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Hi,
I've got a question.I use WARPLab Reference Design. The speed of DAC is 40Mhz. If I set rx_lpf_corn_freq = 3, then what will happen to the signal between [36MHz 40MHz] ? If I want this signal, what can I do?
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The rx_lpf_corn_freq command sets the MAX2829 baseband filter 3dB corner frequency. Refer to the MAX2829 datasheet (pg. 23, plot titled "RX BB FREQUENCY RESPONSE vs. COARSE SETTING") for the actual amplitude vs frequency curves for each filter setting.
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murphpo wrote:
The rx_lpf_corn_freq command sets the MAX2829 baseband filter 3dB corner frequency. Refer to the MAX2829 datasheet (pg. 23, plot titled "RX BB FREQUENCY RESPONSE vs. COARSE SETTING") for the actual amplitude vs frequency curves for each filter setting.
So even if we have 40MHz DAC, we still can't use all the 40MHz bandwidth.
However, when I test will sinusoïd signal. The error occur when frequency near 20MHz. I am not sure if the reason is I set rx_lpf_corn_freq = 3 and the approximate core frequenct 18MHz
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Please see this thread for a related discussion.
The MAX2829 baseband interface is a complex baseband signal. If the ADCs/DACs sample the analog baseband signal at 40MSps, the useable bandwidth is 40MHz, spanning -20MHz to +20MHz. This is consistent with the Nyquist requirement that the highest-frequency component be less than half the sampling rate. The MAX2829 Tx baseband path includes a low-pass filter, which provides the reconstruction filter for the DAC output. This is the low pass filter whose response I mentioned above, and whose response Chris plotted in the other thread I linked to.
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murphpo wrote:
Please see this thread for a related discussion.
The MAX2829 baseband interface is a complex baseband signal. If the ADCs/DACs sample the analog baseband signal at 40MSps, the useable bandwidth is 40MHz, spanning -20MHz to +20MHz. This is consistent with the Nyquist requirement that the highest-frequency component be less than half the sampling rate. The MAX2829 Tx baseband path includes a low-pass filter, which provides the reconstruction filter for the DAC output. This is the low pass filter whose response I mentioned above, and whose response Chris plotted in the other thread I linked to.
You said 'the useable bandwidth is 40MHz, spanning -20MHz to +20MHz'. So if I use channel 3 which central frequency is 2422MHz, my signal wil spaning over 2402~2442MHz?
Suppose I have a 25MHz sinsiod signal in baseband, where should I see it in passband? 2427MHZ or 2407MHz or somewhere else?
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So if I use channel 3 which central frequency is 2422MHz, my signal wil spaning over 2402~2442MHz?
Correct.
Suppose I have a 25MHz sinsiod signal in baseband, where should I see it in passband? 2427MHZ or 2407MHz or somewhere else?
A 40MSps signal can represent sinusoids only up to 20MHz.
If you have 5MHz sinusoid Tx baseband signal and set the MAX2829 to channel 3 (2422MHz), the transmitted RF signal will have a dominant sinusoid at 2427MHz (2422 + 5).
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murphpo wrote:
So if I use channel 3 which central frequency is 2422MHz, my signal wil spaning over 2402~2442MHz?
Correct.
Suppose I have a 25MHz sinsiod signal in baseband, where should I see it in passband? 2427MHZ or 2407MHz or somewhere else?
A 40MSps signal can represent sinusoids only up to 20MHz.
If you have 5MHz sinusoid Tx baseband signal and set the MAX2829 to channel 3 (2422MHz), the transmitted RF signal will have a dominant sinusoid at 2427MHz (2422 + 5).
The original code in example is 'payload = .6*exp(t*j*2*pi*5e6);' what if I use 'payload = .6*exp(t*j*2*pi*25e6);' instead? where should I see the peak?
another question is that how can I use the negative frequency domain?
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As murphpo said, the highest frequency sinusoid that can be represented at 40MSps is only 20MHz. A 25MHz sinusoid is indistinguishable from a -15MHz sinusoid at this sampling rate. In your example, "payload = .6*exp(t*j*2*pi*25e6)" is equivalent to "payload = .6*exp(t*j*2*pi*-15e6)" when the elements in the "t" vector are spaced by 25ns. You can verify this in MATLAB. This is the Nyquist Sampling Theorem.
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