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Determine the \(z\) -transform for each of the following sequences. Sketch the pole zero plot and indicate the region of convergence. Indicate whether or not the Fourier transform of the sequence exisis. (a) \(\delta[n+5]\) (b) \(\delta(n-5]\) (c) \((-1)^{n} u[n]\) (d) \(\left(\frac{1}{2}\right)^{n+1} u[n+3]\) (e) \(\left(-\frac{1}{3}\right)^{n} u[-n-2]\) \((f)\left(\frac{1}{4}\right)^{n} u[3-n]\) (g) \(\left.2^{n} u |-n\right]+\left(\frac{1}{4}\right)^{n} u[n-1]\) (h) \(\left(\frac{1}{3}\right)^{n-2} u[n-2]\)

Short Answer

Expert verified
Most sequences have a non-existent Fourier transform due to ROC not including the unit circle, except \(\delta[n+5]\) and \(\delta[n-5]\).

Step by step solution

01

Understanding Delta Functions

For the sequence \( \delta[n+5] \), the \( z \)-transform is evaluated directly at \( n = -5 \). Since the \( \delta[n+5] \) is a shifted impulse function, its \( z \)-transform is \( z^5 \). The pole-zero plot has a pole at the origin and no zeros. The region of convergence (ROC) is the entire \( z \)-plane, except at the origin. The Fourier transform exists because the impulse is absolutely summable.
02

Impulse Shift Analysis

For \( \delta[n-5] \), it's an impulse shifted to \( n = 5 \). The \( z \)-transform is \( z^{-5} \). The pole-zero plot has a pole at origin and no zeros. The ROC is the entire \( z \)-plane, except at zero. The Fourier transform exists here as well, since it's just a finite impulse at one point.
03

Alternating Unit Step

Analyzing \((-1)^{n} u[n]\), it's an alternating sequence for \( n \geq 0 \). The \( z \)-transform is \( \frac{1}{1 + z^{-1}} \), with a zero at \( -1 \) and a pole at the origin. The ROC is \(|z|>0\), and it does not include the unit circle, meaning the Fourier transform does not exist.
04

Exponential Shifted Step

For \(\left(\frac{1}{2}\right)^{n+1} u[n+3]\), rewrite the sequence as \(\left(\frac{1}{2}\right)^{n+1} z^3 u[n]\). The \( z \)-transform is \(z^3 \cdot \frac{1}{1 - \frac{1}{2} z^{-1}}\) with a pole at \( z = \frac{1}{2} \) and no zeros. The ROC is \( |z| > \frac{1}{2} \), and the Fourier transform does not exist as the unit circle is not included in the ROC.
05

Reversed Exponential Sequence

For \(\left(-\frac{1}{3}\right)^{n} u[-n-2]\), shift \(n\) to positive, rewriting as \((-\frac{1}{3})^{-n-3} u[n]\). The \( z \)-transform is \( \frac{(z^{-1})^3}{1 - \left(-\frac{1}{3}\right)z^{-1}}\) with a pole at \(-3\) and a zero at the origin. The ROC is \(|z| < |-\frac{1}{3}| \), and the Fourier transform does not exist.
06

Upper Bounded Exponential

For \(\left(\frac{1}{4}\right)^{n} u[3-n]\), reverse the sequence as \((\frac{1}{4})^{4-n} u[n]\), giving \(z^{-4} \cdot \frac{1}{1 - \frac{1}{4}z}\). There are zeros at the origin and a pole at \(\frac{1}{4}\). ROC is \(|z| < \frac{1}{4}\) and the Fourier transform does not exist.
07

Compound Exponential and Unit Step

For \(2^n u[-n] + \left(\frac{1}{4}\right)^n u[n-1]\), separate terms to analyze independently. The \( z \)-transform of \(2^n u[-n] \) is zero. For \( \left(\frac{1}{4}\right)^n u[n-1]\), shift \(n\) giving \(z \cdot \frac{1}{1 - \frac{1}{4}z^{-1}}\), with poles at infinity and zero. ROC is \(|z| > \frac{1}{4}\) and the Fourier transform does not exist.
08

Delayed Unit Step

Finally for \(\left(\frac{1}{3}\right)^{n-2} u[n-2]\), shift \( n \) back to give \((\frac{1}{3})^{n-2}z^2 u[n]\). The \( z \)-transform is \(z^2 \cdot \frac{1}{1 - \frac{1}{3}z^{-1}}\), with a pole at \(\frac{1}{3}\) and zeros at the origin. ROC is \(|z| > \frac{1}{3}\) and the Fourier transform does not exist.

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Key Concepts

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

Pole-Zero Plot
Understanding pole-zero plots is crucial for analyzing functions in the z-transform domain. A pole-zero plot visually represents the poles and zeros of a transfer function. Here, poles are values where the function's magnitude goes to infinity, and zeros are where it becomes zero. In the z-domain, these plots are used to determine system characteristics.

