Introduction
This is a summary of the topic Equivalent complex representation. In this we discuss why we need such a representation. It is then followed by some more mathematical manipulation which makes it more convenient to generate and transmit signals. Finally we conclude the section with ECB signal and its interaction with Filters.
Motivation
When analyzing any digital communication modulation schemes it will be useful we can separate the modulation or the mixing of the carrier from the information itself. This allows us to analyze the modulation independent of the center frequency. This motivation drives us to use the Equivalent complex baseband representation of the transmitted RF signals.
To develop an intuition about this process it is easy to think of the transmitted RF signal as just the up-converted baseband signal. So, ideally a down-conversion should bring us back to the baseband. This means that if we can write the baseband signal for any RF signal then we can neglect the carrier frequency and study the modulation at the baseband. In this case the baseband signal is the ECB and of the transmitted RF signal.
However, the details reveal a subtle complication. Now let us consider the reverse situation and ask this question. What if the RF signal was not generated in baseband and up-converted? Instead if RF frequency was generated directly, can we now say for sure that the down-converted signal has the same information as the RF signal? This is the reason behind the detailed analysis that follows.
Mathematical Reasoning
ECB is an equivalent representation of the signal which gets transmitted (RF signal) . The important thing to notice is that the only real signals can be transmitted physically. This is an important fact around which the entire ECB representation works.
Now for real signals we have
$$
X(f) = X^*(-f)
$$
[!Note] Proof: \(X(f) = X^*(-f)\)
$$
\begin{aligned}
\mathcal{F}(x(t)) = X(f) &= \int_{-\infty}^{\infty}x(t)e^{-j\omega t} dt \\
\mathcal{F}(x^*(t)) &= \int_{-\infty}^{\infty}x^*(t)e^{-j\omega t} dt \\
&= \left(\int_{-\infty}^{\infty}x(t)e^{j\omega t}\right)^* dt \\
&= X^*(-f)
\end{aligned}
$$
Since x(t) is real =>
$$
x(t) = x^*(t)
$$
Thus,
$$
X(f) = X^*(-f)
$$
The above statement means that the negative frequency holds the same information as the positive. Hence we can get rid of one side. We shall be using this fact in the next steps.
Analytic Signals from RF
Before we get to the baseband representation (ECB) we encounter what is called the analytic representation. The analytic signal is the RF signal without the negative part of the frequency spectrum. This still has the same information as the RF signal as seen in the previous section.
Already at this point the signal seizes being a real signal and it can’t be transmitted directly anymore . The analytic signal is represented by \(x^+(t)\) and its Fourier transform \(X^+(f)\). In this section we get to the time and frequency domain representation of analytic signals.
The idea is straight forward. We just need to keep the positive side and subtract the negative. Mathematically it can be achieved using the signum function (also called sign function) defined as
$$
\operatorname{sgn}(x) =
\begin{cases}
-1, & x < 0 \\ 0, & x = 0 \\ 1, & x > 0
\end{cases}
$$
$$
X^+(f) = c[1+sgn(f)]X_{RF}(f)
$$
we use the constant c since we would like to keep the energy of both the signals same we can compute c. c can be computed as follows
$$
\begin{aligned}
E_{X_{RF}} &= E_{X^+}\\
E_{X_{RF}} &= \int_{-\infty}^{\infty}\left|c[1+sgn(f)]X_{RF}(f) \right|^2\\
&= \int_{0}^{\infty}\left|c2X_{RF}(f) \right|^2\\\
&= 2c^2E_{X_{RF}}\\
c &= \frac{1}{\sqrt{2}}
\end{aligned}
$$
Taking the inverse Fourier transform we get
$$
\begin{aligned}
x^+(t) &= \frac{1}{\sqrt{2}}[x_{RF}(t) + x_{RF}(t)*\mathcal{F}^{-1}(x_{RF}(t))]\\
&= \frac{1}{\sqrt{2}}[x_{RF}(t) + j(x_{RF}(t)*\frac{1}{\pi t})] \quad \text{[Since $\mathcal{F}^{-1}(x_{RF}(t)) = \frac{j}{\pi t}$]}\\
&=\frac{1}{\sqrt{2}}[x_{RF}(t) + j\mathcal{H}(x_{RF}(t))]
\end{aligned}
$$
where
\(\mathcal{H}(x(t)) = x_{RF}(t)*\frac{1}{\pi t}\) is called the Hilbert transform
[!note] Proof: \(\mathcal{F^{-1}}(sgn(f)) = \frac{j}{\pi t}\)
$$
\mathcal{F^{-1}}(sgn(f)) = \int_{-\infty}^{0}e^{j2 \pi f t}df + \int_{0}^{\infty}e^{j2 \pi f t}df
$$
This clearly is not integrable as the two integrands are sinusoids. Hence, we need a trick.
