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For each of the following Fourier transforms, use Fourier transform properties (Table 4.1) to determine whether the corresponding time-domain signal is (i) real, imaginary, or either and (ii) even, odd, or neither. Do this without evaluating the inverse of any of the given transforms. (a) \(X_{1}(j \omega)=u(\omega)-u(\omega-2)\) (b) \(X_{2}(j \omega)=\cos (2 \omega) \sin \left(\frac{\omega}{2}\right)\) (c) \(X_{3}(j \omega)=A(\omega) e^{j B(\omega)},\) where \(A(\omega)=(\sin 2 \omega) / \omega\) and \(B(\omega)=2 \omega+\frac{\pi}{3}\) (d) \(X(j \omega)=\sum_{k=-\infty}^{\infty}\left(\frac{1}{2}\right)^{|k|} \delta\left(\omega-\frac{k \pi}{4}\right)\)

Short Answer

Expert verified
X1: imaginary, neither; X2: imaginary, odd; X3: neither, neither; X: neither, neither.

Step by step solution

01

Determine the nature of X1(jω)

The given Fourier transform is \(X_1(j \omega) = u(\omega) - u(\omega - 2)\), where \(u(\omega)\) is the unit step function. The unit step makes \(X_1(j \omega)\) a rectangular function in the frequency domain. Since this function is not conjugate symmetric (i.e., \(X_1(j \omega) eq X_1^*(-j \omega)\)), the time-domain signal is not purely real. Also, the lack of symmetry implies the signal is neither even. The signal is, therefore, imaginary and neither even nor odd.
02

Examine X2(jω) for symmetry

For \(X_2(j \omega) = \cos(2 \omega) \sin(\frac{\omega}{2})\), note that \(\cos(2 \omega)\) is an even function and \(\sin(\frac{\omega}{2})\) is an odd function. Multiplying an even function by an odd function results in an odd function. Since the expression is neither real nor imaginary conjugate symmetric, \(X_2(j \omega)\) corresponds to a time-domain signal that is purely imaginary and odd.
03

Analyze X3(jω) via polar form

Consider \(X_3(j \omega) = A(\omega) e^{j B(\omega)}\), where \(A(\omega) = \frac{\sin 2 \omega}{\omega}\) and \(B(\omega) = 2\omega + \frac{\pi}{3}\). The modulus \(A(\omega)\) is an odd function (since it is a sinc function). The phase \(B(\omega)\) has linear and constant terms, which do not impact symmetry directly. Since \(X_3(j \omega)\) is not conjugate symmetric, the corresponding time-domain signal is neither real nor imaginary and is neither even nor odd.
04

Evaluate X(jω) using its expansion

The function \(X(j \omega) = \sum_{k=-\infty}^{\infty} (\frac{1}{2})^{|k|} \delta(\omega - \frac{k \pi}{4})\) is a series of weighted impulses. The series is symmetric with respect to \(\omega = 0\) and the weights \((\frac{1}{2})^{|k|}\) are real and positive, but the spectrum is not symmetric about the origin. Thus, the time-domain signal is neither purely real nor imaginary and is neither even nor odd.

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Key Concepts

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

Fourier Transform Symmetry
The concept of symmetry in Fourier Transforms is incredibly helpful when identifying the characteristics of time-domain signals.
Fourier transform symmetry is mainly about how functions behave when their variables are negated.
When a function is symmetrical about the vertical axis or time axis, this symmetry reveals properties about the signal's realness and evenness.
  • Even Function: A function is even if it is identical when the variable is negated. Mathematically, for a function to be even: \[f(t) = f(-t)\]
  • Odd Function: Conversely, if a function negates its sign with the negation of its variable, it is odd. Thus: \[f(t) = -f(-t)\]
  • Conjugate Symmetry: Calls for the following condition: \[X(-\omega) = X^*(\omega)\] This feature, when present in Fourier transforms, states that the signal is real.
Time-Domain Signal
Understanding how signals behave in both the time and frequency domain is crucial in signal processing. The Fourier transform allows us to move from time-domain to frequency-domain, uncovering different insights about the signal.
  • **Real Signals**: These are signals where each value at a distinct time is real.
  • **Imaginary Signals**: These have time-domain functions that consist of imaginary numbers.
When the Fourier transform of a function is neither real nor imaginary, it hints that the time-domain signal holds both real and imaginary components. The symmetry, or lack thereof, around the origin of the frequency function can disclose additional information.
The nature of the signal, whether real or complex, even or odd, dictates various applications, such as filtering and modulation.
Unit Step Function
The unit step function, often represented as \(u(t)\), is a simple, yet significant function in signal processing. It acts as a building block for piecewise functions, and it plays a crucial role in defining other functions through signal multiplication.
  • **Definition**: The unit step function is defined as: \[ u(t) = \begin{cases} 0, & \text{if } t < 0 \1, & \text{if } t \geq 0 \end{cases} \]
  • **Characteristics**: This function abruptly switches from 0 to 1, marking a shift or threshold in time.

This function is widely utilized in creating signals with sudden starts, stops, or shifts. In Fourier terms, its spectral representation often creates rectangular signals or windows in the frequency domain, similar to what was seen in the original exercise with the expressions involving unit steps. This results in specific characteristics, such as a lack of symmetry in their Fourier transform, which we noted leads to complex or unbalanced signals in the time domain.
Conjugate Symmetry
In the realm of Fourier transforms, conjugate symmetry is a core property, especially visible in real-valued time-domain signals. This property states that the Fourier transform \(X(\omega)\) of a real signal satisfies: \[X(-\omega) = X^*(\omega)\] This denotes that the negative frequency component is the complex conjugate of the positive frequency component.
  • **Application**: Recognizing conjugate symmetry is profoundly beneficial because it implies the realness of its inverse time-domain signal.
  • **Example**: Consider any sinusoidal signal or a cosine wave—these demonstrate conjugate symmetry, illustrating simplicity in decomposition and analysis.
  • **Impact**: If a spectrum lacks conjugate symmetry, it indicates the presence of imaginary components in the time-domain signal, underscoring the importance of careful consideration of these properties in signal processing applications.

Hence, identifying whether a signal's Fourier transform possesses conjugate symmetry simplifies and optimizes the analysis and manipulation of real-time-domain signals.

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Most popular questions from this chapter

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 \(1 / 4\).

Consider an LTI system whose response to the input $$x(t)=\left[e^{-t}+e^{-3 t}\right] u(t)$$ is $$y(t)=\left[2 e^{-t}-2 e^{-4 t}\right] u(t)$$. (a) Find the frequency response of this system. (b) Determine the system's impulse response. (c) Find the differential equation relating the input and the output of this system.

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 \(B\).

(a) Show that the three LTI systems with impulse responses $$\begin{aligned}&h_{1}(t)=u(t)\\\&h_{2}(t)=-2 \delta(t)+5 e^{-2 t} u(t)\end{aligned}$$ and $$h_{3}(t)=2 t e^{-t} u(t)$$ all have the same response to \(x(t)=\cos t\). (b) Find the impulse response of another LTI system with the same response to \(\cos t\). This problem illustrates the fact that the response to cos \(t\) cannot be used to specify an LTI system uniquely.

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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