Is spectral inversion possible in zero IF receivers? If yes, then can somebody explain with an example how that happens and what can be done to correct it.

The following problem can be taken as reference:

A 40Mhz baseband signal is upconverted to fc of 1 GHz and transmitted.


Short Answer

In order to do a complete Zero-IF transmit and receive without inversion, the operations must be done as follows (or otherwise have the signs equally flipped on Q for both transmit and receive):

Zero-IF (Direct Conversion) Transmit:

$$I_{RF}(t) = I_s(t)\cos(\omega_{LO}t) - Q_s(t)\sin(\omega_{LO}t)$$

Zero-IF Receive:

$$I(t) = \text{LPF}\{I_{RF}(t)\cos(\omega_{LO}t)\}$$ $$Q(t) = \text{LPF}\{-I_{RF}(t)\sin(\omega_{LO}t)\}$$

Yes spectral inversion can easily occur by having a sign change on Q on one side and not the other, showing that the way to correct it is to simply change the sign of the Q output.


$I_{RF}(t)$: Transmitter RF signal
$I_s(t)$: Baseband waveform I input at transmitter
$Q_s(t)$: Baseband waveform Q input at transmitter
$\cos(\omega_{LO}t)$: I LO
$\sin(\omega_{LO}t)$: Q LO
$\text{LPF}\{\}$: represents a low pass filtering operation.

Detailed Explanations

Consider the plots below that demonstrate exactly how this occurs:

For context first consider the typical Zero-IF Receiver architecture where we have a real signal that gets multiplied with a complex Local Oscilator, as implemented with two multipliers, one for the real or "in-phase" component as "I" and the other for the imaginary or "quadrature" component as "Q". In the diagram the LO is shown to be fed into a quadature splitter (indicated by 90°) which produces two sinusoidal outputs in quadrature. The two outputs form a single complex source as real and imaginary components. One is designated to be "cosine" while the other is "sine" with "cosine" typically representing the real component. This is consistent with the Euler relationships listed further below.

Zero-IF Receiver

Below shows the spectrums during this downconversion process. Note the Euler identities in the creation of the complex LO:

$$e^{+j\omega t} = \cos(\omega t) + j\sin(\omega t)$$

$$e^{-j\omega t} = \cos(\omega t) - j\sin(\omega t)$$

Multiplication in the time domain is convolution in the frequency domain, so the result of multiplying the real input with the complex LO (which is a single sided impulse in frequency) is to just shift the spectrum accordingly. So when we multiply by $e^{-j\omega t}$ the spectrum slides to the left as a result (and then low pass filter to get the resulting baseband signal), and when we multiply by $e^{+j\omega t}$, the spectrum slides to the right. Notice how the first case results in an inverted spectrum.

Spectrum Plots

To reverse this inversion, we could simply change the sign of the sine (Q) output of the LO resulting in a negative frequency LO instead of a positive frequency. Alternatively and equivalently change the sign of the Q output of the receiver. To see this simply, consider a single baseband positive tone which in the time domain can be represented as a phasor rotating counter-clockwise on a complex IQ plane. If you negate Q (the imaginary component of the phasor), it would then rotate clock-wise instead, representing a single baseband negative tone of the same frequency.

The OP had additional questions specific to the transmitter side (direct-conversion transmitters for which I add the details below:

To explain the graphic we see how the transmitter structure can be viewed as the real component of a full complex multiplication; starting with the baseband Tx waveform on the top spectrum, the complex LO as $e^{+j\omega t}$ specifically to shift the spectrum to the right (for no inversion), resulting in what would be a complex RF signal on the third spectrum (which would require a real and imaginary datapath). By taking the real part of this, we get the conjugate symmetric real spectrum as on the bottom spectrum that can be fed to a single antenna.


To realize this, consider the complex baseband as $I_s+jQ_s$ multiplied with the complex LO as $e^{+j\omega_{LO}t}= \cos(\omega_{LO}t)+j\sin(\omega_{LO}t)$. The result of a full complex multiplication (requiring 4 real multipliers) would be:

$$(I_s(t)+jQ_s(t))\bigl(\cos(\omega_{LO}t)+j\sin(\omega_{LO}t)\bigr) $$ $$ = \bigl(I_s(t)\cos(\omega_{LO}t) - Q_s(t)\sin(\omega_{LO}t)\bigr)+j\bigl(I_s(t)\sin(\omega_{LO}t)+Q_s(t)\cos(\omega_{LO}t)\bigr)$$

With the real part representing the transmitted RF signal specifically as:

$$I_{RF}(t) = I_s(t)\cos(\omega_{LO}t) - Q_s(t)\sin(\omega_{LO}t)$$

Notice the minus sign on the Q LO input as required to not have an inverted spectrum. If one were to make the mistake in using $+\sin(\omega_{LO}t)$ for Q then a spectral inversion would result! This could very well be a source of confusion since $e^{+j\omega t} = \cos(\omega t)+j\sin(\omega t)$, but one needs to review carefully the operation above when taking the real of the full complex multiplication to see why the negative sign is required to actually realize the positive complex LO!

Similarly with the Rx, in order to not introduce and inversion, we must multiply the received real RF signal by $e^{-j\omega_{LO}t}$ as shown earlier in the first spectrum plot on the left hand side. But in this case the RF signal is completely real, so we can show as in the product done out for the transmitter but in this case use $I_{RF}$ as the signal multiplied with the complex LO as $e^{-j\omega_{LO}t}= \cos(\omega_{LO}t)-j\sin(\omega_{LO}t)$ to get the complex baseband output (prior to low pass filtering). This results in:

$$(I_{RF}(t))\bigl(\cos(\omega_{LO}t)-j\sin(\omega_{LO}t)\bigr) $$ $$ = I_{RF}(t)\cos(\omega_{LO}t) -jI_{RF}(t)\sin(\omega_{LO}t)$$

With the final zero-IF receiver output as:

$$I(t) = \text{LPF}\{I_{RF}(t)\cos(\omega_{LO}t)\}$$ $$Q(t) = \text{LPF}\{-I_{RF}(t)\sin(\omega_{LO}t)\}$$

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  • $\begingroup$ Could you show the baseband spectrum on transmitter side before upconversion to RF? $\endgroup$ – Dsp guy sam Jul 4 at 3:07
  • $\begingroup$ Sure, that’s a different question isn’t it? Or did you mean to ask “in zero IF transmitters”? It’s the same thing for the transmitter except we go from complex to real instead of real to complex for the receiver. Change the sign on Q and you invert the baseband spectrum and therefore the spectrum at RF since the transmitter moves the baseband signal to RF as a complex up conversion and then selects the real part. (Look at a full complex multiplier with four real mults and then take the real part and you get the Tx structure above) $\endgroup$ – Dan Boschen Jul 4 at 10:32
  • $\begingroup$ I mean you've shown the RF spectrum, which I suppose is the upconverted spectrum, so I am asking could you show the baseband spectrum, which when upconverted gives the RF spectrum shown in the figure $\endgroup$ – Dsp guy sam Jul 4 at 11:07
  • $\begingroup$ The baseband is on the last line centered at 0 and the RF is the top line $\endgroup$ – Dan Boschen Jul 4 at 11:17
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    $\begingroup$ @Dspguysam See my update where I added Tx spectrums and further explanation on inverstion at the Tx side that could occur if not done properly. $\endgroup$ – Dan Boschen Jul 5 at 0:06

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