We will define 4
$$
\begin{aligned}
sgn(x) &= \lim_{a \to 0} e^{-a|f|} sgn(f) \\
&= \lim_{a \to 0}[e^{-af}u(f) – e^{af}u(-f)] \\
\end{aligned}
$$
Now taking the inverse Fourier transform
$$
\begin{aligned}
\mathcal{F^{-1}}(X(f)) &= \int_{-\infty}^{\infty}X(f)e^{j2 \pi f t} df \\
&= \lim_{a \to 0}\left[\int_{0}^{\infty}e^{-af}e^{j2 \pi f t}df – \int_{-\infty}^{0}e^{af}e^{j2 \pi f t}df \right]\\
&= \lim_{a \to 0}\left[\int_{0}^{\infty}e^{-(a – j2 \pi t)f}df – \int_{-\infty}^{0}e^{(a + j2 \pi t)f}df \right]\\
&= \lim_{a \to 0} \left[ \frac{e^{-(a – j2 \pi t)f}}{-(a – j2 \pi t)} \right]_{0}^{\infty} – \lim_{a \to 0} \left[ \frac{e^{(a + j2 \pi t)f}}{a + j2 \pi t} \right]_{-\infty}^{0} \\
&= – \frac{1}{j2\pi t} – \frac{1}{j 2 \pi t}\\
&= \frac{j}{\pi t}
\end{aligned}
$$
Transformation: RF to ECB
Now that we have the analytic signal, ie. the signal where the negative frequencies are 0 it is a matter of simple downshift to get the baseband.
Thus we have
$$
X(f) = X^+(f+f_0)
$$
or equivalently we have in time domain
$$
\begin{aligned}
x(t) &= x^+(t)e^{-j2\pi f_0t}\
&= \frac{1}{\sqrt{2}}[x_{RF}(t) + j\mathcal{H}(x_{RF}(t))]e^{-j2\pi f_0t}
\end{aligned}
$$
Finally we have the important relation that converts RF to ECB
$$
x(t) = \frac{1}{\sqrt{2}}[x_{RF}(t) + j\mathcal{H}(x_{RF}(t))]e^{-j2\pi f_0t}
$$
Transformation: ECB to RF
This is the reverse transformation of what we did so far. We would like to get the RF signal from the ECB. This part is relatively straight forward. We just need to take the baseband signal and up-convert and keep only the real part. The real signal is the original RF signal we started with
$$
x_{RF}(t) = \sqrt{2}Re(x(t)e^{j2\pi f_0t}) = \frac{x(t)e^{j2\pi f_0t} + x^*(t)e^{-j2\pi f_0t}}{\sqrt{2}}
$$
or the Fourier transform
$$
X_{RF}(f) = \frac{1}{\sqrt{2}}(X(f-f_0) + X(-(f+f_0))
$$
*The above expression takes the ECB and shifts by up by \(f_0\). Also it makes another copy which reflects around 0 and shifts down by \(f_0\)*
ECB in quadrature form
Motivation
We saw in [[Equivalent Complex Baseband]] that the RF signal can be represented by ECB. ECB is a complex signal. So the question remains on how to generate these signals in the real world. This section provides a practical way of producing these signals and using it.
ECB Quadrature to RF
For practical use we would like to generate these ECB signals at baseband and have a way to modulate this to the RF frequency. We start with the describing expanding the ECB into real and imaginary part. And then we proceed by substituting these expressions to get the relevant RF.
We can split the ECB signal into real and imaginary part and call it in phase and quadrature components. The reason for the naming becomes apparent when the derivation ends with multiplication of cos and sin waves.
$$
x(t) = x_I(t) + jx_Q(t)
$$
Then we know to get RF frequency it needs to be multiplied by the center frequency and considering only the real part
$$
\begin{aligned}
x_{RF}(t) &= \sqrt{2}Re[x(t)e^{j2\pi f_0t}] \\
&= \sqrt{2}Re((x_I(t) + jx_Q(t))(\cos{2\pi f_0t} + j \sin{2\pi f_0t})\\
&= \sqrt{2}(x_I(t)\cos{2\pi f_0t} – x_Q(t) \sin{2\pi f_0t})
\end{aligned}
$$
$$
\boxed{x_{RF}(t)= \sqrt{2}(x_I(t)\cos{2\pi f_0t} – x_Q(t) \sin{2\pi f_0t})}
$$
RF to ECB quadrature
The section derives the general conversion first. It is then followed by the more practical conversion as well.
