Theoretically, the passband $y(t)$ signal equivalent into baseband signal $x(t)$ is given by:

$y(t) = \Re \begin{bmatrix} x(t) e^{(-i2 \pi f_ct )} \end{bmatrix}$

where $\Re$ represents the real part.

My question how can we recover the complex signal $x(t)$ from the real part of $y(t)$. I know it's possible by multiplying by the conjugate of same value but I need to understand an example.

let's have $x(t) = 0.2 + 0.6i$ and let's assume also the value of $f_ct$ known, let's put it $f_ct = 1$ so, the results signal $y(t) = 0.2$. So, how can we recover now the complex signal $x(t)$ from $y(t)$ ?

NP: The answer here explains that process well HERE , but I couldn't understand that by an example, for example I couldn't recover the complex signal based on the real passband signal using the signal in the above example.

  • $\begingroup$ Do you know the concept of I/Q downconversion? This + lowpass filtering solves your problem. Btw fc*t = 1 is not possible since t is variable and fc is a fixed carrier frequency. $\endgroup$
    – Don
    Commented Apr 16, 2022 at 8:22
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    $\begingroup$ Does this answer your question? IQ modulation: baseband to bandpass transformation $\endgroup$ Commented Apr 16, 2022 at 9:28
  • $\begingroup$ Your example $y(t)$ is wrong; if $x(t)$ is a constant, then $y(t)$ must be something that oscillates with $f_c$, and not a constant. You can't say "$f_ct = 1$", that is non-sensible, as $t$ is the running time variable and $f_c$ is a constant. So, I think getting that part correct will explain the other direction (passband->baseband) as well . $\endgroup$ Commented Apr 16, 2022 at 9:34
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    $\begingroup$ Does this answer your question? Real and complex low pass equivalent of band-pass signal $\endgroup$ Commented Apr 16, 2022 at 12:25
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    $\begingroup$ Ok - I think I understand where you are stuck; I’ll post an example shortly that should help you assuming you read those other links. $\endgroup$ Commented Apr 16, 2022 at 16:28

1 Answer 1


In order for $x(t)$ as a passband signal to be equivalent to a complex signal, it must be a passband signal. The example the OP gave is ultimately multiplying by a constant, which is DC, which is a real baseband signal. The OP is considering it just one sample of the passband signal, but in this case what is being neglected is the requirement for a low pass filter or the creation of the analytic signal with the Hilbert transform to restore the baseband signal, both of which require more than one sample to recover. I explain this in greater detail below where I demonstrate how the baseband signal is recovered.

I believe the following graphics will clear this up, given the mathematical explanation is already quite clear from MattL's nice answer at this post.

There are some key points that need to be understood first.

Point 1: Any individual impulse in the frequency domain at a frequency $\omega$ is a component of $e^{j\omega t}$ not cosines or sines. $e^{j\omega t}$ is a complex signal with real and imaginary components: $e^{j\omega t} = I + jQ$. $e^{j\omega t}$ has a magnitude of $1$ and a phase given by $\omega t$.

Point 2: All real signals are complex conjugate symmetric (also called "Hermitian symmetric) in the frequency domain: the magnitudes of all signals on the negative frequency axis will be equal in magnitude and opposite in phase to the signals in the positive frequency axis, at the same value for $|f|$ . This means we already know what the negative frequency will be from the positive frequency alone (and vice versa). Euler's Formula is a great example for both of these points, showing the relationship between a sinusoid and the two impulses we would get with it's Fourier Transform, demonstrating how each impulse in the frequency domain is $e^{j\omega t}$ in the time domain:

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

Point 3: "Baseband Signals" are the signal of interest with regards to modulation and demodulation. We use Baseband Signals to modulate a carrier frequency and the result of this is a "Passband Signal". "Modulation" is multiplying a signal by a carrier frequency. Baseband Signals are typically centered about DC ($f=0$) but that is not a hard requirement, while Passband Signals shift the baseband signal to the carrier by translating $f=0$ to $f_c$. ($\omega= 2\pi f$). The baseband signal MUST be complex if the spectrum is not Hermitian symmetric (point 2), but the passband signal can be real or complex depending on how we modulate (and both carry all the same information as the baseband signal:

