# Digital to analog aliasing or mirror query... DAC can output negative frequency?

Using a clock Fs = 1 MHz.

The Digital to Analog (DAC) can make frequencies upto 500 KHz.

500 kHz to 1 MHz is an alias? Or is it called a mirror?? Is aliasing a concept only on the ADC and sampling process.

Does -500 to 0 exist as an alias or mirror too... But surely a DAC can't make negative frequencies? Texas Instruments tutorial on DAC has the attached screen shot showing a DAC outputting a negative frequency.

Is anything different for a DAC with complex samples such as on an ADC complex sampling allows frequencies upto Fs instead of Fs/2...!

Block diagram in this link

## 3 Answers

I think you are confused by negative frequencies and what they mean so let me add this explanation.

When you see a spectrum that contains "positive" and "negative" frequencies, each of the frequencies are of the form:

$$e^{j\omega t}$$

Where $$\omega$$ is the frequency (in this case angular frequency as $$2\pi f$$ with f being the frequency in Hz.

The general form of $$e^{j\phi}$$ is a phasor on the complex plane with magnitude one and angle $$\phi$$, so $$e^{j\omega t}$$ is simply a phasor with unit magnitude rotating with rate $$\omega$$. If $$\omega$$ is positive, the rotation is counter-clockwise: a positive frequency, and if the rotation is negative, the rotation is clockwise and represents a negative frequency. So the concept of positive and negative frequencies only applies to frequencies of the form of the complex exponential and is the direction of rotation.

Consider Euler's identity for a cosine that demonstrates the concept of negative frequencies well:

$$cos(\omega t) = \frac{1}{2}e^{j\omega t} + \frac{1}{2}e^{-j\omega t}$$

Here we get two phasors on the complex plane, one rotating clockwise, one counter-clockwise, and adding the two results in a signal that always stays on the real axis going back and forth sinusoidally.

So therefore a simple cosine wave with a constant frequency would contain one positive (exponential) frequency and one negative (exponential) frequency. We also see from this that ANY real signal must have positive and negative frequencies, AND the negative frequencies MUST be of the opposite phase. Like the cosine example above, each phasor component spinning in a counter-clockwise direction (the positive frequencies) MUST have an equal magnitude phasor spinning in the opposite direction (negative frequency component) in order for the sum to stay on the real axis (which would then be a real signal with no imaginary components. This is said to be "complex conjugate symmetric" when the negative frequency components have equal magnitude and opposite phase compared to the positive frequency components.

In order to IMPLEMENT a signal that does not have a spectrum with this property (positive and negative frequencies not complex conjugate symmetric), you would need to have TWO values in implementation, one to represent the real number and one to represent the imaginary number (or you could do magnitude and phase, either way you need two signals). You see this in the baseband part of your diagram where the real is represented as "I" for "in-phase" and the imaginary is represented as "Q" for quadrature. So in that part of your diagram we are indeed representing a complex signal, and the spectrum at that point can be asymmetric.

I notated your diagram to demonstrate this further. I will call the IQ signal the "baseband signal" and because it is indeed complex, it's spectrum can be assymetric as I have drawn. This complex baseband signal is then digitally upconverted (frequency translated) by multiplying it by $$e^{-j\omega t}$$ since $$e^{-j\omega t} = cos(\omega t)-jsin(\omega t)$$ and then taking the real part:

$$(I_1+jQ_1)(I_2-jQ_2)$$ $$= (I_1I_2+Q_1Q_2)+j(I_1Q_2-I_2Q_1)$$

Taking the real part of this results in the form in your block diagram prior to the DAC:

$$(I_1I_2+Q_1Q_2)$$

Given as:

$$I cos(2\pi f_{IF}t) + Q sin(2\pi f_{IF}t)$$

With $$I_1$$ and $$Q_1$$ as the I and Q inputs to the digital multiplier and $$I_2$$ is the cosine waveform into the upper digital multiplier and $$Q_2$$ is the sine waveform input into the lower digital multiplier in the diagram)

So the signal where we maintain two datapaths (truly a complex signal) will have an asymmetric spectrum while the signal where this is only one datapath must be real and therefore must have a spectrum that is complex conjugate symmetric. I like this diagram as shown as it clarifies that the first signal I circled is indeed complex and has an asymmetric spectrum, while the second one I circled, while it looks like the spectrum is asymmetric if you just considered the positive frequencies alone, is not as we are referring to symmetry between the positive and negative frequency axis.

So we see how negative frequencies exist throughout the entire block diagram, and it is not that the DAC is creating them. The baseband I and Q signal coping out of the encoder is complex and has an assymetric spectrum, this gets digitally upconverted (and is still complex until summed), the output of the summer is now one signal path so must be real and therefore now has a symmetric (complex conjugate) spectrum. The DAC will convert this to the analog domain and the bandpass filter will filter out the images of the DAC (positive and negative--it is a real filter so it too is complex conjugate symmetric in its passband). So assuming the BPF is centered on $$f_{IF}$$, the analog spectrum would look identical to the spectrum prior to the DAC, extend to $$\pm \infty$$ while the unique DAC input spectrum extends only to $$\pm f_s/2$$.

The "image" shown in the diagram are NOT the images due to digital sampling, but the image due to the analog upconversion. This is a real multiplier and as such its output will contain the sum and the difference of the input (review the expansion of $$cos(\alpha)cos(\beta)$$ to see this), so we get the terms RF+IF and RF-IF at both postiive and negative frequency locations. (However you can also see that we are just frequency translating the same spectrum at the output of the BPF by convolving it with the two frequencies of the RF sinusoid.)

