase_detector#Analog_phase_detector
there is a way to measure the phase difference using IQ demoludator.in this post, https://www.physicsforums.com/showthread.php?t=336619
f95toli mentioned that:
the magnitude is just given by the norm (sqrt(I^2+Q^2)) and the phase by the phase angle (arctan I/Q)

So is this method bellow correct?
to get the phase difference of both my RF signal,
i put one the signal into RF input1, and the second signal into the LO,
then i display both the I1, and Q1 in the oscilloscope, and get an get the peak to peak magniture for both the I1 and Q1, then calculate arctan I1/Q1, to get the phase difference?
 
Engineering news on Phys.org
mtxx said:
So is this method bellow correct?
Yes, almost. But think about it differently.

Select one signal as your reference and apply it as the local oscillator. Connect the other signal to the RF input, which will then be down-converted to DC at the IQ output. The I and Q outputs can drive the x and y axes respectively of the oscilloscope to give you a polar plot. Or you can digitise the I and Q, then compute the amplitude = Sqrt(I^2+Q^2) and the phase = Atan2(I/Q).

You will have a problem with the AD8333 because it divides the LO input by 4 to generate the internal LO quadrature signal. In order to use that chip you might run a PLL VCO at 4 times the frequency of your input signals as your LO. Then apply your two input signals to the two RF channels, use the I output from one channel, through a LP filter, to control the PLL VCO frequency. The other channel outputs, I and Q will then give you the phase difference.
 
I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

Similar threads

  • · Replies 10 ·
Replies
10
Views
6K
Replies
26
Views
3K
  • · Replies 23 ·
Replies
23
Views
3K
  • · Replies 21 ·
Replies
21
Views
25K
Replies
5
Views
2K
  • · Replies 8 ·
Replies
8
Views
15K
Replies
6
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
10
Views
4K
  • · Replies 12 ·
Replies
12
Views
2K