INC 212 Signals and systems Lecture#2: Response of system Assoc. - - PowerPoint PPT Presentation

INC 212 Signals and systems

Lecture#2: Response of system

• Assoc. Prof. Benjamas Panomruttanarug

benjamas.pan@kmutt.ac.th

Response of a system

Impulse response: h(t)

( ) for t t   

( ) 1 t dt 

 





Step response: s(t)



1, ( ) 0, t u t t   

Relationship bt. Impulse response and step response

( ) ( ) d h t s t dt 

St Step ep Re Response

• The step response is defined as the output due to a unit step input

signal

• Can be solved by using convolution integral (next time)
• Can be solved by solving the differential equation

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Solving differential equation

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Complete solution: y = y(h) + y(p) y(h) = homogeneous solution, y(p) = particular solution = natural response = forced response

Ex Exam ample ple 2.7 2.7 RC Cir Circuit uit Output Output

Assume that the initial output, y(0) = 1. Show that the impulse response of this circuit is h(t) = e  t u(t).

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1 ( ) ( )

t RC

h t e u t RC





Ex Exam ample ple 2.14 2.14 RC Cir Circuit: uit: St Step ep Respo sponse se

The impulse response of the RC circuit Find the step response of the circuit.

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1 ( ) ( )

t RC

h t e u t RC





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Solving differential equation

     

d y t RC y t x t dt  

Homogeneous Eq:

   

d y t RC y t dt  

Diffential equation:

   

1

V

t h RC

y t c e





Homogeneous solution: Particular solution: y(t) = k=1 Total solution: y(t) = c1e(‐t/RC)+1 , giving the initial condition y(0)=0

Ex Exam ample ple 2.22 2.22 RC Circuit (continued): Complete Response

Find the complete response of the RC circuit to an input x(t) = cos(t)u(t) V, assuming normalized values R = 1  and C = 1 F and assuming that the initial voltage across the capacitor is y(0) = 2 V.

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• 1. Homogeneous sol.:

   

V

t h RC

y t ce





• 2. Particular solution:

   

       

2 2

1 cos sin V 1 1

p

RC y t t t RC RC    

• 4. Coefficient c1 determined by I.C.:
• 3. Complete solution:

 

1 1 cos sin V 2 2

t

y t ce t t



  

0 = 1

R = 1 , C = 1 F

1 1 1 2 cos0 sin 0 2 2 2 ce c



  

    

c = 3/2

• 5. Final solution:

 

3 1 1 cos sin V 2 2 2

t

y t e t t



  

Matlab: impulse input

• function dy = rc(t,y)
• R = 1; C = 1;
• x = 0;
• dy = (x‐y)/(R*C);
• %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
• t = 0:0.01:10;
• R = 1; C = 1;
• [T,y] = ode45(@rc,t,1);
• yt = exp(‐t/(R*C));
• figure;plot(T,y,'b',T,yt,'r‐‐‘);

) ( ) ( ) ( t x t y dt t dy RC  

Differential Equation: h = impulse([1],[R*C 1],t);

Matlab: unit step input

• function dy = rc(t,y)
• R = 1; C = 1;
• x = 1;
• dy = (x‐y)/(R*C);
• %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
• t = 0:0.01:10;
• R = 1; C = 1;
• [T,y] = ode45(@rc,t,0);
• yt = ‐exp(‐t/(R*C))+1;
• x = ones(size(y));
• figure;plot(T,y,'b',T,yt,'r‐‐',T,x,'g:')

) ( ) ( ) ( t x t y dt t dy RC  

Differential Equation: s = step([1],[R*C 1],t);

Matlab: unit step input

2 4 6 8 10 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Time (sec) y(t) y yt input

Matlab: sinusoidal input

• function dy = rc(t,y)
• R = 1; C = 1;
• x = cos(t);
• dy = (x‐y)/(R*C);
• %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
• t = 0:0.01:50;
• R = 1; C = 1;
• [T,y] = ode45(@rc,t,2);
• yt = 3/2*exp(‐t)+1/2*cos(t)+1/2*sin(t);
• x = cos(t);
• figure;plot(T,y,'b',T,yt,'r‐‐',T,x,'g:');
• xlabel('Time (sec)'); ylabel('y(t)');

) ( ) ( ) ( t x t y dt t dy RC  

Differential Equation:

Matlab: sinusoidal input

10 20 30 40 50

• 1
• 0.5

0.5 1 1.5 2 Time (sec) y(t) y yt input