Signals and Systems: Response of LTI systems to Complex Exponentials

[mayan]

How will we evaluate the integral for calculation of Amplitude Factor for an LTI system for which the input and output are related by a time shift of 3, i.e.,

y(t) = x(t - 3)


The answer is: H(s) = e^(-3s)

I want to understand the Mathematics behind the evaluation of the integral.

Thanks.

 

[rbj]

i'm not sure how the title of your post is related to the content.

by definition, on the left side you have,

$$Y(s) = \mathcal{L} \left\{y(t)\right\} = \int_{-\infty}^{+\infty} e^{-st} y(t) \,dt$$

and, also by definition,

$$X(s) = \mathcal{L} \left\{x(t)\right\} = \int_{-\infty}^{+\infty} e^{-st} x(t) \,dt$$

and, it turns out that for LTI systems, after you show that the convolution operator and some impulse response is what relates the input x(t) to the output y(t), then

$$Y(s) = H(s) X(s)$$

where

$$H(s) = \mathcal{L} \left\{h(t)\right\} = \int_{-\infty}^{+\infty} e^{-st} h(t) \,dt$$

and h(t) is the impulse response of the LTI system and completely characterizes the input/output relationship of the LTI system.

now, in your specific case,

$$Y(s) = \mathcal{L} \left\{y(t)\right\} = \mathcal{L} \left\{x(t-3)\right\} = \int_{-\infty}^{+\infty} e^{-st} x(t-3) \,dt$$

which, after you do a trivial substitution of variable of integration is

$$Y(s) = \int_{-\infty}^{+\infty} e^{-s(t+3)} x(t) \,dt = e^{-s \cdot 3} \int_{-\infty}^{+\infty} e^{-st} x(t) \,dt = e^{-s \cdot 3} \cdot X(s)$$

which means that

$$H(s) = e^{-s \cdot 3}$$.

 

[ravenprp]

To understand the mathematics behind the evaluation of the integral for the amplitude factor of an LTI system with a time shift of 3, we need to first understand the concept of complex exponentials in the context of signals and systems.

Complex exponentials are functions of the form e^(st), where s is a complex number and t is the independent variable. These functions are important in the field of signals and systems because they can model a wide range of physical systems, including electrical circuits, mechanical systems, and even biological systems.

In the context of an LTI (linear time-invariant) system, the input-output relationship can be described by a transfer function H(s), which is a function of the complex variable s. This transfer function describes the system's response to a complex exponential input, and it is typically represented in terms of its magnitude and phase response.

Now, let's consider the specific case of a time shift of 3 in the input signal. This means that the input signal x(t) is delayed by 3 units of time, resulting in the output signal y(t) = x(t-3). In the frequency domain, this time shift corresponds to a phase shift of -3s, which can be represented as e^(-3s).

To evaluate the integral for the amplitude factor, we use the property of linearity of the integral. This means that we can split the integral into two parts: one for the input signal and one for the output signal. Since the input signal is a complex exponential, the integral can be easily evaluated using the definition of the complex exponential function.

The integral for the input signal can be written as:

A = ∫ e^(st) dt = (1/s) * e^(st)

Similarly, the integral for the output signal can be written as:

B = ∫ e^(-3s) * e^(st) dt = (1/s) * e^(st-3s)

Using the property of linearity, we can combine these two integrals to get the overall amplitude factor:

A/B = (1/s) * e^(3s)

Substituting the transfer function H(s) = e^(-3s), we get the final expression for the amplitude factor as:

A/B = H(s)/s

This is the mathematical explanation behind the evaluation of the integral for the amplitude factor of an LTI system with a time shift of 3. It is important to note that this is just