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

Short Answer

Expert verified
The impulse response is obtained by inverse Fourier transforming the frequency response term by term and combining the results through convolution and superposition principles.

Step by step solution

01

Understand the Frequency Response

The frequency response of the system is given as \( H(j \omega) = \frac{(\sin^{2}(3\omega)) \cos \omega}{\omega^{2}} \). This is a frequency domain representation of the system's behavior. To find the impulse response, we need to determine the corresponding time domain representation.
02

Analyze Frequency Response Components

The frequency response can be separated into constitutive parts: \( \sin^{2}(3\omega) \) and \( \cos \omega \). Determine how each of these functions behaves and affects the frequency response. \( \sin^{2}(3\omega) \) is a nested sine function which creates a frequency modulation, while \( \cos \omega \) potentially adds another modulation layer.
03

Fourier Transform Properties

Recall that the impulse response function \( h(t) \) is the inverse Fourier transform of \( H(j\omega) \). Use properties like convolution, modulation, and known Fourier transform pairs that relate \( \sin \), \( \cos \), and their products to time domain functions.
04

Convolution in Frequency Domain

Recognize that the expression involves products that imply convolution in the time domain. Specifically, convolution of the inverse transforms of \( \sin^{2}(3\omega) \) and \( \cos \omega \), divided by \( \omega^{2} \), likely involves known convolution pairs.
05

Apply Inverse Fourier Transform Techniques

Determine the inverse Fourier transform of each term separately using Fourier transform tables or known pairs. Use transforms like \( sinc(t) \) for \( \frac{1}{\omega^2} \) and standard results for trigonometric functions in the Fourier domain. Apply linearity and convolution theorems to combine results.
06

Combine Results for Final Impulse Response

Combine the results obtained from the inverse transforms of each component. Utilize convolution in the time domain and superposition to ultimately express \( h(t) \), the impulse response in terms of recognized time domain functions.

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Key Concepts

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

Understanding Frequency Response
The frequency response of a system, given by a function like \( H(j \omega) = \frac{(\sin^{2}(3\omega)) \cos \omega}{\omega^{2}} \), is crucial for analyzing how a system responds to different frequencies of input signals. Essentially, it tells us how each frequency component of an input signal will be affected by the system. This creates a picture of how the system behaves in the frequency domain.

Such responses are often used to design filters and to understand the gain and phase of systems over a range of frequencies. To closely examine this, one could separate it into its parts: \( \sin^{2}(3\omega) \) and \( \cos \omega \). The term \( \sin^{2}(3\omega) \) indicates frequency modulation by a factor of three, while \( \cos \omega \) can imply further modulation. These constituents play vital roles in shaping the system's overall frequency response, making it either resonant or filtering certain frequencies.
  • \( \sin^{2}(3\omega) \): A symbol of frequency doubling and sharpening in the spectral domain.
  • \( \cos \omega \): Possibly affects phase and slightly attenuates certain frequencies.
Unpacking Fourier Transform
The Fourier transform bridges the frequency and time domains. When given a frequency response \( H(j\omega) \), converting it to a time domain representation \( h(t) \) involves using the inverse Fourier transform. This mathematical tool decomposes functions into frequencies and is widely used to solve linear time-invariant systems.

Fourier transform allows us to dissect complex periodic structures into basic components and analyze them individually. For example, trigonometric functions, like sine and cosine in \( H(j\omega) \), are particularly relevant because their transforms are well-documented. When applying Fourier transforms, properties such as linearity and shifting are crucial: they simplify complex operations like multiplication and convolution into manageable expressions.

In our specific exercise's case, we would use well-known pairs and properties to identify their corresponding time domain impulses. This is important because it provides insights into how continuous frequencies interrelate within a defined timeframe. The essence here is using Fourier transform tables to translate expressions like ours back to comprehensible time signals.
Exploring the Convolution Theorem
The convolution theorem is a cornerstone concept linking frequency and time domains. It states that multiplication in the frequency domain is equivalent to convolution in the time domain and vice versa. This theorem is highly beneficial because it converts complex multiplication operations into additions, making calculations manageable.

For our frequency response \( H(j\omega) = \frac{(\sin^{2}(3\omega)) \cos \omega}{\omega^{2}} \), this signifies that the inverse transform—the impulse response \( h(t) \)—involves convolving the inverse transforms of each separate term. The inverse of \( \frac{1}{\omega^2} \) typically represents a \( sinc \) function in time, showcasing how continuous frequency elements map directly to time impulses.

Convolution requires understanding of fundamental properties:
  • Associativity: Reduces complex multi-step convolutions into more manageable operations.
  • Commutativity: Order of functions being convolved does not affect outcome.
  • Distributivity: Allows convolution across summed components, providing analytical clarity.
By applying these characteristics, we effectively blend each component in frequency to reveal the time domain's accurate impulse behavior.
Time Domain Representation
The time domain representation of a system, denoted as \( h(t) \) for its impulse response, shows how the system responds over time to an impulse input. For systems described in frequency terms like \( H(j\omega) \), translating them into time requirements involves clearly defining \( h(t) \) via inverse Fourier transforms.

