Signals and Systems EE235 Leo Lam Leo Lam © 2010-2011

Today’s menu Convolution examples Exponential response of LTI system Leo Lam © 2010-2011

* Convolution examples 3 Approach? How to break it down? 1 t x(t) * 2 h(t) -1 y(t)=x(t)*h(t) Approach? How to break it down? System will start having non-zero output at time t = -1 The signal y(t) can be expressed in terms of 3 time regions: t<-1 (where y(t)=0), -1<t<1, t>1 3 Leo Lam © 2010-2011

* Convolution examples 4 Two non-zero regions: 1 t x(t) * 2 h(t) -1 y(t)=x(t)*h(t) Two non-zero regions: If you flip x(t), you’d get: If you flip h(t), you’d get: Identical? 4 Leo Lam © 2010-2011

Another example y(t)= 5 Approach? What does each part “look” like? = 1 if 3 - > 0 = 1 if t - > 0 5 Leo Lam © 2010-2011

Another example (complicated) y(t)= = 1 if 3 - > 0 = 1 if t - > 0 Need to satisfy both: That is & Two cases to consider then: or 6 Leo Lam © 2010-2011

Another example y(t)= 7 For = 1 if t - > 0 = 1 if 3 - > 0 Leo Lam © 2010-2011

Another example 8 Combining two, with only one active at each t For Then integrate… 8 Leo Lam © 2010-2011

Superposition If the input x(t) can be broken up into smaller, easier to convolve, components Output can be expressed as the sum of the easier convolutions 9 Leo Lam © 2010-2011

Superposition example If the step function is: x(t) = -u(t+2) +2u(t+1) +u(t) -2u(t-1) y(t) = -s(t+2) +2s(t+1) +s(t) -2s(t-1) “Divide and conquer” 1 2 -2 -1 x(t) 10 Leo Lam © 2010-2011

Summary Convolution examples Leo Lam © 2010-2011