lecture5

Dirac Delta Functions I Consider a family of functions Δ( t , T ) , as shown below T T t 2 2 1 T − Δ ( t , T ) I Define the Dirac delta function δ ( t ) as the limit δ ( t ) = lim T → 0 Δ( t , T ) Also called a Dirac impulse or a unit impulse function

Impulse Response Functions I Dirac delta functions are usually depicted as arrows: t 0 δ ( t ) I Definition : Let L : u 7→ y be a LTI system and let u ( t ) = δ ( t ) . Then y is called the impulse response of L t t u y 0 0 L

Some Physical Examples I Mass balance: d dt ( ρ Ay ( t )) = u ( t ) - w ( t ) = u ( t ) - ρ g R y ( t ) So d dt y + a 0 y = b 0 u where a 0 = g RA and b 0 = 1 ρ A

Some Physical Examples I Consider the situation in which u ( t ) = δ ( t ) I Recall that u ( t ) is the mass flow rate of water into the tank I R t -∞ u ( τ ) d τ is the total mass in the tank at time t I In this case Z t -∞ u ( τ ) d τ = 0 : t < 0 1 : t > 0

Some Physical Examples I So the impulse response of the tank is: I y ( t ) = 0 for all t < 0 I y ( 0 + ) = b 0 I For t > 0, d dt y ( t ) + a 0 y ( t ) = b 0 u ( t ) = 0 which is the homogeneous differential equation . I For t > 0, impulse response is the homogeneous response, i.e., y ( t ) = ce λ t , λ = - a 0 where c is chosen such that y ( 0 + ) has the correct value: y ( 0 + ) = ce 0 = c = b 0 0 0 b 0 y ( t ) t b 0 e - a 0 t

Some Physical Examples I Example : Mass-spring-dashpot system y c u k M Suppose u ( t ) = δ ( t ) , find resultant response y ( t ) I u ( t ) = 0 for all t < 0 ⇒ y ( t ) = 0 for all t < 0 I At t = 0, impulse occurs: I Use impulse-momentum principle from physics: m dy dt ( 0 + ) = m dy dt ( 0 - ) + impulse intensity = m · 0 + 1 = 1 So dy dt ( 0 + ) = 1 m and y ( 0 + ) = 0

Some Physical Examples I So we choose c 1 and c 2 such that y ( 0 + ) d dt y ( 0 + ) = 0 1 m = 1 1 λ 1 λ 2 c 1 c 2 implying that c 1 c 2 = 1 1 λ 1 λ 2 - 1 0 1 m = 1 λ 2 - λ 1 λ 2 - 1 - λ 1 1 0 1 m So c 1 = 1 m - 1 λ 2 - λ 1 c 2 = 1 m 1 λ 2 - λ 1

Some Physical Examples I So we choose c 1 and c 2 such that y ( 0 + ) d dt y ( 0 + ) = 0 1 m = 1 1 λ 1 λ 2 c 1 c 2 implying that c 1 c 2 = 1 1 λ 1 λ 2 - 1 0 1 m = 1 λ 2 - λ 1 λ 2 - 1 - λ 1 1 0 1 m So c 1 = 1 m - 1 λ 2 - λ 1 c 2 = 1 m 1 λ 2 - λ 1 I Conclusion : The impulse response for t > 0 is y ( t ) = 1 m 1 λ 1 - λ 2 e λ 1 t - e λ 2 t

Some Physical Examples I Recall that λ 1 = - ξ + ψ λ 2 = - ξ - ψ where ξ , c 2 m and ψ , q ( c 2 m ) 2 - k m

Some Physical Examples I Recall that λ 1 = - ξ + ψ λ 2 = - ξ - ψ where ξ , c 2 m and ψ , q ( c 2 m ) 2 - k m I As such, we have that y ( t ) = 1 m 1 λ 1 - λ 2 e λ 1 t - e λ 2 t = 1 2 m ψ e ( - ξ + ψ ) t - e ( - ξ - ψ ) t = 1 2 m ψ e - ξ t e ψ t - e - ψ t

