![]() ![]() ![]() Furthermore, such a factorization directly implies a minimal (i.e. ![]() with R(s) and P R(s) relatively right (left) prime polynomial matrices in the Laplace operator's and P R( s) column (row) proper. More specifically, a significant number of investigators have now noted and employed the fact that the ( p x m) transfer matrix, T(s), of a linear, time-invariant, multivariable system can be factored as R( s) P -1 R( s). This is due in large part to the development of an alternative (to the state-space approach) time domain technique, namely the differential operator approach for the analysis and synthesis of linear multivariable systems. More recently, however, there has been a resurgent interest in frequency domain methods. With this assumption, y(-1) = 0, leaving y(0) = b.The time domain, state-space approach to control systems analysis and design has dominated the control literature since the early 1960's. Certainly, the difference equation would not describe a linear system if the input that is zero for all time did not produce a zero output. What is the value of y(-1)? Because we have used an input that is zero for all negative indices, it is reasonable to assume that the output is also zero. ![]() In more detail, let's compute this system's output to a unit-sample input:īecause the input is zero for negative indices, we start by trying to compute the output at n = 0. To compute the output at some index, this difference equation says we need to know what the previous output y(n-1) and what the input signal is at that moment of time. Let's consider the simple system having \(p = 1\) and \(q = 0\). The reason lies in the definition of a linear system: The only way that the output to a sum of signals can be the sum of the individual outputs occurs when the initial conditions in each case are zero. One choice gives rise to a linear system: Make the initial conditions zero. These values can be arbitrary, but the choice does impact how the system responds to a given input. The way out of this predicament is to specify the system's initial conditions: we must provide the p output values that occurred before the input started. To compute these values, we would need even earlier values, ad infinitum. To compute them, we would need more previous values of the output, which we have not yet computed. What input and output values enter into the computation of y(1)? We need values for y(0), y(-1)., values we have not yet computed. A MATLAB program that would compute the first 1000 values of the output has the form for n=1:1000 y(n) = sum(a.*y(n-1:-1:n-p)) sum(b.*x(n:-1:n-q)) end An important detail emerges when we consider making this program work in fact, as written it has (at least) two bugs. We simply express the difference equation by a program that calculates each output from the previous output values, and the current and previous inputs.ĭifference equations are usually expressed in software with for loops. We have thus created the convention that \(a_0\) is always one.Īs opposed to differential equations, which only provide an implicit description of a system (we must somehow solve the differential equation), difference equations provide an explicit way of computing the output for any input. We have essentially divided the equation by it, which does not change the input-output relationship. There is an asymmetry in the coefficients: where is \(a_0\)? This coefficient would multiply the \(y(n)\) term in the above equation. ![]()
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