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System in Frequency Domain

Now let's talk about systems.

System Response Function in Frequency Domain​

Previously, we know that all LTI system can be described in the time domain with convolution.

y(t)=x(t)∗h(t)y(t) = x(t) \ast h(t)

We can take this into frequency domain, we get,

Y(ω)=X(ω)H(ω)Y(\omega) = X(\omega) H(\omega)

Here, H(ω)H(\omega) is called the system response function in frequency domain. It is the fourier transform of the impulse response of the system.

Solving System in Differential Equation by Fourier Transform​

Now let's consider solving a LTI system described in differential equation.

P(D)y(t)=x(t)P(D)y(t) = x(t)

Obviously, if we apply fourier transform,

P(−jω)Y(ω)=X(ω)P(-j\omega)Y(\omega) = X(\omega)

So the system response function is,

H(ω)=1P(−jω)H(\omega) = \frac{1}{P(-j\omega)}

Example​

D2y+4Dy+3y=x(t)D^2y+4Dy+3y = x(t)

We apply fourier transform on both sides,

(−ω2−4jω+3)Y(ω)=X(ω)(-\omega^2 - 4j\omega +3)Y(\omega) = X(\omega)

So the system response function is,

H(ω)=1(3−jω)(1−jω)=12(11−jω−13−jω)H(\omega) = \frac{1}{(3 - j\omega)(1 - j\omega)} \\ = \frac{1}{2} (\frac{1}{1 - j\omega} - \frac{1}{3 - j\omega})

Now, we can know that the impulse response of this system is,

h(t)=F−1(H(ω))h(t) = \mathcal{F}^{-1}(H(\omega))

Because we know that,

e−αtu(t)→F1jω+αe^{-\alpha t}u(t) \xrightarrow{\mathcal{F}} \frac{1}{j\omega + \alpha}

Thus,

etu(t)→F1jω−1e^{t}u(t) \xrightarrow{\mathcal{F}} \frac{1}{j\omega - 1}

And,

e3tu(t)→F1jω−3e^{3t}u(t) \xrightarrow{\mathcal{F}} \frac{1}{j\omega - 3}

Due to the linearity of fourier transform, we can get,

F−1(H(ω))=−12(et−e3t)u(t)\mathcal{F}^{-1}(H(\omega)) = -\frac{1}{2}(e^{t} - e^{3t})u(t)

However, you can tell that we still need to calculate the jump value. This issue will be solved with laplace transform.