# User:CSeguinot

This tutorial originates from discussions on discuss-gnuradio@gnu.org. We will explain why simulating digital communications requires equivalent baseband representation of signal which in fact are complex signals. For this unique reason, complex signals are essential in GNURadio.

This tutorial is intended for non specialist, it involves as less maths as possible and present most results with GNURadio flowgraph. Some examples involving simple modulation scheme used in HAM radio are presented. While introducing complex signal can be seen as increasing complexity, we will see that it drastically simplify some impairment such as synchronization.

If you are searching for more detailed information please refer to corresponding literature such as references [1],[2],[3].

## Some maths

Complex number z=a + jb

A complex number is a number of the form a + jb, where a and b are real numbers, and j is an indeterminate satisfying j2=-1 (Mathematician prefer using i instead of j used by physicist and radio engineers). For example, z1=2+3j is a complex number. The real part Re{z} of z1 is 2 and its imaginary part Im{z} is 3.

$\displaystyle z=a+jb$
$\displaystyle \text{Re}\{z\}=a$
$\displaystyle \text{Im}\{z\}=b$

Complex numbers can be represented in the complex plane as vectors. The modulus or magnitude r of a complex number z = a + jb is

$\displaystyle r=|z|=\sqrt{a^2+b^2}$

The phase φ of z mathematically referred to as the argument is the angle of the radius Oz with the positive real axis.

$\displaystyle \phi=\arg(z)=\arctan(b/a)$ (for a≠0)

Together, r and φ give another way of representing complex numbers, the polar form and the exponential form.

$\displaystyle z=r \left(cos(\phi) + j sin(\phi) \right) = r e^{j\phi}$
The complex plane

The exponential form is convenient for computing the multiplication of two complex numbers.

$\displaystyle z_1= r_1 e^{j\phi_1}$
$\displaystyle z_1= r_2 e^{j\phi_2}$
$\displaystyle z=z_1z_2= r_1r_2 e^{j(\phi_1+\phi_2)}$

Following complex number have a unit magnitude r=1 :

$\displaystyle +1=e^{j0}$
$\displaystyle +j=e^{j\pi/2}$
$\displaystyle -1=e^{j\pi}$
$\displaystyle -j=e^{j3\pi/2}$

A complex signal c(t) can be seen as two real signal i(t), q(t) combined to create a complex signal. It can also be represented by its time varying amplitude a(t) and its phase φ(t)

$\displaystyle c(t) = i(t) + j q(t) = a(t) e^{j\phi (t)}$

## Why we need complex and IQ signals

GNURadio software is mainly used to design and study radio communications. Making high frequency transmission requires modulating a high frequency carrier at frequency F0. The most common modulation for analog transmissions are Amplitude modulation (AM) Phase modulation (PM) and Frequency modulation (FM).

AM spectrum

For analog AM, the modulated signal m(t) is simply the mathematical product of the carrier c(t) and the baseband signal to transmit a(t). The corresponding hardware is a mixer whose scheme and mathematical representation is a multiplier.

$\displaystyle m(t) = a(t) c(t) = a(t) \cos(2\pi f_0t)$

We call a(t) a baseband signal since its spectrum is in a low frequency range starting near 0 Hz (For example [0-20kHz] for an HIFI audio signal).

The spectrum of an AM modulated signal M(f) is the translation or the audio spectrum A(f) around ±F0 with A(f) being the entire spectrum of the modulating signal, using both positive and negative frequencies

$\displaystyle M(f) = \frac{1}{2}\big(A(f-f_0) + A(f+f_0)\big)$

N.B. Negative frequencies are often omitted in spectrum representation since, for real signal (a(t), m(t) are real) the power spectrum are symmetric around zero as will be detailed later.

Up to now we have been dealing with real signal. The need for complex signal appears in the next step. Simulation requires sampled signal. Sampling is the operation of observing a continuous signal and taking a finite number of sample at a given sampling rate fs (i.e; one sample each 1/fs second). simulator can only make calculations on a finite number of samples, they require sampled signal. Nyquist Sampling theorem states that the sampling rate must be greater than twice the maximum frequency FMax to be able to reconstruct the original signal from the sampled signal.

$\displaystyle f_s > F_{Max}$

For an HIFI audio signal, maximum audio frequency $\displaystyle F_{Max Audio}$ is closed to 20 kHz, sampling rate must be higher then 40 kHz (44.8 kHz is often used in computer sound card, 8 kHz is used for mobile phone since voice has a lower frequency range then HIFI audio).

