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# Signals and Systems I ECE 350

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This 104 page Class Notes was uploaded by Fredy Okuneva on Thursday October 22, 2015. The Class Notes belongs to ECE 350 at University of Idaho taught by Dennis Sullivan in Fall. Since its upload, it has received 135 views. For similar materials see /class/227726/ece-350-university-of-idaho in ELECTRICAL AND COMPUTER ENGINEERING at University of Idaho.

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1 Introduction 11 Complex Numbers Why do we use complex numbers One of the main reasons for using complex numbers is simply that it is easier than writing two separate numbers In Fig 1 if we consider the line to be a vector in the X Y rectangular coordinate system we could say the vector is A 3 2 Or if we consider the x coordinate to be real and the y coordinate to be imaginary we can write it as one complex number A3j2 111 Y 1mag1nary A3i2 2 3 X real Figure 1 Suppose I have another vector B 2 1 I can write this as the complex number B 2j 112 Now I can start doing operations on them like addition and subtraction A B 3 1392 2j 2 1 1393 A B 3 1392 2 1 2 r We can also do multiplication and division with this form of complex numbers but it is not so straightforward 1 Introduction 1 7102010 AB3j2 2j 3 2j221j312 2 6 2j3 48j39 113 B 3j392 52 B BBquot 2j 2j 114 232j 3 4 22 12 5 It turns out that if we are going to multiply and divide there is an easier way Instead of the rectangular coordinates that we have been using it is worth while to change to polar coordinate where instead of real and imaginary parts or X and Y coordinates we have an amplitude and phase Fig 2 7 I 7 0 B2j A73J2736337 2236153 Figure 112 A and B in polar coordinates We make the transition in the following way A3j2AzA A 3222 1 1312236 AA tan 1 337 Similarly I Introduction 2 7102010 B 221212512 2236 AB tan 1 266 WRONG I got the value AB 2660 by punching l 2 in my calculator and then taking the arctan But look at Fig 2 Does LB look like 7266 degrees The problem comes about because my calculator cannot distinguish between 1 divided by 72 and 71 divided by 2 In actuality the answer is 13 1800 2660 15340 You always have to be aware of which quadrant you are in Now that we have A 1340 B 115340 multiplication and division are much easier To multiply two numbers in polar coordinates we multiply the magnitudes and add the phases ABJEJ z34 1534 J511874 8061 17260 Note that we usually prefer the angles between 180 and 7180 degres Compare this to our answer above AB 8 j 82 12Ztan j J5 180712 28061 17288 We get the division of two complex numbers by dividing magnitudes and subtracting phases B 5115340 J3 Compare this to our result above 34 1534 1611 1194quot 1 Introduction 3 7102010 2 2 12 A 4 1397 4 7 1 7 tan B 5 5 5 4 064196 Z z 180 tan 1 26121 1197quot 1611 1197quot There is another way of writing angles that is more common in electrical engineering We often write the angle as an imaginary exponential For example the above vector can be written 5 1618711970 B There is a set of equations that are a cornerstone in complex analysis They are called the Euler eguations e cos6 jsin6 e9e39 cos6 2 19 19 e e sm6 2 For instance we might have a vector A A 36 We can use Euler s equations to expand A A 36 36cos34 j39sin34quot 360829 10574 3 1392 In using the polar coordinates remember that angles can be in degrees or radians This should not pose a problem as long as you remember that 7139 1800 For instance if I have X 05 but I know that my calculator uses degrees I can make the following substitution X Z 0 583 2 0 58313004 05e1 0 035 1035 1 Introduction 4 7102010 Complex Numbers and Sinusoidal functions Besides the Euler equations above the following trigonometric identities will prove to be very useful sinai sincos i cos asin c0sai cosacos isinasin sinai90 icosa c0sai90 sina Suppose we are now talking about a sinusoidal function over some finite interval T as shown in Fig 4 The functions are y1 I cos 27rI y2 I sin 27rI y3 I c03 2 I j 5 Y1cos2 n t Figure 4 This hints at a way of writing the sinusoidal signals as complex numbers y1 I cos 27rI Re 62 y2 I sin 27rI cos 27rI 90quot Re enH900 1 Introduction 5 7102010 I can use phasor notation to write 900 Y 1 Y2 e j 1 2 I In the phasor notation the t1meharmon1c part Re 5 is understood cosnt sinnt Figure 5 Unit vectors on the unit circle if cosine is the reference What about 13 I Well we can rewrite it as y3 I cos 2m cos 27Icos sin 27rIsin 0866c0s27rI055in27rI ie it can be written as the sum of our two functions sint and cost If that s the case why don t we try the same trick and say cost is the real direction and sint is the imaginary direction and write 13 I as a complex number where the sinusoidal part at frequency a 2H1 is assumed This is the phasor notation Y3 0866 105 1w I can do operations on Ypy 213 I can add and subtract them Y4 Y1 Y2 11 JigW I know that Y4 actually means y4 I Re JigWW J5 cos27rI 754 What if x1 I cos 47rI x2 I cos 67rI 10quot Can I add these using phasors The answer is no because implicit to the phasor notation is that all phasors represent timeharmonic signals of the same frequency 1 Introduction 6 7102010 Phasors to Solve Differential Equations with Sinusoidal Inputs Suppose I m given a differential equation of the form 6 12 t cos 3t dt ylt gt lt gt If I have a linear system and I put in a sinusoidal function of frequency a 3 I know the solution will also be a sinusoid of the same frequency One way I can solve this is to say that the solution must have the form y t A cos 3t B sin3t I know that d t 3A sin 3t 33 c0s3t Now my equation is 3Asin 3t 3B c0s3t12A cos 3t B sin 3 cos 3t For the equation to be true all the cosine and sine terms have to balance 3B c0s3t12A cos 3t cos 3t 3A sin3t 12B sin 3t 0 or 3B 12A 1 3A 12B 0 The second gives me A B 04B So the first becomes 3B 1204B 1 3B048B 1 B 0287A 0115 And the solution is yt 0115c0s3t 0287sin3t 0309 c0s3t 68 I am using the trigonometric identity c0sa c0sac0s sinasin That was straightforward but very laborious Notice that em c0s3t jsin3t So I can say that c0s3t Re em I wonder if I could use the exponential term as my input Then when I have a solution I ll just take the real part If digit 12yt em c0s3t jsin3t 1 Introduction 7 7102010 Linearity tells me that output will be the sum of two parts one due to the real cosine part and one due to the imaginary sine part So if I do the problem with the exponential forcing function and then take the real part of the result I should get the same thing that I got for just the cosine input For the exponential forcing function I know the output will be the form y t Y e 3 where Y is a complex number Now if I put this in the equation I get 3te3 12Ye339 em In fact I can rewrite this as 3jYl2Y 1 from which I get Y 03096680 12 3 3234680 My answer is Re yt Re Ye3 Re 0309e 168 e3 0309 cos 3t 680 What if the input had been cos 3t 100 Re e wwo I could go back and redo everything but that isn t really necessary My phasor equation is 432410 3te3 12Ye339 6 And so 110 110 8 e 03096580 123j 3234680 What if the input had been x t sin 3t Then I m going to use Imle3 sin 3t But the mathematics will all be the same Now suppose I have an input of the form 60 0 dt Linearity tells me I can just solve two separate problems and then put them together yt 0309cos3t10 7 680 5309sin3t 300 680 l2yt cos3tlO5sin3t 300 0309cos3t 580 l545sin3t 980 But in general I would only want to solve for one input The following two formulas are often very helpful sina cosa 90 1 Introduction 8 7102010 cos a sina 90 I could convert either but I d rather work with cosines so 5sin3t 300 5 cos 3t 300 900 5 cos3t 900 and the input is cos 3t 10 5cos3t 600 1e1 56600 89 1390 17 25 1433 161 145 477am Therefore the output will be Y 0309e39580447e 138e52U yt l38cos3t 520 Example d t 2 yd 4yt10sin2t First divide through by 2 so the leading coefficient is l d t 2W 5mm In phasor notation the original equation becomes j2Y 2Y 5 Now Ijust solve for Y y A 4 2 12 1414450 yt 355sin2t 450 3556145 Example 6 dt Clearly this is the same problem with a different magnitude an phase of the forcing function But linearity dictates that the answer is the same as above with the amplitude changed by 2yt 3sin2t 63 7 06 and phase changed by 63 or yt 06355sin2t 450 630 213sin2t 1080 Check 1 Introduction 9 7102010 32 630 32 630 Y 21361080 2 1392 1412450 Adding Sines and Cosines There are many occasions where we nd ourselves adding expressions like the following xt 25 cosat 25 7 5sinat 500 This is where we can use the two formulas cosp sin p 90 7 sinp cos p 900 I can rewrite the above equation as xt 25 cosat 250 5 cosat 50 7 900 Now I can write it in phasor notation X 25e25quot Se wquot 025 cos 25 7 jsin 25 7 05cosl40 7 jsinl40 7 0226 10 106 0383 10321 0609 10215 06466 xt 646 cosat 190 Example Solve for yt The answer should be only sines 6 dt 3yt cos3tsin3t Solution cos 3t sin 3t sin 3t 90 7 sin 3t Im ewe900 em Im1 ew I assume my solution is of the form yt ImYe3 So the original equation becomes j3Ye3 3Y6 1 j 131 3Y 1j 1 Introduction 10 7102010 j3Y3Y1j 1j 1 Y 31j 3 Therefore the answer is ytsin3t 12 Linear Systems and Signals W Time seconds Figure 121 A continuous signal The functionf in Fig 1 is a signal It could represent anything a current a voltage the water ow from a pipe the mean temperature at the North Pole t is continuous ie it is a mathematical function of an independent variable t Note that continuous also means it is uniquely de ned in t except for maybe a nite number of points as shown in Fig 2 Tvme Seumds Figure 122 A continuous signal with a nite number of discontinuities The function g I 5 does not qualify as a signal because the square root has two values Note that signals that represent real quantities in the time domain are real We deal with complex numbers only a er making a transformation to the