Signal Energy in Modulated Waves

Understanding the energy content of signals is fundamental in communication systems and signal processing. When a signal is modulated, its energy distribution can change. This problem asks us to find the energy in a modulated signal \(m(t) \cos \omega_0 t\), given the energy of the original band-limited signal \(m(t)\).

Band-Limited Signal Fundamentals

A band-limited signal \(m(t)\) has its spectrum \(M(\omega)\) confined to a finite frequency range. For this problem, the bandwidth is \(B\), meaning \(M(\omega) = 0\) for \(|\omega| > B\).

• The energy \(E\) of the signal \(m(t)\) is given in the time domain as: \[E = \int_{-\infty}^{\infty} |m(t)|^2 dt\]
• By Parseval's theorem, the energy can also be calculated from its Fourier Transform \(M(\omega)\) in the frequency domain: \[E = \frac{1}{2\pi} \int_{-\infty}^{\infty} |M(\omega)|^2 d\omega\]

Modulated Signal Analysis

We are given the modulated signal \(y(t) = m(t) \cos \omega_0 t\). To find its energy, we first need to determine its Fourier Transform \(Y(\omega)\).

• The Fourier Transform of \(\cos(\omega_0 t)\) is \(\pi[\delta(\omega - \omega_0) + \delta(\omega + \omega_0)]\).
• Using the multiplication property of Fourier Transforms, if \(y(t) = m(t) \cdot f(t)\), then \(Y(\omega) = \frac{1}{2\pi} [M(\omega) * F(\omega)]\), where \( * \) denotes convolution.
• Applying this to \(y(t) = m(t) \cos \omega_0 t\): \[Y(\omega) = \frac{1}{2\pi} \left[ M(\omega) * \pi(\delta(\omega - \omega_0) + \delta(\omega + \omega_0)) \right]\] \[Y(\omega) = \frac{1}{2} \left[ M(\omega) * \delta(\omega - \omega_0) + M(\omega) * \delta(\omega + \omega_0) \right]\] \[Y(\omega) = \frac{1}{2} \left[ M(\omega - \omega_0) + M(\omega + \omega_0) \right]\] This equation shows that the spectrum of the modulated signal \(Y(\omega)\) consists of two shifted copies of the original signal's spectrum \(M(\omega)\), one centered at \(+\omega_0\) and one at \(-\omega_0\).

Calculating Modulated Signal Energy

Now, we use Parseval's theorem to find the energy \(E_y\) of the modulated signal \(y(t)\):

\[E_y = \frac{1}{2\pi} \int_{-\infty}^{\infty} |Y(\omega)|^2 d\omega\] Substitute \(Y(\omega)\): \[E_y = \frac{1}{2\pi} \int_{-\infty}^{\infty} \left| \frac{1}{2} [M(\omega - \omega_0) + M(\omega + \omega_0)] \right|^2 d\omega\] \[E_y = \frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{1}{4} |M(\omega - \omega_0) + M(\omega + \omega_0)|^2 d\omega\] \[E_y = \frac{1}{8\pi} \int_{-\infty}^{\infty} |M(\omega - \omega_0) + M(\omega + \omega_0)|^2 d\omega\] Expand the squared term: \[|A + B|^2 = (A + B)(A^* + B^*) = AA^* + BB^* + AB^* + A^*B = |A|^2 + |B|^2 + 2\text{Re}(AB^*)\] So, \[E_y = \frac{1}{8\pi} \int_{-\infty}^{\infty} \left( |M(\omega - \omega_0)|^2 + |M(\omega + \omega_0)|^2 + 2\text{Re}[M(\omega - \omega_0) M^*(\omega + \omega_0)] \right) d\omega\]

Spectral Separation Impact

A crucial piece of information is \(\omega_0 = 10B\). This condition tells us about the separation of the two spectral components \(M(\omega - \omega_0)\) and \(M(\omega + \omega_0)\):

• The spectrum \(M(\omega - \omega_0)\) exists in the range \((\omega_0 - B, \omega_0 + B)\). Since \(\omega_0 = 10B\), this is \((10B - B, 10B + B) = (9B, 11B)\).
• The spectrum \(M(\omega + \omega_0)\) exists in the range \((-\omega_0 - B, -\omega_0 + B)\). Since \(\omega_0 = 10B\), this is \((-10B - B, -10B + B) = (-11B, -9B)\).

Since the highest frequency of the negative component is \(-9B\) and the lowest frequency of the positive component is \(9B\), there is no overlap between the two spectral components. This means that for any \(\omega\), either \(M(\omega - \omega_0)\) is zero or \(M(\omega + \omega_0)\) is zero, or both are zero.

Therefore, the cross-product term \(2\text{Re}[M(\omega - \omega_0) M^*(\omega + \omega_0)]\) is zero because the non-zero parts of \(M(\omega - \omega_0)\) and \(M(\omega + \omega_0)\) do not coincide.

The energy expression simplifies to:

\[E_y = \frac{1}{8\pi} \int_{-\infty}^{\infty} \left( |M(\omega - \omega_0)|^2 + |M(\omega + \omega_0)|^2 \right) d\omega\] We can split this into two integrals:

\[E_y = \frac{1}{8\pi} \int_{-\infty}^{\infty} |M(\omega - \omega_0)|^2 d\omega + \frac{1}{8\pi} \int_{-\infty}^{\infty} |M(\omega + \omega_0)|^2 d\omega\] For the first integral, let \(u = \omega - \omega_0\), so \(du = d\omega\): \[\int_{-\infty}^{\infty} |M(\omega - \omega_0)|^2 d\omega = \int_{-\infty}^{\infty} |M(u)|^2 du\] For the second integral, let \(v = \omega + \omega_0\), so \(dv = d\omega\): \[\int_{-\infty}^{\infty} |M(\omega + \omega_0)|^2 d\omega = \int_{-\infty}^{\infty} |M(v)|^2 dv\] Both integrals are identical. Let's call the value of one integral \(I\): \[I = \int_{-\infty}^{\infty} |M(\omega)|^2 d\omega\] From Parseval's theorem for \(m(t)\), we know that \(E = \frac{1}{2\pi} \int_{-\infty}^{\infty} |M(\omega)|^2 d\omega\). Therefore, \(\int_{-\infty}^{\infty} |M(\omega)|^2 d\omega = 2\pi E\).

Substituting this back into the expression for \(E_y\): \[E_y = \frac{1}{8\pi} (2\pi E) + \frac{1}{8\pi} (2\pi E)\] \[E_y = \frac{2\pi E}{8\pi} + \frac{2\pi E}{8\pi}\] \[E_y = \frac{E}{4} + \frac{E}{4}\] \[E_y = \frac{2E}{4}\] \[E_y = \frac{E}{2}\]

Thus, the energy in the signal \(m(t) \cos \omega_0 t\) is \(\frac{E}{2}\).

Summary of Energy Calculation

The table below summarizes the key steps and concepts involved in determining the energy of the modulated signal.

The final answer is \(\frac{E}{2}\).