# lec09 - 6.003 Signals and Systems Second-Order Systems...

This preview shows pages 1–14. Sign up to view the full content.

6.003: Signals and Systems Second-Order Systems October 8, 2009

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Last Time We analyzed a mass and spring system. x ( t ) y ( t ) F = K ( x ( t ) y ( t ) ) = M ¨ y ( t ) + K M A A 1 x ( t ) y ( t ) ˙ y ( t ) ¨ y ( t ) Y X = K M A 2 1 + K M A 2
Last Time We also analyzed a leaky tanks system. r 0 ( t ) r 1 ( t ) r 2 ( t ) h 1 ( t ) h 2 ( t ) τ 1 ˙ r 1 ( t ) = r 0 ( t ) r 1 ( t ) τ 2 ˙ r 2 ( t ) = r 1 ( t ) r 2 ( t ) + 1 τ 1 A + 1 τ 2 A r 0 ( t ) r 2 ( t ) ˙ r 1 ( t ) r 1 ( t ) ˙ r 2 ( t ) R 2 R 0 = A 1 1 + A 1 × A 2 1 + A 2

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Second-Order Systems Today: Look more carefully at growth and decay of oscillatory re- sponses by studying an analogous electrical circuit. v i v o R L C
But First . .. The canonical forms for CT and DT diﬀer. + A s 0 X Y + Delay z 0 X Y H = Y X = A 1 s 0 A H = Y X = 1 1 z 0 R h ( t ) = e s 0 t u ( t ) h [ n ] = z n 0 u [ n ] A → 1 s R → 1 z s -plane s -plane z -plane z -plane

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Check Yourself What if we had used the DT canonical form for CT? + A s 0 X Y What is the impulse response of this system? 1. s 0 e s 0 t u ( t ) 2. s t 0 u ( t ) 3. 1 + s 0 e s 0 t u ( t ) 4. δ ( t ) + s 0 e s 0 t u ( t ) 5. none of the above
Check Yourself What is the impulse response of this system? + A s 0 X Y Y X = 1 1 s 0 A = 1 + s 0 A 1 s 0 A Therefore the impulse response is h ( t ) = δ ( t ) + s 0 e s 0 t u ( t )

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Check Yourself Check by tracing signal ﬂow paths. + A s 0 X Y Y X = 1 1 s 0 A = Ø k =0 ( s 0 A ) k
Check Yourself Check by tracing signal ﬂow paths. + A s 0 X Y Y X = 1 1 s 0 A = Ø k =0 ( s 0 A ) k If x ( t ) = δ ( t ) then y ( t ) = δ ( t ) + ··· t y ( t ) 0 s 0

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Check Yourself Check by tracing signal ﬂow paths. + A s 0 X Y Y X = 1 1 s 0 A = Ø k =0 ( s 0 A ) k If x ( t ) = δ ( t ) then y ( t ) = δ ( t ) + s 0 u ( t ) + ··· t y ( t ) s 0 0
Check Yourself Check by tracing signal ﬂow paths. + A s 0 X Y Y X = 1 1 s 0 A = Ø k =0 ( s 0 A ) k If x ( t ) = δ ( t ) then y ( t ) = δ ( t ) + s 0 u ( t ) + s 2 0 tu ( t ) + ··· t y ( t ) s 0 0

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Check Yourself Check by tracing signal ﬂow paths. + A s 0 X Y Y X = 1 1 s 0 A = Ø k =0 ( s 0 A ) k If x ( t ) = δ ( t ) then y ( t ) = δ ( t ) + s 0 u ( t ) + s 2 0 tu ( t ) + 1 2 s 3 0 t 2 u ( t ) + ··· t y ( t ) s 0 0
Check by tracing signal ﬂow paths. + A

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 47

lec09 - 6.003 Signals and Systems Second-Order Systems...

This preview shows document pages 1 - 14. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online