frequency domain eg the Laplace Fourier or Z domain Suppose we have to process t in a computer To do this we have to sample the signal ie se1ect values at a speci c time interval T Then we get a signal which we might designate as kT or just k where it is understood that the integer k Fig 123 1 Introduction 11 7102010 Figure 123 The signal of Fig 1 sampled at an interval T A system can be anything It can be a circuit a factory an ecosystem etc Physically systems are described by differential equations We use the physical properties of the components of the system along with the laws governing their interaction to put together differential equations For instance look at the simple RL circuit in Fig 124 Figure 124 An RL circuit The physical principles are the relationships between the voltage and the current for the individual components vR z39R 121 a dit V 7 121 b Kirchov s voltage law tells us that the sum of the voltages around the circuit must be zero so dit VmtRltL7 122 Now we have a differential equation that describes this system a simple electric circuit with input VmU We could choose 11 as the output What if we want to solve this in a computer First we have to sample the signals at an interval T Then we have to convert the derivatives to difference equations 1 Introduction 12 7102010 d i N ik ik 1 dz T 39 The sampling rate T must always be chosen small enough that it does not play a direct role in the dynamics ofthe system being modeled Now Eq 122 becomes 123 vlnkRikLiUC 124 T RT vaUc Tikik ik l L39k l T k 125 L T R Equation 5 calculates the new value of ik from ik l and vk In this class we will deal with continuous and discrete time dynamic systems with the following two important characteristics 1 Linear 2 Timeinvariant These two characteristics can primarily be seen in the differential equations describing them Linear means the differential equation has constant coefficients Time invariant means that the coefficients do not change in time As an example Eq 126a is linear time invariant Equation 126b is not linear because of the 3y coefficient in front of the second term Equation 126c is not time invariant because of the 3t term in front of the second term d2 i dz dip igt2yfff 126a d2 i d I if 3y2yfi 126 b d2 i dz d 32yfz 126 c Let s take a look at what is necessary to get the differential equation describing a system We will us the simple circuit in Fig 125 1 Introduction 13 7102010 t0 Figure 125 A simple RC circuit The switch closes at time t0 The source is a constant current source The output is the voltage across the capacitor We will take 10 the current source to be the input and V0 the voltage across the capacitor to be the output Once again we use the physical properties of the components to build a differential equation V00 R dvct dt and we use Kirchoff s current law to sum the currents at one of the nodes dv t v t C lt gt 1021 dt R Normally we want the highest order derivative to have a coef cient of one so I will rewrite this as he 41 C Mm tt dt CR C 39 The solution to VCU will be the sum ofthe solutions due to the two sources of energy 127 provided to the circuit namely the input current source 1 and whatever energy is already in the circuit at time zero when the switch is closed ie the initial conditions In the usual approach to solving differential equations we rst determine the forced response This is where the solution will eventually go due to the forcing function In this simple circuit it is easy to see that as time goes on the output will just be the voltage drop due to the current across the resistor ie vFtRIS 128 a Then we have to solve the homogeneous solution which for this simple circuit is of the form 7 RC vHtAe 128 b where A is a constant that we will determine from the initial conditions The initial condition that I need is the voltage on the capacitor right after the switch has closed But 1 Introduction 14 7 102010 physics tells me that the voltage across a capacitor cannot change instantaneously because that would require in nite current I will assume that there is some charge on the capacitor that is sitting there before the switch is closed and this results in a voltage Wm5V The total solution will be the sum of the homogeneous and forced responses 128 a and 128 b vTtAe RC SR 129 Now we use the initial condition to determine the unknown constant A We know the value of V70 for certain at one time only ie right after the switch is closes so we can set WmU5VA mmLRALR Therefore A5 4R Our total solution is now W05 ARV CLR 121m Let s try a little different approach using Laplace transforms If I take the Laplace transform of Eq 127 I get 1 I SVSV0 VS S 1211 RC SC Note that I have assumed that the input current actually has the time domain formulation 10 1110 The reason for adding the step function is the following even though the current source is a constant the output doesn t see it until t 0 I can rewrite Eq 1211 1 S VcssEj v0 SC Note that the two terms on the right side represent the two sources of energy the initial 1212 condition V07 and the forcing function due the current source First we solve for VC 3 V0 1 Vs 1 1 1213 447 C 47 s RC 3HRC The inverse Laplace transform of Eq 1213 is I rRC 71126 vTrgt voaz WHEOe ua 1214 This is of course the same as our previous solution in Eq 1210 As was pointed out earlier we can think of this solution as due to two sources 1 The zero input response due to the initial condition and 1 Introduction 15 7102010 2 The zero state response due to the forcing function I could even write the output as vTt v21 tvzst 1215 a v21t 2 1200 RCut 1215 b 5 71110 vzt RCl e 1215c These are similar to but not the same as the homogeneous and forced responses It also illustrates the very important principle of linearity System Linearity and Time Invariance Let s go back to our RC circuit of Fig 125 only I ll draw it as a block diagram I S Vc gt Figure 126 Block diagram of the RC circuit Now I just show a block with an input I and an output V60 We will assume that the system is at rest ie the initial conditions are all zero If I further assume that C 11F and R 1kQ then I can speci cally write using Eq 1215 c vzxt 215103 Ht103 1216 If 1mI4 then v t1 e 03V 1217 What if 1 5 M Then v t 51 e l103 V We know this by Virtue of linearity In fact can we eXpress linearity in a more general fashion If the input f1 produces the out put yl and f2 produces the outputyz then the input 0 f1 produces the output 001 yz Recall that the input f1 M0 gave the output in Eq 1217 What is the output if f2 15u1 1 1 Introduction 16 7 102010 Timeinvariance dictates that the output must be V110 510317e7071 103util Example Find the response to the RC circuit ofFig 125 for the input in Fig 127 if the circuit is initially at rest Figure 127 Input Solutio The input can be expressed mathematically as M 1 1404104 Equation 1214 and linearity say that the response to this input is WU CliertRCM07C179trlRCut71 However now what happens if 1 say the input is 13t Blt This is also a causal function However it is not a linear combination of the previous input ut That means we have to go back and solve the differential equation all over again Wouldn t it be nice if we had some mathematical description of the system which we could use to nd the response to any input We do It is called the impulse response But rst we have to stop and look at some mathematical preliminaries 13 Special Signals We are all familiar with signals like sin5t 8 3 etc These are ordinary signals These are signals that we can go into a lab and generate with a function generator But there is another group of signals called distributions or singular mctions The most common example is the delta or impulse These are o en ideal functions that we may or may not be able to generate 1Step Function The rst of these new functions is called the step function It is one that we have used already ut 0 tlt 0 1 20 131 1 Introduction 17 7102010 This is one that we can t truly generate because we can t turn anything on infinitely fast but we can come very close The step function is important in its role de ning causa functions Causal functions are those which are zero before time t 0 This is often an important distinction For instance suppose we were given the following differential equations 05ysin3t 132 a 05ysin3tut 132 b Equation 132a would be solved with Fourier transforms whereas Eq 132 b would best be solved with Laplace transforms There is an alternate to the step function called the Heavyside function which is sometimes used u0 lt0 2 05 0 133 1 gt0 This has certain mathematically desirable properties that are desirable when dealing with things like inverse Fourier transforms However the vast majority of the time U 1 may be considered interchangeable with H0 2 Sigmun function This is closely related to the step function sgn l lt 0 1 gt0 134 In fact the Heavyside function is often defined as l l uhEEsgn 135 3 Ramp function The ramp function is given by r 0 lt 0 l 20 136 a And may be thought of as the integral of the step function Z 41 Wk 136b foo 1 Introduction 18 7102010 The tau 239 in the integral illustrates the use of a dummy variable Like I it is a function of time but it is not the real t that appears in rt It is just a parameter being used to do the integration 4 Rectangular pulse p I pftl t2 tSt2 O elsewhere 1 37 5 Triangular pulse A I A t O t S T 2 ll2IZ39 r2 t 0 l2lZ OStSIZ 138 2 O 7 2 St 6 Sinc function sinct sin72t 51nct 7 139 sint Be aware that often the sinc function is de ned as 51mm t However our author uses Eq 139 and this is the one that will be used in this class Example Represent the following signals with mathematical expressions 27 Represent sign39ils given in Figure 217 in terms of step and ramp signals u h 0 FIGURE 217 Continuousrlimc signals 7 The impulse delta function The delta function is defined as having a value at one point only 5z1 20 2 O elsewhere 13933910 1 Introduction 