Some Physical Examples I Recall that λ 1 = - ξ + ψ λ 2 = - ξ - ψ where ξ , c 2 m and ψ , q ( c 2 m ) 2 - k m I As such, we have that y ( t ) = 1 m 1 λ 1 - λ 2 e λ 1 t - e λ 2 t = 1 2 m ψ e ( - ξ + ψ ) t - e ( - ξ - ψ ) t = 1 2 m ψ e - ξ t e ψ t - e - ψ t I The dynamic behavior of y ( t ) looks qualitatively different, depending on whether ψ is real, or imaginary

Some Physical Examples I From previous slide, ξ , c 2 m ψ , q ( c 2 m ) 2 - k m If ψ is real, then it must be the case that ξ > ψ

Some Physical Examples I From previous slide, ξ , c 2 m ψ , q ( c 2 m ) 2 - k m If ψ is real, then it must be the case that ξ > ψ I Consequently, both exponentials in the homogeneous response decay to 0 as t → ∞ y ( t ) = 1 2 m ψ e ( - ξ + ψ ) t - e ( - ξ - ψ ) t

Some Physical Examples I From previous slide, ξ , c 2 m ψ , q ( c 2 m ) 2 - k m If ψ is real, then it must be the case that ξ > ψ I Consequently, both exponentials in the homogeneous response decay to 0 as t → ∞ y ( t ) = 1 2 m ψ e ( - ξ + ψ ) t - e ( - ξ - ψ ) t 0 t

Some Physical Examples I From previous slide, ξ , c 2 m ψ , q ( c 2 m ) 2 - k m If ψ is real, then it must be the case that ξ > ψ I Consequently, both exponentials in the homogeneous response decay to 0 as t → ∞ y ( t ) = 1 2 m ψ e ( - ξ + ψ ) t - e ( - ξ - ψ ) t 0 t y ( t )

Some Physical Examples I If ψ is imaginary, then let ψ = j ω where ω is real. y ( t ) = 1 2 m ψ e ( - ξ + ψ ) t - e ( - ξ - ψ ) t = 1 2 mj ω e - ξ t ( e j ω t - e - j ω t ) = 1 2 mj ω e - ξ t ((cos ω t + j sin ω t ) - (cos ω t - j sin ω t )) = 1 m ω e - ξ t sin ω t

Some Physical Examples I If ψ is imaginary, then let ψ = j ω where ω is real. y ( t ) = 1 2 m ψ e ( - ξ + ψ ) t - e ( - ξ - ψ ) t = 1 2 mj ω e - ξ t ( e j ω t - e - j ω t ) = 1 2 mj ω e - ξ t ((cos ω t + j sin ω t ) - (cos ω t - j sin ω t )) = 1 m ω e - ξ t sin ω t 0 t y ( t )

Exercise A car obeys the following differential equation m d 2 dt 2 y ( t ) + b d dt y ( t ) = u ( t ) where y ( t ) is the car’s displacement, u ( t ) is the engine force, m is the car’s inertia, and b is the viscous drag. Find the response y ( t ) due to an impulsive engine force, i.e., u ( t ) = δ ( t ) .

Exercise A car obeys the following differential equation m d 2 dt 2 y ( t ) + b d dt y ( t ) = u ( t ) where y ( t ) is the car’s displacement, u ( t ) is the engine force, m is the car’s inertia, and b is the viscous drag. Find the response y ( t ) due to an impulsive engine force, i.e., u ( t ) = δ ( t ) . Answer: STEP 1: Find initial conditions: y ( 0 + ) = 0 , d dt y ( 0 + ) = 1 m STEP 2: Find the roots of the characteristic equation λ 2 + b m λ = ( λ + b m ) λ = 0 ⇒ λ 1 = 0 , λ 2 = - b m STEP 3: Find general homogeneous solution, i.e., y ( t ) = c 1 + c 2 e - b m t STEP 4: Choose c 1 and c 2 to get initial conditions: c 1 = 1 b , c 2 = - 1 b

