91Ó°ÊÓ

Let \(h_{d}[n]\) denole the unit sarple response of a desired ideal system with frequency response \(H_{d}\left(e^{j \omega}\right),\) and let \(h[n]\) denote the unit sample response for an FIR system of length \(N\) and with frequency response \(H\left(e^{j \omega}\right) .\) In this problem, we show that a rectangular window of length \(N\) samples applied to \(h_{d}[n]\) will produce a unit sample response \(h[n]\) such that the mean square error \\[\boldsymbol{\epsilon}^{2}=\frac{1}{2 \pi} \int_{-\pi}^{\mathbb{V}}\left|H_{d}\left(e^{j \omega}\right)-H\left(e^{j \omega}\right)\right|^{2} d \omega\\] is minimized. (a) The error function \(E\left(e^{t a}\right)=H_{d}\left(e^{t \omega}\right)-H\left(e^{j \omega}\right)\) can be expressed as the power series \\[E\left(e^{j \omega}\right)=\sum_{n=-\infty}^{\infty} e|n| e^{-j \omega n}\\] Find the coefficients \(e[n]\) in terms of \(h_{d}[n]\) and \(h[n]\) (b) Using Parseval's relation, express the mean square extor \(\epsilon^{2}\) in terms of the coefficients \(e[n]\) (c) Show that for a unit sample response \(h[n]\) of length \(N\) samples, \(\epsilon^{2}\) is minimized when \\[h[n]=\left\\{\begin{array}{ll}h_{d}[n], & 0 \leq n \leq N-1 \\\0, & \text { otherwise }\end{array}\right.\\] That is, simple truncation gives the best mean square approximation to a desired frequency response for a fixed value of \(N\).

Step by step solution

01

Understanding the Error Function

The error function \(E\left(e^{j \omega}\right)\) is given by \(E\left(e^{j \omega}\right) = H_{d}\left(e^{j \omega}\right) - H\left(e^{j \omega}\right)\). This represents the difference between the desired and actual frequency responses.

02

Identifying Coefficients in the Power Series

The power series expansion of the error function is \(E\left(e^{j \omega}\right) = \sum_{n=-\infty}^{\infty} e[n] e^{-j \omega n}\). The coefficients \(e[n]\) can be found as the inverse Fourier transform of the error function: \[ e[n] = h_d[n] - h[n] \].

03

Applying Parseval's Theorem

Parseval's theorem relates the energy in the time domain to the energy in the frequency domain. The mean square error can be expressed as \( \epsilon^2 = \sum_{n=-\infty}^{\infty} |e[n]|^2 \). Here, the energy of the error signal in the time domain corresponds to the mean square error in the frequency domain.

04

Minimizing Mean Square Error with Truncation

To minimize \( \epsilon^2 \) for a finite response \(h[n]\) of length \(N\), we set \( h[n] = h_d[n] \) for \( 0 \leq n \leq N-1 \) and \( h[n] = 0 \) elsewhere. This simple truncation of \( h_d[n] \) results in the minimized mean square error, as all \( e[n]\) outside the window are eliminated.

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Key Concepts

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

Frequency Response

When discussing digital signal processing, the term 'Frequency Response' is crucial. It describes how a system reacts to different frequencies of input signals. Typically, engineers use frequency response to assess the performance of systems, especially filters, in filtering specific frequency bands.

The frequency response function is often denoted as \(H(e^{j\omega})\), where \(\omega\) stands for the angular frequency. In an ideal scenario, the desired frequency response could be represented as \(H_{d}(e^{j\omega})\). This denotes the response we want from a filter to pass through frequency \(\omega\) unaffected. Consequently, figuring out how closely a practical system's frequency response \(H(e^{j\omega})\) aligns with the ideal \(H_{d}(e^{j\omega})\) is essential for assessing any discrepancies or errors.

In the setup provided in the original exercise, we compare \(H_{d}(e^{j\omega})\) with \(H(e^{j\omega})\) to measure the mean square error. This error reflects how badly the actual system deviates from the ideal. The quantity is calculated over a frequency range, capturing the variance between desired and actual behavior.

Fourier Transform

The Fourier Transform is a mathematical tool used extensively in signal processing. It translates a time-domain signal into its frequency-domain representation. This transformation is key in analyzing how signals behave in the frequency domain, which is often easier to interpret and manipulate compared to the time domain.

In many signal processing tasks, especially filtering, it is crucial to understand how signals decompose into various frequency components. The Fourier Transform of a discrete signal is termed the Discrete-Time Fourier Transform (DTFT). For a signal \(x[n]\), the DTFT is given by:

• \(X(e^{j\omega}) = \sum_{n=-\infty}^{\infty} x[n] e^{-j\omega n}\).

In the context of the original problem, we use the inverse Fourier Transform to find the coefficients \(e[n]\) for expressing the error function between the desired and actual frequency responses. Specifically, the error coefficients are calculated using the inverse Fourier Transform of the error function \(E(e^{j\omega})\):

• \(e[n] = h_d[n] - h[n]\).

This step is pivotal in establishing the groundwork for later applying Parseval's Theorem to find the overall mean square error.

Parseval's Theorem

Parseval's Theorem is a powerful concept in signal processing, linking the energies in the time and frequency domains. Simply put, it states that the total energy (or power for infinite signals) of a signal remains the same in both domains.

In mathematical terms, for a signal \(x[n]\), Parseval's Theorem is expressed as:

• \(\sum_{n=-\infty}^{\infty} |x[n]|^2 = \frac{1}{2\pi} \int_{-\pi}^{\pi} |X(e^{j\omega})|^2 d \omega \).

Applied to the problem at hand, the theorem allows us to equate the mean square error in the frequency domain with its counterpart in the time domain. That is, the error energy given by the integral over the frequency spectrum explains the same energy calculated by summing the squared magnitudes of error coefficients \(e[n]\) in the time domain. Therefore,

• \(\epsilon^2 = \sum_{n=-\infty}^{\infty} |e[n]|^2 \).

By employing this theorem, we can confirm that simply truncating the desired sample response \(h_d[n]\) to the length \(N\) effectively minimizes the mean square error, as all non-required parts of \(h[n]\) are set to zero. This guarantees the smallest possible error given the length constraints of the response.