19 7102010 There are several different mathematical techniques that are used to de ne the impulse For instance we could say it s the derivative of the step function d 50 uf dt Or we could de ne it from the pulse function in a limiting sense l 5 11111 pr 0 raO 7 One of the most important uses of the delta function is in integrals no 0 j 51 dz 0 51 dz 1 In mathematics it is often defined in the following way Jiwff5f fod 131 1311 ie because 50 0 is only defined at one point 0 the entire integral is only evaluated at one point This is often called the sifting property There is a similar mathematical definition that is used 0 1 Loft5ttod 5111 This is a classic case of making a definition consistent with other things since ower r0 dr from r0 dz I raw r0 dz 0 When a narrow irregular function appears in an integral it can often be approximated with a delta function For instance take a narrow triangle pulse function A010 What would the following integral give w l j A01t 5dtE01005 What if instead I approximate the pulse function as A010 2 005 5z By doing this I am assigning the area of the triangle function to the amplitude of the delta function Then the integral is 00551 5 dz 005 Time Scaling Property w 1 t0 1 0 I ft5at t0dt f 35at 0 t 1312 lal a a a 39 39 Proof First suppose agt0 and introduce a change of variables 1 Introduction 20 7102010 ot0 do oat t0 dz 7 Z Z grower z dz j f 6ltagtd 1 a ll Z 40 Z I For alt0 the integral above is the same except for the limits of integration aw a 0 w e do 1 t I ft at t0 dz I 7 f o39 fi 4 7 Z Z Z I which proves Eq 1312 Derivative Property If I de ne a new function the derivative of the impulse d 5 I 5I dz then Iff ltodf10 1313 Proof Integration by parts fuw lkz zdd ft5tt0lf1t5t t0d e f lkzo The function f5 0 is zero at plus or minus in nity Similar proofs lead to similar formulas lff5quotf fodf 1quotfquotf0 1314 Examples 5zdz 2 dz 0 5zdz 2 dz 1 5 z dz 5z This result is purely from the definition of 5 e 5z 3 dz e6 e z lt 3 dz e 2quot 3 1 Introduction 21 7102010 J e 2t54l 3 dt 2 14 3 e2t13 2 a 16 Example Problem 223 00 2 4 2cosm6z 1 5735123ui 1 5225 3dt tan2t6t 3dt 56t 2lt 7m 2 3 4 2 3 1 736 7 73 1 7a 72c03ur t7171 95 36 U57726l l9c 35 2 5 Exercises 1 Complex Numbers 111 Using these complex numbers A3j4 B3 j5 C05450 D20e j 6 Find 1 CD 2 CDx 3 AB 4 BA 5 141x 6ACj Show each step of the calculation Any addition or subtraction must appear in rectangular coordinates and any multiplication or division must appear in polar coordinates Note means the complex conjugate ie if x 4 j2 then xi 4 j2 112 Express the following asjust one sine or cosine by rst converting to phasors a x10 3sin2t45quot2sin2t 35 b x2 t 3cos2t45quot 2sin2t 35quot 1 Introduction 22 7102010 113 Solve for yt for the given inputsft WW d it dtz 2 dt 4yt ft a ft 3cos5t 10quot b ft5sin5t45 c ft2cos2t178 114 Prove the following using Euler s equations sin2 6 c0s26 115 A system is described by the following differential equation d3 d2 d Ey 2Wyt 9Eyt 2yt ft Solve for yt if ft 3sint 2cos 31 Your answer should consist ofone sine term and one cosine term 12 Linear Systems 121 Determine Vania fort 2 0 v0w0 5 V R 1M9 C 1nF v1quot sin103t The switch closes at t 0 Use the methods discussed in this chapter not Laplace transforms or another method First nd the forced response then the homogeneous response and then solve for any unknowns using the initial condition R t0 Vi C A 122 Consider the following situation The population ofa small country is given at any time by n t We will assume time is measured in months The birth rate is 10 per month The death rate is 4 The emigration rate people leaving is 7 The immigration rate people coming into the country varies and is described by a function it Write a differential equation to describe the population in the country 1 Introduction 23 7102010 131 Special Signals 131 Evaluate the following Write the answer in the most concise form possible a flu sin j cos 5amp72 dt 3 5 5 2 b sin c0s j 5t7 5 5 2 132 Write each of the following in the simplest form a Inte 3H5rdr b cos10 4tsin102t5t c e39z rt5t72dt d 5t3 5t73equotdt e t5t76 f 51017 2tzdt 133 Describe the following function mathematically 134 Sketch the following function Label the axes f r 1rt27rt73 1 Introduction 24 7102010 MAT LAB Exercises A Write a MATLAB program to do problem 1 above Write all the answers in polar coordinates Hints MATLAB parameters can be complex Furthermore you can enter them in rectangular or polar coordinates eg A 3 i4 C 05expipi4 In expressing the output the following may be useful The magnitude of a complex quantity W is absW the angle can be determined by imagGOgO VD B Redo problem 3 using MATLAB 1 Introduction 25 7102010 21 Introduction to Convolution Earlier we decided that we used differential equations to model physical systems Then we looked for solutions to these equations for various inputs ie forcing functions Let s think about some examples where maybe nding a differential equation isn t so easy Suppose I am hired as a systems engineer for a lumber mill The owner wants to know how to model the output of his mill in boardfeet of lumber for a given input logs Well this doesn t look like something that s going to lend itself so easily to mathematical modeling so I decide to try something different I ll just feed a log in and see what comes out So I feed one log in first thing in the morning and about an hour later lumber starts coming out of the mill It comes out at say 100 boardfeet per hour for three hours I decide that s a good place to start so I make a graph like the one in Fig 1 Input logs I Output boardfeet of lumber 0 l 2 3 4 Time hours Figure l Sawmill input and output So now if they feed one log per hour in over three hours I have a graphical way of estimating what will come out This is shown in Figure 2 Note that this is valid only because I am assuming the mill is linear and time invariant By linear I m assuming that it doesn t matter how many logs are coming in By timeinvariant I m assuming that the time of the day doesn t matter This probably won t hold true at noon time Let s see if we can move towards a little more mathematically precise way of modeling What if the input is ipped around as in Fig 3 I ve already got a function that represents the output of the mill in Fig 1 so I ll just call that Mt Apparently the way I get the output is invert x relative to t0 and move is slowly in the positive direction and look at the overlap with Mt to tell me the output What if the input is somewhat irregular as in Fig 4 I calculate the output in the same way I reverse xtFig 5 and see how much it overlaps with h as I move it to the right Fig 6 Mathematically this convolution is expressed by the convolution integral ytJhrxt rdr 2 Convolution 1 7132010 Input logs I I I Output boardfeet of lumber 0 l 2 3 4 Time hours 0 l 2 3 4 Time hours 0 l 2 3 4 5 6 Time hours 300 200 100 I I 0 l 2 3 4 5 6 Time hours Figure 2 Sawmill output for three logs 2 Convolution 2 7132010 Figure 3 Sawmill output for three logs using convolution Figure 4 Irregular log input response I Input ipped around 3 2 0 l 4 Figure 5 Response with the log input ipped around 2 Convolution 3 7 132010 response Input ipped around t I gt Figure 6 If I move to ipped input to the right to represent time the overlap of the input and the response will give me the output hlexp al 0 1 2 3 4 Figure 7 Two functions to be convolved hlexp al hl xl o 1 2 3 4 Figure 8 Convolution of the two functions in Fig 7 2 Convolution 4 7132010 Mathematical Derivation of the Impulse Response All this is well and good Let s go back to the problem of the RC circuit We had an equation relating the output voltage V I to the input current 5 dvt 1 1 c t dt RCVCO CS 1 We solved this for a step response input but suppose we want to solve it for several different inputs 5 Why not try the same trick we used with the saw mill ie just put something in and see what comes out How about a delta function 50 So we will try to solve dv t CLVCI5I 2 at RC The first thing to do is determine what kind of response we re likely to get To do this we look at the characteristic equation l s 0 RC I got his by ignoring the forcing function and putting in an s for the derivative The solution gives L RC S So I take my solution as having the form v t Ae HRCuU We already knew that first order equations have exponential solution Note the ut which means our function is causal But we expect that Nothing s going to come out until I put something in So if I plug this guess at the solution back in Eq 2 d l v t 2A e ZRcii t Ae ZRc5t dx c RC So Eq 2 becomes 1 rtRC rtRC 1 rtRC Ache utle 5tRCle 5t 3 Equation 3 has two different kinds of functions in it For the equality to hold the following two equations must hold At t 0 The 50 terms give A50 5t and the MU terms give 2 Convolution 5 7132010 A 1 errRCLAerzRC 0 RC RC The only solution to both these equations is A 1 This solution of V60 for the particular input of 5U is so important that it has its own name the impulse response ht gitRC In extrapolating from what we learned from the saw mill to get the output for any input Xt we invert X around t move it in the positive direction and look at the overlap This is expressed mathematically by the convolution integral vows Jiwhltrgtxltr rgtdr 5 Note that the T in Eq 5 is what s called a dummy variable It is also a time variable but it is not the real time in VMW In the most general case the integration limits are plus and minus in nity But if Mr is causal the lower level is zero and if Xt is causal the upper limit is t Instead of solving a differential equation separately for every input xt we solve it once for the input 50 to get the impulse response Mr and then determine the output from the convolution integral Eq 5 Example Solve dv 1 C v u dt RC Solution Since I already have the impulse response the solution for the forcing function xm I 0 no amt Law I 1 d7 If e rRCMI ut Td139 