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( b ) Let us now consider causality. For an input x ( n ) , the response is as given in Eq. (1.22). In order for a system to be causal, the output y ( n ) at time no cannot depend on the input x ( n ) for any n > no. Therefore, Eq. (1.22) must be of the form y(n) = k=-m

Two triangles are similar if an angle of one triangle is congruent to an angle of the other and the sides including these angles are in proportion.

hk(n)x(k)

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which is the desired test for causality. Determine whether o r not the systems defined in Prob. 1 .I5 are (a)stable and (b) causal.

PRINCIPLE 5:

(b) Let us generalize to the case of the function f (x) = Ax + B, where A and B are constants. Then, f (x + h) f (x) [A(x + h) + B] (Ax + B) Ax + Ah + B Ax B Ah = = = =A h h h h f (x + h) f (x) = lim A = A f (x) = lim h h 0 h 0

- k)u(n - k ) , note that h k ( n )grows linearly with n . Therefore, this system cannot be stable. For example, note that if x ( n ) = S(n), the output will be

PRINCIPLE 6:

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which is unbounded. Alternatively, we may use the test derived in Prob. 1 .I6 to check for stability. Because

Two right triangles are similar if an acute angle of one is congruent to an acute angle of the other (corollary of Principle 3). A line parallel to a side of a triangle cuts off a triangle similar to the given triangle.

this system is unstable. On the other hand, because h,(n) = 0 for n < k , this system is causal. ( b ) For the second system, h k ( n ) = S(2n - k), note that h l ( n ) has, at most, one nonzero value, and this nonzero value is equal to I . Therefore,

PRINCIPLE 7:

for all n , and the system is stable. However, the system is not causal. To show this, note that if x ( n ) = &(n- 2 ) , the response is y ( n ) = h 2 ( n )= 6(2n - 2 ) = &(n- I) Because the system produces a response before the input occurs, it is noncausal.

Thus, we have proved:

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Rewriting this system function as a ratio of polynomials, it follows that the locations of the zeros of H ( z ) will depend on the locations of poles as well as the zeros of H,,(s), and there is no way 10 guarantee that the zeros lie inside the unit circle. A simple example showing that a minimum phase continuous-time filter will not necessarily be mapped to a minimum phase discrete-time tilter is the following:

Fig. 7-11

z = -(be-'

Fig. 7-12

7e-2) % - 1.256

PRINCIPLE 8: PRINCIPLE 9:

Theorem 12.1: Dx (Ax + B) = A By letting A = 0 in Theorem 12.1, we obtain: Corollary 12.2: Dx (B) = 0; that is, the derivative of a constant function is 0. Letting A = 1 and B = 0 in Theorem 12.1, we obtain: Corollary 12.3: Dx (x) = 1 By the computations in Problems 11.1, 11.2, and 11.3(a), we have: Theorem 12.4: (i) Dx (x 2 ) = 2x (ii) Dx (x 3 ) = 3x 2 (iii) Dx 1 x = 1 x2

If Ha(s) is minimum phase, the poles and zeros of H,,(s) are in the left-half s-plane. In other words, if H,(s) has a pole or a zero at s = sk, where sk = UL j Q n ,

Triangles similar to the same triangle are similar to each other. The altitude to the hypotenuse of a right triangle divides it into two triangles which are similar to the given triangle and to each other. Triangles are similar if their sides are respectively parallel to each other. Triangles are similar if their sides are respectively perpendicular to each other.

Therefore,

PRINCIPLE 10:

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