Question

Let $m(t)$ be a strictly band-limited signal with bandwidth $B$ and energy $E$. Assuming $\omega_0=10B$, the energy in the signal $m(t)\cos \omega_0 t$ is

• A. $\dfrac{E}{4}$
• B. $\dfrac{E}{2}$
• C. $E$
• D. $2E$

Hint:

Multiplying by $\cos\omega_0 t$ splits the spectrum into two non-overlapping sidebands (for large $\omega_0$). Each carries half the energy $\Rightarrow$ total energy $=E/2$.

Answer 1

The Correct Option is B

Step 1: Modulation in frequency domain.
Let $y(t)=m(t)\cos\omega_0 t$. Then \[ Y(\omega)=\tfrac{1}{2}\big[M(\omega-\omega_0)+M(\omega+\omega_0)\big]. \] Since $m(t)$ is strictly band-limited to $|\omega|\le B$ and $\omega_0=10B\gg B$, the two shifted spectra are disjoint. 
Step 2: Energy via Parseval.
\[ E_y=\int_{-\infty}^{\infty}|y(t)|^2\,dt=\frac{1}{2\pi}\int_{-\infty}^{\infty}|Y(\omega)|^2\,d\omega. \] With disjoint supports, cross-terms vanish: \[ |Y(\omega)|^2=\tfrac{1}{4}\big(|M(\omega-\omega_0)|^2+|M(\omega+\omega_0)|^2\big). \] Thus \[ E_y=\frac{1}{2\pi}. \tfrac{1}{4}\!\left(\!\int |M(\omega-\omega_0)|^2 d\omega + \int |M(\omega+\omega_0)|^2 d\omega\!\right) =\tfrac{1}{4}(E+E)=\frac{E}{2}. \] \[ \boxed{\dfrac{E}{2}} \]