In practical terms, an impulse response \( h(t) \) characterizes how input signals are manipulated over time. This representation is crucial, especially in signal processing and control systems, for identifying the transient and steady-state behavior of signals passing through a system.

Steps to find \( h(t) \) include:
  • Break down the frequency response into simpler known terms, each corresponding to a time domain expression.
  • Apply inverse Fourier transforms to each part, relying on known pairs like \( sinc \) for transformation.
  • Combine results using convolution, which layers time responses ensuring accurate system performance reflection.
  • Utilize superposition principles, adding together convolved results for a complete system reaction profile.
In essence, understanding \( h(t) \) equips us with tools to predict and optimize how systems handle diverse inputs over time.

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Most popular questions from this chapter

Consider a signal \(x(t)\) with Fourier transform \(X(j \omega) .\) Suppose we are given the following facts: 1\. \(x(t)\) is real and nonnegative 2\. \(\mathfrak{F}^{-1}\\{(1+j \omega) X(j \omega)\\}=A e^{-2 t} u(t) .\) where \(A\) is independent of \(t\). 3\. \(\int_{-\infty}^{\infty}|X(j \omega)|^{2} d \omega=2 \pi\). Determine a closed-form expression for \(x(t)\).

Inverse systems frequently find application in problems involving imperfect measuring devices. For example, consider a device for measuring the temperature of a liquid. It is often reasonable to model such a device as an LTI system that, because of the response characteristics of the measuring element (e.g., the mercury in a thermometer \(),\) does not respond instantaneonsly to temperature changes. In particular, assurne that the response of this device to a unit step in temperature is $$s(t)=\left(1-e^{-t / 2}\right) u(t)$$. (a) Design a compensatory system that, when provided with the output of the measuring device, produces an output equal to the instantarieous temperature of the liquid. (b) One of the problems that often arises in using inverse systems as cornpensators for measuring devices is that gross inaccuracies in the indicated temperature may occur if the actual output of the measuring device produces errors due to small, erratic phenomena in the device. since there always are such sources of error in real systems, one must take them into account. To illustrate this, consider a measuring device whose overall output can be modeled as the sum of the response of the measuring device characterized by eq. (P4.52-1) and an interfering "noise" signal \(n(t) .\) Such a model is depicted in Figure \(\mathbf{P 4 . 5 2}(\mathrm{a})\) where we have also included the inverse system of part (a), which now has as its input the overall output of the measuring device. Suppose that \(n(t)=\sin \omega t\). What is the contribution of \(n(t)\) to the output of the inverse system, and how does this output change as \(\omega\) is increased? (c) The issue raised in part (b) is an important one in many applications of LTI system analysis. Specifically, we are confronted with the fundamental trade-off between the speed of response of the system and the ability of the system to attenuate high-frequency interference. In part (b) we saw that this trade-off implied that, by attempting to speed up the response of a measuring device (by means of an inverse system), we produced a system that would also amplify corrupting sinusoidal signals. To illustrate this concept furthes, consider a measuring device that responds instantaneously to changes in temperature, but that also is corrupted by noise. The response of such a system can be modeled, as depicted in Figure \(\mathrm{P} 4.52(\mathrm{b}),\) as the sum of the response of a perfect measuring device and a corrupting signal \(n(t) .\) Suppose that we wish to design a compensatory system that will slow down the response to actual temperature variations. but also will attenuate the noise \(n(t)\). Let the impulse response of this system be $$h(t)=a e^{u t} u(t)$$ Choose \(a\) so that the overall system of Figure \(\mathbf{P} 4.52(\mathrm{b})\) responds as quickly as possible to a step change in temperature, subject to the constraint that the amplitude of the portion of the output due to the noise \(n(t)=\sin 6 t\) is no larger than \(1 / 4\).

Let \(x(t)\) be a signal whose Fourier transform is $$X(j \omega)=\delta(\omega)+\delta(\omega-\pi)+\delta(\omega-5)$$ and let $$h(t)=u(t)-u(t-2)$$. (a) Is \(x(t)\) periodic? (b) Is \(x(t) * h(t)\) periodic? (c) Can the convolution of two aperiodic signals be periodic?

Determine the Fourier transform of each of the following periodic signals: (a) \(\sin \left(2 \pi t+\frac{\pi}{4}\right)\) (b) \(1+\cos \left(6 \pi t+\frac{\pi}{8}\right)\)

The input and the output of a causal LTI system are related by the differential equation $$\frac{d^{2} y(t)}{d t^{2}}+6 \frac{d y(t)}{d t}+8 y(t)=2 x(t)$$ (a) Find the impulse response of this system. (b) What is the response of this system if \(x(t)=t e^{-2 t} u(t) ?\) (c) Repeat part (a) for the causal LTI system described by the equation $$\frac{d^{2} y(t)}{d t^{2}}+\sqrt{2} \frac{d y (t)}{d t}+y(t)=2 \frac{d^{2} x(t)}{d t^{2}}-2 x(t)$$

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