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For each of the following statements, determine whether it is true or false: (a) If \(x[n]=0\) for \(nT_{1}\) and \(h(t)=0\) for \(t>T_{2},\) then \(x(t) * h(t)=0\) for \(t>\) \(T_{1}+T_{2}\)

Short Answer

Expert verified
(a) True, (b) False, (c) True, (d) True

Step by step solution

01

Analyze statement (a)

Statement (a) claims that if the signals \(x[n]\) and \(h[n]\) are zero for indices less than \(N_1\) and \(N_2\) respectively, then the convolution \(x[n] * h[n]\) is zero for indices less than \(N_1 + N_2\). Convolution involves overlapping and summing over the product of two signals, and if \(x[n]\) and \(h[n]\) are zero for indices less than \(N_1\) and \(N_2\), there should be no contribution to the convolution sum for \(n < N_1 + N_2\). Hence, this statement is **true**.
02

Analyze statement (b)

Statement (b) checks the equivalence of delayed signals. Given \(y[n] = x[n] * h[n]\), the convolution of \(x[n]\) and \(h[n]\) does not simply translate to \(y[n-1] = x[n-1] * h[n-1]\) because convolution does not maintain simple translations unless both signals are shifted in the same manner simultaneously. Thus, the statement is **false**.
03

Analyze statement (c)

Statement (c) relates the time-reversal property of continuous-time convolution. For continuous signals, if \(y(t) = x(t) * h(t)\), then using the property of convolution under time reversal, \(y(-t)\) is indeed equal to \(x(-t) * h(-t)\). This is true because convolution is a symmetric operation in time. Thus, this statement is **true**.
04

Analyze statement (d)

In statement (d), the limits of \(x(t)\) and \(h(t)\) are specified for \(t > T_1\) and \(t > T_2\). Convolution involves the integral of overlapping sections of the signals. Thus, if both signals are zero beyond their respective limits, the convolution of these signals beyond \(T_1 + T_2\) will also be zero since there will be no overlaps beyond this limit. Therefore, this statement is **true**.

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Key Concepts

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

Convolution
Convolution is a mathematical operation used to combine two signals to produce a third signal. This operation is fundamental in the analysis of linear systems and helps determine how a system responds to an input signal. In both discrete-time and continuous-time systems, convolution describes how the shape of one signal is modified by the other.

In discrete-time signals, convolution is represented by summing the product of the input signal and the impulse response signal, both shifted in time. Mathematically, if we have two signals, \( x[n] \) and \( h[n] \), the convolution is given by:
\[ y[n] = (x * h)[n] = \sum_{k=-\infty}^{\infty} x[k] h[n-k] \]
For continuous-time signals, convolution involves integrating the product of the two signals over time shifts. Given signals \( x(t) \) and \( h(t) \), convolution is described by:
\[ y(t) = (x * h)(t) = \int_{-\infty}^{\infty} x(\tau)h(t-\tau)\,d\tau \]
Convolution is an essential tool in signal processing, allowing us to understand and predict how signals interact within a system.
Time-Reversal Property
The time-reversal property is an important feature of convolution. It refers to the scenario where reversing the time axis affects the signals. In signal processing, this property is crucial for analyzing symmetric properties of systems.

For discrete-time signals, reversing the time indices essentially flips the signal around the vertical axis. If a signal \( x[n] \) is time-reversed, it becomes \( x[-n] \).

In continuous-time systems, if a signal function \( x(t) \) is reversed, it corresponds to \( x(-t) \). This reversed signal is often used in conjunction with convolution to test the symmetrical properties and behavior of the system. The time-reversal property under convolution for continuous-time signals, as demonstrated, means that if \( y(t) = x(t) * h(t) \), then \( y(-t) = x(-t) * h(-t) \).

This property highlights the robustness and flexibility of convolution, allowing for easier manipulation and analysis of signals.
Discrete-Time Signals
Discrete-time signals are functions defined only at discrete points in time. These signals are vital in digital signal processing where systems operate on sampled data rather than continuous signals.

For example, digital audio or video signals are discrete-time signals where data is processed at specific intervals. Mathematically, a discrete-time signal is represented as a sequence, \( x[n] \), where \( n \) is an integer.

When dealing with discrete-time systems, techniques like sampling, quantization, and discretization are commonly used. The primary goal is to convert continuous signals into a form that can be processed by digital systems. Convolution in discrete-time is especially significant as it allows these signals to be manipulated and analyzed in ways that resemble continuous interactions, even in a discrete setting.

These signals are utilized extensively in technologies such as telecommunications, audio processing, and data transmission, emphasizing their importance in modern electronics.
Continuous-Time Signals
Continuous-time signals are functions that exist and are defined for all values of time. These signals are fundamental in analog signal processing where the data can be any value at any time.

