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First, in the table, item (1) says that if F z is the z-transform of sequence v nT , and if a is constant, then the z-transform of av nT is a times the z-transform of v nT . Next, item (2) says that the z-transform of the sum of two or more sequences equals the sum of the transforms of the individual sequences. This property applies here because we are dealing with linear time-invariant conditions. Next, item (3) can be established as follows. Since, by eq. (573), Zv nT then, Zan v nT

// Turn crawling flag off. crawling = false; // Close matches log file. try { logFileWriter.close(); } catch (Exception e) { showError("Unable to close matches log file."); }

that is, if F z is the z-transform of a sequence v nT , and a is constant, then the ztransform of an v nT is found by replacing z with z=a in the transform of v nT . Problem 299 The notation U t denotes a continuous-time (CT) function called the unit-step function, de ned as equal to 1 for all positive time (including t 0) but equal to zero for all negative time, as illustrated below.

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Given that a CT function v t 3U t 20t is being sampled 100 times per second, what is the z-transform of v nT (Answer: F z z 3z 3:2 = z 1 2 Problem 300 Given v nT U nT U nT T U nT 2T , nd F z . Problem 301 Find F z for the v nT of problem 298.

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The crawling flag is set to false to indicate that crawling is no longer underway. Next, the matches log file is closed since crawling is finished. Similar to opening the file, if an exception is thrown while trying to close the file, an error dialog box will be shown with a call to showError( ). Because crawling is finished, the search controls are reactivated by the following code:

We have rightly said that the solution to a DT problem can be simpli ed if the mathematical work is carried out in the z-domain. After a solution is obtained in the z-domain, the nal step is to inverse transform the answer back into the time domain. In this

section we ll consider how an answer in the z-domain can be inverse-transformed into the time domain. We begin our discussion as follows. Going back to eq. (573), we see that a function of z, say Y z , can be expressed in the basic form Y z y 0 y T z 1 y 2T z 2 y 3T z 3 y nT z n 583

// Mark search as done. crawlingLabel2.setText("Done"); // Enable search controls. startTextField.setEnabled(true); maxComboBox.setEnabled(true); limitCheckBox.setEnabled(true); logTextField.setEnabled(true); searchTextField.setEnabled(true); caseCheckBox.setEnabled(true);

where y 0 sampled value of y t at t 0, y T sampled value of y t at t T, y 2T sampled value of y t at t 2T, and so on. Then the corresponding answer to the above equation in the time domain is given by eq. (571); thus (now writing y instead of v ) ys t y 0 t y T t T y 2T t 2T y nT t nT 584

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