ode 2011 unit3 unit step and unit impulse response

8/11/2019 ODE 2011 Unit3 Unit Step and Unit Impulse Response

1/33

Part II Problems

Problem 1: [Step and delta responses](a) Find the unit impulse response w for the LTI operator 2 D2 + 4D + 4I .(b) Find the unit step response v for the same operator.

. .(c) Verify that v = w (as it should be, since u = ).(d) For each of the following functions, nd the LTI differential operator p(D ) having it asunit impulse response.

(i) 2u( t).(ii) u( t) t .

(iii) u(t)t2.


2/33

Part II Problems and Solutions

Problem 1: [Step and delta responses](a) Find the unit impulse response w for the LTI operator 2D2 + 4D + 4I .(b) Find the unit step response v for the same operator.

. .(c) Verify that v = w (as it should be, since u = ).(d) For each of the following functions, nd the LTI differential operator p(D) having it asunit impulse response.

(i) 2u( t).(ii) u( t) t.

(iii) u(t)t2.

Solution: (a) The roots of the characteristic polynomial are 1 i, so the general solutionto the homogeneous equation is e t(a cos t + b sin t). The unit impulse response for thissecond order operator has w(0) = 0 and w

. (0+) = 12 . The rst forces a = 0 and the secondgives b = 2

1 : w( t) = 12 u(t)e t sin t.

(b) For t > 0, the unit step response is a solution to p(D)x = 1. In our case, x p = 41 is such

a solution, and the general solution is then x = 41 + e t(a cos t + b sin t). We require rest

initial conditions: 0 = x(0) = 14 + a or a = 14 . x

. = e t(( a + b) cos t + ( a b) sin t), so0 = x. (0) = a + b and b = 4

1 as well: v = 41 u( t)( 1 e t(cos t + sin t)) .

(c) v .

= 41 e

t(( 1 + 1) cos t + ( 1 1) sin t) = 12 e

t sin t.

(d) (i) This function has a jump in value, so the operator must be of rst order. (aD + bI )( 2u) = 2a (t) + 2bu(t), so b = 0 and a = 2

1 : p(D) = 12 D.(ii) This function has no jump but its derivative does, so the operator must be of second.. .order. For t > 0, w(t) = t is the solution to a2 = 0 with x(0) = 0 andx + a1x + a0x

x. (0) 1 1

t= 0 a2dt

d t. Plug in: a1 + a0t = 0 implies a1 = a0 = 0, and 1 = implies that= = a2 . ..= 1. So p(D) = D2. Or you can argue that w( t) = u( t) andu( t)t, w(t) w( t) = (t),a2 = so a2 (t) = ( t) and a2 = 1.(iii) This function w( t) has no jump in value or derivative, but its second derivative does jump: w

..(t) = 2u( t). So w(3)( t) = 2 (t). This means that we are looking for a thirdorder operator, a3D3 + a2D2 + a1D + a0I . t2 is a solution to the homogeneous equation,so a2 2 + a1 2t + a0t2 = 0, which implies that a0 = a1 = a2 = 0. w

..(0) = 2 implies thata3 = 2

1 and p(D) = 12 D3. Or you can argue that w(t) = u(t) t2, w. (t) = u( t)2t, w..( t) = 2u(t),

w(3) (t) = 2 ( t), so a3w(3) (t) = (t) implies that a3 = 21 .


3/33

Part I Problems

In the next two problems, nd the unit impulse and the unit step response for

Problem 1: D + kI

Problem 2: D 2 + 02 I


4/33

Part I Problems and Solutions

In the next two problems, nd the unit impulse and the unit step response for

Problem 1: D + kI

Solution: ( t ) = u ( t )e kt

v ( t ) = u ( t ) 1 1 e ktk

Problem 2: D 2 + 02 I

Solution: ( t ) = u ( t ) 10 sin ( 0t )

v ( t ) = u ( t ) 1 (1 cos( 0t )) 20


5/33

18.03SC Practice Problems 25

Step and delta responses

1. Find the unit step and unit impulse responses for the operator 2 D + I , and graph

them.