Is J39 errRC all 0 I can change the limits of integration because both the impulse response and the forcing functions are causal irRC irRC rtRC VDWM J08 dr RCe I RCl e Z 0 2 Convolution 6 7132010 I added the step function on the end because clearly VMWU has no value before t 0 Check 0 1 7 7 EVWWU 2 RC e my I 6 Wu I So the original equation is RC 1 itRC e utERCl e utut Summary System Analysis Here then is a summary of the approach we have developed to analyze a system 1 Use physical properties to develop a differential equation that relates output to input N In the differential equation use 50 as the forcing function to get a solution called the impulse response Mt 3 Now the response to any input xt can be found by convolving h and xt Example Suppose I want to develop an analysis method for the following situation I have a vat whose total water content is described by vt I can control the water I am putting into the system I use the parameter t However there is a loss that is proportional to the volume already in the vat given by 01vt What is the volume of water as a function of time for the following cases a xt l b xtl e SWU Solution The rate of change in volume is dvt dz from which I get my differential equation 01m x ft 01vt But I want the impulse response so I solve dv 0 1w 50 t This is the same form as Eq 2 above so my impulse response must be 2 Convolution 7 7132010 ht 6 0 1WU The response to input a is vt Ihrut Td139 foo so J 102601147 ut rdr The MT means that I can start my integration at zero instead of in nity and H0 139 means I need only take my integration up to t instead of in nity so it simpli es to Z l 7 r VJ1028 01rdrIOZ je 0 1 0 0 102 If lut 103 1 e u z I added the My on the end because vt is only de ned starting at t 0 It turns out that it doesn t matter which lnction I ip I can also write 00 v j 102e 1lt uZ 1W d7 foo Z 3102670104 d7 lozerouj39eou d7 0 0 2 701 1 01r 10 e 6 0 10350 1 60 1 103 1 5 1I ut The response to input b is Z vt 110260104 d7 0 201 04 108 e ra r 7047 Z 1 2102601 0 250e 1 50 1 250e 0 1 e 05 uz Z 0 I could have ipped this one around to get 7132010 2 Convolution Z Va 1028701r87057r d7 21026705160 d7 0 0 21026705 i 6041 04 0 2505 W lut 25050 e uz Note Any physical system with a loss rate proportional to the total entity will invariably give an impulse response with an exponentially decaying term Example In a semiconductor the number of excess electrons is given by nt 0m3 Excess 73 71 electrons are generated from a 11ght source at a rate G 0m S However 1f there are excess electrons they will recombine with the excess holes that were also created at a rate given by n t R 3 cm 331 T where T 1076 S is the lifetime of the electron What is nt for the following light sources a Gt 1 6w b Gt103sin106t 00 lt z lt 00 Solution We start by nding a differential equation to describe this system from which we can determine an impulse response The rate of change of nt is the rate at which electrons are generated minus the rate at which they recombine M dz G T d To find the impulse response we solve for nt using 50 I In GU M50 The mathematics to find the impulse response is now the same as above so we can just write 2 Convolution 9 7132010 ht em u and therefore the solution for the input a will be n e 1 396uz 1 8010 Since both ht and Gt are causal functions I can write I if W Grw 7 d7 108 elm d7 0 o 1086106r6i396106r d7 2 10367106 610 102 1ei106u What if I switched Gt and ht around nlG 71Tdr 104 dr 0 0 108 el r 1021 e40 u The response to b will be a different story because the input is not causal n j ht TGT d7 foo DO 6 2 Jam 0420 I108 sin 106 Td139 6 6 21086101610 r sin1067 d7 em 106 sin106t 106 cos106z 1012 1012 102 s1n106 00310 7 2 Note that this is not causal Of course it s much easier to do it using the phasors The phasor equation is ij 106N l l 1076 1076 N 110610 1 4450 From which we get the answer 76 2108610 2 Convolution 7132010 108104 quot 5 We will see that response to a can be much more easily calculated with Laplace transforms sin106t 45 7 Example Suppose we have to nd the impulse response of a second order system all t dt 043y 5z 6 The characteristic equation is s2 4s3 s3sl 0 so the homogenous solution will be of the form y A873 Be ut 7 a The first derivative is 9 3Ae393 Bequot u A B5t 7 b l and the second derivative is 2 75 2 9A 35 u 3A B 5z A B5lt z 7 0 Putting these back into Eq 6 gives 9A 35 u 3A B5t A B5lt z 4 3Ae393 Be u A B5t 3Ae393 Be u 5z Putting Eq 7 in Eq 6 we will wind up with three types of functions If Eq 6 is to hold true then the coefficients for the different types of functions must satisfy Eq6 so we get three equations The 51 0 terms give AB5l 0 or A B The 5 terms give 3lB 4 4B 50 l l 3ll l 3A B or 2 2 The u ems 9A Be 4 3Ae3 Be393le393 Be39 0 But this is redundant because our choice of the homogeneous equation insured it 2 Convolution 11 7132010 So we can conclude l ht 6 e 3 u 2 What would be the response for the input 3C0 MU yltrgthltrgturgt 071mm 2 6T e 3dr e l 6 3ut 2 This problem was not so difficult because the characteristic equation separated into two simple real roots In general it will be much easier to use Laplace transforms 22 Convolution Properties There are several important properties associated with convolution many of which are obvious by the graphical methods They can also be proven just from the mathematical de nition The formal definition of convolution is ythtxtIiwht rxrdr 1 but we have already said that if both ht and Xt are causal Eq 1 can be written 1 ht x J h TCTd139 2 The following are properties related to convolution 1 Commutativity htxtxtht r r 3 J h TCTd139 J x rhr d7 This one is easy to prove mathematically Start with a change of variables 039 t T 4 This implies T t 039 d7 2 da Remember that inside the integral t is a parameter not a variable or dummy variable That is the reason that d d0 0 Notice also how the limits of integration change when using Eq 4 2 Convolution 12 7132010 T 0 gt a T t gt a 0 so the rst integral in Eq 3 becomes J h TCTd139 J0 haxt a d0 haxt 0 do which is the same as the second integral in Eq 3 except the dummy variable is 039 instead of T This is easy to see graphically Suppose Ihave two functions in Fig 1 If I graphically do the convolution indicated by the rst integral of Eq 3 it s telling me to invert the ht function around the origin and slide it to the right to nd the overlap of the two areas as a function of time If I use the second integral of Eq 3 in rst invert the Xt function and move it Obviously both procedures result in the same graph h l l Xt Figure l Graphical convolution 2 Distributivity This property says that for three functions f1tfz Ifsf f t t t 5 The mathematical proof is trivial because this is just a statement of linearity However this property is very useful Look at the convolution in Fig 2 1 H H X0 2 Convolution 13 7132010 Figure 2 Graphical convolution By far the easiest way to do this convolution is to break ht into two separate function do the convolution with these separate pieces and then add them together at the end 3 Associativity This simply says that when convolving three or more functions the order doesn t ma may ltrgt22ltrgtr1 WW lt6gt This involves convolution with double integrals which is outside the scope of this class I am just going to show you what it looks like 12tf3tfo12t r rdrFt So now Eq 6 becomes f1tFt 0 flt I Fr dr 7 where T l is just a different dummy variable Now Eq 7 becomes tf3t J U T39 or39fZUV TMZU d7 dr 2 I t r Io fzrf3r 139 d7 dr 4 Duration If fl 1 only exists on the interval 111 And f2 I only exists on the interval 212 Then the convolution integral is only defined for M f1tfztl Jaw mm 1H2 This is particularly use ll in doing graphical convolution 2 Convolution 14 7132010 39 m1 y1lt5gtnhltzgtwltrgt Then yt 0ht 0xt And yt 01 02ht 0xt 02 Special Convolutions There are a few functions whose convolutions yield special results that are worth noting Delta function amp NV 60 Jfr6ltr r on ft 8 The integral in Eq 8 only has a value when t r so in effect the 5 just traces out the function Unit step function amp wf139LtI T d7 9 J f 139 d7 The upper limit is changed to t because the ipped function ut only has a value up to fltrgtultrgtj mpg m t But in the second integral HU 139 can just be replaced with one Graphical Convolution Sometimes convolution can most easily be performed graphically The following steps are a guide to graphical convolution Step 1 Use the duration property 4 above to find the interval on which the integral is nonzero Step 2 Flip one signal about the vertical axis usually the simpler one Step 3 Vary the parameter 139 looking for the overlap Exercises 2 Convolution 15 7132010 1 A system is describe by the following equation dyl d1 011 is the out put and fl is the input 20yltrgtfltrgt a What is the impulse response of this system b What is the response toft ul 7 uli 01 c What is the response toft 3cos1OIi 300 2 1n the following problem use the methods from this class Do not use Laplace transforms For the following RC circuitR 100 kg and C 1uF At time time t 0 vc0390V c Vuut a Write a differential equation describing the circuit with vm 141 the input and v out gt the voltage across the capacitor b What is the forced response c What is the homogeneous response d Using parts a and b and the initial condition what is the total response Now solve this using the convolution method e What is the impulse response f What is the response to the inputvm ul g What would be the response if vm e s u I 3 A system is described by the following differential equation 2 Convolution 16 7132010 dyt dt 03yt ft a What is the impulse response to this system b If ft 5t5t71 5t72what is yt c If ftut whatis yt d Find yt ifft 4cos3t foolt tlt oo 4 The impulse response of a system is given by ht 6 7 e39 ut Determine the response to the following input xtut7ut72 5 It is found that the total number of bacteria in a dish n I grow in