For an AM signal modulated by an audio signal, maximum frequency of the modulated spectrum is $\displaystyle F_{Max}=F_0+F_{Max Audio}$ . Direct sampling of such signal is not possible with conventional hardware such as low cost SDR dongle. If the carrier frequency is close to 1 GHz, the sampling rate should be at least 2 GHz. This is obviously too much then computer can handle (higher then some computer clock).

Flowgraph IQ_tutorial_AM_TX_real.grc illustrates amplitude modulation using only real blocks (excepted for bits source). As a consequence, the maximum carrier frequency is limited to several tens of kHz.

• study the modulator part which simply multiply the baseband signal and the sine carrier
• look at the influence of the carrier frequency on the modulated signal spectrum (carrier frequency must stay lower than half the sampling rate)
• look at the spectrum shape for sawtooth input and random bit sequence (QT Gui chooser and Selector)
• When transmitting random bits, you can desactivate the interpolating FIR Filter and replace it by a root raised cosine filter

## Complex enveloppe, equivalent baseband signal

Baseband signals have a spectrum at low frequency near 0 Hz. Audio, video and NRZ line code are baseband signals.

Bandpass signals have no energy near 0 Hz and a spectrum located near a high frequency (generally the carrier frequency). Analog and digital AM PM and FM modulated signals are bandpass signals.

A theorem ([1],[2],[3]) states that any high frequency bandpass signal having a limited bandwidth B can be represented by a baseband equivalent signal having the same bandwidth. This baseband equivalent signal also called the complex envelope is used in simulators since its allow to lower sampling rate as compare to directly sampling the bandpass signal.

The equivalent baseband representation help us for the simulation of bandpass signals. At this step we need some math. We will consider a carrier modulated in phase and/or amplitude (in the sake of simplicity, Frequency modulation is not considered but it can be related to phase modulation.). Such a modulated signal m(t) and it's complex representation $\displaystyle \tilde{m}(t)$ is :

$\displaystyle m(t)=a(t) \cos(2\pi F_0t + \phi(t))$
$\displaystyle \tilde{m}(t)=a(t) e^{(j(2\pi F_0t + \phi(t)))} = a(t) e^{j \phi(t)} e^{j2\pi F_0t} = m^{bb}(t) e^{j2\pi F_0t}$

For modulated signal, the complex representation is obtained by replacing the cos function by an exponent function . For a more general definition see [1],[2],[3] . The real signal correspond to the real part of the complex signal $\displaystyle m(t)=\text{Re}(\tilde{m}(t))$ .

AM spectrum

One important property of $\displaystyle \tilde{m}(t)$ spectrum $\displaystyle \tilde{M}(f)$ is that it has only energy in the positive frequency range and verify.

$\displaystyle \tilde{M}(f)=M^+(f)$

where $\displaystyle =M^+(f)$ denotes the restriction of $\displaystyle =M(f)$ to the positive frequency range.

Let us now look at the complex envelope or equivalent baseband signal $\displaystyle m^{bb}(t)$ of bandpass signal $\displaystyle m(t)$ defined by :

$\displaystyle m^{bb}(t)= \tilde{m}(t) e^{+j2\pi F_0t} = a(t) e^{j \phi(t)}$

Multiplying a signal by $\displaystyle e^{+j2\pi F_0t}$ correspond to a frequency translation of the spectrum so that we have:

$\displaystyle M^{bb}(f)=M+(f+F_0)$

So the spectrum of the complex enveloppe is a baseband signal whose spectrum has the same shape as $\displaystyle =M+(f)$ . For this reason, knowing $\displaystyle \tilde{m}(t)$ or $\displaystyle M^{bb}(f)$ is sufficient to reconstruct $\displaystyle m(t)$ or $\displaystyle M(f)$ .

## IQ modulator and demodulator

Complex signal in the IQ plane
ID modulator and demodulator

Let us now come to hardware and SDR and first rewrite the equivalent baseband signal and the modulated signal m(t) :

$\displaystyle m^{bb}(t)=a(t) e^{j \phi(t)} =i(t) + j q(t)$
$\displaystyle m(t)=\text{Re} \left[ \big(i(t)+jq(t)\big) e^{j2\pi F_0t} \right]$
$\displaystyle m(t)=i(t) \cos(2\pi F_0t) -q(t)) \sin(2\pi F_0t)$

The phase φ(t) of the modulated signal m(t), is identical to the phase of the complex signal c(t)=i(t)+jq(t). The equivalent baseband signal c(t) is represented in a complex plane also refeered to as the IQ plane. The resulting m(t) can be any modulated in AM, PM or even FM signal.

