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A discrete-time system has transfer function \(H\left(e^{j 2 \pi f}\right)\). A signal \(x(n)\) is passed through this system to yield the signal \(w(n) .\) The time-reversed signal \(w(-n)\) is then passed through the system to yield the time-reversed output \(y(-n)\). What is the transfer function between \(x(n)\) and \(y(n)\) ?

Short Answer

Expert verified
The transfer function is \( |H(e^{j 2 \pi f})|^2 \).

Step by step solution

01

Analyze the System's Operations

First, let's understand the described system operations. The signal \( x(n) \) is passed through a system with transfer function \( H\left(e^{j 2 \pi f}\right) \), resulting in \( w(n) \). This operation can be described in the frequency domain as \( W\left(e^{j 2 \pi f}\right) = H\left(e^{j 2 \pi f}\right)X\left(e^{j 2 \pi f}\right) \).
02

Describe Time-Reversal in Frequency Domain

Next, we reverse the signal \( w(n) \) to obtain \( w(-n) \). In the frequency domain, this operation translates to taking the complex conjugate of \( W\left(e^{j 2 \pi f}\right) \), leading to \( W^*\left(e^{j 2 \pi f}\right) \).
03

Pass Time-Reversed Signal through System

The time-reversed signal \( w(-n) \) is passed through the same system, resulting in \( y(-n) \). In the frequency domain, this corresponds to \( Y^*\left(e^{j 2 \pi f}\right) = H\left(e^{j 2 \pi f}\right)W^*\left(e^{j 2 \pi f}\right) \).
04

Relate Initial Input to Final Output

We want the transfer function from \( x(n) \) to \( y(n) \). Given \( Y^*\left(e^{j 2 \pi f}\right) = H\left(e^{j 2 \pi f}\right)\left(H\left(e^{j 2 \pi f}\right)X\left(e^{j 2 \pi f}\right)\right)^* \), we find that \( Y\left(e^{j 2 \pi f}\right) = H^*\left(e^{j 2 \pi f}\right)H\left(e^{-j 2 \pi f}\right)X\left(e^{j 2 \pi f}\right) \).
05

Simplify Transfer Function

The transfer function between the input \( x(n) \) and the output \( y(n) \) is therefore \( H^*\left(e^{j 2 \pi f}\right)H\left(e^{-j 2 \pi f}\right) \). This simplifies to \( |H\left(e^{j 2 \pi f}\right)|^2 \).

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Key Concepts

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

Transfer Function
The concept of a transfer function is essential in systems engineering, particularly within the analysis of linear time-invariant (LTI) systems. For a discrete-time system, the transfer function, denoted as \( H(e^{j 2 \pi f}) \), describes how the system modifies the input signal in the frequency domain.

The transfer function is usually obtained by taking the Fourier transform of the system's impulse response. For any given input \( x(n) \), the output \( y(n) \) in the frequency domain is given by \( Y(e^{j 2 \pi f}) = H(e^{j 2 \pi f}) X(e^{j 2 \pi f}) \), where \( X(e^{j 2 \pi f}) \) is the Fourier transform of the input.

Thus, the transfer function itself represents the system’s frequency response and tells us how each frequency component of the input signal is altered by the system. It provides insight into the system's behavior and helps determine its stability and performance.
Frequency Domain
The frequency domain analysis is a cornerstone of signal processing and provides a powerful means of analyzing signals and systems. Instead of representing signals as they change over time, we focus on how much of each frequency is present.

In our exercise, the initial step involves transforming the time-domain signal \( x(n) \) into its frequency-domain counterpart \( X(e^{j 2 \pi f}) \). This is achieved via the Fourier transform, which decomposes a signal into its sinusoidal components.

This frequency domain representation allows us to understand and characterize the system's effect (described by \( H(e^{j 2 \pi f}) \)) on the signal without solving differential/difference equations. Specifically, it simplifies the process of convolution in the time domain to multiplication in the frequency domain, offering an elegant and efficient means to process signals.
Time Reversal
Time reversal is a less intuitive, yet important operation in signal processing. When a signal is time-reversed, it involves changing the direction of time, mathematically represented as \( x(-n) \) for a signal \( x(n) \).