proportion to a light source shined on the dish This rate is I 106 minquot However at any given time 5 of the total bacteria die per minute a Write a differential equation describing the total number of bacteria if the light source is turned on at time t0 At t 0 assume n 0 b If I were to model this as a system with light being the input and the bacteria population being the output what would be the impulse response 6 Graphically convolve the following two functions 7 Graphically convolve the following two functions I 4 ulzfAfsy 2 Convolution 17 7132010 Appendix The following is a MATLAB program to illustrate convolution m Griconvm A program to illustrate convolution m NN 200 NN2 NNZ delt 1 TT zeros lNN 39 Xl zeroslNN39 X2 zeroslNN XFZ zeroslNN The time buffer for n TTn neNN2delt end The impulse response X1NN2 1 for nNN2lNN Xln expe5deltneNN2 end The forcing function X2NN2 l for nNN2lNN230 X2n 1 end X2NN2 l X2NN220 l X2NN240 1 Flip the forcing function for nlNN XF2n X2NNenl end Plot the impuse response and forcing function subplot4ll plotTTXl plotTTX239r39 AA text4539Original two functions39 setAA39fontsize3912 setgca39XTick39 0 l 2 3 4 5 6 7 8 9 10 axis 5 10 0 11 setgca39fontsize3912 title39Myeconv39 Plot the impuse response and flipped forcing function subplot4l2 2 Convolution 18 7132010 plotTTX1 hold on plotTTXF239r7W AA text4539One function flipped39 5etAA39font5ize3912 5etgca39XTick39 0 1 2 3 4 5 6 7 8 9 10 axis 5 10 0 11 5etgca39font5ize3912 This part does the convolution zeros1NN n zeros1NN ift zeros1NN T 0 Y zeros1NN n5tep51 m Y Yc X5 HD O while n5tep5 gt 0 n5tep5 input39n5tep5 77gtW nin 1n5tep5 T l Shift the flipped function for nT 1NN X5hiftn XF2n7T end Multiply and integrate for nT1NN Yn X1nXF2n7T YconNN2T YconNN2T Yn end end 5ubplot413 plotTTX5hift39rW hold on plotTTX1 barTTY hold off AA text4539T 39num25trTdelt 5etAA39font5ize3912 5etgca39font5ize3912 5etgca39XTick39 0 1 2 3 4 5 6 7 8 9 10 axis 75 10 0 11 ymax maxYcon 5ubplot414 plotTTYcon AA text47ymax39The Convolution function39 5et AA39font5ize3912 5et gca39font5ize3912 set gca39XTick39 0 1 2 3 4 5 6 7 8 9 10 axis 75 10 0 11tymax 5avea5gcf39convbmp39 end Example 1 Convolution of a continuous function and a discrete functin 2 Convolution 19 7132010 vcunv Example 2 Convolution of two continuous functions 2 Convolution 20 7132010 2 Convolution Dneluncuon moved a 7132010 51 Discrete Signals If I start with a continuous signal t and sample at constant intervals T I get a discretetime signal ft gtfkTEfk Instead of differential equations we have difference equations yk2a1yk1a2yk fk With initial conditions yk0yk0 1 Like before we can talk about zeroinput and zerostate response yk ya k ys k ya k is due to IC ya k is due to the forcing function There are many similar functions page 5 1 k0 6k1 0 k 0 1 kZO uk 0 k0 k kZO rk1 0 k0 There is a discrete rectangular pulse page 38 1 m 2 S k S m 2 pm k 0 k elsewhere There are similar theorems Siftz39ng theorem 2 fk15k kofko 511 kroo Noticethat 1 kk 5 k k 0 0 k 0 And therefore fk5kko fko5kko Notice that summation takes the place of integration Convolution of Discrete Time Signals Convolution in the discretetime domain is ZI 1 7292010 fklf1klfzkl w 512 f1mf2kml It has many of the same properties CommutatiVity lkl kl leklquotf1kl DistributiVity fllklquotf2klfskl lkl kl kf3k AssociatiVity f1 k f2 k fslklf1kl fsz kl f3 k Duration If f1 k is duration M1 And f2 k is duration M 2 Then f1k f2 k is duration M1 M2 Note Duration is NOT the number of terms it is the distance between the rst and last terms TimeShifting lkkllquotfzklfkkll f1kk1f2kk2fkk1k2 Example 612 page 298 The convolution of the unit step with any causal signal produces a summation fkukl ifm1uk m rm The convolution of two causal signals or be simpli ed to W0 mklwki mmmk M momkm112k 11mk1m01 Sliding Tape Method page 298 This is a graphical convolution method Step 1 the signals are recorded on two tapes f1 l loll llll lzll l3llm f2 m021m2123t Step 2 The second is ipped 0l 1l 2l 3l 2f21l120ll ZI 2 7292010 Step 3 The second is shifted to the right summing the product of the overlap Example Convolve the following two functions with the sliding tape method Step one 1uu 12 0 1 2 Step Two 1 12 0 1 2 Step Three Dt39 erence Equations to describe digital systems Look at the following difference equation yk 7 05yk7 l xk Suppose the system is at rest ie no initial conditions Take the case where xlkl 5 k l Zl 3 7292010 ls QM yJ1k1 yJLL 0 1 0 1 1 0 1 05 2 0 05 025 3 0 025 125 I surmise that the output can be written k ylklhk05 k Now I want the output for some other input say xlk u k We know that the output is the convolution of the impulse response and the input y1k111k1w1k1 gawk m1 h1m1 Suppose I were given a system described by the differential equation 6W 02 t x t dt y To put this in a computer I have to discritize it I will have to approximate the derivative W t g M dt At Then the differential equation becomes a difference equation szqum yk1 ykAt02ykAtxk yk1 1At02ykAtxk At1 ykl08ykxk 52 Introduction to Z Transforms We will consider the onesided Z transforms FzZfkkfkz k 521 In a way the Ztransform is the simplest because it has the effect of adding 2391 for each delay fk 45k 225k 45k 6 ZI 4 7292010 Fz 42392 22394 2396 A valuable series formula to remember is lqq2q31 zfqltl 522 q Example Zuklz391z392z393z394 1 L 1 2391 2 1 Similarly Zakukla21azz392a32393a42394 171 Z l az z a What is the Z transform of 5 k Obviously this only has one term Z6k iamz k 1 Properties of Z Transforms 1 Linearity Zaf1k fzk Fl z M 2 The proof comes directly from the definition Zaf1kl fz kl afl kl fz klz k gar k gm k oF1z Fzzl 2 Right Time Shift Z fk k0u k k0 z39k Fz k0 positive integer Proof Zfk k0uk k0Zz39kfk k0uk k0 k0 Substitiue i k k0 or k z39k0 0 zquotz k fiui 27k zquotfi z k Fz Because the terms between k0 and 0 fall out ZI 5 7292010 Example The Z transform of the rst is just 2712 1 12 3 14 A 0 1 2 3 1 12 Figure 521 Clearly the second function is just the rst delayed by two 2 271 F F 7 22 2 12 2712 There is another version of this theorem 1 Zfkk ukgtzit jFzzik gfktlz Be careful m Zfkkulukl Amiuikw Change variables i k kn Zfkkuukx fiuiknz39 k fizquotz39 i fizquot i Mi k xzu pkg zFltzgtz filzquot ZI 7292010 Figure 522 Example Look at the previous example fk105kuk And calculate EZZfk2uk Solution 71 Zfk k0uk Z 2Fz 2 22 fiz l 172 Z 2 2 PM 2f2zzf1121 2 2 2 12 72 Z 42z 1z 12 2 12 2 12 2714Z Z 1 42 2 12 2 12 4 22 1 2 2 Example Solve for the impulse response yk 05yk 1 5k Start by taking the Z transform of each term Since the Z transform starts at n0 I am implicitly taking the Z transforms of yku k 05yk lu k 5ku k ZI 7 7292010 Yz705zquot1z7y711 Since we are looking for an impulse response we assume all initial conditions are zero which leaves Yz705271Y zl Yzl705zquotl 1 z Y 17052 1 2705 So hk 05k 14k 3 Le Time Shift Derivative zfk1uk17 zFz7zfOl z fimiuiki zzFz7zzf0l7zf1l Example For fk05kuk Find FzzZfk2uk 1 12 114 13 2 1 o 1 2 Figure 523 Solution z z z l Ha z 2705 2 22 z3 7 z2 52z 7 05 7 z7 05 3 7 3 7 25 z z z 025 Z i Z i Example Solve for the impulse response 8 7292010 21 ykl 05yk 5k1 Notice that this is the same as the previous example with each function shifted to the left by one time step zYz zy0 05Yzz z0 Obviously this will not lead to the same solution as the previous example When you shifted the delta function to the left you shifted it off the radar of the Z transform 7 Convolution Z hh h M hikimikh niminik mi i mmkmk k 160 quot10 nimii nik mw quot10 160 ZMRVEMh no If f is causal then 1 can write kmz k f2k mukmz k And use the rst version of the right shift theorem i kmukmz k 1zz Then the original equation becomes Z k k nimmzw 8 o m 2 E f F1 Example Convolve the following two signals ZI 9 7292010 0 1 2 Figure 524 117121z391z2 F2 2 05 2391 0522 E 21172 z 1 2 1 z zo5 2 1 052 05105z39105051z392 105z 3052quot 0515z 12z 2 152 3 052 Figure 525 Convolution of the two functions in Fig 524 Notice that with that with the one basic transform 2 Z aku k 1 z 7 a We can begin to generate many more For instance there is nothing that says 3 can t be a complex number So Ze kwuk z ziex mT Zequotkwuk 276 ZI 10 7292010 Z cos kwTu Z 056ng equotk 7u 1 Z L Z 1 zz equot 7zz e 7 Zz e quot7 I z equot 7 Z ZZ ZZCOSQT1 z2 zcosaT z2 ZZcosaTl Similarly Z sin kwTu Z 05 e m e39m T u j z z 2 2 6 2 6 zsinaT z2 ZZcosaTl What is the Z transform offk ak coskaTu k fk ak coskaTu k l 5akeka7 ake new lLi F 2zae quot7 zae quot7 z2 zacosaT z2 aZCosaTa2 Property 5 Frequency Scaling Property ZakfkFS m Zakfk akfkz k walla Example Find the Z transform of fMkaY H Solution ZI 7292010 By frequency scaling z e Fe T Z 2 05639WT Property 4 Time Multiplication d Z k F WH2amplt4 Proof Start with the de nition FZ Z flklz k n0 Take the derivative of both sizes with respect to z dF 2 4H k dz kf 12 Multiply both sides by 72 CF Z dz Z ngvazwvn Example Problem 55 in the book f k 2k k Find the Z transform of kfk3kfkfk 2uk 2fkluk Solution Fz The rst term is solved by time multiplication Zkfk i z prm m car 2W ZI 12 7292010 The second term comes from frequency scaling F E z 3 z 3 z 3 2 z 6 The third term form the right shift theorem 71 Z k 2 k 2 4 Z M u 1 2 H H And the last from the left shift theorem 2 Z k 1 k 0 f u 1 2H 2m 2 tmzi z 2 2 2 2 3 Inverse Z Transforms The inverse Z transform is defined as follow 1 k fk j rj Fzz ldz 523 This can be done using contour integration Needless to say what we prefer to do is get every term in a form that we can look up in the table Example What is the inverse Z transform of Fz 1 22392 32394 Solution fk 5k25k 235k 4 Or we might say f01 f22 f43 Example Solve yk 03yk luk y00 m First take the Z transforms of both sizes Yz 03z391Yz i ZI 13 7292010 z 22 Y Z To get back to the sampled time domain we have to transform each term back We use a version of partial fraction expansion similar to what we used for the Laplace transforms Start by dividing everything by z Y z z A B T 2 032 1 2 