Analysis of LTI System with Fourier Transform

For a continuous-time function 𝑥(𝑡), the Fourier transform of 𝑥(𝑡) can be defined as,

$$\mathrm{X\left ( \omega \right )=\int_{-\infty }^{\infty }x\left ( t \right )e^{-j\omega t}dt}$$

System Analysis with Fourier Transform

Consider an LTI (Linear Time-Invariant) system, which is described by the differential equation as,

$$\mathrm{\sum_{k=0}^{N}a_{k}\frac{\mathrm{d}^{k}y\left ( t \right ) }{\mathrm{d} t^{k}}=\sum_{k=0}^{M}b_{k}\frac{\mathrm{d}^{k}x\left ( t \right ) }{\mathrm{d} t^{k}}}$$

Taking Fourier transform on both sides of the above equation, we get,

$$\mathrm{F\left [ \sum_{k=0}^{N}a_{k}\frac{\mathrm{d}^{k}y\left ( t \right ) }{\mathrm{d} t^{k}} \right ]=F\left [ \sum_{k=0}^{M}b_{k}\frac{\mathrm{d}^{k}x\left ( t \right ) }{\mathrm{d} t^{k}} \right ]}$$

By using linearity property $\mathrm{\left [ i.e.,\: ax_{1}\left ( t \right )+bx_{2}\left ( t \right )\overset{FT}{\leftrightarrow}aX_{1}\left ( \omega \right )+bX_{2}\left ( \omega \right ) \right ] }$ of Fourier transform, we get,

$$\mathrm{\sum_{k=0}^{N}a_{k}F\left [ \frac{\mathrm{d}^{k}y\left ( t \right ) }{\mathrm{d} t^{k}} \right ]=\sum_{k=0}^{M}b_{k}F\left [ \frac{\mathrm{d}^{k}x\left ( t \right ) }{\mathrm{d} t^{k}} \right ]}$$

Now, by using time differentiation property $\mathrm{\left [ i.e.,\frac{\mathrm{d} x\left ( t \right )}{\mathrm{d} t}\overset{FT}{\leftrightarrow}j\omega X\left ( \omega \right ) \right ] }$ of Fourier transform, we get,

$$\mathrm{\sum_{k=0}^{N}a_{k}\left ( j\omega \right )^{k}Y\left ( \omega \right )=\sum_{k=0}^{M}b_{k}\left ( j\omega \right )^{k}X\left ( \omega \right )}$$

$$\mathrm{\Rightarrow Y\left ( \omega \right )\sum_{k=0}^{N}a_{k}\left ( j\omega \right )^{k}=X\left ( \omega \right )\sum_{k=0}^{M}b_{k}\left ( j\omega \right )^{k}}$$

Therefore, the transfer function of the given LTI system is,

$$\mathrm{H\left ( \omega \right )=\frac{Y\left ( \omega \right )}{X\left ( \omega \right )}=\frac{\sum_{k=0}^{M}b_{k}\left ( j\omega \right )^{k}}{\sum_{k=0}^{N}a_{k}\left ( j\omega \right )^{k}}}$$

Here, 𝐻(𝜔) is known as frequency response of the LTI System.

Now, when the impulse input is applied to the system, i.e.,

$$\mathrm{x\left ( t \right )=\delta \left ( t \right )}$$

Then,

$$\mathrm{X\left ( \omega \right )=1\; and\; Y\left ( \omega \right )=H\left ( \omega \right )}$$

Also,

$$\mathrm{F^{-1}\left [ H\left ( \omega \right ) \right ]=h\left ( t \right )}$$

Here, the ℎ(𝑡) is called the impulse response of the LTI system. The Fourier transform of the impulse response is called the frequency response or transfer function of the system.

The frequency response of the system is represented as,

$$\mathrm{H\left ( \omega \right )=\left | H\left ( \omega \right ) \right |e^{j\angle H\left ( \omega \right )}}$$

Where,

• $\mathrm{\left | H\left ( \omega \right ) \right |}$ is called the magnitude response of the system, and

• $\mathrm{e^{j\angle H\left ( \omega \right )}}$ is known as phase response of the system.