[SOLVED] How to find image rejection ratio of a complex filter in cadence

I have designed a complex filter in cadence and want to find its image rejection ratio. Can anyone please tell me how to do it??
Thanks in advance

Perform two ac analyses and feed both inputs simultaneously:

1.) Vin=V1+jV1 (90 deg lead)
2.) Vin=V1-jV1 (90 deg lag).

The ratio of both output signals gives you the image rejection.

Dear LvW
Can you please elaborate the method.
1) I am not able to understand what you mean by by perform two anlysis , do you mean that i should do two differnt analysis while giving normal/ usual values in the first, giving the changes values in the second analysis, then compare the results of the two??
2) About the input, Vin=V1+jV1 (90deg lead), Vin=V1-jV1 (90deg lag): I must give AC magnitude value 1 and change the phase to 90 and -90 degress only???
Thanks in advance

richaphy said:

Dear LvW
Can you please elaborate the method.
1) I am not able to understand what you mean by by perform two anlysis , do you mean that i should do two differnt analysis while giving normal/ usual values in the first, giving the changes values in the second analysis, then compare the results of the two??
2) About the input, Vin=V1+jV1 (90deg lead), Vin=V1-jV1 (90deg lag): I must give AC magnitude value 1 and change the phase to 90 and -90 degress only???
Thanks in advance

Click to expand...

Not exactly as proposed by LvW.

There are two methods to verify the IMRR: 1. AC & 2. Transient

For both methods you have to run two complex stimulations which should represent a complex wanted signal and a complex unwanted, or image signal.

If the complex signal path surpress frequency components with "negative" frequencies you first stimulate positive frequencies with two AC sources:

Positive Frequency Stimulation (phasor rotates counterclockwise)
Mag 1.0 Phase 0.0 for the in-phase component & (signal name "I")
Mag 1.0 Phase 90.0 for the quadrature-phase component (signal name "Q")

the second

Negative Frequency Stimulation (phasor rotates clockwise)
Mag 1.0 Phase 0.0 for the in-phase component & (signal name "I")
Mag 1.0 Phase -90.0 for the quadrature-phase component (signal name "Q")

Now you get a frequency response for postive and negative frequencies. The negative is typical the image.

For transient verfication the method is the same but using I=cos,Q=sin for positive frequencies and I=cos,Q=-sin for negative frequencies.

You have to wait the initial settling before measuring the transient image rejection to let decay the initial settling effect one or two orders below the rejection.

rfsystem said:

Positive Frequency Stimulation (phasor rotates counterclockwise)
Mag 1.0 Phase 0.0 for the in-phase component & (signal name "I")
Mag 1.0 Phase 90.0 for the quadrature-phase component (signal name "Q")

the second

Negative Frequency Stimulation (phasor rotates clockwise)
Mag 1.0 Phase 0.0 for the in-phase component & (signal name "I")
Mag 1.0 Phase -90.0 for the quadrature-phase component (signal name "Q")

Click to expand...

Hi rfsystem,

for my understanding that is exactly the method - consisting of two succeeding ac simulation runs - as I have proposed.

Hi LvW,

there is a small risk of misunderstanding:


Perform two ac analyses and feed both inputs simultaneously:

1.) Vin=V1+jV1 (90 deg lead)
2.) Vin=V1-jV1 (90 deg lag).

If 1.) and 2.) meaning is Wanted and Image there should minimum 2 sources for each test. Otherwise a free interpretation would be that "+" means applying two sources. I recognize that this was the meaning but try to explain that in more detail.

I would add that there exist also analog circuits which have image rejection which use 3 inputs. For instance a mixer with 3 phases and a 3-path image reject baseband filter.

rfsystem, thanks for clarification.
I agree - it was not clear that for both simulations under 1.) and 2.) TWO signal sources (one of which with + or - 90 deg phase shift) are necessary.