03W Multiplying through by 203 and evaluating at z 03 gives us A B z 3 03 0428 ltz 1gtZ lltz 1gt 07 Multiplying through by 21 and evaluating at z 1 gives us B A z l B l 0888 2 032 Z1z 03Z1 07 Now we multiply back through by the 2 that we saved to give us terms that we can nd in the tables 04282 08882 Y Z 2 03 2 1 And we can easily go to the sampled timedomain yk 0888 042803kuk Example Find the inverse Z transform of 2 025 Fz ZZ z 022 03 Solution We start by dividing out a zso we will have it at the end M 2z o25 A B z 2 022 03 2 02 2 03 And solve for A and B ZI 14 7292010 A 2z o25 202 025 2 05 1 2 03 HZ 02 03 1 B 2z o25 203 025 205 1 2 02 2 03 02 T Fz Z Z 2 02 3 2 0 fk 02Z 03ku k Example Find the inverse Z transform of 1 F 2 22 1 First I will rewrite it as 05 F 2 z 05 But remember I need to divide out a Zterm so I can get the inverse Fz 05 A B z zz 05 z 2 05 A 05 1 z 05 H B 2 1 205 Fz7z Z 2 05 f k5 k05kuk f0 105 0 f105 f2025 Notice that this is 05k u k with the rst term missing Could I have done this another way Suppose I say ZI 7292010 Iknow Z l 0505kuk then the right shift theorem tells me fk 0505Huk 1 05kuk 1 Notice that f 0 0 f1 05 f2 025 8 Initial Value Theorem no mm This comes about from the de nition of the Z transform and the fact that every term 2quotquot will go to zero 9 Final Value Theorem z 1 1 flk117133i7FltZgti Proof First note that Z we fk 1 Fltzgt 21Fltzgt Fltzgt Then take the following limit as z gt 1 fkfk1Z39k fkfk1 f0f1 f0f2 f1foo f Which is what we were trying to prove ZI 16 7292010 Note that the nal value theorem is only applicable to functions that actually have a limit at in nity For instance Z Fz Does not because the corresponding inverse is k flk 2 k Another way to say this is that it has a pole outside the unit circle In fact the formal test for the applicability of the nal value theorem is the z lF 2 can have no poles on or outside the unit circle Example I can use the nal value theorem because 2 lF 2 has no poles outside the unit circle Example I cannot use the nal value theorem on z 1 z 2 Because it has a pole outside the unit circle Fz Partial Fraction expansion of multiple poles Find the inverse Z of W 1 lz 25 The rst step is to save a z Fz z l A B I C z z 1z 252 z 139z 252 39 2 25 Multiply through by 21 and evaluate at z 1 21 2 A 2 2 356 2 25 75 Multiply through by z 252 21 21 z 252 BCz 25 i ZI 17 7292010 And then evaluate at 2 25 2 1 Z 1 To get C I rst take the derivative of Eq i with respect to 2 1 12 1 iz 252 C d2 2 1 d2 2 1 And then evaluate at 2 25 which will eliminate the term with the A d 12 1 121 125 5 4 5 167 B 75 4 3 Z 25 z 2 1 z12 2 2 22 356 2 1 Z 75 21 3562l 1672 I 3562 FZ2 12 252 2 1 39 z 252 z25 The second term I rewrite to match a formula in Table 52 167 2 2 167 0252 r6 0252 39 2 252 25 2 252 39U2 252 Therefore fk356 668k25k 35625kuk What if I had had FZ 21 12 253 F2 21 A L B L C L D 2 253 2 252 2 25 z 2 12 253 2 1 I will solve for B C and Dby multiplying through by 2 253 21 A Z1 Z12 253 13C2 251D2 252 To get C I take the rst derivative d 21 i13z 252 CD22 25 612 2 1 2 ZI 7292010 And to get DI have to take the second derivative 2 d Z 21 i32z 251CD2 dz 2 1 2 1 So mggl iiiil Notice that I would now have a term Bz z 253 The closest transform in Table 52 is l z39quot1 kk1k2kmaku k Zimm To get it in this form 2 2 23 2 253 Z 2 253 1 23 l k Z Ekklk2025 uk Z391zz 23 k 2k 1k025H uk 2 42k 2k 1k025k uk 2 ZI 19 7292010 34 Inverse of Transforms with Complex Roots So far we have just been dealing with the inverse of real roots Let s look at another possibility Use Laplace transforms to solve the following DE with initial conditions dzym 2 4yt ue y0 0 y390 0 341 dtz Taking the Laplace s2Ys 2sYs 4Ys s 2 or Ys 342 32 2s4s2 There is second order term but when we try to factor it 2i22 44 2i2j1 4 172 z zfz lij we have a complex conjugate pair which I will write 1724ng P l f 343 There is nothing that says that I can t use the same partial fraction expansion for complex numbers so write Eq 342 as A k kquot 344 1 Y S 322s4s2 32 s p s p Notice that I am assuming the coefficient of the k is the complex conjugate of the k We will see that it proves to be true Now we proceed as before The first part is easy 2 32234 2 4 44 4 5 Now solve for k 1 1 1 1 i H 2 32 S1J39 1 1 1 1 1 4450 1 N3 2 N3 22600 2J329 00 443 k s2s p s41J3 Laplace II 1 7212010 k 1 1 Z 1 1 323 p5p 323139J srlrj3 1 1 1 1 41500 l j 2 2 24 600 wig 9 00 443 or we could write Notice that I didn t have to explicitly solve for k So far we have 14 k kquot s2 s p s p m Going to the time domain gives 71150 12 6 t e e y 4 443 4J3 l 2 6 4527150 74527150quot 46 4J e e 345 1 722 1 it 0 Ze me cos t 150ut We probably want to look for a more systematic way If I take 7 e1 1 e115 In3 p a ja I can generalize this by writing yt 6722 2kle cos wt 4kut For this problem a 1 a J3 1 k 4k 1500 I I 4J3 I wonder if there might not be an even easier way to solve a problem like this Look at how we calculated the k in Eq 3411 1 1 1 1 kZES A jxi 1 221 s71 Laplace II 7212010 Notice that the term on the far right could be written 1 ZJ E 2 346 That s because L s p FF p p 2jImp 21w There will always be a term like this because whenever I solve for the coef cient of the s p root the s p will be in the denominator If I know that why not just solve for the simpler term 1 g 1 14600 347 s2 sp 1j39J3 2 In relation to our previous way of solving for the coef cients k g 1 g 348 21w 2ja So now Eq 344 becomes Ys14 g g s2 21w s p s p 4g dj ol 4g milo yr1e urE e e e e ut 4 a 2 j ie ztut em sinm4gut 349 a 1 722 e ut 4 l 2 J e s1n 3t6 6 u t 2J3 since sin6 i900 icos6 7 sin3t 150 7 900 cos3t 1500 Eq 349 is the same answer as Eq 345 Example Find the inverse Laplace of s 2 F s sz 33 5 Solution Sine method i First nd the root 3 32 45 p 32J 15j39166 2 Laplace II 3 7212010 ii Now determine the complex coefficient g gs2 05j166173473 7 571 51 66 iii The corresponding timedomain term is 173 ft e39 s1n166t73 7ut 104e3915 sin166t 730ut Solution Cosine method i The root is the same ii The k coefficient is s 2 k s 15 j166 571 51 66 05j166 1734730 j2166 3324900 iii The timedomain function is ft 2052e cos166t 170ut To check use the trigonometric identity cosa 900 sina ft 104e391539cos166t 17 7 900 900ut 0524 17 7 104e 15 sin166t63 7ut So far we have two approaches to nding the inverse of transforms of the form F S S p S p I The cosine method This consists of the following steps 1 Write the roots as paja2 pa jw Remember p always contains the positive imaginary part 2 Calculate F 39S S Pquot 3 The part due to the complex roots is yt ZIkIe I co ltwt 4kut k k 4k 11 The Sine method Laplace II 4 7212010 This consists of the following steps 1 Write the roots as paja2 pa ja Remember p always contains the positive imaginary part 2 Calculate g F 39S 3 The part due to the complex roots is lg yt e 1 sinatzgut a g4g Example Find the inverse Laplace of 5s 13 ss2 4s13 Xs Solution I Cosine method S25S13 Lk F ss 4s13 S s p s p Ss13 s24s13 Step 1 4 42 413 42x4 13 p 2 2 2j3 Step 2 5s13 5 2j313 ss psp 2j3 2j3 2 13 3j15 1534790 270841350 2j3j6 3641240 71900 Step 3 x5 I 2707e392 cos3t 135 7ut And xt 1141e cos3t 1350ut Solution 11 Sine method Laplace II 5 7212010 Step 1 same Step 2 g 5s13 315 w4254 450 S H 2 1393 364124 Step 3 xx Sin3r 455W x0 1141e sin3t 450 JHU Example problem 445 Find the zerostate response of yquot 3y 2y sin ut Solution Zerostate means no initial conditions Taking the Laplace Yss2 3s2 1 WW szl 1 12 1 s1s2 1 p 2 15 5 572 3 a0a1 1 51 i1z 2 2 g s1s2 l l 4 720 4450 434270 J5 ySt 602 sin I 720u I So yt Eaquot e 0316sint 720ut Laplace II 6 7212010 Question What would be different if the input were ft sint 2ut 2 Then the Laplace transform of the original DE is 8725 sz 1 I Do not try to incorporate this into the mathematics It doesn t work Instead we say that the answer is Yss2 3s2 yt Ed gall 0316sint 2 720 t 2 Question How would I solve it if ft sintut 2 sint 2 2ut 2 Be careful the 2 in the sin2 is in radians 0 2 radians L 1140 7r radians So actually I could write ft sint 2ll4 7ut 2 Then I would solve the problem using f t sin t 1 140ut sin t cos 1 14 7 ut costsin l 14 7 ut 04sintut 09lcostut as my input and then shift the answer by two seconds when I finish l s 3 44 F 04 091 091 S szl szl szl Example Solve yquot2y 5y ut y0 ly 01 Solution The Laplace transform is l slys sy0 y3902Ys y05YsE 1 Y 2 2 5 3 ss 3 3 3 sz3sl A g g Y S ss22s5 Ss ps p Laplace II 7 7212010 A1 5 72J22745 p 7l2 2 7 s23sl s 1747139473j61 ilj2 syn2 7 5 7 5394158 71j2 2244117 1 yl 7equot sin214l ul 5 02l2equot sin21 41 ul 241441 35 Block Diagrams Suppose I have a system described by the differential equation d2yt dyU 0 5 21 3 1 00 3900 351 5112 d1 m d f m m lt gt I can take the Laplace transform to get s2Ys5sY s2Ys sFs3Fs 352 From this I can determine the transfer functions Y Hs S 3 353 Fs s2 5s2i I can take the inverse Laplace of the transfer function to get the impulse response htL 1Hs 354 Notice thatI can make a diagram of this Xs Ys Figure 371 This is referred to as a Block diagram Block diagrams are extremely useful When putting together systems from subsystems The following theorems are helpful 1 Series connections Laplace II 8 7212010 Figure 352 Series connection of two systems For the system in Fig 352 the total transfer functions is HTS H1SH2S 2 Parallel connections Figure 353 Parallel connections of two systems For the parallel connections of Fig 353 the total transfer function is given by HTs H1sH2 s 3 Negative Feedback Figure 354 A negative feedback system The total transfer function of the simple negative feedback circuit can easily be calculated The system consists of the openloop transfer function Has and the feedback transfer function H j s In determining the total transfer function of Fig 354 it is expedient to define an error function right after the summations We solve by noticing Es Fs7H sY s Ys H0 sEs H0SFSHSYS We then group together the Ys terms YSH0SH SW3 H0SFS