Complex Baseband Signal with AM ( real $A(t)$) and PM ($\phi(t)$)

$$m(t) = A(t)e^{j\phi(t)}$$

Complex Passband Signal: modulate carrier $x(t) = e^{j\omega_c t}$:

$$s_1(t) = m(t)x(t) = A(t)e^{j\phi(t)}e^{j\omega_c t} = A(t)e^{j(\omega_c t + \phi(t))}$$

Real Passband Signal: Take the real component of the above modulated result (This is IQ modulation as we typically see it)

$$s_2(t) = Re\{s_1(t)\} = Re\{A(t)e^{j(\omega_c t + \phi(t))}\}$$

$$ =Re\{A(t)(\cos(\omega_c t + \phi(t) + j \sin (\omega_c t + \phi(t))\} $$

$$ = A(t)\cos (\omega _ct + \phi(t))$$

Given those main points, observe the modulation process in which we can create a real passband signal from a complex baseband signal and having all the signal of interest $m(t)$ intact. T

passband and baseband signal

The OP's example starts with a single complex impulse at DC in the frequency domain (a constant in the time domain) with a magnitude and phase as given by $0.2 + j0.6$. This is the "Complex Baseband Signal". To create the passband signal we need to multiply it by $e^{j\omega_c t}$ and then take the real part as done graphically above.

More Mathematical Details and Demonstrating Recovery of Baseband Signal

Given baseband signal:

$$x(t) = 0.2 + j0.6 \tag{1}\label{1}$$

The passband signal at carrier $\omega_c$ is

$$y(t) = Re\{x(t)e^{j\omega_c t}\}$$:

Which will reduce to the following:

$$ = Re\{ (0.2 + j0.6) (\cos(\omega_c t) + j \sin(\omega_c t) \}$$

$$ = 0.2 \cos(\omega_c t) - 0.6\sin(\omega_c t) \tag{2} \label{2}$$

To then recover the baseband signal, we simply multiply by $2e^{-j\omega_c t}$ and then low pass filter:

$$y(t)2e^{-j\omega_c t} = (0.2 \cos(\omega_c t) - 0.6\sin(\omega_c t))(2\cos(\omega_c t) - j2\sin(\omega_c t)) \tag{3} \label{3}$$

One tedious approach is to use the sine and cosine product rules to complete the above cosine and sine products which will result in restoring the baseband signal and also have a high frequency copy of at $2\omega_c$ which we remove with a low pass filter. The cosine and sine product rules are:

$$\cos(a)\cos(b) = \frac{1}{2}\bigg(\cos(a+b)+\cos(a-b))\bigg)$$ $$\sin(a)\sin(b) = \frac{1}{2}\bigg(\cos(a-b)-\cos(a+b))\bigg)$$ $$\sin(a)\cos(b) = \frac{1}{2}\bigg(\sin(a+b)+\sin(a-b))\bigg)$$ $$\cos(a)\sin(b) = \frac{1}{2}\bigg(\sin(a+b)-\sin(a-b))\bigg)$$

Multiplying out \ref{3} and grouping real and imaginary terms together we get:

$$y(t)2e^{-j\omega_c t} = 0.2 \cos(\omega_c t)2\cos(\omega_c t) - 0.6\sin(\omega_c t)2\cos(\omega_c t) - 0.2 \cos(\omega_c t)j2\sin(\omega_c t) + 0.6\sin(\omega_c t)j2\sin(\omega_c t)$$

$$= 0.2 \bigg(\cos(2\omega_c t) + \cos(0)\bigg) - 0.6\bigg(\sin(2\omega_c t)+\sin(0)\bigg) - 0.2j\bigg(\sin(2\omega_c t)-\sin(0)\bigg) + 0.6j\bigg( \cos(0) - \cos(2\omega_c t)\bigg)$$

$$ = 0.2 \bigg(\cos(2\omega_c t) + 1 \bigg) - 0.6\bigg(\sin(2\omega_c t)+0\bigg)- 0.2j\bigg(\sin(2\omega_c t)-0\bigg) + 0.6j\bigg( 1 - \cos(2\omega_c t)\bigg)$$

$$ = 0.2 + j0.6 - 0.2\sin(2\omega_c t) - j0.6\cos(2\omega_c t)$$

$$ = x(t) - (0.6\sin(2\omega_c t) + j0.6\cos(2\omega_c t)) \tag{4} \label{4}$$

Thus we see how each individual sample of $y(t)2e^{-j\omega_c t}$ is the sum of $x(t)$ and a high frequency signal at frequency $\omega_c$. To recover $x(t)$ from the passband signal, a filter is required to separate these, which necessitates multiple samples (not just one). Trying to recover $x(t)$ from one sample of the product is the same as asking what the variables $a$ and $b$ are in the equation $a+b=5$!

