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Dsp implementation

As you begin to implement your PLL on the DSP, it is highly recommended that you implement and test your NCO block firstbefore completing the rest of your phase-locked loop.

Sine-table interpolation

Your NCO must be able to produce a sinusoid with continuously variable frequency. Computing values of θ n on the fly would require a prohibitive amount of computation and program complexity; a look-up table is a betteralternative.

Suppose a sine table stores N samples from one cycle of the waveform: k k 0 N 1 2 k N . Sine waves with discrete frequencies ω 2 N p are easily obtained by outputting every p th value in the table (and using circular addressing). The continuously variable frequency of yourNCO will require non-integer increments, however. This raises two issues: First, what sort of interpolation should be used to get the in-betweensine samples, and second, how to maintain a non-integer pointer into the sine table.

You may simplify the interpolation problem by using "lower-neighbor" interpolation, i.e., by using the integerpart of your pointer. Note that the full-precision, non-integer pointer must be maintained in memory so that thefractional part is allowed to accumulate and carry over into the integer part; otherwise, your phase will not be accurateover long periods. For a long enough sine table, this approximation will adjust the NCO frequency with sufficientprecision. Of course, nearest-neighbor interpolation could be implemented with a small amount ofextra code.

Maintaining a non-integer pointer is more difficult. In earlier exercises, you have used the auxiliary registers( ARx ) to manage pointers with integer increments. The auxiliary registers are not suited for thenon-integer pointers needed in this exercise, however, so another method is required. One possibility is to performaddition in the accumulator with a modified decimal point. For example, with N 256 , you need eight bits to represent the integer portion of your pointer. Interpret the low 16 bits of theaccumulator to have a decimal point seven bits up from the bottom; this leaves nine bits to store the integer partabove the decimal point. To increment the pointer by one step, add a 15-bit value to the low part of the accumulator,then zero the top bit to ensure that the value in the accumulator is greater than or equal to zero and less than256. How is this similar to the addition modulo 2 discussed in the MATLAB Simulation ? To use the integer part of this pointer, shift the accumulator contents seven bits to theright, add the starting address of the sine table, and store the low part into an ARx register. The auxiliary register now points to the correct sample in thesine table.

As an example, for a nominal carrier frequency ω 8 and sine table length N 256 , the nominal step size is an integer p 8 N 1 2 16 . Interpret the 16-bit pointer as having nine bits for the integer part, followed by a decimal point and sevenbits for the fractional part. The corresponding literal (integer) value added to the accumulator would be 16 2 7 2048 . If this value were 2049, what would be the output frequency of the NCO?


You may want to refer to Proakis and Blahut . These references may help you think about the following questions:

  • How does the noise affect the described carrier recovery method?
  • What should the phase-detector look like for a BPSK modulated carrier? (Hint: You would need to considerboth the in-phase and quadrature channels.)
  • How does α affect the bandwidth of the loop filter?
  • How do the loop gain and the bandwidth of the loop filter affect the PLL's ability to lock on to acarrier frequency mismatch?

Questions & Answers

can someone help me with some logarithmic and exponential equations.
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I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
I'm not sure why it wrote it the other way
I got X =-6
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
Commplementary angles
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or infinite solutions?
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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Differences Between Laspeyres and Paasche Indices
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In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
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Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
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Source:  OpenStax, Dsp laboratory with ti tms320c54x. OpenStax CNX. Jan 22, 2004 Download for free at http://cnx.org/content/col10078/1.2
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