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Signals and Systems – Transfer Function of Linear Time Invariant (LTI) System
The transfer function of a continuous-time LTI system may be defined using Laplace transform or Fourier transform. Also, the transfer function of the LTI system can only be defined under zero initial conditions. The block diagram of a continuous-time LTI system is shown in the following figure.
Transfer Function of LTI System in Frequency Domain
The transfer function 𝐻(𝜔) of an LTI system can be defined in one of the following ways −
The transfer function of an LTI system is defined as the ratio of the Fourier transform of the output signal to the Fourier transform of the input signal provided that the initial conditions are zero.
Or, the transfer function is defined as the ratio of output to input in frequency domain when the initial conditions are neglected.
Or, the transfer function of the LTI system is the Fourier transform of its impulse response.
Mathematically, the transfer function of LTI system in frequency domain is defined as,
$$\mathrm{\mathit{H\left ( \omega \right )\mathrm{=}\frac{Y\left ( \omega \right )}{X\left ( \omega \right )} }}$$
The transfer function 𝐻(𝜔) is a complex quantity. Therefore, it has both magnitude and phase.
$$\mathrm{\mathit{H\left ( \omega \right )\mathrm{=}\left | H\left ( \omega \right ) \right |e^{j\theta \left ( \omega \right )} }}$$
The transfer function 𝐻(𝜔) in frequency domain is also known as frequency response of the LTI system. The frequency response of LTI system has magnitude response and the phase response, i.e.,
$$\mathrm{Magnitude\; response\; of\; LTI\; system \mathrm{=}\mathit{\left | H\left ( \omega \right ) \right | }}$$
$$\mathrm{Phase\; response\; of\; LTI\; system \mathrm{=}\mathit{\theta \left ( \omega \right )\mathrm{=}\angle H\left ( \omega \right ) }}$$
Since the frequency spectrum of the output is,
$$\mathrm{\mathit{Y \left ( \omega \right )\mathrm{=} H\left ( \omega \right )\cdot X\left ( \omega \right ) }}$$
Therefore,
$$\mathrm{Magnitude\:of\: output, \mathit{\left | Y \left ( \omega \right ) \right |\mathrm{=} \left | H\left ( \omega \right ) \right |\cdot \left | X\left ( \omega \right ) \right | }}$$
$$\mathrm{Phase\:of\: output, \mathit{\angle Y \left ( \omega \right )\mathrm{=} \angle H\left ( \omega \right )\mathrm{+} \angle X\left ( \omega \right ) }}$$
Also, the transfer function of the LTI system has conjugate symmetry property, i.e.,
$$\mathrm{\mathit{H\left ( -\omega \right )\mathrm{=}H^{*}\left ( \omega \right )}}$$
$$\mathrm{Magnitude,\mathit{\left | H\left ( -\omega \right ) \right |\mathrm{=}\left | H\left ( \omega \right ) \right |}}$$
And
$$\mathrm{Phase,\mathit{\angle H\left ( -\omega \right )\mathrm{=}-\angle H\left ( \omega \right )}}$$
The impulse response ℎ(𝑡) of the LTI system is the inverse Fourier transform of its transfer function 𝐻(𝜔), i.e.,
$$\mathrm{\mathit{h\left ( t \right )\mathrm{=}F^{-1}\left [ H\left ( \omega \right ) \right ]}}$$
Transfer Function of LTI System in s-Domain
The transfer function 𝐻(𝑠) of an LTI system in s-domain can be defined in one of the following ways −
The transfer function of the LTI system can be defined as the ratio of the Laplace transform of the output signal to the Laplace transform of the input signal when the initial conditions are zero.
Or, the transfer function is defined as the ratio of output to input in sdomain when the initial conditions are neglected.
Or, the transfer function of the LTI system is the Laplace transform of its impulse response.
Mathematically, in Laplace domain or s-domain, the transfer function of LTI system is defined as,
$$\mathrm{\mathit{H\left ( s \right )\mathrm{=}\frac{Y\left ( s \right )}{X\left ( s \right )}}}$$
And, the impulse response ℎ(𝑡) of the LTI system in s-domain is the inverse Laplace transform of the transfer function 𝐻(𝑠), i.e.,
$$\mathrm{\mathit{h\left ( t \right )\mathrm{=}L^{-1}\left [ H\left ( s \right ) \right ]}}$$
Note – Once the transfer function of an LTI system in s-domain H(s) is known, then the transfer function in frequency domain 𝐻(𝜔) can be determined just by replacing s in 𝐻(𝑠) by 𝑗𝜔, i.e.,
$$\mathrm{\mathit{H\left ( \omega \right )\mathrm{=}H\left ( s \right )|_{s\mathrm{=}j\omega } }}$$
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