First the signal difference equation of general filter is given by Equation. To solve the difference equation for the output signal y(n)y(n) size 12{y \( n \) } {} when the input signal x(n)x(n) size 12{x \( n \) } {} is applied, we need to know the intial conditions (or boundary conditions), i.e. the state of the filter just before the application of the signal. For this, the filter comprises two types : Relaxed , i.e y(n)=0y(n)=0 size 12{y \( n \) =0} {} at n<0n<0 size 12{n<0} {} , and nonrelaxed , i.e. y(n)y(n) size 12{y \( n \) } {} has some nonzero values at n<0n<0 size 12{n<0} {}. Actually, the solution of digital signal difference equations is quite similar to that of the differential equations of continuous time. This section only presents a brief discussion, interested readers can refer to various references. Now let’s consider a simple equation (Example)

y ( n ) = 0 . 8y ( n − 1 ) − x ( n ) y ( n ) = 0 . 8y ( n − 1 ) − x ( n ) size 12{y \( n \) =0 "." 8y \( n - 1 \) - x \( n \) } {}

For a signal x(n)x(n) size 12{x \( n \) } {} applied at n=0n=0 size 12{n=0} {}, the output at that instant is

y ( 0 ) = 0 . 8y ( − 1 ) − x ( 0 ) y ( 0 ) = 0 . 8y ( − 1 ) − x ( 0 ) size 12{y \( 0 \) =0 "." 8y \( - 1 \) - x \( 0 \) } {}

If the system is not relaxed, y(n−1)y(n−1) size 12{y \( n - 1 \) } {} is differs from zero and it affects values of y(n)y(n) size 12{y \( n \) } {} at successive times. Thus y(n)y(n) size 12{y \( n \) } {} has two components, one due to the applied signal at n=0n=0 size 12{n=0} {}, and the other due to intial condition y(−1)y(−1) size 12{y \( - 1 \) } {} . On the other hand, even for relaxed systems the applied of input signal will cause two responses, the transient response and the steady-state response. The transient response, if any, is the homogeneous root , and steady-sate response is the particular root of the difference equation. The two roots form the total solution of the equation.

The time convolution of input signal x(n)x(n) size 12{x \( n \) } {}with impulse response h(n)h(n) size 12{h \( n \) } {} of the system gives the output signal which is the total solution, from this usually we cannot separate the two homogeneous and particular roots.

We know that in analog world the differential equations can be conveniently solved by using the Laplace transform . Similarly, in the digital domain, we can use the z-transform (chapter 4) to solve the difference equations.