5.11 Discrete-time systems in the time-domain  (Page 6/2)

A discrete-time signal $s(n)$ is delayed by $n_0$ samples when we write $s(n-n_0)$ , with $n_0> 0$ . Choosing $n_0$ to be negative advances the signal along the integers. As opposed to analog delays , discrete-time delays can only be integer valued. In the frequency domain, delaying a signal corresponds to a linear phase shift ofthe signal's discrete-time Fourier transform: $\leftrightarrow (s(n-n_0), e^{-(i\times 2\pi f{n}_0)}S(e^{i\times 2\pi f}))$ .

Linear discrete-time systems have the superposition property.

We want to concentrate on systems that are both linear and shift-invariant. It will be these that allow us the full power offrequency-domain analysis and implementations. Because we have nophysical constraints in "constructing" such systems, we need only a mathematical specification. In analog systems, thedifferential equation specifies the input-output relationship in the time-domain. The corresponding discrete-time specification is the difference equation .

As 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 simply express the difference equation by a program thatcalculates each output from the previous output values, and the current and previous inputs.

Difference equations are usually expressed in software with for loops. 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. What inputand output values enter into the computation of $y(1)$ ? We need values for $y(0)$ , $y(-1)$ , ..., values we have not yet computed. To compute them, we would need more previous values of the output, which wehave not yet computed. To compute these values, we would need even earlier values, ad infinitum. The way out of thispredicament is to specify the system's initial conditions : we must provide the $p$ output values that occurred before the input started. These values can bearbitrary, but the choice does impact how the system responds to a given input. One choice gives rise to a linear system: Make the initial conditions zero. The reasonlies in the definition of a linear system : The only way that the output to a sum of signals can be the sum of theindividual outputs occurs when the initial conditions in each case are zero.