Impulse Responses for General Rational Systems I In the two physical examples, we’ve seen that when u ( t ) = δ ( t ) , the response y ( t ) has the following features: I y ( t ) = 0 for t < 0 I The impulse establishes initial conditions for at t = 0 + I The system responds to these initial conditions for t > 0, based on homogeneous differential equation

Impulse Responses for General Rational Systems I In the two physical examples, we’ve seen that when u ( t ) = δ ( t ) , the response y ( t ) has the following features: I y ( t ) = 0 for t < 0 I The impulse establishes initial conditions for at t = 0 + I The system responds to these initial conditions for t > 0, based on homogeneous differential equation I It turns out that this is true generally, for any rational system that is strictly proper, i.e., any system governed by d n dt n y + a n - 1 d n - 1 dt n - 1 y + ... + a 1 d dt y + a 0 y = b n - 1 d n - 1 dt n - 1 u + ... + b 1 d dt u + b 0 u The effect of the impulse is to instigate a vector of initial conditions h y ( 0 + ) dy dt ( 0 + ) · · · d n - 1 dt n - 1 y ( 0 + ) i T

Impulse Responses for General Rational Systems I Theorem : Let L : u 7→ y be a rational, strictly proper system, and let u ( t ) = δ ( t ) . Then            b 0 b 1 b 2 . . . b n - 3 b n - 2 b n - 1            =            a 1 a 2 a 3 · · · a n - 2 a n - 1 1 a 2 a 3 a 4 · · · a n - 1 1 0 a 3 a 4 a 5 · · · 1 0 0 . . . . . . . . . . . . . . . . . . . . . a n - 2 a n - 1 1 · · · 0 0 0 a n - 1 1 0 · · · 0 0 0 1 0 0 · · · 0 0 0                       y ( 0 + ) y ( 1 ) ( 0 + ) y ( 2 ) ( 0 + ) . . . y ( n - 3 ) ( 0 + ) y ( n - 2 ) ( 0 + ) y ( n - 1 ) ( 0 + )            where y ( k ) ( t ) , d k dt k y ( t )

Impulse Responses for General Rational Systems I Theorem : Let L : u 7→ y be a rational, strictly proper system, and let u ( t ) = δ ( t ) . Then            b 0 b 1 b 2 . . . b n - 3 b n - 2 b n - 1            =            a 1 a 2 a 3 · · · a n - 2 a n - 1 1 a 2 a 3 a 4 · · · a n - 1 1 0 a 3 a 4 a 5 · · · 1 0 0 . . . . . . . . . . . . . . . . . . . . . a n - 2 a n - 1 1 · · · 0 0 0 a n - 1 1 0 · · · 0 0 0 1 0 0 · · · 0 0 0                       y ( 0 + ) y ( 1 ) ( 0 + ) y ( 2 ) ( 0 + ) . . . y ( n - 3 ) ( 0 + ) y ( n - 2 ) ( 0 + ) y ( n - 1 ) ( 0 + )            where y ( k ) ( t ) , d k dt k y ( t ) I So, by inverting the matrix above, we get the initial conditions in terms of the a i and b i coefficients

Impulse Responses for General Rational Systems I Proof: I First, for convenience define the integration operator ( Iy )( t ) , Z t -∞ y ( τ ) d τ Similarly ( I 2 y )( t ) is the twice-integration of y , i.e., ( I 2 y )( t ) , Z t -∞ ( Iy )( τ ) d τ and in general, ( I k y )( t ) is the k th -integration of y , i.e., ( I k y )( t ) , Z t -∞ ( I k - 1 y )( τ ) d τ Use similar notation for u ; i.e., ( Iu )( t ) = Z t -∞ u ( τ ) d τ, ( I k u )( t ) = Z t -∞ ( I k - 1 u )( τ ) d τ

