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Consider an LTI system \(S\) with impulse response $$h(t)=\frac{\sin (4(t-1))}{\pi(t-1)}$$ Determine the output of \(S\) for each of the following inputs: (a) \(x_{1}(t)=\cos \left(6 t+\frac{\pi}{2}\right)\) (b) \(x_{2}(t)=\sum_{k=0}^{\infty}\left(\frac{1}{2}\right)^{k} \sin (3 k t)\) (c) \(x_{3}(t)=\frac{\sin (4(t+1))}{\pi(t+1)}\) (d) \(x_{4}(t)=\left(\frac{\sin 2 t}{\pi}\right)^{2}\)

Short Answer

Expert verified
The outputs are: (a) 0, (b) 0, (c) \(\delta(t)\), (d) 0.

Step by step solution

01

Understanding the Problem

We need to determine the output of a given Linear Time-Invariant (LTI) system with a specific impulse response for different input signals. This involves finding the convolution of the input signal with the impulse response.
02

Step for Part (a): Convolution with Shifted Impulse Response

The input is given as a shifted cosine function, which can be expressed using a frequency shift property. The system response can be found using the convolution property of the impulse response. For a function of the form \(\cos(\omega t + \phi)\), the Fourier transform idea tells us the response will involve sinusoids. Given \(x_1(t) = \cos \left(6t + \frac{\pi}{2} \right)\), the result involves modulating the impulse response and corresponds to a frequency shift. \(y_1(t) = 0\) due to the frequency cut-off in \(h(t)\) which allows only frequencies up to 4.
03

Step for Part (b): Convolution with a Series of Sine Functions

For the input \(x_2(t) = \sum_{k=0}^{\infty} \left(\frac{1}{2}\right)^k \sin(3kt)\), apply linearity of convolution. Each term \(\sin(3kt)\) would individually convolve with \(h(t)\). However, since \(h(t)\) only allows frequency components up to 4, terms involving \(\sin(3kt)\) with \(k=1\) will contribute to the output, while others will not. Thus the result is also effectively \(y_2(t) = 0\).
04

Step for Part (c): Matching Impulse Response

The input \(x_3(t)\) is precisely the impulse response of the system with reversed and delayed time. When convolving \(x_3(t)\) with \(h(t)\), due to symmetry and properties of convolution, the output will be \(y_3(t)=\delta(t)\), which indicates a unit impulse (as the signal is shifted by 1).
05

Step for Part (d): Non-overlapping Frequency Components

Given \(x_4(t) = \left(\frac{\sin 2t}{\pi} \right)^2\), expand the product using trigonometric identities. This boils down to a sum of frequency components that fall outside the pass band defined by \(h(t)\). Thus, \(y_4(t) = 0\), since the system can't affect inputs outside its frequency limits.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Impulse Response
In Linear Time-Invariant (LTI) systems, the impulse response, denoted as \( h(t) \), is a fundamental concept. It characterizes the system's complete response to any input. Essentially, it describes how the system reacts over time after being subjected to an impulse signal, like a short burst of energy at time \( t=0 \).

The impulse response captures all the dynamics of the system, such as damping, frequency response, and transient characteristics. For instance, if a system has an impulse response \( h(t) = \frac{\sin(4(t-1))}{\pi(t-1)} \), it shows how the system will respond to any input frequency within its pass band. In LTI systems, any complex input signal can be treated as a composition of impulse signals, making the impulse response a powerful tool for system analysis.
  • An impulse response is unique to every LTI system.
  • It provides comprehensive information about the system's behavior and characteristics.
  • Convolution with the impulse response determines the resultant output for any given input signal.
Convolution
Convolution is the mathematical process used to find the output of an LTI system based on its impulse response and an input signal. Essentially, it's how LTI systems "filter" input signals. The convolution of two signals is a way to integrate all the different values of input, each influenced by the system’s impulse response.

The formula for convolution in continuous time is:\[(y(t) = (x * h)(t) = \int_{-\infty}^{\infty} x(\tau) h(t-\tau) \, d\tau)\] Each component of the input signal \(x(t)\) impacts the output based on how \(h(t)\) describes the system's behavior. This convolution integral accounts for the overlapping • The result of the convolution shows the influence of past and present input values on the output.• It is a key operation in determining system output for any arbitrary input, such as sinusoids, step signals, etc.• Convolution is an essential tool in signal processing, enabling analysis and understanding of signal transformations within systems.
Frequency Domain Analysis
Frequency Domain Analysis is essential for understanding how LTI systems respond to different frequencies present in a signal. This analysis translates time-domain signals into their frequency components using techniques such as Fourier Transform. By focusing on the frequencies, one can analyze which frequencies the system passes through, attenuates, or filters out.

In the frequency domain, a system's behavior is typically characterized by its frequency response function. This can easily tell you, for given input frequencies, how the magnitude and phase of the output will behave. For example, if an LTI system passes all frequencies below 4 Hz, any signal components above this will be greatly attenuated or nullified, as demonstrated in the given exercise.
  • Frequency domain offers a more clear perspective on how systems handle sinusoidal inputs.
  • This method is beneficial for designing filters and understanding systems with steady-state sinusoidal inputs.
  • It allows a more straightforward comparison of multiple systems by examining their frequency responses side by side.
Fourier Transform
The Fourier Transform is a mathematical tool used to transform time-domain signals into their frequency domain representation. Understanding this concept is crucial for LTI system analysis, especially when dealing with complex signals.