2. Find the unit impulse response for the operator D 2 + 2D , and graph it.

3. From your answer to 2., nd the solution to x + 2x = 3 ( t 1) with rest initialconditions.


6/33

18.03SC Practice Problems 25

Step and delta responses

Solution suggestions

1. Find the unit step and unit impulse responses for the operator 2D +

I, and graph them.

The unit step response h = h(t) is the continuous solution that is zero for t < 0,and is a solution of

2x + x = 1 for t > 0. (1)

This equation has particular solution x p = 1. The homogeneous system 2 x + x = 0has general solution ce t/2 , so the general solution of (1) is x = 1 + ce t/2 .

Because there is no impulse at t = 0 the pre and post-initial conditions are thesame, i.e. x(0 ) = x(0+ ) = 0. We need to choose the constant c to t the post-initial condition: x(0+ ) = 1 + c = 0. Thus, c = 1 and the unit step response his

h( t) = 1 e t/2 u(t).

The unit impulse response w = w( t) is the solution that is zero for x < 0, a solutionof 2x + x = 0 for x > 0, and satises x(0+ ) = 1/ a1 = 1/2, where a1 is the coefcientof x. Or, alternatively, but equivalently, the unit impulse response is the derivativeof the unit step response, so, using the product rule

1

1w( t) = h( t) = e t/2 u(t) + 1 e t/2 ( t) = e t /2 u(t).

2 2

The term (1 e t/2 ) ( t) = 0 because at t = 0 the coefcient of (t) is 0. The graphsof both are given below.

Figure 1: The unit step response h(t) and unit impulse response w(t) for 2D + I .

2. Find the unit impulse response for the operator D 2 + 2D, and graph it.

The unit impulse response for this operator is the function w( t) that is zero for t < 0and satises the equation

x + 2x = 0


7/33


8/33

Unit Step and Unit Impulse Response: Introduction

In real life, we often do not know the parameters of a system (e.g. thespring constant, the mass, and the damping constant, in a spring mass-dashpot system). We may not even know the order of the system. Forexample, there may be many interconnected springs or diodes. Instead, weoften learn about a system by watching how it responds to various inputsignals.

In this session we will study the response of a linear time invariant (LTI)system from rest initial conditions to two standard and very simple signals:the unit impulse ( t) and the unit step function u( t). Reasonably enoughwe will call these responses the unit impulse response and the unit step response.

The theory of the convolution integral studied in the next session willgive us a method of dertemining the response of a system to any input oncewe know its unit impulse response.

Because both ( t) and u(t) are discontinuous at t = 0 we will haveto be careful with our denition of initial conditions. The most sensiblemathematical and physical way to do this is to dene our initial conditionsat 0 . As input an impulse causes a jump when it is applied. This meansthat the conditions at 0 + will be different than those at 0 . To distinguishthese two cases we will use the terms pre-initial conditions (at 0 ) and post-

initial conditions

(at 0

+

). We will be able to state precisely the effect of a unitimpulse on these conditions.


9/33


10/33

..

Initial Conditions OCW 18.03SC

It is easy to see thatx(0+ ) = 0, so the post initial condition is the same

as the pre initial condition. This should not surprise us. Although therate of input jumps from 0 to 1, it is still only inputting an innitesimalamount at a time. So, the responsex( t) should be continuous. But, note. .that x(0 ) = 0 = x(0+ ) = 1.Example 2. Consider the initial value problem

.x = (t), x(0 ) = 0.

We know how to integrate (t) to getx(t) = u( t).

t

1x = u (t )

Here the pre initial conditionx(0 ) = 0 does not match the post initialcondition x(0+ ) = 1. The impulse causes a jump in the value ofx.Example 3. Consider a second order IVP

.. .x = u(t), x(0 ) = 0, x(0 ) = 0.

0 fort < 0Integrating twice we get x(t) = t2/2 for t > 0. t

1 x (t )

. .Again, its easy to check thatx(0 ) = x(0+ ) and x(0 ) = x(0+ ). That..is, the pre and post initial conditions are the same. (But,x(0 ) = 0 = x(0+ ) = 1.)Example 4. Consider the initial value problem

.. .x = (t), x(0 ) = 0, x(0 ) = 0.