Impulse Responses for General Rational Systems I Proof (cont’d): I The system differential equation is y ( n ) ( t ) + a n - 1 y ( n - 1 ) ( t ) + ... + a 1 y ( 1 ) ( t ) + a 0 y ( t ) = b n - 1 u ( n - 1 ) ( t ) + ... + b 1 u ( 1 ) ( t ) + b 0 u ( t ) Integrating both sides of the differential equation gives y ( n - 1 ) ( t ) + a n - 1 y ( n - 2 ) ( t ) + ... + a 1 y ( t ) + a 0 ( Iy )( t ) = b n - 1 u ( n - 2 ) ( t ) + ... + b 1 u ( t ) + b 0 ( Iu )( t ) Now consider evaluating this equation at t = 0 + , and consider the following observations: 1. We know that y ( t ) stays finite as it jumps from y ( 0 - ) to y ( 0 + ) . Because it stays finite, ( Iy )( 0 + ) = 0 2. For u ( t ) = δ ( t ) , u ( t ) and all its derivatives are 0 at t = 0 + 3. For u ( t ) = δ ( t ) , ( Iu )( 0 + ) = 1

Impulse Responses for General Rational Systems I Proof (cont’d): I With these observations the once-integrated differential equation at t = 0 + is y ( n - 1 ) ( 0 + ) + a n - 1 y ( n - 2 ) ( 0 + ) + ... + a 1 y ( 0 + ) = b 0 which is the first line of the matrix equation in the theorem I Similarly, integrating the differential equation a second time gives y ( n - 2 ) ( t ) + a n - 1 y ( n - 3 ) ( t ) + ... + a 1 ( Iy )( t ) + a 0 ( I 2 y )( t ) = b n - 1 u ( n - 3 ) ( t ) + ... + b 1 ( Iu )( t ) + b 0 ( I 2 u )( t ) Applying the same three observations, we have that for t = 0 + , the above equation is y ( n - 2 ) ( 0 + ) + a n - 1 y ( n - 3 ) ( 0 + ) + ... + a 2 y ( 0 + ) = b 1 which is the second line of the matrix equation in the theorem

Impulse Responses for General Rational Systems I Proof (cont’d): I Following the same procedure, integrating the differential equation k times gives the k th row of the matrix equation in the theorem

Impulse Responses for General Rational Systems I Proof (cont’d): I Following the same procedure, integrating the differential equation k times gives the k th row of the matrix equation in the theorem I Example : Suppose the differential equation is d 3 dt 3 y + 2 d 2 dt 2 y + 2 d dt y + y = 2 d 2 dt 2 u - 2 d dt u + u Then   y ( 0 + ) d dt y ( 0 + ) d 2 dt 2 y ( 0 + )   =   2 2 1 2 1 0 1 0 0   - 1   1 - 2 2   =   0 0 1 0 1 - 2 1 - 2 2     1 - 2 2   =   2 - 6 9  

Impulse Responses for General Rational Systems I Once we have the initial conditions we can get the impulse response by solving the homogeneous differential equation

Impulse Responses for General Rational Systems I Once we have the initial conditions we can get the impulse response by solving the homogeneous differential equation I If the roots are all distinct then we know from Lecture 4 that y ( t ) = n X i = 1 c i e λ i t where      c 1 c 2 . . . c n      =      1 1 · · · 1 λ 1 λ 2 · · · λ n . . . . . . . . . λ n - 1 1 λ n - 1 2 · · · λ n - 1 n      - 1      y ( 0 + ) y ( 1 ) ( 0 + ) . . . y ( n - 1 ) ( 0 + )     

