Half IF spurious response and the second order intercept point

An irksome 2nd-order spurious response called the half-IF (1/2 IF) spurious response, is defined for the mixer indices of (m = 2, n = -2) for low-side injection and (m = -2, n = 2) for high-side injection. For low-side injection, the input frequency that creates the half-IF spurious response is located below the desired RF frequency by an amount fIF/2 from the desired RF input frequency. The desired RF frequency is represented by 2400 MHz, and in combination with the LO frequency of 2200 MHz, the resulting IF frequency is 200MHz. For this example, the undesired signal at 2300 MHz causes a half-IF spurious product at 200MHz. For high-side injection, the input frequency that creates the half-IF spurious response is located above (by fIF/2) the desired RF. Note that high side injection implies that the LO frequency is above the RF frequency and low side injection implies that the LO frequency is below the RF frequency.

The second order intercept point is used to predict the mixer performance with respect to the half IF spurious response. For further details please see the article under engineer's corner/engineering pages in our website at www.signalpro.biz.

More on cable modeling

In our efforts to understand cable models a little better we took some measured data from Analog Devices ( Analog Dialogue, Vol 38, July 2004, "An adjustable cable equalizer combines a wideband differential receiver with analog switches", Johnathan Pearson) and an approximate expression provided by David A. Johns and Daniel Essig,( "Integrated circuits for data transmission over twisted pair channels", IEEE Journal of Solid State circuits, March 1997. A little dated but very useful, since this is one paper that actually had something we could get our hands around and use practically.)We used MATLAB to generate the loss characteristics as per David Johns et al's paper and plotted it along with the Analog Devices data. From this it appears that at small cable lengths there is generally good agreement between the two sets of data. As cable lengths increase the correspondence gets worse and worse. If one has to use the approximate expression, then a correction factor of from 2.5dB to 7.5dB may be needed to get closer estimates using the theoretical approximate expression. The results are in the "Engineer's Corner" on our website at www.signapro.biz.

Bandwidth requirement to pass fast rising digital signals

How wide must the bandwidth be to pass a fast rising digital pulse so that at the output it can still be recognized as a pulse and detected? A common enough question. However sometimes the answer is not so obvious. Common wisdom says a minimum 3dB point of the filter or medium through which the pulse has to transition must be at least 1/pi*tr where tr is the risetime and pi is 3.1415 etc. Upon simulation using a simple RC filter, the result is: (a) The rule is correct. (b) The pulse width and period must be such as to accomodate the rise and fall time of the pulse. (c) The bandwidth may be narrower if the detection threshold can be set higher. (d) If the detection threshold is low then detection errors may occur if the above rules are disobeyed!