.Integrating once gives

0 fort < 0x(t) = t for t > 0.

x( t) = u( t). Integrating a second time gives

t

x (t )

Checking the pre and post initial conditions gives

x(0 ) = 0 = x(0+ ) . .x(0 ) = 0 = x(0+ ) = 1

2


11/33

Initial Conditions OCW 18.03SC

In other words, x(t) itself is continuous, but for the second order equation

the input ( t) caused a jump in the rst derivative.If we continued these examples wed nd that for annth order equation

an input of ( t) causes a jump in the derivative of ordern 1.

3. Rest Initial Conditions

The case wherex(t) = 0 fort < 0 is calledrest initial conditions . If wehave a DE of ordern this translates into pre initial conditions

x(0 ) = 0, x. (0 ) = 0, . . . ,x(n 1)(0 ) = 0.

4. Conclusion

A unit step inputu( t) causes a smooth response with matching pre andpost initial conditions. For a unit impulse input (t) the pre and post initialconditions match except for the derivative one less than the order of theequation.

3


12/33

First order Unit Step Response

1. Unit Step Response

Consider the initial value problem.x + kx = ru (t), x(0 ) = 0, k , r constants.

This would model, for example, the amount of uranium in a nuclear reactorwhere we add uranium at the constant rate of r kg/year starting at timet = 0 and where k is the decay rate of the uranium.

As in the previous note, adding an innitesimal amount ( r dt) at a timeleads to a continuous response. We have x( t) = 0 for t < 0; and for t > 0

we must solve .x + kx = r, x(0) = 0.The general solution is x( t) = ( r/ k ) + ce kt. To nd c, we use x(0) = 0:

r r0 = x(0) =

k + c c =

k .

Thus, in both cases and u-format

x( t) = 0k r (1 e kt)

for t < 0 = r (1 e kt)u( t). (1)for t > 0 k

With r = 1, this is the unit step response , sometimes written v( t). To

be more precise, we could write v(t) = u( t)( 1/ k )( 1

e

kt).

The claim that we get a continuous response is true, but may feel a bitunjustied. Lets redo the above example very carefully without makingthis assumption. Naturally, we will get the same answer.

The equation is

. 0 for t < 0x + kx = r for t > 0, x(0 ) = 0. (2)

Solving the two pieces we get

c1e kt for t < 0x(t) =

k r

+ c2e kt

for t > 0.This gives x(0 ) = c1 and x(0+ ) = r/ k + c2. If these two are different thereis a jump at t = 0 of magnitude

x(0+ ) x(0 ) = r/ k + c2 c1.


13/33

First order Unit Step Response OCW 18.03SC

The initial condition x(0 ) = 0 implies c1 = 0, so our solution looks

like

0 for t < 0x(t) = k r + c2e kt

for t > 0.To nd c2 we substitute this into our differential equation (2). (We must usethe generalized derivative if there is a jump at t = 0.) After substitution theleft side of (2) becomes

. 0 for t < 0x + kx = ( r/ k + c2) (t) + kc2e kt + r + kc2e kt

for t > 0

0 for t < 0= ( r/ k + c2) (t) + r for t > 0.

Comparing this with the right side of (2) we see that r/ k + c2 = 0, or c2 = r/ k . This gives exactly the same solution (1) we had before.

Figure 1 shows the graph of the unit step response ( r = 1). Notice thatit starts at 0 and goes asymptotically up to 1/ k .

t

1/k

v(t)

.Figure 1. Unit step is the response of the system x + kx = f (t) when f ( t) = u(t).

The Meaning of the Phrase Unit Step Response In this note looked at the system with equation

.x + kx = f (t)

and we considered f ( t) to be the input. As we have noted previously, itsometimes makes more sense to consider something else to be the input.For example, in Newtons law of cooling

.T + kT = kT e

it makes physical sense to call T e, the temperature of the environment, theinput. In this case the unit step response of the system means the responseto the input T e( t) = u(t), i.e. the solution to

.T + kT = ku(t).

2


14/33

Unit Step Response: Post-initial Conditions

Quiz: Consider the equation.v + kv = u ( t )

with rest initial conditions, v (0 ) = 0..

For the solution v ( t ) what is v (0+ )?

Choices:

.a)

v(0

+

) = 0. b) v (0+ ) = 1/ k

.c) v (0+ ) = 1

.d) v (0+ ) = k

e) None of these.