For a given sequence, once you find its z-transform, identifying poles and zeros is the next step. Poles are generally represented by an 'X' mark, while zeros are shown by an 'O' mark on the plot. This method helps illustrate the behavior of the system over various values of the z variable.
  • To plot, calculate the z-transform of the sequence.
  • Identify where the function is undefined (poles) and zero (zeros).
  • Plot these on the complex plane, with the horizontal axis as the real part, and the vertical axis as the imaginary part.
In example sequences, like \((-1)^n u[n]\), poles and zeros indicate system stability and frequency response features.
Region of Convergence
The Region of Convergence (ROC) is a fundamental concept in z-transforms that tells us where the z-transform of a sequence converges. The ROC depends on the sequence and helps deduce if the z-transform exists for specific values.

Understanding ROC is essential to determine the stability and causality of a system. It is usually a ring or an annulus on the z-plane, dictated by the poles of the sequence:
  • The ROC does not contain any poles.
  • For causal sequences, the ROC is outside of the outermost pole.
  • For anti-causal sequences, it is inside the innermost pole.
For a sequence like \(\left(\frac{1}{2}\right)^{n+1} u[n+3]\), the ROC is outside \(z = \frac{1}{2}\) since it is causal. Different ROCs provide information on whether the Fourier transform exists.
Fourier Transform
The Fourier Transform is a powerful tool used to convert a time-domain signal into its frequency domain representation. It works by decomposing a signal into its constituent sinusoids, allowing for analysis in the frequency domain.

When dealing with z-transforms, checking for the Fourier transform's existence is necessary. This is possible if the ROC includes the unit circle \(|z| = 1\).
  • If the ROC includes \(|z| = 1\), the Fourier transform exists.
  • If it does not, the Fourier transform does not exist, often indicating non-absolutely summable sequences.
For example, in the sequence \(-(-1)^{n} u[n]\), the ROC \(|z| > 0\) doesn't include the unit circle, so the Fourier Transform is not feasible.
Delta Function
Delta functions, also known as impulse functions, are important in analyzing sequences. The delta function, \(\delta[n]\), is zero everywhere except at = 0\, where it equals one. This makes it invaluable for "sampling" a sequence at specific points.

In the z-domain, shifts in delta functions correspond to simple algebraic manipulations. For example, \(\delta[n-a]\) transforms directly to \(z^a\):
  • Shifting the impulse results in powers of \(z\) in the transform.
  • The z-transform of \(\delta[n+5]\) is \(z^5\).
Delta functions help in representing systems in discrete domains, triggering specific reactions at desired points.
Exponential Sequences
Exponential sequences in discrete systems often appear in the form \(a^n\). They are fundamental in modeling growth or decay processes in signals.

Assessing exponential sequences in z-transforms involves:
  • Identifying if the sequence has a growing or decaying nature.
  • Finding the z-transform using \( \frac{1}{1-az^{-1}}\), provided \(|z| > |a|\).
Examples include \(\left(\frac{1}{4}\right)^n u[n-1]\); here, the decay factor of \(\frac{1}{4}\) indicates a shrinking sequence. Understanding these allows for the ease of handling systems exhibiting exponential behavior.

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Most popular questions from this chapter

Find the inverse \(z\) -transform of \\[ X(z)=\frac{1}{1,024}\left\\{\frac{1,024-z^{-10}}{1-\frac{1}{2} z^{-1}}\right\\},|z|>0 \\].

Consider the following system functions for stable LTI systems. Without utilizing the inverse \(z\) -transform, determine in each case whether or not the corresponding system is causal. (a) \(\frac{1-\frac{4}{3} z^{-1}+\frac{1}{2} z^{-2}}{z^{-1}\left(1-\frac{1}{2} z^{-1}\right)\left(1-\frac{1}{z} z^{-1}\right)}\) (b) \(\frac{z-\frac{1}{2}}{z^{2}+\frac{1}{2} z-\frac{3}{16}}\) (c) \(\frac{z+1}{z+\frac{4}{3}-\frac{1}{2} z^{-2}-\frac{2}{3} z^{-3}}\)

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?

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 \(\cos \left(\frac{\pi}{3} n\right) ?\)

By means of a specific filter design procedure, a nonideal continuous-time lowpass filter with frequency response \(H_{0}(j \omega),\) impulse response \(h_{0}(t),\) and step response \(s_{0}(t)\) has been designed. The cutoff frequency of the filter is at \(\omega=2 \pi \times 10^{2}\) rad/sec, and the step response rise time, defined as the time required for the step response to go from \(10 \%\) of its final value to \(90 \%\) of its final value, is \(\tau_{r}=10^{-2}\) second. From this design, we can obtain a new filter with an arbatrary cutoff frequency \(\omega_{c}\) by the use of frequency scaling. The frequency response of the resulting filter is then of the form \\[ H_{\mathrm{I} \rho}(j \omega)=H_{0}(j a \omega) \\] where \(a\) is an appropriate scale factor. (a) Determine the scale factor a such that \(H_{1 p}(j \omega)\) has a cutoff frequency of \(\omega_{c}\) (b) Determine the impulse response \(h_{| p}(t)\) of the new filter in terms of \(\omega_{c}\) and \(h_{0}(t)\) (c) Determine the step response \(s_{1 p}(2)\) of the new fiter in terms of \(\omega_{c}\) and \(s_{0}(f)\) (d) Determine and sketch the rise time of the new filler as a function of its cutoff frequency \(\omega_{c}\) This is one illustration of the trade-off between fine-domain and frequency- domain characteristics. In particular, as the cutoff frequency decreases, the rise time tends to increase

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