Leo Lam © Signals and Systems EE235

Leo Lam © Today’s menu Homework 2 due now Convolution!

y(t) at all t Leo Lam © At all t t<0 The product of these two signals is zero where they don’t overlap ShiftMultiplyIntegrate

y(t) at all t Leo Lam © At all t 0≤t<1 ShiftMultiplyIntegrate

y(t) at all t Leo Lam © At all t 1≤t<2 y(t)=2-t for 1≤t<2 ShiftMultiplyIntegrate

y(t) at all t Leo Lam © At all t t≥2 y(t)=0 for t≥2 (same as t<0, no overlap) ShiftMultiplyIntegrate

y(t) at all t Leo Lam © Combine it all –y(t)=0 for t 2 –y(t)=t for 0≤t<1 –y(t)=2-t for 1≤t<2

Another example Leo Lam © At all t t<0 The product of these two signals is zero where they don’t overlap ShiftMultiplyIntegrate

Another example Leo Lam © At all t 0≤t<0.5 ShiftMultiplyIntegrate h(t) moving right

Another example Leo Lam © At all t 0.5≤t<1 h(t) moving right ShiftMultiplyIntegrate

Another example Leo Lam © At all t 1≤t<1.5 ShiftMultiplyIntegrate h(t) moving right

Another example Leo Lam © At all t 1.5≤t? ShiftMultiplyIntegrate y(t)=0 because there is no more overlapping

Another example Leo Lam © At all t Combining Can you plot and formulate it?

Another example Leo Lam © At all t

Few things to note Leo Lam © Three things: –Width of y(t) = Width of x(t)+Width of h(t) –Start time adds –End time adds –y(t) is smoother than x(t) and h(t) (mostly) Stretching the thinking –What if one signal has infinite width?

From yesterday Leo Lam © Stretching the thinking –What if one signal has infinite width? Width = infinite (infinite overlapping) Start time and end time all infinite

One more example Leo Lam © For all t: x(t) 2 1 t FlipShift Can you guess the “width” of y(t)?

One more example Leo Lam © For all t: x(t) 2 1 t Multiply & integrate