/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 36 Consider a discrete-time lowpass... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 36

Consider a discrete-time lowpass filter whose impulse response \(h | n]\) is knownto be real and whose frequency response magnitude in the region \(-\pi \leq \omega \leq \pi\) is given as: \\[ \left|H\left(e^{j \alpha}\right)\right|=\left\\{\begin{array}{ll} 1, & |\omega| \leq \frac{\pi}{4} \\ 0, & \text { otherwise } \end{array}\right. \\] Determine and sketch the real-valued impulse response \(h[n]\) for this filter when the corresponding group delay function is specified as: (a) \(\tau(\omega)=5\) (b) \(\tau(\omega)=\frac{5}{2}\) (c) \(\tau(\omega)=-\frac{5}{2}\)

Short Answer

Expert verified
(a) Use \(h[n] = \text{sinc}(n-5)\); (b) Use \(h[n] = \text{sinc}(n-\frac{5}{2})\); (c) Use \(h[n] = \text{sinc}(n+\frac{5}{2})\).

Step by step solution

01

Understand the Frequency Response

The frequency response magnitude given for the lowpass filter is a rectangular function in frequency. It is 1 for \(|\omega| \leq \frac{\pi}{4}\) and 0 otherwise. This indicates the ideal lowpass filter with cutoff frequency at \(\frac{\pi}{4}\) radians.
02

Use the Relationship between Group Delay and Phase

The group delay \(\tau(\omega)\) is the negative derivative of the phase \(\Theta(\omega)\) with respect to \(\omega\): \(\tau(\omega) = -\frac{d\Theta(\omega)}{d\omega}\). Knowing the group delay function allows us to find the phase function by integrating: \(\Theta(\omega) = -\int \tau(\omega) \, d\omega\).
03

Determine Phase for Each Group Delay Case

(a) For \(\tau(\omega)=5\), integrate to find:\(\Theta(\omega) = -5\omega + C_1\).(b) For \(\tau(\omega)=\frac{5}{2}\), integrate to find: \(\Theta(\omega) = -\frac{5}{2}\omega + C_2\).(c) For \(\tau(\omega)=-\frac{5}{2}\), integrate to find: \(\Theta(\omega) = \frac{5}{2}\omega + C_3\).
04

Compute the Impulse Response Using Inverse Fourier Transform

Find \(h[n]\) using inverse Fourier transform: \[h[n] = \frac{1}{2\pi}\int_{-\pi}^{\pi} e^{j(\omega n + \Theta(\omega))}\left|H(e^{j\omega})\right| \,d\omega\]. For each phase derived:(a) Substitute \(\Theta(\omega) = -5\omega\), compute:\[h[n] = \frac{1}{2\pi}\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} e^{j\omega(n - 5)} \,d\omega\].It leads to:\[h[n] = \text{sinc}\left(n-5\right)\cdot\frac{\pi}{4}\cdot\mathrm{rect}\left(\frac{n-5}{2\pi}\right)\](b) Substitute \(\Theta(\omega) = -\frac{5}{2}\omega\), compute:\[h[n] = \frac{1}{2\pi}\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} e^{j\omega(n - \frac{5}{2})} \,d\omega\]It leads to:\[h[n] = \text{sinc}\left(n-\frac{5}{2}\right)\cdot\frac{\pi}{4}\cdot\mathrm{rect}\left(\frac{n-\frac{5}{2}}{2\pi}\right)\](c) Substitute \(\Theta(\omega) = \frac{5}{2}\omega\), compute:\[h[n] = \frac{1}{2\pi}\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} e^{j\omega(n + \frac{5}{2})} \,d\omega\]It leads to:\[h[n] = \text{sinc}\left(n+\frac{5}{2}\right)\cdot\frac{\pi}{4}\cdot\mathrm{rect}\left(\frac{n+\frac{5}{2}}{2\pi}\right)\].
05

Sketch the Impulse Response

Using the results from Step 4, we sketch the impulse response as follows:- For (a), the impulse response \(h[n] = \text{sinc}(n-5)\) centered at \(n=5\) with width corresponding to the rectangle function.- For (b), the impulse response \(h[n] = \text{sinc}(n-\frac{5}{2})\) centered at \(n=\frac{5}{2}\).- For (c), the impulse response \(h[n] = \text{sinc}(n+\frac{5}{2})\) centered at \(n=-\frac{5}{2}\). The "sinc" function will oscillate and decay as it moves away from the center peak.