An example of a continuous-time signal is sound waves, which naturally occur in the physical world. These signals, denoted usually by \( x(t) \), where \( t \) is the time variable, are often analyzed using calculus-based approaches.

In continuous-time systems, the convolution operation is integral. Given that these signals can change at any moment, convolution helps in understandings how systems like filters or amplifiers reshape these signals through their impulse responses. This understanding is crucial as it informs how we design and predict the behavior of a wide range of electronic devices and systems.

The importance of continuous-time signals lies in their ability to represent real-world phenomena accurately, making them essential for the modeling and analysis of systems in fields such as control engineering and communications. They serve as a bridge between the physical world and analytical models.

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Most popular questions from this chapter

Draw block diagram representations for causal LTI systems described by the follwing differential equations: (a) \(y(t)=-\left(\frac{1}{2}\right) d y(t) / d t+4 x(t)\) (b) \(d y(t) / d t+3 y(t)=x(t)\)

Determine whether each of the following statements concerning LTI systems is me or false. Justify your answers. (a) If \(h(r)\) is the impulse response of an LTI system and \(h(t)\) is periodic and nonzero. the system is unstable. (b) The inverse of a causal LTI system is always causal. (c) \(\mathbf{f}|h[n]| \leq K\) for each \(n\), where \(K\) is a given number, then the \(L T\) is system with \(h[n]\) as its impulse response is stable. (d) If a discrete-time LTI system has an impulse response \(h[n]\) of finite duration. the system is stable (e) If an LTI system is causal, it is stable. (f) The cascade of a noncausal LTI system with a causal one is necessarily noncausal. (g) A continuous-time LTI system is stable if and only if its step response \(s(t)\) is absolutely integrable - that is, if and only if $$\int_{-\infty}^{+\infty}|s(t)| d t<\infty$$ (h) A discrete-time LTI systew is causal if and only if its slep response \(s[n]\) is zero for \(n<0\)

Compute and plot \(y[n]=x[n] * h[n],\) where $$\begin{array}{l} x[n]=\left\\{\begin{array}{ll} 1, & 3 \leq n \leq 8 \\ 0, & \text { otherwise } \end{array}\right. \\ h[n]=\left\\{\begin{array}{ll} 1, & 4 \leq n \leq 15 \\ 0, & \text { otherwise } \end{array}\right. \end{array}$$

Another application in which matched filters and correlation functions play an important role is radar systems. The underlying principle of radar is that an electromaguetic pulse transmitled at a target will be reflected by the targe and will subsequently return to the sender wath a delay proportional to the distance to the target Ideally, the received sigral wil? simply be a shifted and possibly scaled version of the original transmitted signal Let \(p(t)\) he the original pulse that is sent out. Show that $$\phi_{\mu p}(0)=\max \phi_{, p}(t)$$ if the waveform that comes back to the sender is $$x(t)=\alpha p\left(t-t_{0}\right)$$ where \(\alpha\) is a positive constant, then $$\phi_{r p}\left(t_{t}\right)=\max _{t} \phi_{r p}(t)$$ (Hint: Use Schwartz's inequality.) Thus, the way in which simple radar ranging systems work is hased on using a matched filter for the transmitted waveform \(p(t)\) and noting the tome at which the output of this system reaches its maximum value.

A \(\$ 100,000\) mortgage is to be retired by equal monthly payments of \(D\) dollars. In terest, compounded monthly, is charged at the rate of \(12 \%\) per annum on the unpaid balance; for example, after the first month, the total debt equals $$\$ 100,000+\left(\frac{0.12}{12}\right) \$ 100.000=\$ 101,000$$ The problem is to determine \(D\) such that after a specified time the mortgage is paid in full, leaving a net balance of zero. (a) To set up the problem, let \(y[n]\) denote the unpaid balance after the \(n\) th monthly payment. Assume that the principal is borrowed in month 0 and monthly payments begin in month 1. Show that \(y[n]\) satisfies the difference equation $$y[n]-\gamma y[n-1]=-D \quad n \geq 1$$ with initial condition $$y[0]=\$ 100,000$$ where \(\gamma\) is a constant. Determine \(\gamma\) (b) Solve the difference equation of part (a) to determine $$y[n] \quad \text { for } n \geqslant 0$$ (Hink: The particular solution of eq. \((\mathrm{P} 2.63-1)\) is a constant \(Y\). Find the value of \(Y,\) and express \(y(n]\) for \(n \geq 1\) as the sum of particular and homogeneous solutions. Deternine the unknown constant in the homogeneous solution by directly calculating \(y[1]\) from eq. \((P 2.63-1)\) and comparing it to your solution.) (c) If the mortgage is to be retired in 30 years after 360 monthly payments of \(D\) dollars, determine the appropriate value of \(D\) (d) What is the total payment to the bank over the 30 -year period? (e) Why do banks make loans?

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