Answer: (c) v ( t ) is continuous so v (0 ) = v (0+ ) = v (0) = 0 Therefore the DE shows .v (0

+) = u (0

+) = 1.


15/33


Quiz: Consider the equation

. v + kv = u ( t )

with rest initial conditions, v (0 ) = 0. .


Choices:

.a)

v(0

+

) =

0

. b) v (0+ ) = 1/ k

.c) v (0+ ) = 1

.d) v (0+ ) = k

e) None of these.

Pick what you think is the correct choice and then look at the answer.


16/33



. v + kv = u ( t )

with rest initial conditions, v (0 ) = 0. .


Think about your answer and then look at the choices.


17/33

First order Unit Impulse Response

1. Unit Impulse Response

Consider the initial value problem.x + kx = (t), x(0 ) = 0, k , r constants.

This would model, for example, the amount of uranium in a nuclear reactorwhere at time t = 0 we add 1 kilogram of uranium all at once and k is thedecay rate of the uranium.

Because of the rest initial conditions we have x( t) = 0 for t < 0. Theeffect of the input is to cause the amount x( t) to jump from 0 to 1 at t = 0.

That is, x(0+ ) = 1. For t > 0 the input (t) = 0 and, therefore, for t > 0 weshould solve .

x + kx = 0, x(0) = 1.

The general solution is x( t) = ce kt. To nd c, we use x(0) = 1, which givesc = 1. Thus, in both cases and u format

x( t) = 0e kt for t < 0 = e ktu( t). (1)for t > 0

This is called the unit impulse response , which we denote w(t). Insome sense it is the simplest nontrivial solution; you just give the system a

unit kick at t =

0, stand back, and watch the result. For t

> 0 it is just thehomogeneous solution with initial condition x(0) = 1.

2. Graph of the Unit Impulse Response w(t)

Figure 1 shows the graph of the unit impulse response. Notice that att = 0 it jumps to x = 1 and then decays exponentially to 0.

t

1w ( t )

.Figure 1. The unit impulse response of the system x + kx.

3. (t) as a limit of box functions

Originally we found (t) as a limit of box functions of area 1. In this sec-tion we will compute the unit impulse response as the limit of the responses


18/33

First order Unit Impulse Response OCW 18.03SC

to these box functions. The main two points in doing this are: rst, to gain

more comfort and facility with this circle of ideas and second, to convinceyou that the delta function is much nicer to work with than box functions.We invite you to compare the amount of work required for solving the unitimpulse with the amount of work needed in the unit step case.

A quick review: Dene the box function as uh(t) =

0 for t < 01/ h for 0 < t < h

0 for h < t.It has total area 1 for all h > 0 and the graph of uh(t) becomes a spike ash 0, i.e.

lim uh(t) = ( t).h 0

For the equation .x + kx = uh( t), x(0

) = 0

the three pieces of the solution are easily found to be

x( t) =

c1e kt for t < 0

1hk + c2e

kt for 0 < t < h

c3e kt for h < t.

Using the initial condition x(0 ) = 0 and matching the value of x at theendpoints of each piece we nd c1 = 0, c2 = 1/ hk , c3 = ( ekh

1)/ hk .

This gives the solution

x(t) =

0 for t < 0

1 kh 1)e kt kh (e for h < t.

kh e 1hk

1hk (1 e

kt) for 0 < t < h

Letting h 0 this becomes (since limh 0

= 1)

0 for t < 0x( t) =

e kt for 0 < t.

This limit is exactly the unit impulse response w( t) we found in a previousnote.

Figure 2 shows this graphically by plotting the input and output forseveral values of h.

2


19/33

First order Unit Impulse Response OCW 18.03SC

t

11

1/1t

2

1

1/2t

3

1

1/3t

1

Figure 2. Responses for h = 1, h = .5, h = .333, and h 0.

The input is plotted in black and the output in red. Notice how the outputrises faster and gets closer to 1 as h 0. Finally, in the limit of small h, it

jumps directly to 1.The Meaning of the Phrase Unit Impulse Response Exactly as in the case of the unit step response, the unit impulse responsemeans the response of the system when the input is a unit impulse. In thisnote we looked at the the system

.x + kx = f (t)

and we considered f (t) to be the input. Suppose, instead, we have thesystem .