• $\displaystyle i(t)=a(t) \cos(\phi(t))$
• $\displaystyle q(t)=a(t) \sin(\phi(t))$

As a result, the modulated signal m(t) is the addition of :

• $\displaystyle i(t) \cos(2\pi F_0t)$ which is an AM modulated signal, the product of i(t) by a signal In phase with the carrier (i stand for In phase)
• $\displaystyle q(t) \sin(-2\pi F_0t)$ which is an AM modulated signal, the product of q(t) by a signal in Quadrature with the carrier (q stand for Quadrature)

The corresponding hardware is called an IQ modulator. Every modern radio communication uses IQ modulator for emitting and IQ demodulator for receiving. The IQ demodulator is able to recover incoming i(t) and q(t). If the amplitude of the recovered carrier is 2, and if modulator and demodulator carrier are synchronous (same frequency and phase) the output of the IQ demodulator correspond to input i(t) and q(t).

$\displaystyle \hat{i}(t)=i(t)$
$\displaystyle \hat{q}(t)=q(t)$

In real hardware, carrier are not synchronous and the receiver must compensate any phase and frequency difference between emitter and receiver. This is done using some hardware and/or software such as polyphase clock sync, and Costas loop.

SDR Module such as USRP N320 block Diagram) and SDR Dongle (RTL-2832) input and output are the 2 real signals i(t) and q(t) combined to form the complex signal i(t) + j q(t) which turns to be the equivalent baseband of the modulated emitted or received signal. These hardware are based on IQ modulator and IQ demodulator associated with mixers when intermediate frequency (IF) is used.

## Some examples of EqBB signals

In order to get familiar with complex signal, let us consider some basic examples.

First, we suppose our emitter carrier is $\displaystyle \cos (2\pi F_0t)$ so that every equivalent baseband signal will be defined according to this reference.

### complex envelope of a pure sine wave

We will consider a pure sine wave, close to the carrier having a Δf frequency shift and φ phase shift as compared to the carrier. After some math we get its complex equivalent signal.

$\displaystyle m(t) = A \cos (2\pi (F_0+\Delta f)t+ \phi)$
$\displaystyle \tilde{m}(t) = A e^{(j(2\pi (F_0+\Delta f)t + \phi))}$
$\displaystyle m^{bb}(t) = A e^{2j(\pi (\Delta ft+\phi)} = A e^{j \Delta ft} e^{j\phi}$

The complex envelope of the carrier itself is found for Δf=0 and φ=0 which yields $\displaystyle m^{bb}(t) = A$ . We conclude that in a GNURadio flowgraph, the carrier is represented by a continuous components, a pure DC signal ; this may seemed counter intuitive. The spectrum $\displaystyle M^{bb}(f)$ is a single peak at f=0 which can be represented using the Dirac δ(f) function (or distribution) : $\displaystyle M^{bb}(f)=\delta(f))$ . Shifting this spectrum of $\displaystyle +F_0$ towards positive frequency we get the positive part of the carrier spectrum (a peak at $\displaystyle +F_0$ ) : $\displaystyle M^+(f)=\delta(f-F_0))$ . The negative part of the spectrum is simply obtained by symmetry of the real part (a peak at $\displaystyle -F_0$ ).

This can be simulated with GNURadio flowgraph IQ_tutorial_eq_bb.grc. The spectrum of our carrier is centered at 0 Hz. The Frequency sink has a feature to shift this spectrum around $\displaystyle F_0$ which is a parameter of the sink. In a simulation, it is not necessary to represent the negative part of the spectrum since m(t) being real, it spectrum is obviously symmetric.

We will now consider Δf ≠0 which simulate a signal not exactly synchronous to the carrier. This yieds

$\displaystyle m^{bb}(t) = A e^{2j\pi \Delta ft}$

The complex enveloppe is rotating vector (you can simulate this with Δf=1 Hz giving a vector which rotate at 60 rpm or 1 turn per second) , its spectrum is a single peak at f=+Δf. This complex signal exhibits a non symmetric spectrum (no peak at f=-Δf). Change Δf to -0.5 Hz, the vector now rotates counter clockwise at 30 rpm or 0.5 turn per second.

Exercice: Open flowgraph IQ_tutorial_eq_bb.grc. Parameter delta_f is set with an increment of 1/12, which correspond to 5 rotation per minute.

• For Δf=1/12=0.0833, what is the speed of rotation of the complex signal?
• What do you observe when Δf=-1/12=-0.0833 ?
• Slowly increase Δf to reach fs/2 and observe the spectrum really has a single peak at Δf. Explain your observation when Δf>fs/2'.
• Do the same for a negative Δf
• For Δf =fs what is the equivelant baseband frequency. What is the carrier frequency ?
• Set Δf=0 and φ≠0. Discusss the simulated equivalent baseband signal.

