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Is there any difference in energy content of baseband and bandpass signals? When we see their representation in frequency domain, it looks like a bandpass signal has double the energy content compared to basband. But my friend says there's no difference. Someone please clarify.

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  • $\begingroup$ does your friend believe in the existence of negative frequencies? if yes, he/she is wrong ("says there's no difference"). if no, then your friend is folding over the "negative frequency" content onto the image that exists in positive frequency and is counting the energy there. $\endgroup$ – robert bristow-johnson May 7 '14 at 15:57
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    $\begingroup$ I don't think something as physical as energy depends on your belief regarding negative frequencies. $\endgroup$ – Jazzmaniac May 7 '14 at 18:09
  • $\begingroup$ What does Parseval's theorem and the Fourier transform pair $$x(t)\cos(2\pi f_ct) \leftrightarrow \frac{1}{2}[X(f+f_c) + X(f-f_c)]$$ have to say about the matter especially when we assume that for all $f$, $X(f+f_c)^{*}X(f-f_c)=0$? $\endgroup$ – Dilip Sarwate May 7 '14 at 18:31
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    $\begingroup$ @Talasila If $X(f)$ is nonzero only when $|f|<W$, then $X(f+f_c)=0$ except when $-f_c-W<f<-f_c+W$ while $X(f-f_c)=0$ except when $f_c-W<f<f_c+W$. If we assume that the carrier frequency $f_c>W$, then $-f_c+W<0$ and $f_c-W>0$ meaning that $X(f+f_c)$ (ditto $X(f+f_c)^*$) and $X(f-f_c)$ both cannot be nonzero for the same value of $f$, i.e., $X(f+f_c)^*X(f-f_c)=0$ for all $f$. Thus, Parseval's theorem gives the energy of the modulated signal as $$\int_{-\infty}^\infty \frac 14|X(f+fc)+X(f−fc)|^2\,df=\int_{-\infty}^\infty\frac 14|X(f+f_c)|^2+|X(f-f_c)|^2\,df=\frac 12|X(f)|^2=\frac 12 E_x.$$ $\endgroup$ – Dilip Sarwate May 8 '14 at 13:13
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    $\begingroup$ @Talasila I stand by the correctness of my formula. The $\frac{1}{2\pi}$ that you want is needed when Fourier transforms are defined as functions of radian frequency $\omega$, but are not there when Fourier transforms are defined as functions of Hertzian frequency $f$. $\endgroup$ – Dilip Sarwate May 8 '14 at 18:35
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To give a very simple example: let us consider random signals and let's assume that $x(t)$ is a zero-mean wide-sense stationary real-valued baseband signal with power $E[x^2(t)]$. Define a modulated bandpass signal

$$y(t)=x(t)\cos(\omega_0t+\theta)$$

where $\theta$ is a random phase uniformly distributed in the interval $[0,2\pi)$, which is independent of the signal $x(t)$.

Then we get for the power of $y(t)$

$$E[y^2(t)]=E[x^2(t)\cos^2(\omega_0t+\theta)]=E[x^2(t)]E[\cos^2(\omega_0t+\theta)]=\frac{1}{2}E[x^2(t)]$$

So for this simple example you were both wrong.

(And the same holds of course also for the energy of deterministic signals, as hinted at in Dilip Sarwate's comment.)

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  • $\begingroup$ The key point is that $\cos(2\pi f_c t)$ is a signal of power $\frac 12$ watts unless you are a power systems engineer who uses only the r.m.s. value for the amplitude so that $P = VI$ or $P = V^2/1\Omega$ works for DC as well as for AC. Communications engineers do not follow this tradition which is the leading cause of all those $\sqrt{2P}\cos(2\pi f_ct)$'s that we see floating around in the literature. $\endgroup$ – Dilip Sarwate May 7 '14 at 20:09
  • $\begingroup$ @Matt so from the answer do you mean that the energy of bandpass signal is neither the same or double the initial energy. So do you say it's not determinable as a general case. $\endgroup$ – Karan Talasila May 8 '14 at 1:20
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    $\begingroup$ @Talasila: What I'm saying is that the energy/power of the bandpass signal is half the energy/power of the baseband signal, if we use the model $y(t)=x(t)\cos\omega_0t$. $\endgroup$ – Matt L. May 8 '14 at 11:16
  • $\begingroup$ @MattL. I got it from clear explanations of yours and Dilip Sarwate. On a more intuitive note, where did half the energy vanish. It's just modulation and shift of frequency. Is the distribution of cos(x) responsible for the reduction of energy. $\endgroup$ – Karan Talasila May 8 '14 at 17:17
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    $\begingroup$ @Talasila: It's the cosine which contributes the factor $1/2$. Note that $\cos^2 x=(1+\cos 2x)/2$. After averaging the double frequency term disappears and you're left with the factor $1/2$. $\endgroup$ – Matt L. May 8 '14 at 17:55
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Let $x(t)$ denote a real-valued finite-energy signal of energy $E_x$, that is,

$$\int_{-\infty}^\infty |x(t)|^2\,\mathrm dt = E_x < \infty \tag{1}$$

and let $X(f)$ denote the Fourier transform of $x(t)$. Parseval's theorem assures us that

$$\int_{-\infty}^\infty |x(t)|^2\,\mathrm dt = \int_{-\infty}^\infty |X(f)|^2\,\mathrm df \tag{2}$$

and thus the energy $E_x$ can also be computed via the right side of $(2)$.