Laplace H 9 7212010 This is so universal it is worth memorizing Example a What is the transfer function of the block diagram shown in Fig 354 if 1 2 H s H s oltgt S m S 3 b What is its impulse response 0 What is the step response Solution 1 J 3 3 S a HTS S 2 1 2 1 ss32 s 3s2 s3 s A B b H s T sz3s2 sl 32 S 1 3 22 321 l sls2 l ht equot 2639 ut c l YstepsHTS s 1 l A L B sz3s2s sz3s2 slls2 in 1 3 s2571 312 ymp Er 63922 equotu t 453 Find the Closed loop system transfer function for the feedback system represented in Figure 415 ms gt f 0 my l4 FIGURE 415 Laplace II 10 7212010 Ho3H1SHzS HfsG1sG2 s Hle 1H1HSG1GZ HTS Em lt7 Law FIGURE 417 Answer rm Fm 1 GMSH1ltSgt Hll H2 39I133G18 63m Write the little loop on the right as H V H3 1 H 361 Now we have M 0 1 HSGI H f G2 G3 So HIHZHS 1 H 361 1 7H1H2H3 G2 G3 1 H 361 1H3G1 H1H2H3 G2 63 T Example Find the impulse and step response of the following system 55 HS 1405 Ys Laplace II 11 7212010 S olution s3 s3 s2s31 s25s7 75JS2747 5 6 p 77J7 2 2 2 5 J5 5 377 3973 2 J 2 537 7 2 J 2 Sine method gs3 1 3 7j 1460 2 2 J5 J5 2 2 ht 6 2 sin it60 1154e 2 sin it60 2 2 2 Check Block Diagram Analysis We have introduced some theorems that can be used is the analysis of block diagrams But an alternative method is this define an auxiliary variable after each adder Write an algebraic expression for each adder Solve to get output as a function only of input E1F7H3Y E2 HlEliHEY YH2E2 H2H1E1 H3Y H2H1F7H3Y7H2H3Y Laplace H 12 7212010 Y 1 H2ng HIHZHEY HIHZF Y HIH HT 7 2 F 1 H1H2 H1H2H3 Alternative Using the Theorems First of all I m going to redraw the diagram E Finally HlHZ H 7 1H2H3 7 HIH2 T 7 1H3 HlH2 lH2H3H1H2H3 1H2H3 Example Find the transfer function of the following block diagram Laplace H 13 7212010 S olution For this one it would be hard to apply the theorems E1 HIHZF 7 Y EZ HIF H3E1 HIF H3H1HZF 71 Y H4EZ H4H1F H3H1HZF 717 H1H4 H1H2H3 F 7 H3H4Y Y 1 H3H4H1H4 H1H2H3 F Y 1 H3H4H1H4 H1H2H3 F Y H1H4 H1H2H3 HT F 1 H3H4 Problems 34 Finding the Inverse ofComplex Roots 341 Solve for yt 22dyz dz dz 5Yl9fl y0 1 y10 0 fle 3tul 342 Find the step response of the system described by the following equation dzygl 3 d 12 dz dz 3yzfz 343 Solve the following equation for yl M dlz 4d75fl3yzfz Laplace H 14 7212010 y0 0 y 0 1 ft sin2tut 344 An RCL series circuit has the transfer function HS SR L sL1sCR s1 sRL1LC R1kQ L1mH C1nF Find the impulse response and the step response 35 Black Diagrams 351 Find the total transfer function of the following block diagram FS Laplace 11 15 7212010 Laplace II 16 7212010 31 Introduction to the Laplace Transform The Laplace transform of a timedomain signal ft is given by FsLftj0 fte39 d 311 For instance the Laplace transform of an exponential function is Fs I e39mute39 dt I elsmyemdt 1 7sat 01 312 sae 0 sa 1 s a Notice that e39mut ut if a 0 so we can conclude Lutl 313 s Here s an interesting one L5tJ5te dt 1 314 We know the LT of ft but what about it s derivative This can be solved by integration by parts 1 de 7 Lf t 10 7e dt 72 OO O is me 0 10m se d 315 f0 SI ftse39 dt sF s fltogt With just these few transforms let s see what we can do Suppose we want the impulse response of the first order differential equation given by 03yt5t 316 The Laplace transform of this equation is just the transform of the individual components so SY s y003Y s 1 Since we are looking for the impulse response we may assume y0 0 so this leaves 1 Y s s03 We don t know how to take inverse Laplace transforms yet but we know from Eq 312 that the time domain version of this function which is our impulse response is ht e 03tut 317 Laplacel 1 7152010 a 75041 J1 xt re alt J1 x039e do e39 Xs LaplaceI 2 7152010 The integral was a Laplace transform using the parameter 039 instead oft so Eq 3114 becomes Ys hre39S Xsdr I Z 3115 Xsj0 hre chquot X sH 3 Can this be true Instead of doing the convolution I can take the Laplace transforms of the two functions and just multiply them Yes it s true And with these simple examples using just a few functions we ve demonstrated the two main reasons for using Laplace transforms 1 Differential equations become algebraic equations 2 Convolution becomes multiplication Let s look at our original differential equation but this time with a different forcing function 03yte ut 3116 Taking the Laplace transform we get 1 sY s 03Y s 3117 s01 Solving for YS 1 1 Y s 3118 s03s01 Ultimately we always want to go back to the time domain There is an inverse Laplace transform given by 1 M m 2 These are solved by techniques referred to as contour integration But as you ve probably noticed our approach is just to get it in a form we recognize and take it back to the time domain term by term So once again we write Ys 1 1 A B 3120 s03 s01 s03 s01 This time we ll be a little more systematic For A and B to be the right values they have to be the right values everywhere We will exploit this To look for A multiply through by s03 Fse ds 3119 1 B Y s s03 A s03 s01 s01 In order to isolate A I evaluate it at s 03 L A B s 03 s01503 s01 5703 The term containing B falls out and we are left with L L 5 s01503 301 2 LaplaceI 3 7152010 A similar procedure gives us B B 1 1 i 5 s 03 570 1 1 03 2 Equation 3120 then becomes 5 5 Y s 3121 s03 s01 And the corresponding time domain function is yt 5e 1 e 3 ut 3122 Let s go back to the original problem m03 te ut 3116 dt y But suppose we also had an initial condition of y0 02 Now when Itake the Laplace transform of Eq 3116 I get 1 sY s 0 03Y s 3123 y s01 Solving for Ys gives Ys 3124 s03 s03s01 We already know what the inverse of the second term on the right is The inverse of the term containing the initial condition just adds an exponential terms yt 0 e20 ut5e e39 3 ut 3125 This demonstrates the full power of the Laplace transform method of solving differential equations The initial condition just looks like another input Hints on taking the inverse Laplace Before we go on let s stop and look at how the properties can be used in taking the inverse of Laplace function Example Find the inverse Laplace of Fs S s 3 This looks easy enough however it s not specifically in my table The reason is the partial fraction expansion cannot be used directly on an equation like Eq 3315 The order of the numerator must be one less than the denominator So instead I just divide s 3 s 3 s 3 Fs The inverse Laplace of this is LaplaceI 4 7152010 f t 5t 3e 3 ut Or how about if I looked at it this way It almost looks like I could use the derivative rule remember L110 sFm f039 3126 What is f039 You might be temped to say f0 e 3 utlt0 1 But be careful f 039 means the value of f right before time t0 Therfore f0 e 3 utt0 0 This will be true for any causal function So if f t is a causal function and its Laplace transform is F s L iFU sFs dt And we have a theorem d L391 SF 3 t d t f So in fact we can use the derivative rule Ell S st 1 J 3 3 3 3 e393 ut 3e393 ut 50 Before going on lets find the Laplace transform off I te ut This is a new one so we just use the definition 0 7a 7 Lte utdt J39O re e dt This is not an integral that most of us carry in our head but it is readily available in most basic tables of integrals My table of indefinite integrals tells me Ixeaxdx e ye l a a Therefore the above integral becomes no no 7sat 1 J te me WI J39 te dt e I j 0 0 s a s a 3127 LaplaceI 5 7152010 Notice that 639 tends to make the limit at t 00 converge to zero This is one of the strengths of the Laplace transform and one of the key reasons it was developed More generally n L n in 7 3128 t e 0 Ha Example Find 1 S s 32 71 1 i 732 L dt re 110 e39mua 3te393 ut te393 5t e39mua 3te393 ut Check I can check my answer by just looking the time domain function up in the table of Laplace transforms 3 Le 3 ut 363940 2 i s3 3 3 332 332 Review What do we have so far Table of Laplace transforms ft Fs 5t l l 0 s e39 ut 1 30 1 a n te ut Sam 6 7152010 Laplace I Properties of Laplace transforms sFs f039 ftgt FS GS 32 Theorems and Properties of the Laplace Transform We have already developed some of the characteristics of the Laplace transform Here are some more Linearity This comes from the de nition La f1 0 m 0 Ha f1 I 3 f2 New I Iowa 1te dt te dt 321 OIF1 s F2 3 Integration I would like the Laplace of the integral of a function The easiest way to do this is remember that the integration of a causal function may be thought of as convolution with the unit step function LU oat m ut And use the convolution theorem Lfltrgt ultrgt Fltsgt Derivative We have already developed the Laplace of the first derivative LEt Fs f0 322 We will almost certainly want to know how to do 2quotd and higher order derivatives 2 LZISZFS Sfl0 f0 323 The proof is left as an exercise The Laplace of higher order derivatives follows in a similar manner Time Scaling LaplaceI 7 7152010 Lfat Iowfate39 dt 039 at 7 d0 1 no 5 l flak 77L 76 dc 324 8 The last step comes about because the integral looks like the Laplace transform except s has been replaced by sa This is only valid for agt0 because Laplace is a causal transform Time Shifting Lfttout t0d IJflgtt0eistd Changing Variables to r t to Lftt0ut I0d tJ 0wfTeisrtndr w 325 as fre39 dr e39 Fs Example Solve the following d t ii02yte 5 1ut l y00 Start by taking the Laplace transform of both sides 7 e SY s 02Y s S 05 Warning The e appears in the numerator because of the delay However do not try to incorporate that into the partial fraction expansion solution Instead just set it aside and make the shift at the end So I will solve l A B YS s02305 s02 s05 333 333 s02 s05 Then when I go back to the time domain I make the shift yt 333e e 5quot1ut 1 Example A Find the Laplace transforms of ft e39 ut to Be careful To use Eq 325 the tterms all have to be in the same form ft e39 e H ut to LaplaceI 8 7152010 Lft e ULe quot ut 10 S n L S a ef ng Example B Find the Laplace transform of ft t1equotut 1 In order to use the time shifting theorem Eq 325 we have to get every time term in the same form as the