Digital Signal Processing Assignment Help

Digital Signal Processing Assignment Help

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Problems Problem 3.1

DISCRETE-TIME SIGNALS AND SYSTEMS

Stability of a system was defined in the, and it was shown that for an LSI system a necessary and sufficient condition for stability is that h(n) be absolutely summable. Show that each of the unit sample responses listed below corresponds to a stable system

Problem 3.2 Causality of a system was defined in the lecture and in section 1.3 of the text. In this problem we wish to show that for an LSI system a necessary and sufficient condition for causality is that the unit sample response be zero for n < 0. (a) Show from the convolution sum that if h(n) = 0 for n < 0 then the system must be causal, i.e. that y(n0) depends only on x(n) for n < n0 where n0 is arbitrary. This then shows that a sufficient condition for causality of an LSI system is that h(n) = 0, n ,< 0. (b) Now, suppose that h(n) is not zero for n < 0. Argue that the system cannot be causal. This can be done simply by showing that there is at least one input for which the output "anticipates" the input i.e. for which y(n0) depends on values of x(n) for n > n0. This then establishes that a necessary condition for causality of an LSI system is that h(n) = 0, n < 0. matlabassignmentexperts.com

Problem 3.3 Determine whether or not each of the following systems is stable and/or causal: (a) y(n) = g(n)x(n) where g(n) is bounded

Problem 3.4 Consider a causal system for which the input x(n) and output y(n) are related by the linear constant coefficient difference equation

(a) Determine the unit sample response of the system. (b) Using your result in (a) and the convolution sum determine the response to the input matlabassignmentexperts.com

x(n) = ejwn (c) Determine the frequency response of the system; (d) Determine the response of the system to the input

Problem 3.5 Consider a system with input x(n) and output y(n). The input-out put relation for the system is defined by the following two properties: (1) y(n) -ay(n -1) = x(n) (2) y(O) = 1 (a) Determine whether the system is shift-invariant. (b) Determine whether the system is linear.

Problem 3.6 Consider a discrete-time system with input x(n) and output y(n). The system transformation y(n) = F [x(n)] is arbitrary and may be nonlinear and time-varying. The only known property of the system is that it is well defined, i.e. the output for any given input is unique. Suppose that the input x(n) is chosen as x(n) = ejon and some parameter P of the output is measured (e.g. the maximum amplitude). P will in general be a function of w. Let us consider P for different excitation frequencies. Show that P is periodic in w and determine the period. matlabassignmentexperts.com

Solutions

Solution 3.1 To correspond to a stable system, the unit sample response must be absolutely sum able. For each system

Solution 3.2 (a) Consider the convolution sum in the form

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and hence depends only on values of x(k) for k < n0 (b) If h(n) is not zero for n < 0, then an example of an input for which the output anticipates the input is, of course, a unit sample. Alternatively, consider the convolution sum

If h(n) is not zero for n < 0 then at least one of the terms in the summation x(k) h(n -k) will be non-zero, i.e. y(n) will depend on at least one value in x(k) for k > n. matlabassignmentexperts.com

Solution 3.3 Not all of these systems correspond to LSI systems. For those that don't we can examine stability and causality by referring back to the basic definitions. (a) Clearly causal. Also stable since |y(n)I = |g(n)| |x(n)I Thus, if x(n) is bounded then y(n) will be bounded (b) Not causal, since for n < n0, y(n) will depend on values in x(k) for k > n. Also, not stable, since if x(n) = 1 for all n then y(n) = (n -n0 + 1) which is unbounded. (c) Clearly stable. It is causal if n0 > 0 and not causal otherwise.

Solution 3.4 (a) Rewriting the difference equation, with x(n) = 6(n) and h(n) denoting the unitsample response

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(b) Substituting x(n) and h(n) into the convolution sum we obtain

(c) The frequency response of the system is the complex amplitude ofthe response with an excitation ejw on. Thus, from part (a)

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Solution 3.5 (a) If x(n) = 6(n) then the system response is y(n) = an u(n). If x(n) = 6(n -1) then for n > 0 the system response will be (an + an-1) Therefore the system is not shift-invariant. (b) The system is not linear, as can be demonstrated in a variety of ways. For example, a linear system has the property that T[a x(n)]= aT[x(n)] . Therefore if the input is doubled (for example) the output must double at each value of n, which for this system cannot happen at n = 0.

Solution 3.6 The system excitation is x(n) = ejw. With w replaced by (w + 2) the excitation becomes ej(w+2 )n ejwn = e e j27rn and hence the input is periodic in o. This means, in essence that in considering complex exponential inputs, a variation in o over a range of 27 generates all of the distinguishable discrete-time complex exponentials. When w varies outside that range we simply see the same ones over again. For the system described in this problem, the output for a given input is unique. Thus since lwn and ej(o+27)n are identical input signals their outputs must be identical, and consequently the output parameter P will be periodic with period 27. matlabassignmentexperts.com