Chapter 2 Discrete-Time System Analysis in the Time Domain.ppt

§2-1 Linear Input/Output Difference Equations with Constant Coefficients
Chapter 2 Discrete-Time System Analysis in the Time Domain
§2-2 Discretization in Time of Differential Equations
Problems
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§2-1 Linear Input/Output Difference Equations with Constant Coefficients
Now consider single-input single-output discrete-time system defined by the input/output difference equation
where n is the integer-valued discrete-time index, x[n] is the input, and y[n] is the output. Here it is assumed that the coefficients a1, a2, …, aN and b0, b1, b2, …, bM are constants.
()
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Since Eq. () is a linear difference equation with constant coefficients, the system defined by the equation is linear, time invariant, and finite dimensional. The integer N in () is the order or dimension of the system. Also, any discrete-time system in the form of Eq. () is causal since the output y[n] at time n depends only on previous values of the output and the current and previous values of the input x[n].
Solution by Recursion
Unlike linear input/output differential equations, linear input/output difference equations can be solved by a direct numerical procedure. More precisely, the output y[n] for some finite range of integer values of n can puted recursively as follows. First, rewrite () in the form
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()
Then setting n=0 in () gives
y[0] a1y[1] a2y[2] · · · aNy[N]
+ b0x[0] b1x[1] · · · bMx[M]
Thus the output y[0] at time 0 is a bination of y[1], y[2], · · ·, y[N] and x[0], x[1], · · ·, x[M].
Setting n=1 in () gives
y[1] a1y[0] a2y[1] · · · aNy[N+1]
+ b0x[1] b1x[0] · · · bMx[M+1]
So y[1] is a bination of y[0], y[1], · · ·, y[N+1] and x[1], x[0], · · ·, x[M+1].
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If this process is continued, it is clear that the next value of the output is a bination of the N past values of the output and M+1 values of the input. At each step of putation, it is necessary to store only N past values of the output (plus, of course, the input