T + kT = kT e ,

where we consider T e to be the input. Then the unit impulse response is theresponse of the system to input T e( t) = (t), i.e. the solution to

.T + kT = k (t).

3


20/33

Unit Impulse Response: Post-initial Conditions

Quiz: Consider the equation.w + kw = ( t )

with rest initial conditions, w (0 ) = 0..

For the solution w ( t ) what is w (0+ )?

Choices:

.a)

w

(0+

) =

0.

b) w (0+ ) = 1/ k

.c) w (0+ ) = 1

.d) w (0+ ) = k

e) None of these.

Answer: (d). .Using the DE we get w (0+ ) + kw (0+ ) = (0+ ). We know w (0+ ) = 1 and. (0

+) = 0. Therefore w (0

+) = k .

We could also look at the solution w ( t ) = e kt for t > 0. Thus w. ( t ) =

ke kt for t > 0. This implies w. (0+ ) = k .Using the solution to the DE probably seems easier than the rst method,

but it is important to be able to draw conclusions without knowing the so-lution.


21/33



. w + kw = ( t )

with rest initial conditions, w (0 ) = 0. .


Choices:

.a)

w(0

+

) =

0

. b) w (0+ ) = 1/ k

.c) w (0+ ) = 1

.d) w (0+ ) = k

e) None of these.



22/33



. w + kw = ( t )

with rest initial conditions, w (0 ) = 0. .




23/33

Second order Unit Step Response

1. Unit Step Response

We will use the example of an undamped harmonic oscillator with in-put f ( t) modeled by ..

mx + kx = f (t).

The unit step response is the solution to this equation with input u(t) andrest initial conditions x(t) = 0 for t < 0. That is, it is the solution to theinitial value problem (IVP)

.. .mx + kx = u(t), x(0) = 0, x(0) = 0.

This could be an undamped spring mass system with mass m and springconstant k . The mass is at rest at equilibrium until time t = 0 when a steadyforce starts to act on it.

Force represents a change in momentum over time. A nite force F(t) can only cause an ininitesimal change in momentum (i.e. F( t) dt) at a time.Therefore, the mass does not change position abruptly, nor does it changevelocity instantaneously. Because of this we should expect a solution whichis continuous with continuous derivative. Only the acceleration experi-ences a discontinuity.

For t < 0 we already know that x( t) = 0. For t > 0 the DE is

..mx + kx = 1.

This has a constant particular solution x( t) = 1/ k , and a general homoge-neous solution

xh( t) = c1 cos( n t) + c2 sin ( n t), where n = k / m.Putting the two together gives the general solution

x(t) = 1/ k + c1 cos( n t) + c2 sin ( n t) for t > 0.. . .

The continuity of x and x implies x(0) = x(0) = 0 and x(0) = x(0) = 0.This allows us to nd c1 and c2.

0 = x. (0) = 1/ k + c1

c1 = 1/ k 0 = x(0) = c2 n c2 = 0.


24/33

Second order Unit Step Response OCW 18.03SC

The unit step response for this system is (in both cases and u format)

0 for t < 0 1x(t) = 1

k (1 cos( n t)) for t > 0.=

k (1 cos( n t)) u( t).

As in the rst order case, we will sometimes denote this v(t).The claim that we get a continuous response is true, but may feel a bit

unjustied. Lets redo the above example very carefully without makingthis assumption. It will take more work, but we will get the same answer.

In cases format the equation for the IVP is

.. 0 for t < 0 .mx + kx = 1 for t > 0,

x(0) = 0, x(0) = 0. (1)

Solving the two pieces we get

c1 cos( n t) + c2 sin ( n t) for t < 0x(t) = 1/ k + c3 cos( n t) + c4 sin ( n t) for t > 0.

.The pre initial conditions x(0) = x(0) = 0 easily imply c1 = c2 = 0. Soour solution looks like

0 for t < 0x(t) =

1/ k + c3 cos( n t) + c4 sin ( n t) for t > 0.

To nd c3 and c4 we substitute x( t) into equation (1)..