Now consider the DSB-AM (Double-Side-Band Amplitude-Modulated) signal $x(t)\cos(2\pi f_c t)$. Its energy is

$$\begin{align} \int_{-\infty}^\infty |x(t)\cos(2\pi f_c t)|^2\,\mathrm dt &= \int_{-\infty}^\infty |x(t)|^2|\cos(2\pi f_c t)|^2\,\mathrm dt\\ &= \frac 12 \int_{-\infty}^\infty |x(t)|^2[1 + \cos(2\pi 2f_c t)]\,\mathrm dt\\ &= \frac 12 E_x + \int_{-\infty}^\infty |x(t)|^2\cos(2\pi 2f_c t)\,\mathrm dt\tag{3} \end{align}$$ Now, if the value of the integral on the right side of $(3)$ is exactly zero, then the modulated signal $x(t)\cos(2\pi f_c t)$ has exactly half the energy of the unmodulated signal $x(t)$. In the time domain, it can be plausibly argued that if the integral is not identically zero, it is at least very small when $x(t)$ is a low-pass signal and $f_c$ is very large compared to the highest frequency in $x(t)$. Denote by $T = (2f_c)^{-1}$ the period of the double-frequency sinusoid in $(3)$ and consider that $$\int_{-\infty}^\infty |x(t)|^2\cos(2\pi 2f_c t)\,\mathrm dt = \sum_{n=-\infty}^\infty \int_{nT}^{(n+1)T}|x(t)|^2\cos(2\pi 2f_c t)\,\mathrm dt. \tag{4}$$ Now, $x(t)$ is a slowly-varying signal that does not change very much during $[nT, (n+1)T]$ and so $$\int_{nT}^{(n+1)T}|x(t)|^2\cos(2\pi 2f_c t)\,\mathrm dt \approx |x(nT)|^2 \int_{nT}^{(n+1)T}\cos(2\pi 2f_c t)\,\mathrm dt \approx 0.$$ Note also that $\displaystyle \int \cos(2\pi 2f_c t)\,\mathrm dt = -\frac{1}{4\pi f_c}\sin(2\pi 2f_c t)$ so that any contribution from $\int_{nT}^{(n+1)T}|x(t)|^2\cos(2\pi 2f_c t)\,\mathrm dt$, which we have already argued is going to be small is further tamped down towards $0$ by that $\frac{1}{4\pi f_c} << 1$ factor.

Calculations such as these are more easily handled in the frequency domain. We have the Fourier transform pair $$x(t)\cos(2\pi f_c t) \leftrightarrow \left.\left.\frac 12\right[X(f+f_c)+X(f-f_c)\right]$$ so that Parseval's theorem gives us that $$\begin{align} \int_{-\infty}^\infty |x(t)\cos(2\pi f_c t)|^2\,\mathrm dt &= \frac 14 \int_{-\infty}^\infty |X(f+f_c)+X(f-f_c)|^2\,\mathrm df\\ &= \frac 14 \left[\int_{-\infty}^\infty |X(f+f_c)|^2\,\mathrm df + \int_{-\infty}^\infty |X(f-f_c)|^2\,\mathrm df \right.\\ &= \qquad \left. + \int_{-\infty}^\infty X(f+f_c)X^*(f-f_c) +X^*(f+f_c)X(f-f_c)\,\mathrm df\right]\\ &= \frac 12 E_x + \frac 14 \left[\int_{-\infty}^\infty X(f+f_c)X^*(f-f_c) +X^*(f+f_c)X(f-f_c)\,\mathrm df\right].\tag{5} \end{align}$$ We now argue (as in the comments) that if the support of $X(f)$ is $[-W,W]$ and $f_c > W$, then the supports of $X(f+f_c)$ and $X(f-f_c)$ are $[-f_c-W, -f_c+W]$ and $[f_c-W, f_c+W]$ respectively, and since $-f_c+W < 0 < f_c-W$, the supports do not overlap so that the integrand in $(5)$ has value $0$ for all $f$. Thus, $f_c > W$ is an explicit sufficient condition for the energy in $x(t)\cos(2\pi f_c t)$ to be half of the energy in $x(t)$. The modulation process "loses" half the energy because $\cos(2\pi f_c t)$ is a carrier signal of power $\frac 12$; if we had used a unit-power signal such as $\sqrt{2}\cos(2\pi f_c t)$, there would be no "loss".

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