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Table 7.3 Iteration 0 1 2 3 4 5 6 7 Total cost
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Use Case 1 Use Case 2 Use Case 3 Use Case 4 Use Case 5
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2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Gain mean 12.00 dB 1.50 dB 10.00 dB 8.00 dB 7.00 dB 0.61 dB 30.00 dB Gain 1.00 dB 0.25 dB 2.00 dB 0.59 dB 2.00 dB 1.82 dB 2.00 dB Gain s 0.50 dB 0.17 dB 1.00 dB 0.41 dB 0.80 dB 1.27 dB 1.30 dB mean NF gain, cum NF, cum at mean gain at filter at filter 2.00 dB 12 2 1.54 dB 0 0 4.00 dB 0 0 8.06 dB 0 0 3.00 dB 0 0 0.93 dB 0 0 14.45 dB 0 0 mean 12.00 dB 10.50 dB 20.50 dB 12.50 dB 19.50 dB 18.89 dB 48.89 dB dB dB dB dB dB dB dB 2.0 dB DERIVED Gain Gain max min 13.00 dB 11.00 dB 1.25 dB 1.74 dB 12.00 dB 8.00 dB 7.41 dB 8.59 dB 9.00 dB 5.00 dB 1.21 dB 2.43 dB 32.00 dB 28.00 dB CUMULATIVE Gain max min 13.00 dB 11.00 dB 11.75 dB 9.26 dB 23.75 dB 17.26 dB 16.34 dB 8.67 dB 25.34 dB 13.67 dB 26.55 dB 11.24 dB 58.55 dB 39.24 dB 1.00 1.25 3.25 3.84 5.84 7.66 9.66 s 0.50 0.53 1.13 1.20 1.45 1.93 2.32 dB dB dB dB dB dB dB NF using mean NFs at mean G 2.00 dB 2.07 dB 2.42 dB 2.54 dB 2.67 dB 2.68 dB 3.42 dB
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Example 8.9 LO Contaminant Equivalent Sideband Leaking into IF Given: Hard LO limiting, 23-dB LO-to-IF balance, and two contaminants as shown in Fig. 8.19. The 87-MHz contaminant is decomposed into sidebands at (87 50 =) 37 MHz offset from the 50-MHz LO. The lower of these sidebands is at (50 37 =) 13 MHz in the IF passband. The level of the contaminant is 50 dBm so the equivalent FM sidebands are at 56 dBm. This is attenuated 23 dB by the LO-to-IF balance, causing it to arrive at ( 56 dBm 23 dB) = 79 dBm in the IF. The 118-MHz contaminant is decomposed into sidebands at (118 50 =) 68 MHz on the 50-MHz LO. The lower sideband is at (50 68 =) 18 MHz. The absolute value is 18 MHz and a sinusoid at this frequency is produced in the IF. Since the LO contaminant is 10 dB larger than the rst one considered, the level in the IF at 18 MHz is ( 79 dBm + 10 dB =) 69 dBm. The negative frequency only affects the phase of the signal. If we used negative frequencies, as in a proper Fourier analysis, we would see that there is a +18-MHz component produced by the negative frequencies corresponding to the positive frequencies shown in Fig. 8.19. Together with the 18-MHz component produced by the positive frequencies, these two impulses at 18 MHz represent an 18-MHz sinusoid.
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For some cases, the transmission matrix method could be advantageous for analyzing multiple layers. This is primarily due to the fact that the overall transmission matrix of two consecutive layers can be obtained by simple multiplication of the individual transmission matrices. Unlike the GSM cascading process, no matrix inversion is required here. This can be proven by direct substitution. However, in many situations, the [T ] matrix formulation suffers from a numerical instability problem that will be explained shortly. Let us represent the [T ] matrix of a layer in terms of its GSM. From (6.1) we can write a = S11 a+ + S12 a+ 1 1 2 a = S21 a+ + S22 a+ 2 1 2 From (6.19) we write a+ = T21 a+ + T22 a 2 1 1 (6.22) (6.20) (6.21)
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PREDICTIVE COSTING, PREDICTIVE ACCOUNTING, AND BUDGETING
MATRIX OPERATIONS USING WIZARD
CHARGING FLEXIBLE CONTRACTS
d Ph 1 Ph 2 Ph 3 Neutral
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