Handout F: Fundamental Theorem of Linear Systems

Theorem: Let a linear time-invariant system $g$ has an $e_f(t)$ denote the complex sinusoid $e^{j 2 \pi ft}$. Then, $g(e_f(.),t)=g(e_f(.),0)e_f(t)=c   e_f(t)$.

Example: Analog RC Lowpass Filter

Figure 1: A First-Order Analog Lowpass Filter
\begin{figure}\epsfxsize = 10cm
\centering\leavevmode
\epsffile{circuit.eps}\end{figure}

The impulse response for the circuit in Fig. 1, i.e. the output measured at $y(t)$ when $x(t) = \delta(t)$, is


\begin{displaymath}
h(t)=\frac{1}{RC}e^{-\frac{1}{RC}t}u(t)
\end{displaymath}

For a complex sinusoidal input, $x(t)=e_f(t)=e^{j2\pi ft}$,


$\displaystyle y(t)$ $\textstyle =$ $\displaystyle \int_{-\infty}^\infty x(t-\lambda)h(\lambda) d\lambda$  
  $\textstyle =$ $\displaystyle \int_{-\infty}^\infty e^{j2\pi f(t-\lambda)}\frac{1}{RC}e^{-\frac{1}{RC}\lambda}u(\lambda) d\lambda$  
  $\textstyle =$ $\displaystyle e^{j2\pi ft} \left[
\frac{1}{RC}\int_{-\infty}^\infty
e^{-j2\pi f\lambda}e^{-\frac{1}{RC}\lambda} d\lambda \right]$  
  $\textstyle =$ $\displaystyle \left[ \frac{\frac{1}{RC}}{{j2\pi f+\frac{1}{RC}}} \right]
e^{j2\pi ft}$  
  $\textstyle =$ $\displaystyle g(e_f(.),0)   e_f(t)$  

So, $g(e_f(.),0)= H(f)$, which is the transfer function of the system.

Placeholder - please ignore.





Brian L. Evans