# A Direct Digital Synthesizer with Arbitrary Modulus

Suppose you have a system with a 10 MHz sample clock, and you want to generate a sampled sinewave at any frequency below 5 MHz on 500 kHz spacing; i.e., 0.5, 1.0, 1.5, … MHz. In other words, f = k*f_{s}/20, where k is an integer and f_{s} is sample frequency. This article shows how to do this using a simple Direct Digital Synthesizer (DDS) with a look-up table that is at most 20 entries long. We’ll also demonstrate a Quadrature-output DDS. A note on terminology: some authors call a DDS a Numerically-Controlled Oscillator (NCO).

Disclaimer: I have not implemented this DDS in hardware, so there could be problems with the scheme that I have not anticipated.

This article is available in PDF format for easy printing.

## Background [1,2]

A continuous-time sinewave with frequency f_{0} is given by y = sin(2πf_{0}t + φ_{0}). For a sampled signal, we replace t by nT_{s}, where n is the sample number and T_{s} is the sample time. Letting φ_{0} = 0, we have:

y = sin(2πf_{0}nT_{s})

The phase of the signal is:

Φ = 2πf_{0}nT_{s} rad (mod 2π),

Or

Φ = f_{0}nT_{s} cycles (mod 1) (1)

The phase wraps every 2π radians = 1 cycle. Equation 1 shows that the phase increases (accumulates) by f_{0}T_{s} every sample. So we can calculate the phase using an accumulator with input = f_{0}T_{s}, as shown in Figure 1a. The value of φ has a range of 0 to 1 (cycles). We generate the sinewave from the phase using a look-up table (LUT). What we’ve just described is a basic DDS. Note that another option to generate the sinewave from the phase not discussed here is the CORDiC algorithm [3].

Figure 1b adds quantization in the accumulator register, the phase, and the LUT entries. The accumulator input has 2^{C} steps over a range of 0 to 1, giving a frequency step Δf = f_{s}/2^{C}, where f_{s} is the sample frequency. The resulting output frequencies are f_{s}/2^{C}, 2f_{s}/2^{C}, 3f_{s}/2^{C} … Given the 2^{C} steps, we can say the DDS has a modulus of 2^{C}. As an example, if C= 24 bits, and f_{s}= 10 MHz, the frequency step is:

Δf = 10E6/2^{24} = 0.59605 Hz.

This frequency step is impressively small. However, if you want to program a frequency that is not on one of the steps, such as f_{s}/10, there will be a small frequency error of up to Δf/2.

If we were to maintain the 24 bits of phase, the LUT size for this example, taking symmetry of the sine into account, would be ¼*2^{24} = 2^{22} = 4,194,304 entries. To avoid such a large LUT, the phase is normally quantized to P < C bits. The phase quantization results in so-called phase truncation spurs in the output spectrum. A typical value of P used in DDS chips is 15 bits, which, taking advantage of the symmetry of the sine, gives LUT size of 2^{13}= 8192 entries.

You can see that a standard DDS is not a perfect solution to our problem of generating f_{0} = k*f_{s}/20: it does not produce the exact frequency; it requires a not-so-small LUT; and it has spurs due to truncation of the phase.(Note that there are techniques for reducing phase-truncation spurs [4]).

Figure 1. a) Implementation of Equation 1. b) DDS with quantization.

## DDS with Arbitrary Modulus

A DDS with modulus other than 2^{C }can address the shortcomings of a conventional DDS for our application.

If we multiply both sides of Equation 1 by an integer L, we get:

LΦ = Lf_{0}nT_{s }(mod L)

This equation can be implemented by modifying the accumulator in Figure 1a as shown in Figure 2. Here we require m to be an integer between 0 and L-1, so there are L entries in the LUT, where L is not restricted to 2^{C}. The input L*f_{0}/fs is an integer:

L*f_{0}/fs = k (2)

or f_{0} = k*fs/L (3)

Since k is an integer, f_{0} has a step size of Δf = fs/L. For a given Δf and fs, we have:

L = fs/Δf (4)

Letting f_{s} = 10 MHz and Δf= 0.5 MHz, we get L= 20. The number of bits required for the accumulator is found by taking log_{2}(L) and rounding up to the next integer. For L= 20, we need 5 bits.

As shown in Figure 2, m = Lφ, so the phase is φ = m/L. Simplistically, the LUT entries are:

u(m) = sin(2πm/L), m= 0: L-1 (5)

However, for fixed point entries, we need to round the values of u(m) and prevent overflow when m = L/4 and u(L/4) = sin(π/2) = 1.0. (For example, if the number of bits D= 8, the largest allowable entry is not 1.0 but (2^{7} -1)/2^{7} = 127/128 = 01111111). We can compute the fixed-point entries as:

u(m) = (1 – ε) * sin(2πm/L), m= 0: L-1

LUT(m) = round(u(m)*2^{D-1})/2^{D-1 }(6),

Where D is the number of bits in the 2’s complement LUT entry and ε << 1. I used ε= 1/2^{D-2}. Multiplication by 1 – ε makes the LUT entry for m = L/4 less than 1.0 after rounding.

