March 7, 2017

Troubleshooting your Oscilloscope -Tektronix by Robert L. Goodman

By Robert L. Goodman

Explains the right way to arrange and use the technician's and complex hobbyist's most useful troubleshooting device - the triggered-sweep oscilloscope. The e-book describes the newest triggered-sweep oscilloscopes and new troubleshooting strategies for quick waveform research.

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Extra info for Troubleshooting your Oscilloscope -Tektronix

Sample text

These are circularly displaced to the left by 3 bits for each PPi. In the case  of radix-8 Booth encoding, the ith partial product can be seen to be PPi ¼ 23i d i X2n À1 , This is   modified to include the bias B as PPi ¼ 23i ðB þ di XÞ2n À1 . 4f. Hence, the correction word will be one n-bit word if k is chosen to be prime to n. e. the delay of the k-bit ripple carry adder). Here, the partial product accumulation by CSA tree has a time complexity of O(log (n + n/k)). The delay of the hard multiple generation, CC generation, partial product generation by BS and BE blocks, and two operand parallel-prefix modulo (2n À 1) adder are respectively O(k), O(1),O(1), O(logn).

Considering multiplication of X ¼ 12 and Y ¼ 17, and base 2, for modulus m ¼ 19, the indices corresponding to X and Y can be seen to be 15 and 10 which in residue form are (3, 1) and (4, 3) corresponding to (m1, m2). Adding the indices corresponding to the product XY, we obtain the indices as (1, 4). Using CRT (which will be introduced later in Chapter 5), the decoded word corresponding to moduli {6, 7} can be shown to be 25 which mod 18 is 7. Thus, the final result can be obtained as 14 since 27 mod 19 ¼ 14.

The first stage is a n-bit by n-bit multiplier whose output 2n-bit word is reduced mod m using a second stage. Hiasat [9] has suggested a modulo multiplier which uses a conventional multiplier to obtain first z ¼ xy followed by a modulo reduction block. In this method, defining a ¼ 2n À m and considering that k bits are used to represent a, the 2n-bit product can be written as Z ¼ D22nÀkÀ1 þ C2n þ B2nÀ1 þ A ð4:2Þ where A, B, C, and D are n À 1, 1, n À 1 À k and k + 1 bit words. It may be noted that (C2n) mod m ¼ Ca.

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