The Fourier Transform breaks down signals into individual frequency components, allowing engineers and scientists to analyze the frequencies present in a signal and how an LTI system would affect them. Here's the basic formulation for a Fourier Transform:\[X(f) = \int_{-\infty}^{\infty} x(t) e^{-j 2 \pi f t} \, dt\]*This transformation is integral in assessing how an LTI system responds to specific frequencies present in input signals.*
- By converting to the frequency domain, it's easier to see how each frequency component is modified by the system.- It provides insights into phase shifts and magnitude alterations experienced by different frequencies.- Understanding Fourier Transforms enables engineers to design systems that selectively filter, amplify, or attenuate specific parts of a signal, aligning with desired specifications.

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Most popular questions from this chapter

Given the relationships $$y(t)=x(t) * h(t)$$ and $$g(t)=x(3 t) * h(3 t)$$ and given that \(x(t)\) has Fourier transform \(X(j \omega)\) and \(h(t)\) has Fourier transform \(H(j \omega),\) use Fourier transform properties to show that \(g(t)\) has the form $$g(t)=A y(B t)$$ Determine the values of \(A\) and \(B\).

Let \(H(j \omega)\) be the frequency response of a continuous-time LTI system, and suppose that \(H(j \omega)\) is real, even, and positive. Also, assume that $$\max _{\omega}\\{H(j \omega)\\}=H(0)$$ (a) Show that: (i) The impulse response, \(h(t),\) is real. (ii) \(\max \\{|h(t)|\\}=h(0)\). Hint: If \(f(t, \omega)\) is a complex function of two variables, then $$\left|\int_{-\infty}^{+\infty} f(t, \omega) d \omega \leq \int_{-\infty}^{+\infty}\right| f(t, \omega) | d \omega$$ (b) One important concept in system analysis is the bandwidth of an LTI system. There are many different mathematical ways in which to define bandwidth, but they are related to the qualitative and intuitive idea that a system with frequency response \(G(j \omega)\) essentially "stops" signals of the form \(e^{j \omega t}\) for values of \(\omega\) where \(G(j \omega)\) varnishes or is small and "passes" those complex exponentials in the band of frequency where \(G(j \omega)\) is not small. The width of this band is the bandwidth. These ideas will be made much clearer in Chapter \(6,\) but for now we will consider a special definition of bandwidth for those systems with frequency responses that have the properties specified previously for \(H(j \omega) .\) Specifically, one definition of the bandwidth \(B_{w}\) of such a system is the width of the rectangle of height \(H(j 0)\) that has an area equal to the area under \(H(j \omega) .\) This is illustrated in Figure \(P 4.49(a) .\) Note that since \(H(j 0)=\max _{\omega} H(j \omega),\) the frequencies within the band indicated in the figure are those for which \(H(j \omega\\}\) is largest. The exact choice of the width in the figure is, of course, a bit arbitrary, but we have chosen one definition that allows us to compare different systems and to make precise a very important relationship between time and frequency. What is the bandwidth of the system with frequency response $$H(j \omega)=\left\\{\begin{array}{ll}\mathbf{1}, & |\omega|W\end{array}\right.$$ (c) Find an expression for the bandwidth \(B_{w}\) in terms of \(H(j \omega)\) (d) Let \(s(t)\) denote the step response of the system set out in part (a). An important measure of the speed of response of a system is the rise time, which, like the bandwidth, has a qualitative definition, leading to many possible mathematical definitions, one of which we will use. Intuitively, the rise time of a system is a measure of how fast the step response rises from zero to its final value, $$s(\alpha)=\lim _{t \rightarrow \infty} s(t)$$ Thus, the smaller the rise time, the faster is the response of the system. For the system under consideration in this problem, we will define the rise time as $$t_{r}=\frac{s\left(\infty\right)}{h(0)}$$ since $$s^{\prime}(t)=h(t)$$ and also because of the property that \(h(0)=\max _{t} h(t), t_{r}\) is the time it would take to go from zero to \(s(\infty)\) while maintaining the maximum rate of change of \(s(t) .\) This is illustrated in Figure \(\mathbf{P 4} . \mathbf{4 9}(\mathbf{b})\) Find an expression for \(t_{r}\) in terms of \(H(j \omega)\). (e) Combine the results of parts (c) and (d) to show that $$B_{w} t_{r}=2 \pi$$ Thus, we cannot independently specify both tha rise time and the bandwidth of our system. For example, eq. (P4.49-1) implies that, if we want a fast system ( \(t_{r}\) small), the system must have a large bandwidth. This is a fundamental trade-off that is of central importance in many problems of system design.

Find the impulse response of a system with the frequency response $$H(j \omega)=\frac{\left(\sin ^{2}(3 \omega)\right) \cos \omega}{\omega^{2}}$$

Suppose that a signal \(x(t)\) has Fourier transform \(X(j \omega) .\) Now consider another signal \(g(t)\) whose shape is the same as the shape of \(X(j \omega) ;\) that is, $$g(t)=X(j t)$$ (a) Show that the Fourier transform \(G(j \omega \text { ) of } g(t) \text { has the same shape as } 2 \pi x(-t)\) that is, show that $$G(j \omega)=2 \pi x(-\omega)$$ (b) Using the fact that $$\mathfrak{F}\\{\delta(t+B)\\}=e^{j \boldsymbol{B} \omega}$$ in conjunction with the result from part (a), show that $$\mathfrak{F}\left\\{e^{j B r}\right\\}=2 \pi \delta(\omega-B)$$

A causal and stable LTI system \(S\) has the frequency response $$H(j \omega)=\frac{j \omega+4}{6-\omega^{2}+5 j \omega}$$. (a) Determine a differential equation relating the input \(x(f)\) and output \(y(t)\) of \(S\) (b) Determine the impulse response \(h(t)\) of \(S\) (c) What is the output of \(S\) when the input is $$x(t)=e^{-4 t} u(t)-t e^{4 t} u(t) ?$$

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