# Signals and Systems – Response of Linear Time Invariant (LTI) System

Signals and SystemsElectronics & ElectricalDigital Electronics

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## Linear Time-Invariant System

A system for which the principle of superposition and the principle of homogeneity are valid and the input/output characteristics do not change with time is called the linear time-invariant (LTI) system.

## Impulse Response of LTI System

When the impulse signal is applied to a linear system, then the response of the system is called the impulse response. The impulse response of the system is very important for understanding the behaviour of the system.

Therefore, if

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

Then,

$$\mathrm{\mathit{\mathrm{Output}, y\left(t\right)=h\left(t\right)}}$$

As the Laplace transform and Fourier transform of the impulse function is given by,

$$\mathrm{\mathit{L\left [\delta\left(t\right) \right ]\mathrm{=}\mathrm{1}\:\:\mathrm{and} \:\:F\left [\delta\left(t\right) \right ]\mathrm{=}\mathrm{1}}}$$

Hence, once the transfer function $\mathit{H\left(\omega\right)}$ of an LTI system is known in frequency domain, then the impulse response of the system can be determined by taking the inverse Fourier transform of $\mathit{H\left(\omega\right)}$, i.e.,

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

Similarly, once the transfer function $\mathit{ H\left(s\right)}$ of an LTI system is known in s-domain, then the impulse response of the system can be determined by taking the inverse Laplace transform of $\mathit{ H\left(s\right)}$, i.e.,

$$\mathrm{\mathit{h\left(t\right)=L^{-\mathrm{1}}\left [ H\left(s\right) \right ]}}$$

Once the impulse response $\mathit{ h\left(t\right)}$ of the linear system is known, then the response of the linear system $\mathit{ y\left(t\right)}$ for any given input signal $\mathit{ x\left(t\right)}$ can be obtained by convolving the input with the impulse response of the system, i.e.,

$$\mathrm{\mathit{ y\left(t\right)=h\left(t\right)*x\left(t\right)=x\left(t\right)*h\left(t\right)}}$$

## Step Response of LTI System

The convolution integral can be used to obtain the step response of a continuous-time LTI system. If the unit step signal $\mathit{u\left(t\right)}$ is an input signal to a system having impulse response $\mathit{h\left(t\right)}$, then the step response of the system is given by,

$$\mathrm{\mathit{s\left(t\right)\mathrm{=}h\left(t\right)*u\left(t\right)}}$$

If the given system is non-causal, then the step response is given by,

$$\mathrm{\mathit{s\left(t\right)\mathrm{=}\int_{-\infty }^{t}h\left(\tau\right)u\left(t-\tau\right)d\tau =\int_{-\infty }^{t}h\left(\tau\right)d\tau }}$$

And, when the given system is causal, then the step response is,

$$\mathrm{\mathit{s\left(t\right)=\int_{\mathrm{0}}^{t}h\left(\tau\right)d\tau} }$$

Hence, the step response of an LTI system is the running integral of impulse response of the system.

Now, when causal and non-causal signals are applied to causal and non-causal systems, we get the following outputs −

• When a non-causal signal is applied to a non-causal system, then,

$$\mathrm{\mathit{y\left(t\right)\mathrm{=}\int_{-\infty }^{\infty }h\left(\tau\right)x\left(t-\tau\right)d\tau \mathrm{=}\int_{-\infty }^{\infty }x\left(\tau\right)h\left(t-\tau\right)d\tau }}$$

• When a causal signal is applied to a non-causal system, then,

$$\mathrm{\mathit{y\left(t\right)\mathrm{=}\int_{-\infty }^{t }h\left(\tau\right)x\left(t-\tau\right)d\tau \mathrm{=}\int_{\mathrm{0} }^{\infty }x\left(\tau\right)h\left(t-\tau\right)d\tau }}$$

• When a non-causal signal is applied to a causal system, then,

$$\mathrm{\mathit{y\left(t\right)\mathrm{=}\int_{\mathrm{0} }^{\infty }h\left(\tau\right)x\left(t-\tau\right)d\tau \mathrm{=}\int_{-\infty }^{t}x\left(\tau\right)h\left(t-\tau\right)d\tau }}$$

• When a causal signal is applied to a causal system, then,

$$\mathrm{\mathit{y\left(t\right)\mathrm{=}\int_{\mathrm{0} }^{t }h\left(\tau\right)x\left(t-\tau\right)d\tau \mathrm{=}\int_{\mathrm{0} }^{t}x\left(\tau\right)h\left(t-\tau\right)d\tau }}$$

Updated on 17-Dec-2021 10:31:53