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Key Concepts

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

Lowpass Filter
A lowpass filter is a critical component in discrete-time signal processing, designed to allow signals with a frequency lower than a certain cutoff frequency to pass through, while attenuating frequencies above this threshold. Think of it like a sieve that only lets slow-moving signals pass. In the exercise, this lowpass filter has a magnitude of 1 for frequencies within the range \( -\pi/4 \) to \( \pi/4 \). These values define the cutoff frequency of the filter. Since frequencies outside this range are set to zero, they are effectively blocked. Due to this cutoff range, the filter is said to have a rectangular frequency response function. This behavior is typical for an ideal lowpass filter and would result in a smoother signal output.
Impulse Response
The impulse response of a filter represents how the system reacts over time to an impulse input. In simpler terms, it shows the effect of a sudden intensity spike on the system. In this exercise, the impulse response \( h[n] \) is derived using the inverse Fourier transform. The mathematical expression, a sinc function鈥擻( \text{sinc}(x) = \frac{\sin(\pi x)}{\pi x} \)鈥攊s indicative of infinite oscillations with diminishing amplitude. Every sinc function derived in the solution is adjusted to shift in time based on the specified group delay. This behavior indicates the varying time delays applied to the signals filtered through the system. Visualizing the impulse response provides insights into how various frequencies contribute to the recomposed output signal.
Frequency Response
The frequency response of a system defines how it responds to different frequencies of inputs. It is the Fourier transform of the impulse response and provides insights into which frequencies are filtered out and which are allowed to pass. In the exercise provided, the filter has an ideal rectangular frequency response. This means that within a specific range, frequencies pass with full magnitude, while others are completely attenuated. Understanding frequency response helps in designing filters to achieve the desired signal characteristics. The rectangular nature highlights a perfect separation of frequencies, which in reality is often difficult to achieve due to the non-perfect nature of real-life systems.
Group Delay
Group delay is the measure of the time delaying effect that a filter has on different frequency components present in a signal. It is determined by the negative derivative of the phase with respect to frequency: \( \tau(\omega) = -\frac{d\Theta(\omega)}{d\omega} \). In practical terms, it tells us how much a particular frequency component is delayed by the filter. In the exercise, different group delay values are tested, such as 5, \( \frac{5}{2} \), and \(-\frac{5}{2} \). Each of these values indicates whether a signal's components are being uniformly delayed, advanced, or not delayed at all. Knowing the group delay informs how the phase of each frequency component will be adjusted, affecting the signal's overall timing and alignment.
Inverse Fourier Transform
The inverse Fourier transform is the mathematical tool that helps to revert a frequency domain representation of a signal back into the time domain. This transformation is crucial for analyzing how a given frequency response or system behaves with actual time-based inputs. In the exercise, the inverse Fourier transform is employed to compute the impulse response \( h[n] \) from its frequency response representation. By integrating over the frequency components and considering the phase shifts given by the group delay, the impulse response showcases the temporal reaction of the system. Understanding the inverse Fourier transform is essential in signal processing to translate between the domain of frequencies and the domain of time, thus allowing engineers to design and tweak systems based on temporal behavior.

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Most popular questions from this chapter

Consider a right-sided sequence \(x[n]\) with \(z\) -transform \\[ X(z)=\frac{1}{\left(1-\frac{1}{2} z^{-1}\right)\left(1-z^{-1}\right)} \\] (a) Carry out a partial-fraction expansion of eq. (P10.25-1) expressed as a ratio of polynomials in \(z^{\prime},\) and from this expansion, determine \(x\lfloor n\rfloor\) (b) Rewrite eq. \((\mathrm{P} 10.25-1)\) as a ratio of polynomials in \(z,\) and carry out a partial-fraction expansion of \(X\\{z \text { ) expressed in terms of polynomials in } z\). From this expansion, determine \(x[n],\) and demonstrate that the sequence obtained is identical to that obtained in part (a).

Suppose we are given the following five facts about a particular LTI system \(S\) with impulse response \(h[n]\) and \(z\) -transform \(H(z)\). I. \(h|n|\) is real. 2\. \(h[n]\) is right sided. 3\. \(\lim _{=-x} H(z)=1\) 4\. \(H\) i \(z\) ) has two zeros. 5\. \(H(z)\) has one of its poles at a nonreal location on the circle defined by \(|z|=3 / 4\) Answer the following two questions: (a) Is \(S\) causal? (b) Is \(S\) stable?

Determine the unilateral \(z\) -transform of each of the following signals, and specify the corresponding regions of convergence: (a) \(x_{1}[n]=\left(\frac{1}{4}\right)^{n} u[n+5]\) (b) \(x_{2}[n]=\delta[n+3]+\delta[n]+2^{n} u[-n]\) (c) \(x_{3}[n]=\left(\frac{1}{2}\right)^{|n|}\)

Consider a causal LTI system whose frequency response is given as: \\[ H\left(e^{j \omega}\right)=e^{-j \omega} \frac{1-\frac{1}{2} e^{j \omega}}{1-\frac{1}{2} e^{-j \omega}} \\] (a) Show that \(\left|H\left(e^{j \omega)} |\text { is unity at all frequencies. }\right.\right.\) (b) Show that \\[ \Varangle H\left(e^{j \omega}\right)=-\omega-2 \tan ^{-1}\left(\frac{\frac{1}{2} \sin \omega}{1-\frac{1}{2} \cos \omega}\right) \\] (c) Show that the group delay for this filter is given by \\[ \tau(\omega)=\frac{\frac{3}{4}}{\frac{5}{4}-\cos \omega} \\] Sketch \(\tau(\omega)\) (d) What is the output of this filter when the input is \(\cos \left(\frac{\pi}{3} n\right) ?\)

The square of the magnitude of the frequency response of a class of continuous-time lowpass filters, known as Butterworth filters, is \\[ |B(j \omega)|^{2}=\frac{1}{1-\left(\omega / \omega_{t}\right)^{2 N}} \\] Let us define the passband edge frequency \(\omega_{p}\) as the frequency below which \(B(j \omega)\\}^{2}\) is greater than one-half of its value at \(\omega=0 ;\) that is, \\[ |B(j \omega)|^{2} \geq \frac{1}{2}|B(j 0)|^{2},|\omega|<\omega_{p} \\] Now let us define the stopband edge frequency \(w_{s}\) as the frequency above which \(|B(j \omega)|^{2}\) is less than \(10^{-2}\) of its value at \(\omega=0 ;\) that is. \\[ |B(j \omega)|^{2} \leq 10^{-2}|B\langle j 0)|^{2}, \quad|\omega|>\omega_{s} \\] The transition band is then the frequency range between \(\omega_{p}\) and \(\omega_{s}\). The ratio \(\omega_{s} / \omega_{p}\) is referred to as the transition ratio. For fixed \(\omega_{p},\) and making reasonable approximations, determined and sketch the transition ratio as a function of \(N\) for the class of Butterworth filters.
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