For our case, with L= 20, the LUT values are plotted in figure 3. The LUT contains one cycle of a sinewave evaluated over L samples. Note that when L is a multiple of 4, it is possible to reduce the LUT size to L/4 entries by taking the symmetry of the sinewave into account.

Figure 2. DDS with arbitrary modulus

Figure 3. Sine look-up table for L= 20

Let’s look at the behavior of our example DDS, with f_{s} = 10 Hz and Δf = 0.5 Hz. The Matlab code is listed in the Appendix. To start out, let the output frequency f_{0}= 0.5 Hz. From equations 2 and 4, k = f_{0}/Δf, so k= 1. As shown in Figure 4, m increments through all the integers from 0 to L-1, then repeats. So the DDS just steps through every entry of the LUT. Also shown in Figure 4 is the phase φ = m/L cycles, and the sampled sinewave output.

Now, if we let f_{0} = 1 Hz, k = 2. Thus m = 0, 2, 4, … and the DDS steps through every 2^{nd} entry of the LUT, as shown in Figures 5a and 5b.

If we let f_{0} = 1. 5 Hz, k= 3. Thus m= 0, 3, 6, … and the DDS steps through every 3^{rd} entry of the LUT, as shown in Figures 5c and 5d. As can be seen in Figure 5c, it takes three cycles for the phase sequence to repeat.

For L= 20, the allowable output frequencies f_{0} that are less than f_{s}/2 are: 0.5, 1, 1.5, 2, 2.5, 3, 3.5, 4, and 4.5 Hz, corresponding to k = 1: 9. For L even, there are L/2 -1 allowable values of f_{0}.

Since accumulator output m is always an integer, there is no phase truncation error. The only error in the output y is due to rounding of the LUT entries. Figure 6 compares spectra for f_{0} = 1.5 Hz of a conventional DDS with 15-bits of phase to our DDS with L= 20 (4.3 bits of phase). Both have 16-bit LUT entries. The modulus 20 DDS has lower spurious, with the worst spur at about -105 dB with respect to the level at 1.5 Hz.

Finally, note that it is also possible to make a DDS with an arbitrary *programmable* modulus.The approach involves using two accumulators [5,6].

Figure 4. DDS with L= 20 and f_{s} = 10 Hz.

a) Accumulator output m for f_{0 }= 0.5 Hz. b) Phase in cycles. c) LUT output y.

Figure 5. DDS with L= 20 and f_{s} = 10 Hz.

a) Accumulator output m for f_{0} = 1.0 Hz, and

b) LUT output y

c) Accumulator output m for f_{0} = 1.5 Hz, and

d) LUT output y

Figure 6. Spectra of conventional DDS and DDS with modulus 20. f_{0} = 1.5 Hz and f_{s} = 10 Hz.

Left: Conventional DDS with 15 bits of phase and 16-bit LUT entries.

Right: DDS with L= 20 (4.3 bits of phase) and 16-bit LUT entries.

## Quadrature Output DDS

A quadrature output DDS has both cosine and -sine outputs. The cosine phase leads sine phase by π/2 radians = ¼ cycle. Given m as the LUT address for a sine, the address for the cosine is:

p = m + L/4 mod(L)

where L is the DDS modulus = LUT length, which must be a multiple of 4. We can modify the Matlab code in the Appendix to compute both sine and cosine. Here is the modified *for* loop:

sine(1)= 0; cosine(1)= 1; m= 0; for n= 2:N r = k + m; m= mod(r,L); % LUT address/ sine p= mod(m+ L/4,L); % LUT address/ cosine sine(n)= lut(m+1); % sine output cosine(n)= lut(p+1); % cosine output end

The Quadrature DDS outputs for L= 20, f_{s}= 10 Hz, and f_{0} = 1 Hz are shown in Figure 7.

Figure 7. Quadrature DDS with L= 20, f_{s} = 10 Hz, and f_{0} = 1 Hz.

a. cosine address p. b) cosine output. c) sine address m. d) -sine output.

## Simplest DDS with L= 4

If we let L= 4, there is only one output frequency below f_{s}/2:

f_{0} = k*f_{s}/L = f_{s}/4 (k= 1)

The LUT sine values from Equation 5 are:

LUT = [0 sin(π/2) 0 sin(3π/2)]

= [0 1 0 -1]

The cosine values are [1 0 -1 0].

A quadrature L= 4 DDS using cosine and -sine can be used to down-convert a signal centered at f_{s}/4 to complex baseband [7,8]. Since all LUT values are 0 or +/-1, no multiplier is needed to perform the frequency conversion.