To measure the jumps we compute x(0+ ) = 1/ k + c3 and x(0+ ) = c4 n .We use this as we compute derivatives of x.

. 0 for t < 0x(t) = ( 1/ k + c3) ( t) +

c3 n sin ( n t) + c4 n cos( n t) for t > 0... 0 for t < 0x(t) = ( 1/ k + c3)( t) + c4 n ( t) +

c3 2 cos( n t) c4 2 sin ( n t) for t > 0.n n Since the right hand side of equation (1) does not have any delta functions

..or any ( t) the coefcients in front of these terms in the formula for x must be 0:

1/ k + c3 = 0 c3 = 1/ k

c4 n = 0 c4 = 0.

2


25/33

Second order Unit Step Response OCW 18.03SC

In the end, we have exactly the same solution as above for the unit step

response.To summarize: the continuity assumptions follow because any jumps.

in x(t) or x(t) would result in delta functions when x is substituted intoequation (1).

The generalized derivative (t) is not something weve seen before. Itis often called a doublet. There is an entire theory of these and other gener-alized functions, but we will only use (t) in this course.

Figure 1 shows the graph of the unit step response (with k = 1 andm = 0.5.

t

2 /k

..Figure 1. The unit step response for the system mx + kx = u( t).

If we added some damping the homogeneous part of the solution wouldgo to 0 and the unit step response would go asymptotically to 1/ k .

The Meaning of the Phrase Unit Step Response

As we noted in the rst order case, the unit step response is the response ofthe system to a unit step input. For example, if our system is.. . .

mx + bx + kx = by

and we consider y to be the input, then the unit step response is the solutionto .. . . .. .

mx + bx + kx = bu(t) equivalently mx + bx + kx = b (t).

3


26/33

Second order Unit Impulse Response

1. Effect of a Unit Impulse on a Second order System

We consider a second order system.. .

mx + bx + kx = f (t). (1)

Our rst task is to derive the following. If the input f (t) is an impulsec ( t a), then the systems response to f (t) has the following properties..1. The momentum mx( t) jumps by c units at t = a. That is,

mx. (a+ ) mx. (a ) = c.

2. The position x(t) is unchanged at t = a. That is,

x(a+ ) = x(a ).

Recall the argument that we used before: If x( t) had a jump at a then. ..x(t) would contain a multiple of (t a). So, mx(t) would contain a multi-ple of the doublet ( t a). This is impossible since the input ( t a) doesnot contain a doublet. This shows point (2) above.

.To show point (1), we note that if mx(t) has a jump of c units at t a ..

then mx( t) contains the term c (t a). This is needed to make the left handside of equation (1) match the right hand side when f (t) = c (t a).

Another way to show points (1) and (2) is a physical argument. A forceacting on the mass over time changes its momentum. In fact, the best wayto state Newtons second law is that

dp = f ( t),dt

where p(t) is the momentum of a system and f ( t) is an external force actingon the system. If a force f (t) acts over the time interval [t1, t2] the totalchange of momentum due to the force is

t2 f (t) dt.

t1

Physicists call this the impulse of the force f ( t) over the interval [t1, t2]. Ifa very large force is applied over a very short time interval and has totalimpulse of 1 the result will be a sudden unit jump in the momentum of thesystem.


27/33


28/33

Second order Unit Impulse Response OCW 18.03SC

w(t) = 0 for t < 0. At t = 0 the input causes a unit jump in momentum,.i.e., 2w(0

+) = 1. So, for t > 0 we have to solve

2w .. + 8w . + 26w = 0, w. (0+ ) = 1/2, w(0+ ) = 0.

The roots of the characteristic polynomial are 2 3i. Which implies

w(t) = c1e 2t cos(3t) + c2e

2t sin (3t), for t > 0.

The initial conditions give

0 = w(0+ ) = c1,.1/2 = w(0+ ) = 2c1 + 3c2 c2 = 1/6.

Thus, the unit impulse response (in both cases and u format) is

0 for t < 0 1w(t) = 16 e

2t sin (3t) for t > 0=

6e 2t sin (3t)u(t). (3)

Figure 1 the graph of the unit impulse response. Notice that at t = 0 the.graph has a corner. This corresponds to the slope w jumping from 0 to 1/2.For t > 0 the graph decays to 0 while oscillating.

t

Figure 1. The unit impulse response for the system 2 .. .x + 8x + 26x.

