5 0 Impulse Response Step Response and Convolution In this chapter we confine ourselves to systems that can be modeled as linear and timeinvariant 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 Functions An impulse or delta function d t is defined as a function that is zero everywhere except at one point and has an area of one Mathematically we can write this as d l 0 l 0 m d l d l 1 m 0 m Note that we do not know the value of d 0 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 1 Figure 5 2 Triangular model of an impulse delta function Figure 5 3 Gaussian model of an impulse delta function 2 Although 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 f t d t t0 f t0 d t t0 b f t d t t0 dt f t0 a t0 b Property 2 Sifting Property a 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 t0 and then at t0 both sides have the same value The second property follows directly from the first property as follows b b b a a a f t d t t0 dt f t0 d t t0 dt f t0 d t t0 dt f t0 a t0 b 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 et d t 1 e1d t 1 t 2d t 2 4d t 2 2 t d t 2 dt 4 0 10 t 1 e d t 1 dt e 0 10 t e d t 20 dt 0 10 d t 1 d t 2 dt 0 3 5 2 Unit Step Heaviside Functions We will define the unit step function as 1 t 0 u t 0 t 0 1 2 The argument of the We will not define u 0 though some textbooks define 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 u 0 Example 5 2 1 The following are some simple examples with unit step functions a u t 1 1 for t 1 0 or t 1 b u 2 t 1 for 2 t 0 or 2 t t t u 4 1 4 0 3 for 3 c 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 u t l u l 1 d l Example 5 2 1 Simplify as much as possible We need both unit step functions to be one or the integral is zero We need then u t l 1 for t l 0 or t l u l 1 1 for l 1 0 or l 1 t The integral then becomes 1 1 d l t 1 However we are not done yet We need to be sure both of the unit step functions are 1 which means we need t l 1 or t 1 So the answer is zero for t 1 and t 1 for t 1 The way we can write this compactly is 1 t 1 u t 1 which is the final answer 4 t 2 t e u l 2 d l Simplify Example 5 2 2 as much as possible We need the step function to be one or the integral is zero We need then u l 2 1 for l 2 0 or l 2 The integral becomes t 2 l e 1 d l e 2 e t 2 e 2 1 e t 2 However the integral will be zero unless t 2 l 2 or t 0 The final answer is then e 2 1 e t u t u t l d l 2 d l Example 5 2 3 Simplify 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 u t l d l 2 d l u t 2 t t l e d l 2 d l Example 5 2 4 Simplify 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 t e t 2 t l e d l 2 d l 0 t 2 t 2 t 2 u t 2 which we can write in a more compressed form as e Example 5 2 5 Simplify property we have t l e t 1 d l 2 d l as much …
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