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the power supply circuitry includes U5 to provide a regulated and smoothed +12 V supply from an external supply such as a 14.5 18 V wall-wart. This is followed by 5V regulator U4 which has its output jacked up using 300 ohm and 47 ohm resistors to provide close to +6 V for the SA605D (U1). Note that if you want to run the receiver from a 12 V battery, this can be quite easily done by replacing REG1 with a wire link. In addition, the 2200 F capacitor should be replaced with a 16 V Zener diode (ZD2) for over-voltage protection.
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If you sit in a room with no radioactive materials present and switch on a radiation counter with the window of a tube closed, you ll notice an occasional click from the device. Some of the particles come from the Earth; there are radioactive elements in the ground almost everywhere (usually in small quantities). Some of the radiation comes indirectly from space. These
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Let s call the five points in set B of Fig. 20-1 by the names b1 through b5, and the five points in set C by the names c1 through c5. Suppose that b1 maps to c5, b2 maps to c4, b3 maps to c3, b4 maps to c2, and b5 maps to c1. How can we state this mapping as a set of ordered pairs
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Let us now consider the gates in the classical FFT circuit after the recursive calls to F F TM/2 : the wires pair up j with M/2 + j, and ignoring for now the phase that is applied to the contents of the (M/2 + j)th wire, we must add and subtract these two quantities to obtain the jth and the (M/2 + j)th outputs, respectively. How would a quantum circuit achieve the result of these M classical gates Simple: just perform the Hadamard gate on the rst qubit! Recall from the preceding discussion (Section 10.5.1) that for every possible con guration of the remaining m 1 qubits x, this pairs up the strings 0x and 1x. Translating from binary, this means we are pairing up x and M/2+x. Moreover the result of the Hadamard gate is that for each such pair, the amplitudes are replaced by the sum and difference (normalized by 1/ 2) , respectively. So far the QFT requires almost no gates at all! The phase that must be applied to the (M/2 + j)th wire for each j requires a little more work. Notice that the phase of j must be applied only if the rst qubit is 1. Now if j is j represented by the m 1 bits j1 . . . jm 1 , then j = m 1 2 l . Thus the phase j can be l=1 l applied by applying for the lth wire (for each l) a phase of 2 if the lth qubit is a 1 and the rst qubit is a 1. This task can be accomplished by another two-qubit quantum gate the conditional phase gate. It leaves the two qubits unchanged unless they are both 1, in which case it applies a speci ed phase factor. The QFT circuit is now speci ed. The number of quantum gates is given by the formula S(m) = S(m 1)+O(m), which works out to S(m) = O(m 2 ). The QFT on inputs of size M = 2m thus requires O(m2 ) = O(log2 M ) quantum operations. 302
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