This operation corresponds to altering the phase of its frequency components. In particular, when you time-reverse a signal \( w(n) \), the frequency domain representation becomes the conjugate of \( W(e^{j 2 \pi f}) \), denoted as \( W^*(e^{j 2 \pi f}) \).

In the problem at hand, after time-reversal, the reversed signal is passed through the same system, resulting in \( y(-n) \). The challenge lies in confirming that the operation not only changes the signal’s temporal characteristics but considerably affects its frequency representation, which ultimately alters the output \( y(n) \). This understanding is crucial for correctly interrelating inputs and outputs through transformations like the one we examined.

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Most popular questions from this chapter

A digital filter is determined by the following difference equation. $$y(n)=y(n-1)+x(n)-x(n-4)$$ a) Find this filter's unit sample response. b) What is the filter's transfer function? How would you characterize this filter (lowpass, highpass, special purpose, \(\ldots) ?\) c) Find the filter's output when the input is the sinusoid \(\sin \left(\frac{\pi n}{2}\right)\). d) In another case, the input sequence is zero for \(n<0,\) then becomes nonzero. Sammy measures the output to be \(y(n)=\delta(n)+\delta(n-1)\). Can his measurement be correct? In other words, is there an input that can yield this output? If so, find the input \(x(n)\) that gives rise to this output. If not, why not?

Sammy loves to whistle and decides to record and analyze his whistling in lab. He is a very good whistler; his whistle is a pure sinusoid that can be described by \(s_{a}(t)=\sin (4000 t)\). To analyze the spectrum, he samples his recorded whistle with a sampling interval of \(T_{S}=2.5 \times 10^{-4}\) to obtain \(s(n)=s_{a}\left(n T_{S}\right)\). Sammy (wisely) decides to analyze a few samples at a time, so he grabs 30 consecutive, but arbitrarily chosen, samples. He calls this sequence \(x(n)\) and realizes he can write it as $$x(n)=\sin \left(4000 n T_{S}+\theta\right), \quad n=\\{0, \ldots, 29\\}$$ a) Did Sammy under- or over-sample his whistle? b) What is the discrete-time Fourier transform of \(x(n)\) and how does it depend on \(\theta ?\) c) How does the 32 -point DFT of \(x(n)\) depend on \(\theta\) ?

A discrete-time system is governed by the difference equation $$y(n)=y(n-1)+\frac{x(n)+x(n-1)}{2}$$ a) Find the transfer function for this system. b) What is this system's output when the input is \(\sin \left(\frac{\pi n}{2}\right) ?\) c) If the output is observed to be \(y(n)=\delta(n)+\delta(n-1)\), then what is the input?

Sampling and Filtering The signal \(s(t)\) is bandlimited to \(4 \mathrm{kHz}\). We want to sample it, but it has been subjected to various signal processing manipulations. a) What sampling frequency (if any works) can be used to sample the result of passing \(s(t)\) through an \(\mathrm{RC}\) highpass filter with \(R=10 \mathrm{k} \Omega\) and \(C=8 \mathrm{nF} ?\) b) What sampling frequency (if any works) can be used to sample the derivative of \(s(t) ?\) c) The signal \(s(t)\) has been modulated by an \(8 \mathrm{kHz}\) sinusoid having an unknown phase: the resulting signal is \(s(t) \sin \left(2 \pi f_{0} t+\phi\right)\), with \(f_{0}=8 \mathrm{kHz}\) and \(\phi=?\) Can the modulated signal be sampled so that the original signal can be recovered from the modulated signal regardless of the phase value \(\phi ?\) If so, show how and find the smallest sampling rate that can be used; if not, show why not.

A problem on Samantha's homework asks for the 8 -point DFT of the discrete- time signal \(\delta(n-1)+\) \(\delta(n-7)\) a) What answer should Samantha obtain? b) As a check, her group partner Sammy says that he computed the inverse DFT of her answer and got \(\delta(n+1)+\delta(n-1) .\) Does Sammy's result mean that Samantha's answer is wrong? c) The homework problem says to low pass-filter the sequence by multiplying its DFT by $$H(k)=\left\\{\begin{array}{ll} 1 & \text { if } k=\\{0,1,7\\} \\ 0 & \text { otherwise } \end{array}\right.$$ and then computing the inverse DFT. Will this filtering algorithm work? If so, find the filtered output; if not, why not?

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