Z-transform of sequence product
- Signal and System Spring 2023 Homework Requirements and Reference Answer Summary
- Signals and Systems 2023 (Spring) Assignment Requirements - Eleventh Assignment
- Signals and Systems 2023 (Spring) Homework Reference Answers - 11th Homework
01 The eleventh homework
1. Introduction to Exercises
Two sequences, each with a corresponding z-transform. The z-transform corresponding to their product is a special convolution result of their respective z-transform. Using this formula can help us calculate the z-transformation of some special sequences. In the eleventh assignment, which consists of two problems, let us relate the z-transform of a product of sequences using convolutions in the z-transform domain. Let us look at the specific solution ideas below.
2. Problem solving
1. The first sub-question
The z-transforms corresponding to the two sequences of the first sub-problem are all rational fractions. The corresponding z-transform of their product is calculated by z-transform transformation domain convolution theorem. Since the expressions of the two sequences are first-order rational fractions, the choice of which convolution theorem formula has little effect on the complexity of the later solution. Simplify the interior of the integral, and finally get the contour integral on the z plane. Next, use the residue theorem of the complex variable function to calculate the contour integral. The expression has two poles, one at third, and one at 3z. It is necessary to use the regions of convergence corresponding to the z-transforms of the original two sequences, and it can be judged that the two poles are located in the contour integral. Therefore, the result of the integration is equal to the sum of the residues of the integrand at these two poles. .Using the residue calculation rule, the integral result can be finally obtained, which is equal to 1.
Let's analyze the rationality of this result. The z-transform corresponding to the product of two sequences is equal to 1, which is related to the particularity of the two sequences. The region of convergence corresponding to the first sequence is outside a circle with a radius of one-third, so this sequence is a right-hand sequence. The region of convergence corresponding to the second sequence is the area inside a circle with a radius of one-third, corresponding to the left sequence. The product of two sequences is actually non-zero only when n is equal to zero. Corresponds to delta [n]. Therefore, the product of these two sequences corresponds to a z-transform equal to 1.
3. The second question
The second sub-problem includes an exponential sequence and a sinusoidal oscillatory sequence. The exponential sequence z-transform contains only one pole, and the sinusoidal sequence contains a pair of conjugate poles. When choosing the convolution formula, we deliberately choose to take z of v for the variable of H(z), which can simplify the complexity of calculating the residue in the later stage. Substituting X(z), H(z) into the integral According to the convergence domains of the two sequences, it can be judged that the pole corresponding to the negative beta of e is within the perimeter integral, and from the convergence domain of H(z), it can be judged that the value of z in v is modulo greater than 1. In this way, the size of the modulus of v and z can be judged. From this relationship, we can know that the conjugate pole corresponding to the following expression is not in the convergence area. Thus, the contour integral can be obtained only by computing the pole residues at negative betas of e. The following simplification is performed to calculate the residue at the pole. This is the result of the calculation. Refer to the typical sequence z-transform table, which corresponds to the z-transform of an exponentially decaying sequence. This is consistent with the fact that the two series are the product of an exponential series and an oscillatory series, respectively. This example shows that when choosing the convolution theorem in the transformation domain of two z-transforms, a reasonable choice of the convolution formula can simplify the steps in the final calculation of the residue.
※ Summary ※
This paper discusses the transform-domain convolution theorem for the z-transform and its application to solving the z-transform of the product of sequences. According to the complexity of the sequence involved in the operation, a reasonable choice of two convolution forms can simplify the later calculation steps.
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