# Heterodyne modeling confusion in SDR

I've been reading this lab sheet which explains the signal processing math of the RTL-SDR radio dongle. http://www.eas.uccs.edu/~mwickert/ece4670/lecture_notes/Lab6.pdf In pages 5 and 6, the local oscillator (LO) is said to be modeled by a multiplication with the complex frequency $$e^{-j2{\pi}f_{c}t}$$ as shown in the image below taken from the lab sheet.

I understand how, mathematically, this multiplication with the negative frequency will shift the signal spectrum to the left by $$f_c$$ Hz, but what I don't understand is how a local oscillator can physically achieve this. The negative frequency in the behavioral model does not have a physical interpretation (does it?), and the physical LO signal in reality is just a sine wave with frequency $$f_c$$. Multiplying $$sin(2{\pi}f_ct)$$ with the radio signal of interest $$s(t)$$ will not remove the carrier frequency $$f_c$$ from the radio signal, for example if the radio signal is also $$sin(2{\pi}f_ct)$$, the resulting product will be a signal with frequency $$2f_c$$. Can someone explain what I am getting wrong?

A negative frequency is given by the complex signal $$e^{-j2 \pi f_c t}$$ for which the physical implementation requires two real signal paths, which is made clearer from Euler's formula:

$$e^{-j2 \pi f_c t} = \cos(2 \pi f_c t) + j \sin(2 \pi f_c t)$$

So if the Local Oscillator is generated with sine and cosine outputs, and the multiplier is a complex multiplier (in this case it may have a real input so would result in two real multipliers) then the receiver can be implemented as described.

Note that a full complex multiplier requires four real multipliers:

$$(I_1+jQ_1)(I_2+jQ_2) = (I_1 I_2 - Q_1 Q_2) + j(I_1Q_2- I_2Q_1)$$

In this case the input from the LNA is real, representing $$I_1$$ in the formula above, while the input from the Local Oscillator is complex, representing $$I_2+jQ_2$$ and the product is:

$$(I_1)(I_2+jQ_2) = (I_1 I_2) + j(I_1Q_2)$$

With the complex output as indicated in the block diagram.

Note I detail this further and the implications of the real versus imaginary input after the LNA in this post: Frequency shifting of a quadrature mixed signal (for imaginary input the output of the LNA would be fed into a Hilbert Transform and then the multiplier would be a full complex multiplier using four real multipliers).

• Thanks Dan, so are complex signals physically accomplished by using two wires? Or are the 90˚ shifted components just added together? In QAM modulation, the IQ components are (I believe) superimposed into one single physical signal, but it can be uniquely split again into IQ components. In the first diagram of the answer that you linked to, you split 2cos(wt) into two signals, and each one is mixed differently, so they must be two different wires? – Carlos Vazquez Sep 3 at 1:40
• @CarlosVazquez Two wires--- it takes "two wires" to realize a complex signal. This can be I and Q in I+jQ or magnitude and phase, either way there are two real numbers to describe a single complex number. We can represent QAM as a real waveform as a passband waveform but not at baseband. Review the link I gave in my answer and that should be clearer. – Dan Boschen Sep 3 at 2:17
• Note that for all real signals, the spectrum will be complex conjugate symmetric. So imagine your baseband waveform with a non-symmetric spectrum (must be complex), and how we can translate that spectrum to a higher carrier friequency at which point it can have a negative complex conjugage symmetric image (and therefore is real). This is a why a real passband can have the same spectrum (just centered at a different carrier frequency) as a complex baseband (where the carrier frequency = 0). – Dan Boschen Sep 3 at 2:23