In order to obtain the ECB from RF we have to: get the analytic function, down-convert and scale
$$
\begin{aligned}
x(t) &= \frac{1}{\sqrt{2}}x^+(t) e^{-j2\pi f_0t} \\
&= \frac{1}{\sqrt{2}} (x_{RF}(t)+ j\mathcal{H}(x_{RF}(t))(\cos {2\pi f_0t} – j\sin{2\pi f_0t})\\
&= \frac{1}{\sqrt{2}}(x_{RF}(t)\cos{2\pi f_0t}+\mathcal{H}(x_{RF}(t)\sin{2\pi f_0t}) + j (\mathcal{H}(x_{RF}(t)\cos{2\pi f_0t} – x_{RF}(t)\sin{2\pi f_0t})
\end{aligned}
$$
Thus we have
$$
\begin{align}
\boxed{x_I(t) = \frac{1}{\sqrt{2}}(x_{RF}(t)\cos{2\pi f_0t}+\mathcal{H}(x_{RF}(t)\sin{2\pi f_0t})}\\
\boxed{x_Q(t) =\frac{1}{\sqrt{2}}(\mathcal{H}(x_{RF}(t)\cos{2\pi f_0t} – x_{RF}(t)\sin{2\pi f_0t})}
\end{align}
$$
The above expressions can convert any RF frequency down to its ECB in quadrature form. This expression is particularly complex due to the presence of Hilbert transformer and so many terms. Firstly a Hilbert transformer which spans the entire frequency range is not practical and these many terms complicate the process.
Intuition behind the simpler version
This can be simplified if we make an assumption that the our signal of interest is usually band limited. This assumption simplifies the case as we don’t have to bother about the entire spectrum from 0 to infinity. In other words, the spectrum is 0 everywhere other than the band of interest. Under this assumption one can avoid the Hilbert transform but instead use a low pass filter.
Mathematical proof
From the basic definition of ECB we have
$$
\begin{align}
X(f) = \frac{1}{\sqrt{2}}(1 + sgn(f+f_0))X_{RF}(f+f_0)
\end{align}
$$
The signum function is 0 below \(-f_0\) and 1 after that. But since the signal is band-limited this implies that the signal after down-conversion resides only between \(-f_0 \leftrightarrow f_0\) . Thus \((1 + sgn(f+f_0)\) can be replaced by a rectangle centered at 0 with a width of \(2f_0\). This in frequency domain is a low pass filter.
Thus we have
$$
\begin{align}
X(f) &= \sqrt{2}.rect\left( \frac{f}{2f_0} \right)X_{RF}(f+f_0) \quad \text{[Since $(1 + sgn(f+f_0)$ scales to 2]}\\
\end{align}
$$
Taking inverse Fourier transform
$$
\begin{align}
x(t) &= \sqrt{2}h_{LP}(t)*(x_{RF}(t)e^{-j2\pi f_0t})\\
&= \sqrt{2}h_{LP}(t)*(x_{RF}(t)(\cos{(-2\pi f_0t)} + \sin{(-2\pi f_0t)})\\
\end{align}
$$
$$
\begin{align}
\boxed{x_I(t) = \sqrt{2}h_{LP}(t)*(x_{RF}(t)(\cos{(2\pi f_0t)}}\\
\boxed{x_Q(t) = -\sqrt{2}h_{LP}(t)*(x_{RF}(t)(\sin{(2\pi f_0t)}}\\
\end{align}
$$
This is the same as what was described in the intuition section.
Analysis on synchronization error
We now have expressions to go from ECB to RF and back. ECB to RF is done at the transmitter and RF to ECB is done at the receiver. Often there will be an offset in the frequency and phase of the oscillators at the TX and RX. This section we shall see how the received signal gets impacted.
Let us assume that ECB signal at the transmitter be
$$
s(t) = s_I(t) + j s_Q(t)
$$
Thus we get the transmitted RF as
$$
s_{RF}(t) = \sqrt{2}Re(s(t)e^{j2\pi f_0t})
$$
The received signal now is given by
$$
\begin{align}
r(t) &= \sqrt{2}h_{LP}(t) * (s_{RF}(t)e^{-j2\pi (f_0 + \Delta f)t + \phi })\\
&= 2 h_{LP}(t)*[(s_I(t)\cos({2\pi f_0 t}) – s_Q(t)sin({2\pi f_0 t}))(\cos({2\pi (f_0 + \Delta f)t + \phi })-j\sin({2\pi (f_0 + \Delta f)t + \phi }))]\\
&=s_I(t)(\cos(2\pi\Delta ft+\phi)+j \sin (2\pi\Delta ft+\phi)) + s_Q(t)(\cos(2\pi\Delta ft+\phi)+j \sin (2\pi\Delta ft+\phi))\\
&= (s_I(t) + js_Q(t))e^{-j2\pi \Delta ft + \phi }\\
&= s(t)e^{-j2\pi \Delta ft + \phi }
\end{align}
$$
Thus we have that the received signal is multiplied by a complex exponential whose frequency is the difference in frequency and phase is the difference in phase
$$
\boxed{r(t) = s(t)e^{-j2\pi \Delta ft + \phi }}
$$
ECB through LTI
Motivation
Any digital system would make use of filters. Using the ECB representation we are able to bring the RF frequency down to the base-band. Now we would like to determine the equivalent filter that should operate in base-band frequencies so that we get the same response as the filter at the HF frequency.