Impulse Responses for General Rational Systems I Once we have the initial conditions we can get the impulse response by solving the homogeneous differential equation I If the roots are all distinct then we know from Lecture 4 that y ( t ) = n X i = 1 c i e λ i t where      c 1 c 2 . . . c n      =      1 1 · · · 1 λ 1 λ 2 · · · λ n . . . . . . . . . λ n - 1 1 λ n - 1 2 · · · λ n - 1 n      - 1      y ( 0 + ) y ( 1 ) ( 0 + ) . . . y ( n - 1 ) ( 0 + )      I In our example, we have that λ 1 = - 1 λ 2 = - 1 2 + j √ 3 2 λ 3 = - 1 2 - j √ 3 2

Impulse Responses for General Rational Systems I So in this example,   c 1 c 2 c 3   =   1 1 1 λ 1 λ 2 λ 3 λ 2 1 λ 2 2 λ 2 3   - 1   y ( 0 + ) y ( 1 ) ( 0 + ) y ( 2 ) ( 0 + )   =    1 1 1 - 1 - 1 2 + j √ 3 2 - 1 2 - j √ 3 2 ( - 1 ) 2 ( - 1 2 + j √ 3 2 ) 2 ( - 1 2 - j √ 3 2 ) 2    - 1   2 - 6 9   =    5 - 3 2 + j 5 2 √ 3 - 3 2 - j 5 2 √ 3   

Impulse Responses for General Rational Systems I The resultant impulse response for t > 0 is y ( t ) = c 1 e λ 1 t + c 2 e λ 2 t + c 3 e λ 3 t = 5 e - t + - 3 2 + j 5 2 √ 3 e ( - 1 2 + j √ 3 2 ) t + - 3 2 - j 5 2 √ 3 e ( - 1 2 - j √ 3 2 ) t = 5 e - t + e - 1 2 t - 3 cos √ 3 2 t - 5 √ 3 sin √ 3 2 t

Impulse Responses for General Rational Systems I Finding impulse response in Matlab (fast way): % Define differential equation coefficients a3 = 1; a2 = 2; a1 = 2; a0 = 1; b3 = 0; b2 = 2; b1 = -2; b0 = 1; % Create a system object L = tf([b3 b2 b1 b0],[a3 a2 a1 a0]); % Specify the time values at which to evaluate y(t) t = linspace(0,10,1000); % Evaluate the impulse response at these times y = impulse(L,t); % Plot figure(1) plot(t,y)

Impulse Responses for General Rational Systems I Finding impulse response in Matlab (fast way): % Define differential equation coefficients a3 = 1; a2 = 2; a1 = 2; a0 = 1; b3 = 0; b2 = 2; b1 = -2; b0 = 1; % Create a system object L = tf([b3 b2 b1 b0],[a3 a2 a1 a0]); % Specify the time values at which to evaluate y(t) t = linspace(0,10,1000); % Evaluate the impulse response at these times y = impulse(L,t); % Plot figure(1) plot(t,y) I Note that if you just type impulse(sys) at the command line, Matlab will automatically plot the impulse response

Impulse Responses for General Rational Systems I Finding impulse response in Matlab (from scratch): % Solve for initial conditions A = [ a1 a2 a3 ; a2 a3 0 ; a3 0 0 ]; B = [b0 ; b1 ; b2]; initial_conditions = A\B; % Solve for roots of the characteristic equation lambda = roots([a3 a2 a1 a0]); % Solve for the "c" values in the homogeneous equation D = [ 1 1 1 ; lambda(1) lambda(2) lambda(3) ; lambda(1)^2 lambda(2)^2 lambda(3)^2 ]; c = D\initial_conditions; % Find the impulse response y = c(1) * exp(lambda(1) * t) + c(2) * exp(lambda(2) * t) ... + c(3) * exp(lambda(3) * t); % Find the impulse response figure(1) plot(t,y)

Impulse Responses for General Rational Systems I What happens if L is proper, but not strictly proper? I.e., if d n dt n y + a n - 1 d n - 1 dt n - 1 y + ... + a 1 d dt y + a 0 y = b n d n dt n u + b n - 1 d n - 1 dt n - 1 u + ... + b 1 d dt u + b 0 u where b n 6 = 0

