Homework Tutorial Video SS2023-HW11: Prove that the z-transform is cumulative

Z-transform of sequence accumulation

 

01 The eleventh homework


1. Introduction to Exercises

  In the eleventh assignment, a proof question is included. After accumulating the sequences, a new sequence is obtained. The corresponding z-transform is equal to the original sequence's z-transform multiplied by z minus 1/z. Let's discuss the idea of ​​this simple proof problem.

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2. Problem solving

  It is relatively simple to prove this conclusion by applying the convolution theorem of the z-transform. Here the conclusion in a convolution needs to be applied. That is, the sequence obtained by accumulating the sequence can be regarded as the convolution of the sequence and u[n]. Thus the z-transform of the accumulated sequence is equal to the z-transform of the sequence multiplied by the z-transform of u[n]. For u[n] sequence, its z transformation is equal to z minus 1/z. This result is the proof requirement of this question. Applying the convolution theorem of z-transform simplifies the proof process.

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  Can this problem be proved by directly applying the definition of z-transform? Let's take a look. Substitute the cumulative sum of the sequence directly into the formula for the z-transform. A double-accumulation expression is obtained. Since the second accumulation has the variable n, they cannot be swapped yet. But if there is any kind of accumulation in the back, add a u[n], so that the upper limit of the second accumulation can be modified to infinity. In this way, the two cumulative sums can be exchanged, all related to k are placed in the outer cumulative sum, and the inner cumulative variable is n, and k is a constant. The inner accumulation is actually the z-transform corresponding to the delay of u[n]. Then the corresponding accumulation result can be obtained according to the time-shift characteristic of the z-transform. Add up the remaining outermost layers. Taking z minus 1 out of the accumulation, what remains is the z-transformation on x[n]. The final calculation result is obtained. You can see that this proof process is actually to prove the convolution theorem of z-transform again.

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Summary  ※


  In this paper , the z-transform proof process of sequence accumulation is given. Applying the convolution theorem simplifies the proof.
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Origin blog.csdn.net/zhuoqingjoking97298/article/details/130701405