There is more not detailed about the DAC upconversion that I didn't cover but you should be aware of including the option to pass a higher Nyquist zone by setting the bandpass filter on one of the DAC images, and that the DAC itself will introduce a Sinc roll--off in frequency due to the staircase reconstruction.

• Hi Dan, I do understand what you are saying (I hope), but my question still is this... the DAC is not creating a negative frequency, how can it? Its a sinusoidal voltage level. So your black coloured spectrums (after the DAC) in the top middle under the first image is okay to show like this if using mathematical terms, but in physical reality terms, there is a one side spectrum of physical frequencies which make up that signal right? So if I put it on an oscilloscope, its one sided... Dec 3, 2019 at 9:56
• and that upconverted image is not actually physically there because the DAC cannot generate a negative frequency, so the -F upconverted in figure 21 is not there in physical terms but in mathematical terms it is. So I don't actually have to physically filter out that 'upconverted image' of the 'negative frequency' because its after the DAC and the DAC physically makes real voltage signals only Dec 3, 2019 at 10:15
• That image is from the mixer: a mixer will put out the sun and difference of the two real frequencies at the input— so you have a real IF and a real LO and the output of it is the sum and the difference. The DAC is not creating the negative frequency: if you notice it exists at the DAC input as well. Do not get caught up with what actually exists versus what is “mathematical”: real and imaginary numbers are both just as mathematical ways in which we describe the physical world. A real cosine has both positive and negative frequencies as I described in detail. Dec 3, 2019 at 12:03
• But you can also see from the diagram that the mixer is moving the frequencies centered about 0 at the mixer input to the LO frequency. Bottom line: the reason negative frequencies are shown is because in that graph each impulse in the frequency domain represents and $\e^{j\omega t}$ NOT $cos(\omega t}$!! Dec 3, 2019 at 12:07
• Completely lost with what you are saying. I understand how upconversion works with sum and difference, but the converter usually has its own filter that removes this. Whereas your diagram labels this an ' upconverted image' which seems to be implying the converter has also upconverted the 'image' or negative frequency, but this doesnt exist after the DAC as what Fat32 says below. Why represent a plot after a DAC with complex exponentials because theres no FFT analysis being applied in that block diagram after the DAC - Its labelled analog section. Dec 3, 2019 at 12:29

Yes you are right.

Aliasing only happens when you sample (either analog to digital conversion or during digital downsampling) a signal and refers to the overlap of signal spectrum due to an inadequate sampling rate; sampling rate below the Nyquist rate.

Aliasing cannot happen at the output of a DAC (digital to analog converter). What happens is, if you do not use a proper lowpass reconstruction (or interpolation) filter at the output of the DAC, then you will have images of the original signal spectrum repeating at multiples of DAC output frqeuencies, 1 MHz in your case. These are called image spectrums and they do not overlapp, but just create imaging distortion at the output signal, if not properly filtered out.

In your case the spectrum from 500 kHz to 1 MHz will be a (conjugate) mirror of the original spectrum from 0 Hz to 500 kHz.

• Fat32, I added a screenshot. This seems contrary to the other answer. The images have nothing to do with negative frequencies? Is it any different for complex or IQ Digital to analog DAC method like the difference on complex sampling is the entire Fs range is filled with unique frequency and not just upto FS/2 with normal sampling Nov 28, 2019 at 3:16
• @NatalieJohnson If you want to have a chat session with me so as to ask your most fundemantal questions on DSP, I would like to help. If yes, please indicate your suitable calendar for this. Dec 3, 2019 at 19:38

Keep in mind that digitized signals always contain images of the original analog signal's spectrum, modulated around multiples of the sample rate.

The graph shows only the positive spectrum, but it extends in the negative direction too. However, you can ignore negative frequencies for real signals because 1) they only appear with a corresponding positive frequency of the same amplitude, and 2) you can't tell the difference between a sine wave of a given positive frequency and the same of negative frequency at the same amplitude.

So, if you want to consider the negative frequency spectrum, consider that when you convert back to analog, -100 Hz and 100 Hz would both be reproduced, each carrying half the amplitude, in phase, and summing to the expected amplitude. But in most cases, except maybe mathematical derivations, we just ignore the negative spectrum.

(You might ask "what about phase? Could positive and negative cancel?"—but that would be a different question. The short answer is they are always in phase like a cosine wave that is symmetrical around zero.)

• earlevel How can -100 to 100 be produced on a DAC, it makes a real signal. Seems contrary to what Fat32 says. Nov 28, 2019 at 2:45
• It's really only useful to consider negative frequencies mathematically. Consider amplitude modulation (AM). If you modulate a 100 Hz sinusoid by a 10 Hz sinusoid you get 110 Hz and 90 Hz, the sum and difference of the two inputs. If the two inputs are 100 Hz and 110 Hz, the resulting frequencies are 210 Hz and -10 Hz. That is the true mathematical result. However, if you draw a -10 Hz sinusoid on paper, or play it out a DAC, there is no way to differentiate it from a 10 Hz sinusoid. Nov 29, 2019 at 19:06
• Plot cos(x): Start with x = 0, and increment the angle by a small amount till you get to 2pi. Below it, start again with x = 0, but use negative increments of the same size. Clearly, you generated a negative frequency in the second case. But that task done, the two plots are identical, and it's just a mathematical argument whether there is a difference. (But if you do the same with sin(x), there is a phase difference. For that reason, you can't outright say there is no difference between a positive and negative frequency, but for real signals they are the same.) Nov 29, 2019 at 19:13
• Could some one explain the reasons for the down-vote for my answer? I don't think I said anything wrong, but I'd like to improve the answer if needed, Nov 29, 2019 at 19:19
• I guess I was looking in the wrong place—my first sentence referred to aliasing instead of images, I fixed it. (An alias is a special case of an image.) Nov 30, 2019 at 7:37