We would like to initially know if such a system exists. If it does then we would like to know how to build such a system.
Proof of existence
We start with a filter in RF frequency with the following setup at the RF frequency
$$
Y_{RF}(f) = X_{RF}(f)H_{RF}(f)
$$
Let us assume that we have a ECB works the same way for H(f) as it works for X(f) and Y(f) we should get
$$
Y(f) = H(f)X(f)
$$
where
$$
\begin{align}
X_{RF}(f) &= \frac{1}{\sqrt{2}}(X(f-f_0) + X^*(-(f+f_0)))\\
Y_{RF}(f) &= \frac{1}{\sqrt{2}}(Y(f-f_0) + Y^*(-(f+f_0)))\\
H_{RF}(f) &= c_h(H(f-f_0) + H^*(-(f+f_0)))\\
\end{align}
$$
If this was true, then we should be able to start with the RHS and get the LHS.
$$
\begin{align}
X_{RF}(f)H_{RF}(f) &= \frac{c_h}{\sqrt{2}}(X(f-f_0) + X^*(-(f+f_0)))(H(f-f_0) + H^*(-(f+f_0)))\\
&= \frac{c_h}{\sqrt{2}}(X(f-f_0)H(f-f_0) + X^*(-(f+f_0))H^*(-(f+f_0)) \\
& +X(f-f_0)H^*(-(f+f_0)) + X^*(-(f+f_0))H(f-f_0))\\
&= \frac{c_h}{\sqrt{2}}(X(f-f_0)H(f-f_0) + X^*(-(f+f_0))H^*(-(f+f_0)) \\
&= \frac{c_h}{\sqrt{2}}(Y(f-f_0)+ Y^*(-(f+f_0)))
\end{align}
$$
this will be equal to \(Y_{RF}(f)\) if \(c_h = 1\)
So we have,
$$
\begin{align}
H_{RF}(f) &= (H(f-f_0) + H^*(-(f+f_0)))\\
\end{align}
$$
[!note]
There is a difference in scaling between the H and X.
The difference in scaling makes the ECB for transfer function different by a scaling factor.
RF to ECB:
$$
\boxed{h(t)= \frac{1}{2}[h_{RF}(t) + j\mathcal{H}(h_{RF}(t))]e^{-j2\pi f_0t}\}}
$$
ECB to RF:
$$
\boxed{
h_{RF}(t) = 2Re(h(t)e^{j2\pi f_0t}) = h(t)e^{j2\pi f_0t} + h^*(t)e^{-j2\pi f_0t}
}
$$
ECB filter action
The ECB representation of both the input and the filter response in complex in the ECB. Hence they can be expanded to see how it can be realized
$$
\begin{align}
h(t) &= h_I(t) + jh_Q(t)\\
x(t) &= x_I(t) + jx_Q(t)\\
y(t) &= h(t)*x(t)\\
&= ( h_I(t) + jh_Q(t))*(x_I(t) + jx_Q(t))\\
&= (x_I(t)*h_I(t)-x_Q(t)*h_Q(t)) + j(x_Q(t)*h_I(t) + x_I(t)*h_Q(t))
\end{align}
$$
Simplification
There are the following steps to get the ECB from RF
- Get the analytic signal
- Translate down
- scale
Since the input signal is already analytic for a ECB system, we can relax the condition on the transfer function. This means for the transfer function we can skip step 1 and the end result would still be the same filter provided that the input is ECB.
They an alternate simpler system
$$
\tilde{H}(f) = H_{RF}(f)e^{-j2\pi f_0t}
$$
will perform the same way as H(f).
Summary
We see that ECB is a convenient representation of the signals in baseband. This helps in studying the modulation independent of the carrier frequency. We also see how the signals can be transformed between these two states. Finally, we also deduce the effects of mismatch frequency and phase between the transmitter and the receiver on the ECB representation.
Reference
- Fischer, R. F. H., & Huber, J. B. (2024). Digital Communications: A Foundational Approach. Cambridge: Cambridge University Press.