step function er 87184271 t1 t 12 So now we have ft t 1 2e 1equotquot ut 1 e 1t 1e 1ut 1 2e 1equotquot ut 1 If you look closely the second term is just the exponential with the time shifting L 4quot 1 875 e I s1 The first is te39mu t with the time shifting L 1 1 4quot 1 1 eis e s12 Combining them and adding the constants L 1 1 equotut 1 e39le39s Ze39le39s 312 H1 Frequency Shifting Lfte Fs t Proof 12 is 4571 J39O fte e dt J39O fte dt F s A Because it looks like I took the Laplace transform using a parameters 1 instead of just s Modulation Notice that the above holds even when A is an imaginary quantity Lfte m m Using Eulers Identity LaplaceI 9 7152010 2ijFSjwoFSij Time Multiplication d quotF s Lrquotfltrgtlt 1gtquot W Start with the de nition Fs Ifte39 dt 0 And take the derivative with respect to s Fs gym Or 41 iFs Tamem ds 0 Which proves it for nl Obviously the higher terms are proven by subsequent derivatives Initial Value Theorem lirnft 1imsFs 326 Example 1 1 139 3 t 1 139 1 33 0 ahaS 1 ulttgt1IES1 limrtlim Si 0 tgt0 Sgtoo 32 139 m39 t 1 1 L 26 s1nau 5gssazwz a ws 11m s 11m 0 H s 2asa a H l2asa a s LaplaceI 10 7152010 A Brief Lecture on Stability We know that the Laplace function 1 2 will become ei2tut a function that s will decay away nicely in time IfI have I H S 1 s2s3 Even before I take the inverse Laplace I know that the time domain function will decay away as some combination of ei2tut and eigtu I would say that HS is stable because I know that any reasonable input ie one that is also stable will decay away If in contrast I had I H S 2 S2373 I would say that it is unstable because it has a time domain term that will explode ie emu I I doesn t even matter that it has one stable term39 if one term is unstable the system is unstable Since the parameter s is complex I can write it as S 7 jw Therefore I can uses a complex graph to show the location of these poles at s2 and s 3 Since we have seen that a pole at s 2 leads to a stable result we say the All the poles must be in the left half plane for stability Complex 8 plane jw sigma 2 3 Final Value Theorem lgg SFS 327 This is only valid if all poles are in the left half plane 2 hm SH j 11m Li s o 0 3a 54 0 3a 54 702 coswtut 0 and in fact Laplace I 11 7152010 Example Use the nal value theorem to nd lim f 0 tgtoo 33 aquot Fmim1 Can t use it 3 3 h 1W1 33 3 my 93SFS 93s2s4 0 s c Fs SZ9 Can t use it 33 Solving Differential Equations with Laplace Transforms Taking what we have learned so far let s see how the Laplace transforms can be used to solve differential equations Example 331 Use Laplace transforms to solve the following DE with initial conditions wan gun dz dtz 3yf W W L y390 1 3 3 1 Taking the Laplace szYs sy0 y 0 4sYs y0 3Ys 1 S Yss2 4s3 sy0y3904y0l S l s 5 s We want to solve for Ys so 1 1 s57 s5 SZSS1 ya3zngnu snynu m LaplaceI 12 7152010 Notice that we factored the denominator into individual terms Once again we use partial fraction expansion to break this down into terms we can look up in the table 2 s 531 A B C YS 13 S S S S S S 332 We want to isolate the terms A B C one at a time To isolate A multiply through by s s2 53 l B s s sls3 sl 33 and then we evaluate at s 0 sz53l L l sls3s0 l3 3 Similarly to nd B we multiply both sides of Eq 332 by sl and evaluate at s l sz53l 1 51 3 3 ss3 Pl 12 12 2 Similarly for C sz53l 9 1 51 5 ssl H 3 2 6 So now we have YS1332 56 3 31 33 and going back to the time domain gives 1 3 7 53 2 e e at y 3 2 6 Example 332 Back in example 34 1 what would be different if we had Ly 43yt amt y0 0 W 0 W 333 The Laplace transform of both sides yields Yss2 43 3 i s which we rewrite as l l Ys sls3s3 31s3Z 334 LaplaceI 13 7152010 Now we have a repeated root This calls for a slightly different approach First we expand 1 A I B I C 3l33Z S1 332 33 The rst one we handle in the same way as previously We crossmultiply by 31 and then evaluate at 3 1 to isolate the A m l l A 2 Z 33 5 1 13 3 32 To get the B we cross multiply by 1 A 2 C 2 33 B 33 31 31 33 and then evaluate at s3 B 1 3 1 573 2 33 Still no problem Now we look for C We start by crossmultiplying by l A B 33 233C 3l33 31 33 However now if we evaluate at s3 we are not going to isolate C there will still be a 2 3 3 term with containing B However if we once again crossmultiply by as above 1 A 33Z BC33 3 l 3 l and then take the derivative with respect to 3 i l i A332 C d3 31 d3 31 we get rid of the B Then if we evaluate at s 3 we get rid of the term containing A to give d l l 1 C7 1 2 7 3 3 573 31 573 335 Now our expansion is YS14 122 14 31 33 33 LaplaceI 14 7152010 So the inverse Laplace is l l l y Zef Ete 3 Ze 3 1110 Example 333 Find the inverse Laplace transform of FSiz 3231 Solution EXpandFsas F 34 2 A B 2 C 3231 32 31 31 4 2 S 2 22 31 572 1 BS4 223 321 1 Cd34 l32 l34 1 3 2 2 d3 32 51 52 771 1 Remember the formula dux dvx i W dx Vxux dx dX vx vx2 Check by crossmultiplying 34 2 L 3 L 2 3l332 32 39 312 39 31 34232231332 232332 32 02 2 31 143 6 3 426 4 Example344 Solve Z 44 0e um y01y 00 3316 Taking the Laplace 15 7152010 Laplace I l 32Y3 3y0 43Y3 y0 4Y3 33 which gives Y33243434S3 34 Y 3 S 32 There is a couple different ways to approach it from here A 34331 32 7313 322 33 322 33 This is relatively straightforward but at some point I m going to wind up having to take a derivative of a fairly complex function of s with polynomials in the numerator and denominator m B 34i Ys 323 H42 21 32 32 32 33 This gives me two different terms but they re simpler YSY1SY2S 34 1 Y 3 Y 3 1 322 2 32233 Firstterm Look at this very carefully before you go wailing away Y0 34 2 32 2 1 322 322 322 l 2 32 322 y1t e392 2te392 u I Now look at the second term YS 1 C L D L E 2 32233 s3 39 322 39Hz 1 1 C 2 1 32 573 12 LaplaceI 16 7152010 l 33H 1 d l l l 1 d333 2 S32S2 1 Check 1 Y2 mm32y 3243433 32536 1 33322 33322 SO y2 t 63932 te39 e39 ut And putting the two solutions together yt y1t y2 t 2te392 e39 l ut 63932 te39 e39 l ut 63932 3t ut Partial Fraction Expansion for Multiple Roots Let s try to generalize the results for partial fraction expansion when we have multiple roots So if we write for instance N 3 A B G3 2 G 3 2 30 30 30 336 I The functionG S is that part that doesn t contain an S 00 A G33 002 root I solve for these by B iGss 002 d3 3tz However what do I do when I have a triple root G3 MS G s A 3 B 2 337 30 30 30 30 3 The rst step is to multiply through by S 0 G33 03 2 G 33 03 14 3 0B 3 02C 338 To get A I evaluate at a 3 LaplaceI 17 7152010 A Gss 003 To get B take the first derivative of Eq 338 with respect to s diGss 23 G39s3s 002 3 2s aC S a 5 57a and evaluate at a 3 to get rid of the terms I don t want d 3 B EGss a Now how do I get C I take another derivative with respect to s 57a 3 Gssa3 G s32sa2C S 7a S and evaluate at a 3 to isolate the C parameter But notice the difference C 2j SGssa3 57a C has acquired a factor 12 Guess what D would be if we increase by one power N A B C D Gltsgtil61sgt 4 3 2 30 30 30 30 Ha 339 You guessed it D Ld 3Gssa4 32 03933 57 A general formula is CS CSNG1S kN N kNilNi1 J k2 2 k1 30 30 30 30 3 3310 1 d N k j Gltsgtsagt N n ds 3311 Check the above notation for N4 Azk 1 dj44Gssa4 4 m E Cssa4 5712 5712 l d 473 4 d 4 B k3 mg Gss a aws 0 1 d H 4 l d 2 4 C k2 mg Gss a 5a CsXs a LaplaceI 18 7152010 1 d 471 4 1 d 3 4 D21 me WW Z 32103 mm Example 1 A B C D E F H3 5 2 3 4 5 331 S 31 31 31 31 31 Solve for B For this problem N 5 and B is the k1 term in Eq 335 So using Eq 336 1 d 5 1 5 1 d 4 1 k1 15 13 1W3 11 1 i4 1 2 3 4 i 1 432 d3 s5 1 SS P1 Example S 32 ABC F 33312 33 312 31 SolveforC Solution S 32 ABC F 33312 33 312 31 1 32 132 133 C al33351 332 1 1 l 332 4 Example A system is described by the following equation m d71fgt4yogtfltrgt dtz 5 Find the impulse response LaplaceI 19 7152010 S olution In finding the transfer function or the impulse response initial conditions play no role Sol start by taking the Laplace SZY s 5sY s 41 s Fs The impulse response means the solution for f I 6 I 71 A B s25s4 s1 s4 13 7 13 7s1 s4 So 1 it 74 hI7 e 7e uI Problems 3 InIraducIian 131 For the following RC circuitR 100 kg and C 1uF At time time t 0 vc0390V c Vuut We had developed a differential equation for this circuit dv I 71 vltrgti1 v0 dI RC RC chit 10 I 10v I a Set vm I 6 I and solve for the impulse response using Laplace transforms b What is the step response The step response is the response to vm I u I Laplacel 20 7152010 c Suppose vm t 0 but there is an initial charge on the capacitor expressed as v 0 5V Whatis v0t 0 312 Find the Laplace transform of emu t Use this and the Euler equations to derive the Laplace transforms of sinat and cos wt dzf I dtz do any integrals Use the fact that you know the answer for the single derivative 313 Find the Laplace transform of given that L f F Hint Do not 314 Write the Laplace transform of the following equation d2 y t d t dtf 5 6 4yt ft If ft ut y039 1 andy39039 2 solve for for Ys in the Laplace domain 32 Properties of Laplace Transforms 321 Use the time delay property and find the Laplace transforms of the following signals a ft t 42ut 4 b ft t2ut 4 322 Take the Laplace transforms of the following functions a ewdr b 1672072611quot 2 c f1 t I sin2t 1 dr 0 d f2 t te ut 1 323 Find the Laplace transforms of the following functions LaplaceI 21 7152010 a f1 t 3tcos3tu t b f2 t tze s ut 1 33 Solving Di erential Equations with Laplace Transforms 331 Take the inverse Laplace transforms of the following functions sZ l s 4s4 332 Solve for B C and D s2ABCDE XSss1quot s n1quot s13 s12 s1 333 Solve for D 2 A B C D ES s1s23 31s23 39 s22 39s2 334 Solve the following differential equation using Laplace transforms dZEI3dyd tt2yidZEI4fi ft ui y0 1 y 0 2 335 Find the inverse Laplace transform of the following functions s2 7s 12 F S 2 s 3s 2 336 Find the inverse Laplace transforms of the following functions LaplaceI 22 7152010 sz3s2 sz231 a F10 b F2 3 e 63925 LaplaceI 23 7152010

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