Sinusoidal input, frequency response

In this lecture, we explore the relationship—which turns out to be pretty chummy—between a system’s frequency response function \(H(j\omega)\) and its sinusoidal forced response.

Let’s build from the frequency response function \(H(j\omega)\) definition: $$\begin{align} y(t) &= \mathcal{F}^{-1}Y(\omega) \\ &= \mathcal{F}^{-1}(H(j\omega)U(\omega)). \label{eq:sinusoidal_response_sol_01} \end{align}$$ We take the input to be sinusoidal, with amplitude \(A\in\mathbb{R}\), angular frequency \(\omega_0\), and phase \(\psi\): \[\begin{aligned} u(t) = A \cos(\omega_0 t + \psi). \end{aligned}\] The Fourier transform of the input, \(U(\omega)\), can be constructed via transform identities from . This takes a little finagling. Let $$\begin{align} p(t) &= A q(t), \\ q(t) &= r(t - t_0),\ \text{and} \\ r(t) &= \cos\omega_0 t,\ \text{where} \\ t_0 &= -\psi/\omega_0. \end{align}$$ The corresponding Fourier transforms, from , are $$\begin{align} P(\omega) &= A Q(\omega), \\ Q(\omega) &= e^{-j\omega t_0} R(\omega),\ \text{and} \\ R(\omega) &= \pi \delta(\omega-\omega_0) + \pi \delta(\omega+\omega_0). \end{align}$$ Putting these together, $$\begin{align} U(\omega) &= A \pi \left( e^{j\psi \omega/\omega_0} \delta(\omega-\omega_0) + e^{j\psi \omega/\omega_0} \delta(\omega+\omega_0) \right) \\ &= A \pi \left( e^{j\psi} \delta(\omega-\omega_0) + e^{-j\psi} \delta(\omega+\omega_0) \right). \tag{because $\delta$s} \end{align}$$

And now we are ready to substitute into ; also applying the “linearity” property of the Fourier transform: \[\begin{aligned} y(t) &= A \pi\left( e^{j\psi} \mathcal{F}^{-1}( H(j\omega) \delta(\omega-\omega_0) ) + e^{-j\psi} \mathcal{F}^{-1}( H(j\omega) \delta(\omega+\omega_0) ) \right). \end{aligned}\] The definition of the inverse Fourier transform gives \[\begin{aligned} {3} y(t) = \dfrac{A} {2}\Bigg( &e^{j\psi} &&\int_{-\infty}^\infty e^{j\omega t} H(j\omega) \delta(\omega-\omega_0) d\omega {}+{} \nonumber \\ {}+{} &e^{-j\psi} &&\int_{-\infty}^\infty e^{j\omega t} H(j\omega) \delta(\omega+\omega_0) d\omega \Bigg). \end{aligned}\] Recognizing that \(\delta\) is an even distribution (\(\delta(t) = \delta(-t)\)) and applying the sifting property of \(\delta\) allows us to evaluate each integral: \[\begin{aligned} y(t) &= \frac{A} {2} \left( e^{j\psi} e^{j\omega_0 t} H(j\omega_0) + e^{-j\psi} e^{-j\omega_0 t} H(-j\omega_0) \right). \end{aligned}\] Writing \(H\) in polar form, $$\begin{alignat}{4} y(t) = \frac{A} {2} \Big( &e^{j(\omega_0 t + \psi)} &&|H(j\omega_0)| &&e^{j\angle H(j\omega_0)} {}+{} \nonumber\\ {}+{} &e^{-j(\omega_0 t + \psi)} &&|H(-j\omega_0)| &&e^{j\angle H(-j\omega_0)} \Big). \end{alignat}$$ The Fourier transform is conjugate symmetric—that is, \(F(-\omega) = F^*(\omega)\)—which allows us to further simply: $$\begin{align} y(t) &= \frac{A |H(j\omega_0)|} {2} \Big( e^{j(\omega_0 t + \psi)} e^{j\angle H(j\omega_0)} + e^{-j(\omega_0 t + \psi)} e^{-j\angle H(j\omega_0)} \Big) \\ &= A |H(j\omega_0)| \dfrac{ e^{j(\omega_0 t + \psi + \angle H(j\omega_0))} + e^{-j(\omega_0 t + \psi + \angle H(j\omega_0))} }{2}. \end{align}$$ Finally, Euler’s formula yields something that deserves a box. For input \(A \cos(\omega_0 t + \psi)\) to system \(H(j\omega)\), the forced response is $$\begin{align} y(t) &= A |H(j\omega_0)| \cos(\omega_0 t + \psi + \angle H(j\omega_0)). \end{align}$$

This is a remarkable result. For an input sinusoid, a linear system has a forced response that

• is also a sinusoid,

• is at the same frequency as the input,

• differs only in amplitude and phase,

• differs in amplitude by a factor of \(|H(j\omega)|\), and

• differs in phase by a shift of \(\angle H(j\omega)\).

Now we see one of the key facets of the frequency response function: it governs how a sinusoid transforms through a system. And just think how powerful it will be once we combine it with the powerful principle of superposition and the mighty Fourier series representation of a function—as a “superposition” of sinusoids!