A Removable Singularity in Lead-Lag Coefficients As discussed in section 2.2.4 of Lead-Lag Algorithms, a recursive simulation of a first-order lead-lag transfer function can be written in the form where the subscripts "c" and "p" denote current and past values respectively.  If the time constants tD and tN are variable, and vary linearly over each time increment Dt, then the coefficients of the recurrence can be written as where On the surface it may appear that B is infinite when , but we can show that this is due to a removable singularity, and B is actually well-behaved at .  To show this, we first re-write A in the form Assuming , we can expand the binomial to write this as Extending the series and simplifying, we have We now re-write B in the form Notice that as  goes to -1, the coefficient A goes to Dt/tDP, so the quantity in the right hand parentheses goes to zero.  In the same condition the quantity in the left hand parentheses goes to infinity.  Our assertion is that the product of these two quantities approaches a finite value in this limit.  To determine this limiting value, we can substitute the series expression for A into the above expression for B to give We can therefore cancel a factor of  from the numerator and denominator, leaving us with Now, setting  to -1 (which makes A equal to Dt/tDP), we have Thus when  to -1 the coefficient B has the value given by Note that the original restriction  again corresponds to the condition of convergence, since  equals -1, so the absolute value of Dt is less than tDP.  To evaluate the infinite sum, let x denote the ratio Dt/tDP, and differentiate the summation as follows. The summation in the right hand expression is the series expansion of -ln(1-x), so we have Making use of the well-known integral with a = -b = 1, we have and therefore setting z = 1 - x (which gives dz = -dx) we arrive at Combining this with the 1/x term in the derivative of our infinite sum, and simplifying, we have the overall integral where the constant of integration is determined by the fact that our original summation vanishes as x = 0, and noting that the product (1/x - 1)ln(1 - x) goes to -1 as x goes to zero.  Thus our summation can be written in closed form as Therefore, in terms of our original problem, the singularity in B when  equals -1 can be removed by assigning B the value which was to be shown. Return to MathPages Main Menu