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This is a rst answer which, upon multiplying the numerator and denominator of the right-hand fraction by z 4 , is thereby put in the equivalent form of eq. (606); thus H z z 1 ; answer: 1 1:4z 1 0:71z 2 0:154z 3 0:012z 4
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314. This is a parallel connection of processors, and thus (section 13.8) H z H1 H2 H3 z z z z 0:2 z 0:4 z2 0:8z 0:15
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which, upon combining the three fractions into a single fraction, gives the equivalent answers H z z4 2z4 1:2z3 0:53z2 0:31z 1:4z3 0:71z2 0:154z 0:012
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2 1:2z 1 0:53z 2 0:31z 3 1 1:4z 1 0:71z 2 0:154z 3 0:012z 4
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The second answer is in the form of eq. (606), and is found by multiplying the numerator and denominator of the rst answer by z 4 . 315. As explained in section 13.6, for unit-pulse input the output Y z of a processor is numerically equal to the transfer function H z . Thus, in this particular case for unitpulse input voltage, we can use either of the equivalent answers found in problem 313. The problem now is to put Y z into the form of eq. (583) in section 13.4. To do this we can apply algebraic long division to either of the equivalent answers found in problem 313. If we choose to use the rst answer, the details of the long division are as follows.
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   z4 1:4z3 0:71z2 0:154z 0:012 
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z 1 1:4z 2 1:25z 3 z3 z3 1:4z2 0:71z 0:154 0:012z 1 1:4z2 0:71z 0:154 0:012z 1 1:4z2 1:96z 0:994 0:2156z 1 0:0168z 2
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1:25z 0:840 0:2036z 1 0:0168z 2 This is as far as we need to go to nd the required answer. If, however, you wish to continue the above division for a couple of more terms, you ll nd that Y z 0 z 1 1:4z 2 1:25z 3 0:91z 4 0:5901z 5 which is the output of the network, in the z-domain, for unit pulse input. By comparison with eq. (583) we see that, y 0 0:000 y 2T 1:400 y T 1:000 y 3T 1:250 thus y 3T 1:25 V; answer: y 4T 0:910 y 5T 0:5901
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316. The notation in Fig. 356 is tied to that of eq. (606). Let us therefore put the given H z into the equivalent form of eq. (606) by multiplying the numerator and denominator of H z by z 3 , thus getting H z 2 1:3z 1 0:9z 2 1 2:2z 1 1:5z 2 0:75z 3
Note that H z now has the form of eq. (606), and thus comparison with eq. (606) shows that b0 2 b1 1:3 b2 0:9 a1 2:2 a2 1:5 a3 0:75
and hence, from comparison with Fig. 356, the required block diagram is as follows:
317. Comparing the given network with Fig. 356 shows that a1 1:66 b0 1:9 b2 1:6
a2 1:5353 and hence, by eq. (606), we have that H z
b1 2:2
1:9 2:2z 1 1:6z 2 1:9z2 2:2z 1:6 2 1 1:66z 1 1:5353z 2 z 1:66z 1:5353
Now setting the denominator equal to zero, then making use of the formula for the roots of a quadratic function, you should nd that z hence q jzj 0:83 2 0:92 2 1:2391* which means that the points 0:83 j 0:92 lie outside the unit circle; thus the given processor is unstable, answer.
* The radius of a circle with center at the origin of the complex plane is r jzj p x2 y2 .
1:66 j1:84 0:83 j 0:92; 0:83 j 0:92; 2
first root; second root
a1 0:54
Solutions to Problems
318. From comparison of the given network with Fig. 356 b0 0:92 b1 0:82 a2 0:7453 Hence, by eq. (606), H z 0:92 0:82z 1 0:92z2 0:82z 2 1 0:54z 1 0:7453z 2 z 0:54z 0:7453
To nd the poles of H z , set the denominator equal to zero and solve for z. Doing this, making use of the quadratic formula, gives two poles, 0:27 j 0:82, both of magnitude 0.8633. Thus all poles lie inside the unit circle, so the processor is stable, answer. 319. First, by eq. (608) H z 1 z 1 then, H r 1  j2pr 1 cos 2pr j sin 2pr that is, H r 1 cos 2pr j sin 2pr or, if we wish to work in degrees instead of radians, H r 1 cos 360r j sin 360r Hence, remembering that sin2 x cos2 x 1, we have that p jH r j 2 1 cos 360r and  
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