Alternatively in implementation and with the mathematical analysis we can recover the baseband signal from the real passband signal using the analtic signal which is given as:

$$y_a(t) = y(t) + j\hat y(t)$$

Where $y_a(t)$ is the "analytic signal" given by the above relationship, and $\hat y(t)$ is the Hilbert Transform of $y$ (which I will note also requires multiple samples to solve for in implementation).

It is easy to see combined with the graphical description above that the analytic signal, which is a complex passband signal, is created by simply removing the negative frequency component and doubling. Note that the complex baseband signal as $0.2+j0.6$ has a magnitude of $\sqrt{0.2^2+0.6^2} = \sqrt{0.4}$ and a phase of $tan^{-1}(.6/.2) \approx 1.249$ radians, and therefore $x(t)=\sqrt{0.4}e^{j1.249}$. From that it's easier to see how we get from baseband to passband just by multiplying by $e^{j\omega_c t}$ in one direction and $e^{-j\omega_c t}$ in the reverse. As mentioned, once it is passband, if it is real we can simply double it and eliminate the negative frequency to restore this complex signal, which is done in practical implementations with the Hilbert Transform. An example of this is Euler's Formula with the relationship between the real $cos(\omega_c t)$ and $e^{j\omega_c t}$.

So doing it this way, with complex exponentials, becomes trivial:

$$x(t)=0.2+j0.6 = \sqrt{0.4}e^{j1.249}$$ $$y(t) = Re\{x(t)e^{j\omega_c t}\} = Re\{\sqrt{0.4}e^{j(\omega_c t + 1.249)}\}$$ $$=\sqrt{0.4}\cos(\omega_c t + 1.249))= 0.2 \cos(\omega_c t) - 0.6\sin(\omega_c t)$$

$$y_a(t) = \sqrt{0.4}e^{j(\omega_c t + 1.249)}$$

$$x(t) = y_a(t)e^{-j\omega_c t} = \sqrt{0.4}e^{j1.249} $$

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    $\begingroup$ Thank you so much for your detailed answer. Sorry for agreeing it late since I was reading it and the links you provided, it seems I understood it. But, is it possible to continue now and get $x(t)$ back from $y(t)$ I put $w_ct$ = $45$. $i.e$ if we had $y(t) = 0.2*cos(pi/4) - 0.6*sin(pi/4) = -0.2828$, so we should be able to recover the complex value of $x(t)$ based on $y(t) = -0.2828$ $\endgroup$
    – Sajjad
    Commented Apr 16, 2022 at 17:59
  • $\begingroup$ I got what you mean, I now understand it well. But when I tried it by matlab (or mathematically), I can't get $x(t)$ back. For example let $w_ct$ be $45$, so $y(t) = 0.2cos(45) - 0.6sin(45) = -0.4055$ . Then to recover $x(t)$, we will have $x(t) = -0.4055 * e^{-j45} = -0.2130 + 0.3450i$. which is not the same of $x(t)$. $\endgroup$
    – Sajjad
    Commented Apr 18, 2022 at 0:21
  • $\begingroup$ @Sajjad What you are doing specifically is multiplying by a constant (which is DC! -- that is still a baseband signal.) To make it a passband signal you need to increment $t$ with time so not multiply by a constant 45 but by a waveform that appropriately changes with time. And then keep in mind to recover you will also need to low pass filter. $\endgroup$ Commented Apr 18, 2022 at 0:29
  • $\begingroup$ But I am considering here $t = 1$ which, I think, should not be an issue. I mean $t$ should normally increment following the length of base-band signal $x(t)$. Unless, could you please edit your answer by giving any example for $w_ct$ and then recover $x(t)$ accordingly ? $\endgroup$
    – Sajjad
    Commented Apr 18, 2022 at 0:42
  • $\begingroup$ In order to low pass filter which requires convolution, meaning memory, you have to consider the waveform over a time span. You are computing one sample, but you are not able to separate the low frequency and high frequency components in that sample without also considering previous time samples. $\endgroup$ Commented Apr 18, 2022 at 0:44

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