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where we made use of the law of exponents, z z z ,z z z and so on. Now multiply both sides of eq. (A) by z 1 , as suggested. Doing this, eq. (A) becomes z 1 F z z 1 z 2 z 3 z 4 z n z2 z n z1 z n z n z 1 B
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Now take the algebraic sum of eqs. (A) and (B), as suggested. Upon doing this, note that on the right-hand side all of the terms except two will cancel out, leaving 1 z 1 F z 1 z n z 1 which upon solving for F z gives eq. (577), as explained in the text. 296. First, for this problem, v nT cos !nT. Then, by Euler s formulas, we have
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 j!nT cos !nT j sin !nT  j!nT cos !nt j sin !nT which, upon taking the algebraic sum of the two equations, shows that cos !nt 1  j!nT  j!nT v nT 2 and upon substituting this value of v nT into the basic eq. (573) we have F z 1 2 or, if we wish, F z 1 2
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n 1 X n 0 n 1 X n 0
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 j!nT z n  j!nT z n
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in which we made use of the fact that since (X ab X a b , then also (X ab X a b . Note, now, that eq. (580) applies to eq. (A), where b j! in the rst term and b j! in the second term. Thus eq. (581) applies to eq. (A), where k j!T for the rst term and k  j!T for the second term. Hence, by eq. (581), eq. (A) becomes ! z 1 1 F z 2 z  j!T z  j!T Now, inside the brackets, combine the two fractions together over the common denominator (the product of the two denominators), then apply Euler s formulas to the result. Carefully doing this, you should nd that the z-transform of v nT cos !nT is truly given by (7) in the table. 297.
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298. The rst two sequences, and their algebraic sum, are shown below.
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Therefore the addition of the delayed step sequence, 3U nT 7T , produces the nal answer shown to the right.
Solutions to Problems
299. IN SAMPLED form U t becomes U nT and t becomes nT. Thus we have that v nT 3U nT 20nT Hence, making use of (1) and (2) in Table 2, then (2) and (4) in Table 1, we have that Zv nT F z 3z 20Tz z 1 z 1 2
By eq. (91), Chap. 5, T 1=f 1=100 0:01. Making use of this fact, then combining the two fractions over a common denominator, should produce the answer given with the problem. 300. Applying (2) of Table 1 and (4) of Table 2, we have F z z z 1 z z 2 z z 1 z 1 = z 1 ; answer: z 1 z 1 z 1
301. Using (4) and (2) of Table 1 and (4) of Table 2, we have F z or, if you wish, F z 302. z 1 0:5z 2 0:25z 3 0:125z 4 0:0625z 5 z 0:5 1 1 0:5z 1
   
Tz z 1
Tz 2 z 1
z 6 ; answer; z 1
Tz z 1
T z2 z 1
z6 z
1 ; answer: 1
0:5z 1 0:5z 1 0:25z 2 0:25z 2 0:25z 2 0:125z 3 0:125z 3 0:125z 3 0:0625z 4 0:0625z 4 hence the answers are y 0 0:000 y 2T 0:500 y T 1:000 y 3T 0:250 y 4T 0:125 y 5T 0:0625
Solutions to Problems
303. For convenience, let s temporarily write a in place of 0.45. Thus we have z F z 2 , and our problem is to put F z into the form of eq. (583). To do this z a we can use algebraic long division; thus,
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