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

3. Oh beer… An infinite amount of mathematicians walk into a bar. The first one orders a beer. The second one orders half a beer. The third one orders one quarter of a beer. The fourth one starts to order, but the bartender interrupts "Here's two beers; you lot can figure the rest out yourself." Leo Lam © 2010-2011

4. Today’s menu To do: –If you still haven’t: join the Facebook Group! –Homework 1 posted, due Tuesday. From Wednesday: Describing Common Signals –Introduced the three “building blocks” –General description for sinusoidal signals Today: Periodicity

5. Periodic signals Definition: x(t) is periodic if there exists a T (time period) such that: The minimum T is the period Fundamental frequency f 0 =1/T Leo Lam © 2010-2012 For all integers n

6. Periodic signals: examples Sinusoids Complex exponential (non-decaying or increasing) Infinite sum of shifted signals v(t) (more later) Leo Lam © 2010-2012 x(t)=A cos(   t+  ) T0T0

7. Periodicity of the sum of periodic signals Question: If x 1 (t) is periodic with period T 1 and x 2 (t) is periodic with period T 2 –What is the period of x 1 (t)+x 2 (t)? Can we rephrase this using our “language” in math? Leo Lam © 2010-2012

8. Rephrasing in math Leo Lam © 2010-2012 Goal: find T such that

9. Rephrasing in math Leo Lam © 2010-2012 Goal: find T such that Need:  T=LCM(T 1,T 2 ) Solve it for r=1, true for all r

10. Periodic sum example If x 1 (t) has T 1 =2 and x 2 (t) has T 2 =3, what is the period of their sum, z(t)? LCM (2,3) is 6 And you can see it, too. Leo Lam © 2010-2012 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 1 2 1 1 T 1 T 2

11. Your turn! Find the period of: Leo Lam © 2010-2012 No LCM exists! Why? Because LCM exists only if T 1 /T 2 is a rational number

12. A few more Leo Lam © 2010-2011 Not rational, so not periodic Decaying term means pattern does not repeat exactly, so not periodic

13. Summary Description of common signals Periodicity Leo Lam © 2010-2011

14. Playing with signals Operations with signals –add, subtract, multiply, divide signals pointwise –time delay, scaling, reversal Properties of signals (cont.) –even and odd Leo Lam © 2010-2011

15. Adding signals Leo Lam © 2010-2011 1 1 123 t t + = ?? x(t) y(t) 1 t 123 x(t)+y(t)

16. Delay signals Leo Lam © 2010-2011 unit pulse signal (memorize) t 0 1 1 What does y(t)=p(t-3) look like? P(t) 0 34

17. Multiply signals Leo Lam © 2010-2011

18. Scaling time Leo Lam © 2010-2011 Speed-up: y(t)=x(2t) is x(t) sped up by a factor of 2 t 0 1 1 t 0 1.5 y(t)=x(2t) How could you slow x(t) down by a factor of 2? y(t)=x(t)

19. Scaling time Leo Lam © 2010-2011 y(t)=x(t/2) is x(t) slowed down by a factor of 2 t 01 t 01 y(t)=x(t/2) 2 -2 y(t)=x(t)

20. Playing with signals Leo Lam © 2010-2011 What is y(t) in terms of the unit pulse p(t)? t 8 3 5 t 0 1 1 Need: 1.Wider (x-axis) factor of 2 2.Taller (y-axis) factor of 8 3.Delayed (x-axis) 3 seconds

21. Playing with signals Leo Lam © 2010-2011 t 8 3 5 in terms of unit pulse p(t) t 8 2 first step: 3 5 t 8 second step:

22. Playing with signals Leo Lam © 2010-2011 t 8 3 5 in terms of unit pulse p(t) t 8 2 first step: 3 5 t 8 second step: replace t by t-3: Is it correct?

23. Playing with signals Leo Lam © 2010-2011 3 5 t 8 Double-check: pulse starts: pulse ends:

24. Do it in reverse Leo Lam © 2010-2011 t Sketch 1

25. Do it in reverse Leo Lam © 2010-2011 t Let then that is, y(t) is a delayed pulse p(t-3) sped up by 3. 1 1 4/3 1 3 4 Double-check pulse starts: 3t-3 = 0 pulse ends: 3t-3=1

26. Order matters Leo Lam © 2010-2011 With time operations, order matters y(t)=x(at+b) can be found by: Shift by b then scale by a (delay signal by b, then speed it up by a) w(t)=x(t+b)  y(t)=w(at)=x(at+b) Scale by a then shift by b/a w(t)=x(at)  y(t)=w(t+b/a)=x(a(t+b/a))=x(at+b)

27. Playing with time Leo Lam © 2010-2011 t 1 What does look like? 2 1 -2 Time reverse of speech: Also a form of time scaling, only with a negative number

28. Playing with time Leo Lam © 2010-2011 t 1 2 Describe z(t) in terms of w(t) 1 -213 t

29. Playing with time Leo Lam © 2010-2011 time reverse it: x(t) = w(-t) delay it by 3: z(t) = x(t-3) so z(t) = w(-(t-3)) = w(-t + 3) t 1 2 1 -21 3 x(t) you replaced the t in x(t) by t-3. so replace the t in w(t) by t-3: x(t-3) = w(-(t-3))

30. Playing with time Leo Lam © 2010-2011 z(t) = w(-t + 3) t 1 2 1 -21 3 x(t) Doublecheck: w(t) starts at 0 so -t+3 = 0 gives t= 3, this is the start (tip) of the triangle z(t). w(t) ends at 2 So -t+3=2 gives t=1, z(t) ends there

31. Summary: Arithmetic: Add, subtract, multiple Time: delay, scaling, shift, mirror/reverse And combination of those Leo Lam © 2010-2011