I also have same doubt. If RF is cos(theta_IF+theta_LO) and LO is cos(theta_LO) then as per Qui output is 1/2[cos(theta_IF)+jsin(theta_IF)]

If RF would be cos(theta_IF-theta_LO) then i can assume output would be

1/2[cos(theta_IF)-jsin(theta_IF)]

Now in this both case cos(theta_IF-theta_LO) and cos(theta_IF+theta_LO) are Image of each other. Thus in the output signal and image are opposite phase, this does not mean image is rejected.

I am typically giving RF 2GHZ and LO 2.07GHz. If Image signal is present at 2.14GHz then should I expect both signal at image present at I and Q at 70MHz?

Hello Dwijen,

In one case where RF = 2 GHz and LO = 2.07 GHz, the IF will be at 70 MHz on both the I and Q channel. For this case the LO signal is higher in frequency than the RF. This is termed high side LO. In the second case, RF= 2.14 GHz and LO = 2.07 GHz. This is low side LO because the RF is higher in frequency than the LO. Regardless, if you have high side or low side LO, the base band output for both cases will be 70 MHz on both the I and Q paths.

I am not sure if I completely understand your concern about images. If you are looking at the complex spectrum ( I + jQ) then you will see the image at -70 MHz in both cases. Images are due to amplitude and phase errors between the I and Q paths. The dominate source of phase error is due to the ADL5380; specifically where the 90 degree phase split happens in the the LO path. You can get additional amplitude and phase errors if the routing and components on the PCB are not matched for the I and Q paths. This is a minor contribution compared to the phase error in the demodulator. Without any additional correction, the ADL5380 will have an image rejection ratio of approx 45 dBc. If more rejection is needed than this, then most customers tend to implement an image reject algorithm digitally in the back end.

Qui

Hello Dwijen,

A point of clarification. When RF = 2.14 GHz and LO = 2.07 GHz, the IF tone will be at 70 MHz and the image will be at -70 MHz. When RF = 2 GHz and LO = 2.07 GHz, then the IF tone will be at -70 MHz and its image will be at +70 MHz. The statement I made above stating that both cases will result in an image at -70 MHz is not correct. The statement I was trying to make is that wherever the IF lands, the image will be at the opposite frequency.

Qui

Hello Qui,

Let me simplify my concern,

I have RF at 2.07GHZ with 0dBm power which i want to convert in 70MHz IF, I have 2.14GHz unwanted signal with -10dBm power which i dont want to appear in IF.

Will ADL5380 or in that sense any I/Q demod be able to reject this 2.14GHz unwanted signal at the output?

From your answer above I can assume that 2.07GHz RF would get converted to 70MHz IF with power 0dBM+Conversion gain. Unwanted 2.14GHz signal would also be converted to 70MHz IF with power -10dBm+conversion gain.

Thus at IF both 2.07GHz and 2.14GHz signals are mixed. To achieve rejection of unwanted 2.14GHz signal I have phase shift I and Q and add in signal processing.

To be precise I/Q demodulator would not in fact reject 2.14GHZ signal (2.14GHZ signal is Image for 2.07GHz signal) but simply convert it into I and Q which would be 90 degree phase shifted with respect to I and Q generated from 2.07GHZ which can be rejected using further signal processing.

Is this understanding correct?

Can you also provide procedure you follow to test Image Rejection spec of ADL5380? That should clarify things better I guess.

Dwijen

Hello Dwijen,

Your desired signal will be at 70 MHz and its image will be at -70 MHz. Lets just assume that the image rejection of the ADL5380 is 45 dBc. This assumes that there are only amplitude and phase errors due to the demod and everything else is perfect. This is a big assumption here but I think the best way to explain this is to put some numbers in. So your desired signal is 0 dBm at 70 MHz and its image is at -45dBm at -70 MHz.

Now the unwanted signal will be at -70 MHz with a signal amplitude of -10 dBm and its image will be at 70 MHz with a signal amplitude of -10 dBm - 45 dB = -55 dBm.

Lets put both pictures together.

To answer one of your questions, the demod will not reject the undesired signal but rather it will mix in the same manner as the desired signal. Therefore, you will need to either filter the undesired signal at RF or try to improve on the image rejection. Since the undesired signal is so close in frequency it may be difficult to filter. The second option may be the best where you try to improve the image rejection. This can be done digitally in the DSP where you monitor the amplitude and phase difference of the I and Q paths and correct for this difference.

To test this using the ADL5380, feed the differential I and Q outputs to a spectrum analyzer that allows complex spectrum. An instrument that does this is the Rhode & Schwartz FSQ8. Another option is to use a dual ADC and process the FFT in the digital domain.

Qui

Hi Qui,

I need one more clarification.

On the datasheet on page 25, following diagram is shown

Here, Is the phase shift of IF signal at I (or Q) and addition to Q (or I) is carried out internally or it is just shown for illustration purpose?

In the functional block shown bellow which is on page 22 and page 1 of datasheet, phase shift and addition process is not shown. Also in the circuit description on page 22 nothing is mentioned about phase shifting and addition. So I guess process shown in fig 85 on page 25 is just for illustration purpose. Is that correct?

Hello Shouxi,

Your analysis is correct. The only mistake I saw is the last equation where -sin(theta_IF) should be +sin(theta_IF). Your final output at the demod should be 1/2[cos(theta_IF)+jsin(theta_IF)]

Qui