## References

- MT-085, “Fundamentals of Direct Digital Synthesis (DDS)”, Analog Devices, 2009, https://www.analog.com/media/en/training-seminars/tutorials/MT-085.pdf
- “A Technical Tutorial on Digital Signal Synthesis”, Analog Devices, 1999, https://www.analog.com/media/cn/training-seminars/tutorials/450968421DDS_Tutorial_rev12-2-99.pdf
- Rice, Michael,
__Digital Communications, A Discrete-Time Approach__, Pearson, 2009, section 9.4. - Rice, section 9.2.
- Gentile, Ken, AN-953, “Direct Digital Synthesis with a Programmable Modulus”, Analog Devices, 2014, https://www.analog.com/media/en/technical-documentation/application-notes/AN-953.pdf
- Hou, Yuqing, et. al., “An Accurate DDS Method Using Compound Frequency Tuning Word and Its FPGA Implementation”, Electronics, Nov, 2018, https://www.mdpi.com/2079-9292/7/11/330
- Harris, Fredric J.,
__Multirate Signal Processing__, Prentice-Hall PTR, 2004, section 13.2.1. - Lyons, Richard G.,
__Understanding Digital Signal Processing__, 3^{rd}Ed., Prentice-Hall, 2011, section 13.1.2.

Neil Robertson June 3, 2019. Revised 6/8/19

## Appendix Matlab Code for DDS with Modulus = 20

% dds_mod20.m 5/30/19 Neil Robertson % DDS with modulus L = 20 % output frequency f0 = k*fs/L % Plot LUT, phase, and output fs= 10; % Hz sample freq df= 0.5; % Hz desired freq step L= fs/df % length of LUT= modulus of accumulator if mod(L,1)~=0 error('fs/fstep must be an integer') end % create LUT with one full cycle of sinewave (not using symmetry) D= 16; % bits LUT entries quantization m= 0:L-1; phi_lut= m/L; % cycles phase epsilon= 1/2^(D-2); u= (1 - epsilon) *sin(2*pi*phi_lut); lut= round(u*2^(D-1))/(2^(D-1)); % quantize lut entries % % DDS N= 30; % number of output samples f0= 0.5; % Hz output frequency (must be multiple of df) k= L*f0/fs; % integer input to DDS y(1)= 0; m= 0; for n= 2:N r = k + m; m= mod(r,L); % LUT address y(n)= lut(m+1); % output phi(n)= m/L; % cycles phase end % % % Plotting % % plot LUT stem(0:L-1,lut),grid axis([0 32 -1 1]) xlabel('m'),ylabel('lut'),figure % %plot m and phi subplot(311),plot(0:N-1,phi*L,'.-','markersize',9),grid axis([0 N 0 20]) xlabel('n'),ylabel('m') subplot(312),plot(0:N-1,phi,'.-','markersize',9),grid axis([0 N 0 1]) xlabel('n'),ylabel('phi (cycles) = m/L') % % plot y along with "continuous" sinewave y2 in grey fs_plot= fs*16; % fs of "continuous" sine Ts= 1/fs_plot; Len= 16*N; i= 0:Len-1; y2= sin(2*pi*f0*i*Ts); % "continuous" sine subplot(313),plot(0:N-1,y,'.','markersize',9),grid hold on plot(i/16,y2,'color',[.5 .5 .5]) axis([0 N -1 1]) xlabel('n'),ylabel('y')

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- Comments
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Hi Neil,

Thanks for the useful subject. I have only few things to add.

1) DDS term was coined for the physical NCO chip plus ADC (actually DAC, as Neil pointed below) and marketed as such. The NCO is more generic term to indicate actual cos/sin function generator at digital level.

2) a dedicated NCO that targets specific frequencies to centre on the target bins is a preferred requirement in some applications over generic NCO (based on modulo 2) for two reasons;a nondedicated NCO can suffer bin shift and phase drift though frequency tends to be accurate over long time window. The phase drift occurs because of rounding of tuning word calculation right at start. The accumulator itself is not truncated and wraps around if modulo 2.

3) The use of LUT here is just another case of using precomputation approach which can be applied to any equation (fully or partially precomputed). The address then represents input to precomputation table.

If lookup table can't cover enough resolution then intermediary values can be derived on the fly by interpolation leading to larger virtual LUT.

The cordic is direct computation of each value of cos/sin at given point of phase.

4) I am not familiar with your matlab code for cos/sin lut generation. I just write this for a full cycle table:

lut = round(2^15*exp(j*2*pi*(0:N-1)/N));

Then I model NCO emulating the accumulator plus addressing.

Some engineers prefer their LUT calculation to target centre of step rather than edges.

5) A generic NCO with small LUT compared to accumulator bitwidth and no intermediate computations may be ok for some applications if phase accuracy is more of interest such as PLLs.

NCO ip vendors specify three elements of resolution:

LUT resolution (size)

phase resolution (accumulator width)

amplitude resolution (bit width)

Regards

Kaz

Thanks Kaz,

Did you mean to say DAC rather than ADC in item 1?

yes DAC indeed

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