lecture 24 continuous-time signals 2

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Fundamentals of Digital Signal Processing Lecture 24 Continuous-Time Signals and LTI Systems 2 Fundamentals of Digital Signal Processing Spring, 2012 Wei-Ta Chu 2012/5/25 DSP, CSIE, CCU 1

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Page 1: Lecture 24 Continuous-Time Signals 2

Fundamentals of Digital Signal Processing

Lecture 24 Continuous-Time Signals and LTI Systems 2

Fundamentals of Digital Signal ProcessingSpring, 2012

Wei-Ta Chu2012/5/25

DSP, CSIE, CCU1

Page 2: Lecture 24 Continuous-Time Signals 2

Time-Invariance� Suppose that a continuous-time system is represented

by

� This system is time-invariant if, when we delay the input signal by an arbitrary amount t0, the result is only to delay the output by that same amount.

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� This condition must be true for any signal x(t) and for any real number t0.

Test of time-invariance for a discrete-time system

Page 3: Lecture 24 Continuous-Time Signals 2

Squaring System is Time-Invariant� When the input is x(t), the output is y(t) = [x(t)]2

� If the input is x(t-t0), the corresponding output will be

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Page 4: Lecture 24 Continuous-Time Signals 2

Integrator is Time-Invariant

� Changing the “dummy variable” of integration to .

� is replaced by , the upper limit becomes

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Page 5: Lecture 24 Continuous-Time Signals 2

A Time-Varying System� Amplitude modulator

� Such a system is a fundamental component of many radio systems.

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� It’s a time-varying system

Page 6: Lecture 24 Continuous-Time Signals 2

Linearity� A continuous-time system is linear if when

and , then

� Principle of superposition

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Test of linearity for a discrete-time system

Page 7: Lecture 24 Continuous-Time Signals 2

Squaring is Nonlinear

� The corresponding output is

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Page 8: Lecture 24 Continuous-Time Signals 2

Integrator is Linear

� The corresponding output is

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� The integrator system is linear

Page 9: Lecture 24 Continuous-Time Signals 2

The Convolution Integral

� This operation is usually written as , meaning x(t) is convolved with h(t).

Every LTI system can be described by a convolution integral

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meaning x(t) is convolved with h(t).

� A system is LTI if and only if its output can be represented as a convolution.

Page 10: Lecture 24 Continuous-Time Signals 2

The Convolution Integral� Assume that

� The corresponding output is

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� Thus, the operation of convolution of an input x(t) with h(t) is a linear operation.

Page 11: Lecture 24 Continuous-Time Signals 2

The Convolution Integral

� Make a substitution

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� Therefore, convolution is time-invariant as well as linear.

For every LTI system, the output y(t) is always equal to, the convolution of the input signal x(t) with

the system impulse response h(t).

Page 12: Lecture 24 Continuous-Time Signals 2

Convolve Unit Steps� When the impulse response is a unit step, h(t) = u(t), and

the input is also a unit step, x(t) = u(t), the convolution integral becomes

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� The upper limit becomes t because when , or

� is called a unit ramp because it is linearly increasing with a slope of one.

Page 13: Lecture 24 Continuous-Time Signals 2

Properties of Convolution� The operation of convolution is commutative,

associative, and distributive over addition.

� Commutativity:

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� Recall that reversing the sign of a definite intergalreverses the order of the limits

Page 14: Lecture 24 Continuous-Time Signals 2

Properties of Convolution� Associativity:

� Distributivity Over Addition:

� Convolution is a linear operation. Therefore, convolution of x(t) with the sum must be the sum of the individual convolutions.

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individual convolutions.

� Identity Element of Convolution: � When will the following equation works?

� The answer is the unit-impulse signal by substituting

Page 15: Lecture 24 Continuous-Time Signals 2

Properties of Convolution

� The impulse is the identity signal for the operation of convolution, much as 1 is the identity element for

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convolution, much as 1 is the identity element for ordinary multiplication.

Page 16: Lecture 24 Continuous-Time Signals 2

Impulse Responses – Integrator � Integrator

� The superscript (-1) means the first anti-derivative, i.e., integral

� The impulse response of this system is the unit step as shown previously in Equation (9.21).

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shown previously in Equation (9.21).

� Thus we can represent the integrator operationally as follows:

A running integral is equivalent to convolution with a unit step.

Page 17: Lecture 24 Continuous-Time Signals 2

Impulse Responses – Differentiator

� It’s an LTI system.