3. Checking Example 1 by Substitution

With any differential equation you can verify a solution by plugging itinto the equation. We will do that with example 1 to gain some more insightinto why we get the solution.

First, we take the derivatives of the solution in equation (3) for t > 0

w. ( t) = 16 e

2t( 2 sin(3t) + 3 cos(3t)) for t > 0w..( t) = 16 e

2t( 5 sin(3t) 12 cos(3t)) for t > 0

Next we look at the jumps at t = 0

w(0 ) = 0, w(0+ ) = 0. .w(0 ) = 0, w(0+ ) = 1/2

3


29/33

Second order Unit Impulse Response OCW 18.03SC

Now we can compute the full generalized derivatives

w. (t) = 16 e 2t( 2 sin (3t) + 3 cos(3t)) u(t)

w.. (t) = 12 (t) +

16 e

2t( 5 sin (3t) 12 cos(3t)) u(t)

Finally we substitute w for x in equation (2)

2w..(t) = ( t) 3

5 e 2t sin (3t) 4e 2t cos(3t)

8w. (t) = 3

8 e 2t sin (3t) + 4e 2t cos(3t)

1326w( t) = 3 e 2t sin (3t)

.. .2w + 8w + 26w = ( t).

The Meaning of the Phrase Unit Impulse Response As weve noted several times already, the response to a given input de-pends on what we in our equation we consider to be the input. For exam-ple, if our system is .. . .

mx + bx + kx = by

and we consider y to be the input, then the unit impulse response is thesolution to

.. . . .. . .mx + bx + kx = b (t) equivalently mx + bx + kx = b ( t)..

(Here, is what weve called a doublet.)

4


30/33


..Quiz: Let w ( t ) be the solution to mx + kx = ( t ) with rest initial condi-.tions. What is w (0+ )?

Choices:

.a) w (0+ ) = 0

. b) w (0+ ) = m

.c) w (0+ ) = k

.d) w (0+ ) = k / m

.e) w (0+ ) = 1/ m

f) None of these.

Answer: (e). The unit impulse input causes a unit jump in momentum. Starting from . .rest this means mw (0+ ) = 1 or w (0+ ) = 1/ m .


31/33



Choices:

.a) w (0+ ) = 0

. b) w (0+ ) = m

.c) w (0+ ) = k

.d) w (0+ ) = k / m

.e) w (0+ ) = 1/ m

f) None of these.



32/33





33/33

Higher Order Unit Impulse Response

We can extend our reasoning in the rst and second order cases to anyorder. Consider an nth order system with DE

an x(n) + an 1x(n 1) + . . . + a1x + a0x = f ( t) , (1)

where we take f ( t) to be the input. The equation for the unit impulse re-sponse of this system is

an x(n) + an 1x(n 1) + . . . + a1x + a0x = ( t), with rest IC. (2)

The effect of the function input is to cause a jump in the n 1st derivativeat time t = 0, while the lower order derivatives do not jump. That is, thesystem is put in the state

x(0+ ) = 0, x(0+ ) = 0, . . . , x(n 2)(0+ ) = 0, x(n 1)(0+ ) = 1/ an .

To show this we use the same reasoning as in the second order case. Sup-pose there was a jump in a lower derivative. For example, suppose

x(n 3)(0+ ) = b = 0.

Then the expression for x(n 2)( t) contains b (t), which implies that x(n 1)(t) contains b( t) and x(n)( t) contains b(t). This is impossible because theright hand side of (2) does not have any derivatives of the delta function.

Since xn 1(t) has a jump of x(n 1) (0+ ) = 1/ an at t = 0, its derivativean x(n)(t) has a unit impulse, ( t), at t = 0.

We conclude that the solution to (2) is 0 for t < 0 and for t > 0 it isexactly the same as the solution to

an x(n) + an 1x(n 1) + . . . + a1x + a0x = 0

with initial conditions

x(0) = 0, x(0) = 0, . . . , xn 2(0) = 0, xn 1(0) = 1/ an .

ode 2011 unit3 unit step and unit impulse response

Documents