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1. |
What is the proper fraction and polynomial form of the improper rational transform |
A. | = (1+3z<sup>-1</sup>+11/6 z<sup>-2</sup>+1/3 z<sup>-3</sup>)/(1+5/6 z<sup>-1</sup>+1/6 z<sup>-2</sup> )? |
B. | 1+2z -1+(1/6 z<sup>-1</sup>)/(1+5/6 z<sup>-1</sup>+1/6 z<sup>-2</sup> ) |
C. | 1-2z -1+(1/6 z<sup>-1</sup>)/(1+5/6 z<sup>-1</sup>+1/6 z<sup>-2</sup> ) |
D. | 1+2z -1+(1/3 z<sup>-1</sup>)/(1+5/6 z<sup>-1</sup>+1/6 z<sup>-2</sup>) |
Answer» B. 1+2z -1+(1/6 z<sup>-1</sup>)/(1+5/6 z<sup>-1</sup>+1/6 z<sup>-2</sup> ) | |