5.0 Impulse Response, Step Response, and ConvolutionIn this chapter we confine ourselves to systems that can be modeled as linear and time-invariant, or LTI systems. For these types of systems, we can determine the output of the system to any input in a very systematic way. We can also determine a great deal about the system just by looking at how it responds to various types of inputs. The most fundamental of these inputs is the impulse response, or the response of a system at rest to an impulse. However, the response of a system to a step is much easier to determine and can be used to determine the impulse response of any LTI system.5.1 Impulse or Delta FunctionsAn impulse, or delta function, , ( )tdis defined as a function that is zero everywhere except at one point, and has an area of one. Mathematically, we can write this as( ) 0, 0( ) 1, 0dmmd l ld l l m-= �= >�Note that we do not know the value of (0)d, it is undefined! We can think of, or model, delta functions as functions that exist in some type of limit. For example, the functions displayed in Figures 5.1, 5.2, and 5.3 can be thought of as different models for delta functions, since the meet our two (simplistic) requirements above.Figure 5.1. Rectangular model of an impulse (delta) function.1Figure 5.2. Triangular model of an impulse (delta) function.Figure 5.3. Gaussian model of an impulse (delta) function.2Although delta functions are really idealized functions, they form the basis for much or the study of systems. Knowing how a system will respond to an impulse (an impulse response) tells us a great deal about a system, and lets us determine how the system will response to any arbitrary input.There following two very important properties of delta functions will be used extensively:Property 1: 0 0 0( ) ( ) ( ) ( )t t t t t tf d f d- = -Property 2 (Sifting Property): 0 0 0( ) ( ) ( )bat t t dt t a t bf d f- = < <�The first property is pretty easy to understand if we think about the definition of a delta function. A delta function is zero everywhere except when its argument is zero, so both sides of the equation are zero everywhere except at 0t, and then at 0tboth sides have the same value. The second property follows directly from the first property as follows:0 0 0 0 0 0 0( ) ( ) ( ) ( ) ( ) ( ) ( )b b ba a at t t dt t t t dt t t t dt t a t bf d f d f d f- = - = - = < <� � �It is very important that the limits of the integral are such that the delta function is within the limits of the integral, or else the integral is zero.Example 5.1.1. You should understand each of the following identities, and how to use the two properties to arrive at the correct solution.122010101010( 1) ( 1)( 2) 4 ( 2)( 2)dt = 4( 1)( 20) 0( 1) ( 2) 0ttte t e tt t tt te t dt ee t dtt t dtd dd ddddd d�-�- �- = -- = --- =- =- - =����35.2 Unit Step (Heaviside) FunctionsWe will define the unit step function as0(01)0uttt>��<=�We will not define (0)u, though some textbooks define 1(0)2u =. The argument of the unit step was deliberately not written as t, since this sometimes leads to some confusion when solving problems. It is generally better to remember that the unit step is one whenever the argument (t) is positive, and then try and figure out what this might mean in terms of t.Example. 5.2.1. The following are some simple examples with unit step functions:a) ( 1) 1u t - = for 1 0t - > or 1t >b) (2 ) 1u t- = for 2 0t- > or 2 t>c) 4 13tu� �- =� �� � for 4 03t- > or 12 t>Unit step functions also show up in integrals, and it is useful to be able to deal with them in that context. The usual procedure is to determine when the unit step function (or functions) are one, and then do the integrals. If the unit step functions are not one, then the integral will be zero. When you are done with the integral, you may need to preserve the information indicating that the integral is zero unless the unit step functions are “on”, and this is usually done by including unit step functions. A few examples will hopefully clear this up.Example 5.2.1. Simplify ) ( 1)(u dut l l l��---� as much as possible. We need both unit step functions to be one, or the integral is zero. We need then( ) 1u t l- = for 0t l- > or t l>( 1) 1u l - = for 1 0l - > or 1l >The integral then becomes 1(1)(1) 1td tl = -�. However, we are not done yet. We need to besure both of the unit step functions are 1, which means we need 1t l> >, or 1t >. So theanswer is zero for 1t < and 1t - for 1t >. The way we can write this compactly is( 1) ( 1)t u t- -, which is the final answer.4Example 5.2.2. Simplify2( 2)tte u dl l+-- �-�as much as possible. We need the step functionto be one, or the integral is zero. We need then ( 2) 1u l - = for 2 0l - > or 2l >The integral becomes 22 ( 2) 221 (1 )( )tt tde e ee ell+- - - + - -= = --�However, the integral will be zero unless 2 2t l+ > > or 0t >. The final answer is then2(1 ) ( )te e u t- --.Example 5.2.3. Simplify ( ) ( 2)u t dl d l l�- �- +� as much as possible. This integral has both an impulse and a unit step function. While we might be tempted to use the unit step function to set the limits of the integral, the best (and easiest) thing to do is to just use the sifting property of impulse functions. This gives the result ( ) ( 2) ( 2)u t d u tl d l l�- �- + = +�Example 5.2.4. Simplify ( )( 2)ttdeld l l- -- �-�as much as possible. We can again use the sifting property with this integral, but we must be careful. If we do not integrate past the impulse function, the integral will be zero. Hence we have( 2)( )20( 2)2tttde tetld l l- -- -�-�>�=�<�-��which we can write in a more compressed form as ( 2)( 2)tue t- --.Example 5.2.5. Simplify 1( 2)tte dld l l�+-+� as much as possible. Using the sifting property we have 5211 2( 2)0 1 2ttte te dtld l l�-+-�- <-�+ =�- >-���which we can write more compactly as 2(1 )te u t--.Finally, if we consider integrating an impulse, ( )tdd l l- ��, we will either get a one (if we integrate …
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