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AD9856/PCB 数据表(PDF) 24 Page - Analog Devices |
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AD9856/PCB 数据表(HTML) 24 Page - Analog Devices |
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24 / 37 page ![]() AD9856 Rev. C | Page 23 of 36 –150 –120 –90 –60 –30 0 16 12 48 020 24 2 DISPLAYED FREQUENCY IS RELATIVE TO I/Q NYQ. BW 8 32 –4.0 –3.5 –3.0 –2.5 –2.0 –1.5 –1.0 –0.5 0 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 DISPLAYED FREQUENCY IS RELATIVE TO I/Q NYQ. BW Figure 39. CIC Filter Frequency Response (R = 2, HBF 3 Selected) Figure 40. Pass-Band Detail (R = 2, HBF 3 Selected) –150 –120 –90 –60 –30 0 0 144 288 432 576 720 864 1008 DISPLAYED FREQUENCY IS RELATIVE TO I/Q NYQ. BW –4.0 –3.5 –3.0 –2.5 –2.0 –1.5 –1.0 –0.5 0 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 DISPLAYED FREQUENCY IS RELATIVE TO I/Q NYQ. BW Figure 41. CIC Filter Frequency Response (R = 63, HBF 3 Active) Figure 42. Pass-Band Detail (R = 63, HBF 3 Active) DIGITAL QUADRATURE MODULATOR Following the CIC filter stage the I and Q data (which have been processed independently up to this point) are mixed in the modulator stage to produce a digital modulated carrier. The carrier frequency is selected by programming the direct digital synthesizer (see the Direct Digital Synthesizer Function section) with the appropriate 32-bit tuning word via the AD9856 control registers. The DDS simultaneously generates a digital (sampled) sine and cosine wave at the programmed carrier frequency. The digital sine and cosine data is multiplied by the Q and I data, respectively, to create the quadrature components of the original data upconverted to the carrier frequency. The quadrature components are digitally summed and passed on to the subsequent stages. The key point is that the modulation is done digitally, which eliminates the phase and gain imbalance and crosstalk issues typically associated with analog modulators. Note that the modulated signal is actually a number stream sampled at the rate of SYSCLK, which is the same rate at which the DAC is clocked (see Figure 23). Note that the architecture of the quadrature modulator results in a 3 dB loss of signal level. To visualize this, assume that both the I data and Q data are fixed at the maximum possible digital value, x. Then the output of the modulator, y, is: y = x × cos(ω) + x × sin(ω) = x × [cos(ω) + sin(ω)] From this equation, y assumes a maximum value of x√2 (a gain of 3 dB). However, if the same number of bits were used to represent the y values, as is used to represent the x values, an overflow would occur. To prevent this, an effective divide-by-two is implemented on the y values, which reduces the maximum value of y by a factor of two. Because division by two results in a 6 dB loss, the modulator yields an overall loss of 3 dB (3 dB − 6 dB = −3 dB, or 3 dB of loss). |
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