QPSK InPhase and Quadrature

Question

I am new to DSP. Please bear with my question. In QPSK why are we multiplying InPhase (Odd Bit) with Cosine and Quadrature (Even Bit) with Sine wave in this QPSK formulae

$$ s \left( t \right) = I \left( t \right) \cos\left(2 \pi f c t \right) − Q \left( t \right) \sin \left( 2 \pi f c t \right)$$

This page explains and illustrates in a clear way the IQ concept: https://www.ni.com/docs/en-US/bundle/ni-rfsa/page/nirfsa/iq-modulation.html — , Feb 04 '23 at 20:08
//why are we multiplying InPhase (Odd Bit) with Cosine and Quadrature (Even Bit) with Sine wave (?)// This sounds to me to be Offset Quadrature Phase Shift Keying. But I think the convention is that the even-indexed bits goes with the in-phase and the odd-indexed bits goes to the quadrature component. We geeks start counting with 0. — robert bristow-johnson, Feb 04 '23 at 20:52
@robertbristow-johnson not necessarily-- we can encode the datastream I Q I Q I Q but would only be offset QPSK if we delay I relative to Q by 1/2 a symbol. So important to define explicitly how those I and Q bits are applied in time. — Dan Boschen, Feb 04 '23 at 22:00
I know, @DanBoschen. I am just saying when we have an ordered pair, when do we conventionally precede the Real part with the Imaginary part ? Also $$ \big( \sqrt[4]{1} \big)^n= e^{j \frac\pi2 n} = ? \qquad n\in\mathbb{Z}$$ when $n$ is even or odd? — robert bristow-johnson, Feb 04 '23 at 22:11

Dan Boschen · Accepted Answer · 2023-02-04T19:28:01.673

This requires understanding complex waveforms in time and frequency and what "negative" and "positive" frequencies are. If that isn't clear, start with these other posts and then read on further below:

Frequency shifting of a quadrature mixed signal

Use of I/Q representation for unmodulated Rx Signal

The QPSK waveform at baseband is a complex time domain waveform, passing through the four complex locations given in a QPSK constellation at the correct sampling location within each symbol.
The expression given by the OP moves this complex QPSK baseband waveform as given by $I(t) + jQ(t)$ with $I(t)$ representing the real component and $jQ(t)$ representing the imaginary component, to a real carrier frequency as described with the following intermediary steps.

Frequency translate the complex baseband signal by multiplying it with a single positive frequency at carrier frequency $\omega_c$ as given by $e^{j\omega t}$:

$y(t) = (I(t) + jQ(t)) e^{j\omega_c t} \tag{1}\label{1}$

Note Euler's formula relating sinusoids with positive and negative frequency components:

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

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

From which we get the relationship for $e^{j\omega t}$:

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

Substituting this into equation \ref{1} results in:

$y(t) = (I(t) + jQ(t)) (\cos(\omega_c t) + j \sin(\omega_c t)) \tag{2}\label{2}$

We must represent a baseband waveform as complex if it represents independent control of magnitude and phase for a real passband waveform. However once this baseband waveform is frequency translated to a passband, we can take the real part which maintains this magnitude and phase relationship versus time. That said, multiply out equation \ref{2} and take the real part to get:

$$Real\{y(t)\}=I(t)\cos(\omega_c(t)) - Q(t)\sin(\omega_c(t))$$

QPSK InPhase and Quadrature

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