How can we find the signal power of $x_{out}(t)=\frac{1}{2}Am(t)\cos \theta$?

I am trying to solve the following question: I have partially worked out the problem and we have: $$x_{out}(t)=\frac{1}{2}Am(t)\cos \theta,$$ which is the output of the LPF.

and $$x(t)=Am(t)\cos (2\pi f_c t)$$

My problem is that how can I find out the power of $x_{out}(t)$ and $x(t)$ because I don't know the amplitude of $m(t)$. Moreover, if $m(t)$ is not a singletone frequency signal, how can we find out the power of $x_{out}(t)$ and $x(t)$ ?

In short my question is:

How can we find the power of

$x_{out}(t)=\frac{1}{2}Am(t)\cos \theta$

and

$x(t)=Am(t)\cos (2\pi f_c t)$?

• Welcome to SE.DSP! The question just asks about the ratio of $\frac{P_{\rm out}}{P_\mu}$, so do you need to know the power in $m(t)$ at all? – Peter K. Apr 9 '18 at 12:48
• @PeterK. The average power of signal $x(t)$ is defined as: $$P=\lim_{T\to 0}\frac{1}{2T}\int_{-T}^T |x(t)|^2dt$$ Therefore: $$P_{\mu}=\lim_{T\to 0}\frac{1}{2T}\int_{-T}^T |x(t)|^2dt$$ $$P_{\mu}=\lim_{T\to 0}\frac{1}{2T}\int_{-T}^T |Am(t)\cos (2\pi f_c t)|^2dt$$ and $$P_{out}=\lim_{T\to 0}\frac{1}{2T}\int_{-T}^T |x_{out}(t)|^2dt$$ $$P_{out}=\lim_{T\to 0}\frac{1}{2T}\int_{-T}^T |\frac{1}{2}Am(t)\cos \theta|^2dt$$ – Anwesa Roy Apr 9 '18 at 13:19
• @PeterK. Therefore: $$\frac{P_{\rm out}}{P_\mu}=\frac{\lim_{T\to 0}\frac{1}{2T}\int_{-T}^T |\frac{1}{2}Am(t)\cos \theta|^2dt}{\lim_{T\to 0}\frac{1}{2T}\int_{-T}^T |Am(t)\cos (2\pi f_c t)|^2dt}$$ Sir, I can't take $m(t)$ out of the integration, as it is dependent on the variable $t$. This is the source of my confusion. – Anwesa Roy Apr 9 '18 at 13:31
• Hint: In general, if $m(t)$ has power $P$, what is the power of $m(t)\cos(2\pi f_c t)$? – MBaz Apr 9 '18 at 13:51
• @MBaz Pardon sir, I am not able to comprehend the answer. – Anwesa Roy Apr 9 '18 at 14:18