� Substitute for x(t)

� is the impulse response of the differentiator system, the following operational definition must be

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system, the following operational definition must be true:

� is called the doublet.

Differentiation is convolution with derivative of an impulse.

Page 18: Lecture 24 Continuous-Time Signals 2

Example: Convolution with Doublet� The convolution of the unit step with the doublet

can be evaluated by using (9.47)

� The derivative property of the unit step (9.22); i.e.,

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Page 19: Lecture 24 Continuous-Time Signals 2

Impulse Responses – Ideal Delay

� Substituting gives the impulse response of the ideal delay system as

� The impulse response of the ideal delay is a delayed impulse

Time shift is the same as

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Time shift is the same as convolution with a shifted impulse.

Page 20: Lecture 24 Continuous-Time Signals 2

Convolution of Impulses� Convolution of impulses

� The convolution of two shifted impulses at t1 and t2gives a shifted impulse located at the sum

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Page 21: Lecture 24 Continuous-Time Signals 2

Convolution of Impulses� When combined with the linearity property, we can

work problems like:

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� So we end up with impulses at

� The area of the impulses at is -3.

Page 22: Lecture 24 Continuous-Time Signals 2

Impulse Convolution Causes Shifting� Consider the pulse input x(t) and the impulse response

� The equation for the output is simply

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� We simply shift a scaled copy ofthe continuous signal to the location of each impulse and sum all the shifted and scaledcopies.

Page 23: Lecture 24 Continuous-Time Signals 2

Evaluating Convolution Integrals� Delayed Unit-Step Input

� Suppose we wish to evaluate a convolution integral

� is what we call the “dummy variable” of integration in the integral because it disappears when we evaluate

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in the integral because it disappears when we evaluate at the upper and lower limits.

Page 24: Lecture 24 Continuous-Time Signals 2

Evaluating Convolution Integrals� t is the independent variable of y(t)

� To compute each value of the function y(t), we must form the product and then evaluate the integral for each different value of t.

� Make substitution

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� This form does not make it obvious how the limits of integration depend on t.

Page 25: Lecture 24 Continuous-Time Signals 2

Evaluating Convolution Integrals� Key idea: draw an auxiliary sketch of the two

functions whose product is the integrand of the convolution integral.

� The sketch of is the same as

� For the sketch of

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� The desired is obtained by first time-reversing

and then shifting the resultby t.

Page 26: Lecture 24 Continuous-Time Signals 2

Evaluating Convolution Integrals� Show how and interact in evaluating the

convolution integral.

� There are two different regions

� On the left, we see a “typical” plot that is representative of all values of t such that t – 1 < 0.

� The nonzero parts of the two

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� The nonzero parts of the twofunctions do not overlap and hence, their product is zerofor t < 1.

Page 27: Lecture 24 Continuous-Time Signals 2

Evaluating Convolution Integrals� On the right, this plot is representative for all values of

t such that t -1>0. Since the flipped and shifted extends infinitely to the left, the nonzero parts of and will overlap for all t such that t > 1.

� The output for t>1 is given by

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Page 28: Lecture 24 Continuous-Time Signals 2

Evaluating Convolution Integrals� This is an example of the more general case where

for and for . (In this case T0 = 1 and T1 = 0.)

� If t < T0 + T1, then the nonzero parts of and do not overlap. It is always true that y(t) = 0 for t < T0

+ T1 .

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+ T1 .

� Even if one or both of the signals involved in the convolution is discontinuous, the result of the convolution will be a continuous function of time.

Page 29: Lecture 24 Continuous-Time Signals 2

Exercise 9.4

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Page 30: Lecture 24 Continuous-Time Signals 2

Evaluation of Discrete Convolution� Recall that the discrete convolution sum is defined as

� Assume that the input is a shifted discrete-time unit-step sequence x[n] = u[n-1] and the impulse response is a discrete-time exponential sequence

� Make substitution

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� Make substitution

Page 31: Lecture 24 Continuous-Time Signals 2

Evaluation of Discrete Convolution� We need to form the product for n fixed and

sum the values of the resulting product sequence over all k. We must do this for all values

� It helps if we plot x[n-k] as a function of k for different values of n. We can do this by first time-reversing x[k] to obtain g[k]=x[-k], and then shifting g[k] by nto obtain

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to obtain

Page 32: Lecture 24 Continuous-Time Signals 2

Evaluation of Discrete Convolution� In Fig. 9-17(a), nonzero parts of the sequences h[k]

and x[n-k] do not overlap if n-1<0.

� In Fig. 9-17(b), for

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� In Fig. 19-7(c)