Impulse Responses for General Rational Systems I What happens if L is proper, but not strictly proper? I.e., if d n dt n y + a n - 1 d n - 1 dt n - 1 y + ... + a 1 d dt y + a 0 y = b n d n dt n u + b n - 1 d n - 1 dt n - 1 u + ... + b 1 d dt u + b 0 u where b n 6 = 0 I Define ˜ y = y - b n u Then the above differential equation is equivalent to d n dt n (˜ y + b n u )+ a n - 1 d n - 1 dt n - 1 (˜ y + b n u )+ ... + a 1 d dt (˜ y + b n u )+ a 0 (˜ y + b n u ) = b n d n dt n u + b n - 1 d n - 1 dt n - 1 u + ... + b 1 d dt u + b 0 u which simplifies to d n dt n ˜ y + a n - 1 d n - 1 dt n - 1 ˜ y + ... + a 1 d dt ˜ y + a 0 ˜ y = ( b n - 1 - a n - 1 b n ) d n - 1 dt n - 1 u + ... + ( b 1 - a 1 b n ) d dt u + ( b 0 - a 0 b n ) u

Impulse Responses for General Rational Systems I From previous slide, we have that d n dt n ˜ y + a n - 1 d n - 1 dt n - 1 ˜ y + ... + a 1 d dt ˜ y + a 0 ˜ y = ˜ b n - 1 d n - 1 dt n - 1 u + ... + ˜ b 1 d dt u + ˜ b 0 u where ˜ b i , b i - a i b n

Impulse Responses for General Rational Systems I From previous slide, we have that d n dt n ˜ y + a n - 1 d n - 1 dt n - 1 ˜ y + ... + a 1 d dt ˜ y + a 0 ˜ y = ˜ b n - 1 d n - 1 dt n - 1 u + ... + ˜ b 1 d dt u + ˜ b 0 u where ˜ b i , b i - a i b n I Noting that the mapping u 7→ ˜ y is strictly proper, we can solve for its impulse response using the technique described in this class.

Impulse Responses for General Rational Systems I From previous slide, we have that d n dt n ˜ y + a n - 1 d n - 1 dt n - 1 ˜ y + ... + a 1 d dt ˜ y + a 0 ˜ y = ˜ b n - 1 d n - 1 dt n - 1 u + ... + ˜ b 1 d dt u + ˜ b 0 u where ˜ b i , b i - a i b n I Noting that the mapping u 7→ ˜ y is strictly proper, we can solve for its impulse response using the technique described in this class. I The impulse response of the original system is then y ( t ) = ˜ y ( t ) + b n δ ( t )

Impulse Responses for General Rational Systems I From previous slide, we have that d n dt n ˜ y + a n - 1 d n - 1 dt n - 1 ˜ y + ... + a 1 d dt ˜ y + a 0 ˜ y = ˜ b n - 1 d n - 1 dt n - 1 u + ... + ˜ b 1 d dt u + ˜ b 0 u where ˜ b i , b i - a i b n I Noting that the mapping u 7→ ˜ y is strictly proper, we can solve for its impulse response using the technique described in this class. I The impulse response of the original system is then y ( t ) = ˜ y ( t ) + b n δ ( t ) I Conclusion : So when a system is proper, but not strictly proper, its impulse response contains an impulse as well

Summary I The Dirac impulse , denoted δ ( t ) , is a signal that has the following properties I δ ( t ) = 0 for t 6 = 0 I δ ( t ) = ∞ for t = 0 I R 0 + 0 - δ ( t ) dt = 1 I For a system L : u 7→ y , if u ( t ) = δ ( t ) then y ( t ) is called the impulse response of the system I Impulse responses can often be found based on physical reasoning , such as in the impulse-momentum principle of physics I Impulse responses for general rational systems can be found systematically, in terms of the coefficients of the differential equation; i.e., the a i and b i values I It turns out that the impulse response plays a central role in the analysis of linear systems. More on that next time