1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

A(6)

2

9.3 • DIGITAL SIGNAL F ILTERS 3 7 5

and x [l ] = J2 has the solution x[n] = 2cos(in). Thus cos(i(n + 2))

is a solution to x[n] - J2x[n - 1] + x(n - 2] = 0 and so are the

signals cos( ~n) and sin(~n). This can also be proven by direct sul:>-

stitution of x [n] = cos(~n), and using the trigonometric identities

x[n - l ] = cos(%n - i) = ~(cos( in)+ sin( in)) and x [n - 2] =

cos(;fn - ~) = sin( ;fn). An easy calculation shows that

y[n] = x(n] - J2x[n - l ] + x[n -2] =cos (in)

- v'2 (~(cos (in)+ sin (in)))+ sin (in) = 0.

Testing x[n] = sin(in) is similar.

For this filter, the amplitude response is A(IJ) = IH(ei^9 )1 where H(z) =

1 - J2z -^1 + z-^2 is the transfer function. Calculation reveals that

A(i) = IH(elf )I =Ii-J2 (elf) -

1

+ (eit) -

2

1

(

1 .)-

1

(1 .)-

2

(1 .)

= 1 - J2 ~i + ~i = 1 - J2 :i?.i -i = o.

The graph of A( IJ) is given in Figure 9.4. Notice that there is a

zero amplitude response at IJ = ;f and that the amplitude response

increases for values of IJ in the interval [i, 11'].

1t 1t

4 2

Figure 9.4 The amplitude response A(8) =II - J2z-^1 + z-^21 for the zeroing out filter
y{n] = x [n l - J2x[n - 1] + x(n - 2).