信号与系统信号f(t)=2cos(t)+4cos(2t)+2cos(4t+π)的功率是多少

要计算一个信号的功率，需要先求出信号的平均功率，即：

$$P = \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} |f(t)|^2 dt$$

对于给定的信号 $f(t) = 2\cos(t) + 4\cos(2t) + 2\cos(4t+\pi)$，我们可以将其拆分为三个简单的信号：

$$f(t) = 2\cos(t) + 4\cos(2t) - 2\cos(4t)$$

然后，我们可以计算每个信号的功率：

$$\begin{aligned}
P_1 &= \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} |2\cos(t)|^2 dt \\
&= \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} 4\cos^2(t) dt \\
&= \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} 2(1+\cos(2t)) dt \\
&= \lim_{T \to \infty} \frac{1}{T} \left[2t + \frac{1}{2}\sin(2t)\right]_{-T/2}^{T/2} \\
&= \lim_{T \to \infty} \frac{1}{T} (T + \sin(T)/2) \\
&= 1
\end{aligned}$$

$$\begin{aligned}
P_2 &= \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} |4\cos(2t)|^2 dt \\
&= \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} 16\cos^2(2t) dt \\
&= \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} 8(1+\cos(4t)) dt \\
&= \lim_{T \to \infty} \frac{1}{T} \left[8t + 2\sin(4t)\right]_{-T/2}^{T/2} \\
&= \lim_{T \to \infty} \frac{1}{T} (4 + \sin(2)) \\
&= 4
\end{aligned}$$

$$\begin{aligned}
P_3 &= \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} |2\cos(4t)|^2 dt \\
&= \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} 4\cos^2(4t) dt \\
&= \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} 2(1+\cos(8t)) dt \\
&= \lim_{T \to \infty} \frac{1}{T} \left[2t + \frac{1}{16}\sin(8t)\right]_{-T/2}^{T/2} \\
&= \lim_{T \to \infty} \frac{1}{T} (T/8) \\
&= 0
\end{aligned}$$

因此，总功率 $P$ 等于这三个功率的和：

$$P = P_1 + P_2 + P_3 = 1 + 4 + 0 = 5$$

所以给定信号的功率为5。