What is use of Quadrature Converter in OFDM?

Question

I'm learning about OFDM Transmitter and see that output of IFFT Block produces complex digital signal and we need DAC to convert it to analog and then use RF Frequency mixing to send it out of transmitter. But this RF Freq mixing is not simply multiplying signal with high frequency, rather I see something called as inphase and Quadrature phase, i don't understand this , can someone pls guide ? Why it is done and what's the use of it? Is it done to send a real signal out of transmitter?

Answer 1

In OFDM, the output of DAC after IFFT is a complex analog signal. In order to transmit this complex signal, two orthogonal RF channels are needed, one to transmit the real part, another one to transmit the imaginary part. So, in practice, the cosine wave and the sine wave of a same RF frequency are used to respectively transmit the real part and imaginary part. Such kind of frequency-converter (also called frequency mixer) converting a baseband complex signal to a two-channel RF signal is called quadrature frequency converter (or mixer).

Expressed in math, if the baseband complex signal is expressed as

$$x(t)=i(t)+jq(t)$$

where $i(t)$ and $q(t)$ are often respectively referred to as the in-phase part and the quadrature-phase part of $x(t)$. The RF signal then is

$$r(t)=i(t)\cos(\omega_0t)-q(t)\sin(\omega_0t)$$

Clearly, $r(t)$ is a real RF signal to be transmitted. Equivalently, $r(t)$ is often expressed as

$$r(t)=\Re e \big\{x(t)e^{j\omega_0t} \big\}$$

for convenience, meaning that $r(t)$ is the real part of the product of $x(t)$ and $e^{j\omega_0t}$. Indeed, this can be easily verified with simple math.

Answer 2

"Mixing" has a different meaning for RF than it does for audio.

In audio mixing is a linear combination of the two (or more) signals. In RF, it's the modulation that bumps a baseband signal up to the RF frequency.

The reason they call it "mixing" in RF is that the way they created the cross-product term, $x(t) \cos(\omega_0 t)$ which is what we want for the RF modulated waveform, they first add the two signals and then pass that sum through a non-linear element that generates the cross-product term.

Now quadrature modulation is a slightly different multiplication. It like first , $$ x(t) = i(t) \, + \, j\, q(t) $$

where $i(t)$ and $q(t)$ are the in-phase and quadrature components and the RF output would eventually be the real part of

$$ x(t) \cdot e^{j \omega_0 t} $$

Now, if your DSP is fast enough, you can do that complex math in a DSP and output the real part as the composite RF signal. I normally think that most modulation to RF is done in the analog domain because the RF frequencies are often so high. 100s of MHz. But a DSP can easily modulate the baseband quadrature signal, $x(t)$, to an Intermediate Frequency (IF) which is much higher than baseband, but much lower than the physical RF frequency it will be bumped to when it's going out a microwave dish.