### AM mod demod example

This example will consider signal baseband signal a(t) modulating a carrier at F0 in AM, and its demodulation. As no phase modulation in used, φ(t)=0 and consequently q(t)=0.

$\displaystyle m(t) = a(t) \cos (2\pi (F_0t)=i(t) \cos (2\pi F_0t) -q(t) \sin (2\pi F_0t)=$
$\displaystyle i(t) = a(t)$
$\displaystyle q(t) = 0$

AM modulation is a special case for which the equivalent baseband complex signal has a null imaginary part and is real. Considering the schematic diagram of an IQ modulator demodulator, when q(t)) is null the diagram is simplified (imaginary path is not used) yielding the well known AM modulation/demodulation scheme.

Flowgraph IQ_tutorial_AM_TX_complex.grc contains two equivalent diagram for an AM modulation with a sawtooth input:

• the upper one uses real signals. It is the exact AM modulator uses at the beginning of this tutorial.
• sampling frequency is 200 kHz, 40 times the input rate which equal 5 kHz.
• allowing a maximum carrier frequency close to 100Khz
File:AM TX complex.png
AM modulator flowgraph
• the lower one uses an equivalent baseband representation.
• sampling frequency is 25 kHz, 5 times the input rate which equal 5 kHz.
• the sawtoooth correspond to identical generator in both modulator
• each blue input or output is the baseband equivalent of the corresponding signal in the upper AM modulator.
• the carrier frequency can be any value compatible with connected Hardware
• the carrier equivalent signal equal 1 (as stated in the previous section) so it has been disable and replaced by a complex constant
• the Hardware input is the complex equivalent baseband signal
• the carrier frequency is not needed and used in complex blocks, excepted in the spectrum to label the center frequency which is 0 but correspond to the F>sub>0</sub>.

Three blocks are unusefull and can be removed from the lower complex modulator:

• remove the complex multiplier, the carrier equivalent baseband (complex constant=1) and the carrier complex source (the disabled one). Reconnect complex to float and throttle blocks to obtain the flowgraph sketched on the right.

You can now run this flowgraph and compare signal and their spectrum in both modulator. Once again, these simulations suppose perfectly synchronous emitter and receiver carrier which is quite far from reality.

#### Further work : construct the demodulator flowgraph

Taking into account that an AM demodulator extract the amplitude (magnitude) of the modulated signal, what would be the equivalent baseband representation of the demodulator (see IQ demodulator) ?

This text is collapsible. Template:Lorem

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### QPSK example

In this section, we will simulate a QPSK without Nyquist filter in order to get simple constellation.

QPSK digital signal exhibit four phase state $\displaystyle \phi \in \{\pi/4, 3\pi/4, -3\pi/4, -\pi/4 \}$ .

$\displaystyle m^{bb}(t)=a(t) e^{j \phi(t)} =i(t) + j q(t)$

And the baseband equivalent signal also exhibit 4 different values each one being used to code a 2 bits sequence 00 01 10 or 11 :

$\displaystyle m^{bb}(t) \in \{1+j , -1+j , -1-j, 1-j \}$

At this step, we consider unfiltered QPSK. The QPSK modulator/demodulator (complex representation) correspond to flowgraph XXX. The upper flowgraph is a classic QPSK. The constellation q(t) versus i(t)) shows that this signal exhibit only the 4 phase state.

The lower diagram is similar, however, only 3 phase states are used as shows the constellation diagram. This QPSK has not practical interest, it will be usefull for further work in this tutorial.

Simulating these diagrams clearly shows that the complex baseband signal spectrum is no longer symmetric as expected for complex signals. As for the previous example, this simple modulation/demodulation scheme is far from reality and we will have to investigate synchronisation later.

## Equivalent baseband scheme

This diagram sketch a classical Tranceiver Receiver (Tx/Rx).in order to simulates this, we need a equivalent representation of

• this IQ modulator and demodulator
• bandpass filters
• channel noise

Note that complex baseband signal don't permit to precisely modelize non linear effects such as intermodulation so they are not considered here.

### Tx/Rx baseband model

The following diagram is the baseband equivalent of the TX/RX given above.

## References

Further reading for complex equivalent baseband signal:

• [1] Proakis J., Digital Communication, McGraw Hill Series in Electrical and Computer Engineering, Singapore, 1989
• [2] Gallager R., Principles of digital communication, Cambridge University Press Cambridge, UK, 2008
• [3] Benedetto S. and Biglieri E., Principles of digital transmission : with wireless applications, Kluwer Academic/Plenum Publishers, NY, 1999