Chapter 3 CT LTI Systems Updated: 9/16/13

A Continuous-Time System How do we know the output? System X(t)y(t)

LTI Systems Time Invariant –X(t)  y(t) & x(t-to)  y(t-to) Linearity –a1x1(t)+ a2x2(t)  a1y1(t)+ a2y2(t) –a1y1(t)+ a2y2(t)= T[a1x1(t)+a2x2(t)] Meet the description of many physical systems They can be modeled systematically –Non-LTI systems typically have no general mathematical procedure to obtain solution What is the input-output relationship for LTI-CT Systems?

An approach (available tool or operation) to describe the input-output relationship for LTI Systems In a LTI system –  t)  h(t) –Remember h(t) is T [  t)] –Unit impulse function  the impulse response It is possible to use h(t) to solve for any input-output relationship One way to do it is by using the Convolution Integral Convolution Integral LTI System X(t)=  (t) y(t)=h(t) LTI System: h(t) X(t)y(t)

Convolution Integral Remember So what is the general solution for LTI System X(t)=A  (t-kto) y(t)=Ah(t-kto) LTI System X(t)y(t) ?

Convolution Integral Any input can be expressed using the unit impulse function LTI System X(t)y(t) Sifting Property Proof: to   and integrate by d 

Convolution Integral Given We obtain Convolution Integral That is: A system can be characterized using its impulse response: y(t)=x(t)*h(t) LTI System X(t)y(t) LTI System: h(t) X(t)y(t) Do not confuse convolution with multiplication! y(t)=x(t)*h(t) By definition

Convolution Integral LTI System: h(t) X(t)y(t)

Convolution Integral - Properties Commutative Associative Distributive Thus, using commutative property: Next: We draw the block diagram representation!

Convolution Integral - Properties Commutative Associative Distributive

Simple Example What if a step unit function is the input of a LTI system? S(t) is called the Step Response LTI System u(t)y(t)=S{u(t)}=s(t) Step response can be obtained by integrating the impulse response! Impulse response can be obtained by differentiating the step response

Example 1 Consider a CT-LTI system. Assume the impulse response of the system is h(t)=e^(-at) for all a>0 and t>0 and input x(t)=u(t). Find the output. h(t)=e^-at u(t)y(t) Draw x(  ), h(  ), h(t-  ),etc.  next slide Because t>0 The fact that a>0 is not an issue!

Example 1 – Cont. y(t) t>0 t<0 Remember we are plotting it over  and t is the variable U(-(  -t)) * t y(t); for a=3 t

Example 1 – Cont. Graphical Representation (similar) a=1 In this case!

Example 1 – Cont. Graphical Representation (similar) Note in our case We have u(t) rather than rectangular function! Note in our case We have u(t) rather than rectangular function!

Example 2 Consider a CT-LTI system. Assume the impulse response of the system is h(t)=e^-at for all a>0 and t>0 and input x(t)= e^at u(-t). Find the output. h(t)=e^-at x(t)y(t) Draw x(  ), h(  ), h(t-  ),etc.  next slide Note that for t>0; x(t) =0; so the integration can only be valid up to t=0

Example – Cont. * h(t)=e^-at u(t) x(t)= e^at u(-t) ?

Another Example notes

Properties of CT LTI Systems When is a CT LTI system memory-less (static) When does a CT LTI system have an inverse system (invertible)? When is a CT LTI system considered to be causal? Assuming the input is causal: When is a CT LTI system considered to be Stable? notes

Example Is this an stable system? What about this? notes

Differential-Equations Models This is a linear first order differential equation with constant coefficients (assuming a and b are constants) The general nth order linear DE with constant equations is

Is the First-Order DE Linear? Consider Does a1x1(t)+ a2x2(t)  a1y1(t)+ a2y2(t)? Is it time-invariant? Does input delay results in an output delay by the same amount? Is this a linear system? notes Sum Integrator a - + X(t) Y(t)e(t)

Example Is this a time invariant linear system? R L V(t) Ldi(t)/dt + Ri(t)= v(t) a= -R/L b=1/L y(t)=i(t) x(t) = V(t)

Solution of DE A classical model for the solution of DE is called method of undermined coefficients –yc(t) is called the complementary or natural solution –yp(t) is called the particular or forced solution

Solution of DE Thus, for x(t) =constant  yp(t)=P x(t) =Ce^-7t  yp(t)= Pe^-7t x(t) =2cos(3t)  yp(t)=P1cos(3t)+P2sin(3t)

Example Solve –Assume x(t) = 2 and y(0) = 4 –What happens if notes yc(t) = Ce^-2t; yp(t) = P; P = 1 y(t) = Ce^-2t + 1 y(0) = 4  C = 3  y(t) = 3e^-2t + 1

Schaum’s Examples Chapter 2: –2, 4-6, 8, 10, 11-14, 18, 19, 48, 52, 53,