91Ó°ÊÓ

A causal LTI system is described by the difference equation $$y[n]-a y[n-1]=b x[n]+x[n-1]$$ where \(a\) is real and less than 1 in magnitude. (a) Find a value of \(b\) such that the frequency response of the system satisfies. $$\left|H\left(e^{j \omega}\right)\right|=1, \text { for all } \omega$$ This kind of system is calkd an all -pass system, as it does not atienuate the input \(e^{j \omega n}\) for any value of \(\omega .\) Use die value of \(b\) that you have found in the rest of the problem. (b) Roughly sketch \(\Varangle H\left(e^{j \omega}\right), 0 \leq \omega \leq \pi,\) when \(a=\frac{1}{2}\) (c) Roughly sketch \(\Varangle H\left(\rho^{\prime \omega}\right), 0 \leq \omega \leq \pi,\) when \(a=-\frac{1}{2}\) (d) Find and plot the output of this system with \(a=-\frac{1}{2}\) when the input is $$x[n]=\left(\frac{1}{2}\right)^{n} u[n]$$ From this example, we see that a nonlinear change in phase can have a significantly different effect on a signal than the time shift that results from a linear phase.

Step by step solution

01

Formulate the System's Frequency Response

The system is described by the equation \(y[n] - ay[n-1] = b x[n] + x[n-1]\). Taking the z-transform on both sides, the transfer function \(H(z)\) can be deduced.\[ H(z) = \frac{b + z^{-1}}{1 - az^{-1}} \]We want \( |H(e^{j\omega})| = 1 \) for all \(\omega\). This implies that the magnitude response of the transfer function should be 1 across all frequencies.

02

Solve for Constant b to Satisfy All-Pass Condition

To satisfy \( |H(e^{j\omega})| = 1 \),\[ |b + e^{-j\omega}|^2 = |1 - ae^{-j\omega}|^2 \]Expanding both side:Left side: \[ |b + e^{-j\omega}|^2 = b^2 + 2b \, \cos(\omega) + 1 \]Right side: \[ |1 - ae^{-j\omega}|^2 = 1 + a^2 - 2a \, \cos(\omega) \]Equating both expressions:\[ b^2 + 2b \, \cos(\omega) + 1 = 1 + a^2 - 2a \, \cos(\omega) \]Simplifying, we need:\[ b^2 + 2b \, \cos(\omega) = a^2 - 2a \, \cos(\omega) \]This equation holds for all \(\omega\) if,\[ b^2 = a^2 \] and \[ b = -a \]Thus, the value of \(b\) is \(-a\) to ensure the system is all-pass.

03

Compute the Phase Response for a Given Value of a

The phase response for an all-pass system is:\[ \angle H(e^{j\omega}) = \angle (b + e^{-j\omega}) - \angle (1 - ae^{-j\omega}) \]Substituting \(b = -a\):\[ \angle H(e^{j\omega}) = \angle (-a + e^{-j\omega}) - \angle (1 - ae^{-j\omega}) \]We can evaluate this phase for specific values of \(a\).

04

Sketch Phase Response for a = 1/2

For \(a = \frac{1}{2}\), we have:\[ \angle H(e^{j\omega}) = \angle \left(-\frac{1}{2} + e^{-j\omega}\right) - \angle \left( 1 - \frac{1}{2} e^{-j\omega} \right) \]This will be an upside-down plot from zero phase to negative phase angle compared to usual linear phase response and can be sketched using specific points considering symmetry on the interval \( [0, \pi]\).

05

Sketch Phase Response for a = -1/2

For \(a = -\frac{1}{2}\), we repeat the process:\[ \angle H(e^{j\omega}) = \angle \left( \frac{1}{2} + e^{-j\omega} \right) - \angle \left( 1 + \frac{1}{2} e^{-j\omega} \right) \]Sketch around the intervals noting how the signs introduce a different symmetry pattern compared to \(a = \frac{1}{2}\).

06

Find Output Signal for Input and a = -1/2

Using the equation:\[ y[n] = x[n] + \left( b + a \right) x[n-1] \]Since \(b = a = -\frac{1}{2}\):\[ y[n] = x[n] \]Since input is \(x[n] = \left(\frac{1}{2}\right)^n u[n]\), output will be the same as the input function. Plot this to get decreasing exponential.

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Key Concepts

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

Differential Equations in LTI Systems

Differential equations are mathematical formulas that relate a function to its derivatives. In the context of Linear Time-Invariant (LTI) systems, they describe how output signals are generated in relation to the input signals. Specifically, the essence of differential equations in LTI systems lies in the difference equation. For instance, in the given system:

• Difference equation: \( y[n] - ay[n-1] = bx[n] + x[n-1] \)

It fundamentally shows how the present output \( y[n] \) is influenced by its prior value \( y[n-1] \) and current as well as past input values \( x[n] \) and \( x[n-1] \). The coefficients \( a \) and \( b \) determine the system's behavior. In this exercise, we seek a condition where the system does not alter the amplitude of the input signal, leading to an all-pass system. This condition is satisfied when \( b = -a \), which ensures uniform amplitude across frequencies.

Understanding Frequency Response

Frequency response is a critical concept in signal processing that illustrates how an LTI system reacts to different frequencies within an input signal. It indicates how much each frequency component of the input gains or loses amplitude and undergoes phase shifts after going through the system. The frequency response is usually represented by a transfer function in the frequency domain.

• Transfer function: \( H(z) = \frac{b + z^{-1}}{1 - az^{-1}} \)
• Condition for frequency response: \( |H(e^{j\omega})| = 1 \) for all \( \omega \)

This condition shows that regardless of the input frequency \( \omega \), the system output will have the same amplitude. An all-pass filter must satisfy this condition to maintain the original signal amplitude across all frequencies. The exercise focused on setting the constant \( b \) such that the magnitude response remains unity, ensuring consistent signal integrity.

The Role of Z-Transform in Analysis

The \( z \)-transform is a powerful tool that converts discrete-time signals into a form that is easier to analyze, akin to the Laplace transform in continuous time signals. By transforming a time-domain difference equation into the \( z \)-domain, we can derive the system's transfer function.

• Relationship: Converts \( y[n] - ay[n-1] = bx[n] + x[n-1] \) to \( H(z) = \frac{b + z^{-1}}{1 - az^{-1}} \)

The \( z \)-transform simplifies the process of determining system response to any arbitrary input, particularly in discrete systems. For the given problem, we utilize the \( z \)-transform to determine the frequency response by evaluating \( H(e^{j\omega}) \), thereby aiding in deducing the needed value of \( b \) for an all-pass system.

Interpreting Phase Response

The phase response of a system gives insights into the phase shift each frequency component of the input signal experiences as it passes through the system. This is crucial for applications where the relative timing of waveform components must be preserved or controlled.

• Phase response formula: \( \angle H(e^{j\omega}) = \angle (b + e^{-j\omega}) - \angle (1 - ae^{-j\omega}) \)

Using different values of \( a \) yields different phase graphs. A positive \( a \) (e.g., \( a = \frac{1}{2} \)) skews the phase response differently than a negative \( a \) (e.g., \( a = -\frac{1}{2} \)). Understanding this phase response behavior is essential as it can dramatically alter the time-domain reconstruction of the signal, a realization flatly expressed in point (d) of the exercise, showing the distinctive effects of differing phase but not amplitude.