2.9: The Angle of a Transfer Function
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A particularly useful property of the Fourier (and Laplace) transform is that the magnitude of the transfer function scales a sinusoidal input, and the angle of the transfer function adds to the angle of the sinusoidal input. In other words,
\[ u(t) = u_o \cos (\omega_o t + \psi) \longrightarrow \]
\[ y(t) = u_o |H(\omega_o)| \cos (\omega_o t + \psi + \arg (H(\omega_o)) \]
To prove the above relations, we'll use the complex exponential:
\[\begin{align} u(t) &= Re \, (u_o e^{i(\omega_o t + \psi)}) \\[4pt] &= Re \, (\tilde{u}_o e^{i \omega_o t}), \end{align}\]
making \(u_o e^{i \psi} = \tilde{u}_o\) complex; then
\[\begin{align*} y(t) &= h(t) * u(t) \\[4pt] &= \int\limits_{-\infty}^{\infty} h(\tau) u(t - \tau) \, d \tau \\[4pt] &= \displaystyle \int\limits_{-\infty}^{\infty} Re(\tilde{u}_o e^{i \omega_o (t - \tau)}) \, d \tau \\[4pt] &= Re \left( \displaystyle \int\limits_{-\infty}^{\infty} h(\tau) e^{-i \omega_o \tau} \, d \tau \,\, \tilde{u}_o e^{i \omega_o t} \right) \\[4pt] &= Re \left( H(\omega_o) u_o e^{i (\omega_o t + \psi)} \right) \\[4pt] &= u_o |H(\omega_o)| \, cos(\omega_o t + \psi + \arg (H(\omega_o))) \end{align*}\]
As an example, let \(u(t) = 4 \, \cos (3t+ \pi /4)\), and \(H(\omega) = 2i\omega /5\). Then \(H(\omega_o) = H(3) = 6i/5 = 1.2 \, \angle \pi /2\). Thus, \(y(t) = 4.8 \, \cos (3t + 3\pi /4)\).