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Problem 1: Listed below are several z-transforms. Determine by "inspection" the sequences to which they correspond.

Solution:

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Problem 2 Listed below are three z-transforms. For each determine the inverse z-transform by using contour integration, and again by using a partial fraction expansion.

Solution: (i) Using Contour Integration

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Since the contour of integration must lie inside the region of convergence, i.e. for For n > 0 there are no poles at Thus, f or n > 0

It encloses the pole at

For n < 0 we use the substitution of variables. Thus

where new the contour of integration must lie inside the region of convergence of X(l/p) i.e. for IpI < 2. Then

For n negative the only pole is at p = -2 which is outside the contour of integration. Thus for n < 0 x(n) = 0.

Combining these two results, then

Since for this example X(z) has only a single pole, the partial fractions expansion method wouldn't apply.

The inspection method would and in fact corresponds to problem 6.1 (iii) for a=-1/2

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(ii) Using contour integration,

where on the contour c, For n > 0

For n < 0

The contour of integration is |p| < 2. The poles of the integrand for n < 0 are at p = -4 and p = -2. Therefore there are no poles inside the contour of integration and thus

Using a partial fractions expansion, Matlab Assignment Help

With X(z) expressed in this form, x(n) can be obtained by inspection. Specifically, since the region of convergence of X(z) is , x(n) is a right-sided sequence and is given by

(iii) Using contour integration,

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for n > 0 there is one pole, at z = 1/2. Thus for n > 0

For n = 0 there is also a pole at z = 0. Hence for n = 0, residue of x(n) at the origin = For n < 0 following the same procedure as in (i) and (ii), x(n) = 0. Thus ,

Using the partial fractions expansion, we note first that the order of the numerator and denominator are equal, and thus the partial fractions expansion of X(z) is in the form

To obtain the constant a we use long division. Thus:

Since the region of convergence is Izi > ½, the inverse z-transform is (by inspection)

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Problem 3 By using a power series expansion determine a sequence x(n) whose z-transform is

Solution: Expanding ez in a power series, we obtain

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Question 4.

Given here are four z-transforms. Determine which ones could be the transfer function of a discrete linear system which is not necessarily stable but for which the unit-sample response is zero for n < 0.

Clearly state your reasons.

Solution: If h(n) = 0 for n < 0, then H(z) must be expressible as a power series of the form

Specifically, it cannot contain any positive powers of z. Consequently, expressed as a ratio of polynomials in z, the order of the numerator must be less than or equal to the order of the denominator. In that problem we showed that if x(n) = 0 n < 0 then lim X(z) must be z+“ finite. Applying either of these conditions, (a) and (c) could correspond to causal systems while (b) and (d) could not.

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Problem 5.

If the contour of integration is the unit circle, for example, this has the effect of mapping the outside of the unit circle to the inside, and vice versa. It is worth convincing your self, however, that without this substitution, we can obtain x(n) for n < 0 by evaluating the residues at the multiple-order poles at z = 0.

Let

where the region of convergence includes the unit circle. (a) Determine x(0), x(-l), and x(-2) by evaluating Eq. (4.67) explicitly, obtaining the residue for the poles at z = 0. (b) Determine x(n) for n > 0 by evaluating expression (4.67) and for n < 0 by evaluating Eq. (4.74).

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Solution:

Since the region of convergence includes the unit circle and the pole is at z = 2, |z| < 2 on the contour c . Therefore the pole at z = 2 is not inside the contour of integration. Hence for n < 0 the only poles inside the contour of integration are at z = 0:

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(b) For n > 0 x(n) = 0 since there are no poles inside the region of convergence. For n < 0

Note in particular that

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