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Leo Lam © 2010-2012 Signals and Systems EE235

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Leo Lam © 2010-2012 Pet Q: Has the biomedical imaging engineer done anything useful lately? A: No, he's mostly been working on PET projects.

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Leo Lam © 2010-2012 Todays menu System properties examples –Invertibility –Stability –Time invariance –Linearity

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Invertibility test Positive test: find the inverse For some systems, you need tools that well learn later in the quarter… Negative test: find an output that could be generated by two different inputs (note that these two different inputs might only differ at only one time value) Each input signal results in a unique output signal, and vice versa Invertible Leo Lam © 2010-2011

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Invertibility Example Leo Lam © 2010-2011 1)y(t) = 4x(t) 2)y(t) = x(t –3) 3)y(t) = x 2 (t) 4)y(t) = x(3t) 5)y(t) = (t + 5)x(t) 6)y(t) = cos(x(t)) invertible: T i {y(t)}=y(t)/4 invertible: T i {y(t)}=y(t/3) invertible: T i {y(t)}=y(t+3) NOT invertible: dont know sign of x(t) NOT invertible: cant find x(-5) NOT invertible: x=0,2 π,4 π,… all give cos(x)=1

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Stability test For positive proof: show analytically that –a bounded input signal gives a bounded output signal (BIBO stability) For negative proof: –Find one counter example, a bounded input signal that gives an unbounded output signal –Some good things to try: 1, u(t), cos(t), 0 Leo Lam © 2010-2012

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Stability test Is it stable? Leo Lam © 2010-2012 Bounded input results in a bounded output STABLE!

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Stability test How about this? Leo Lam © 2010-2012 Stable Let for all t

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Stability test How about this, your turn? Leo Lam © 2010-2012 Not BIBO stable Counter example: x(t)=u(t) y(t)=5tu(t)=5r(t) Input u(t) is bounded. Output y(t) is a ramp, which is unbounded.

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Stability test How about this, your turn? Leo Lam © 2010-2012 Stable NOT Stable Stable

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System properties Leo Lam © 2010-2012 Time-invariance: A System is Time-Invariant if it meets this criterion System Response is the same no matter when you run the system.

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Time invariance Leo Lam © 2010-2012 The system behaves the same no matter when you use it Input is delayed by t 0 seconds, output is the same but delayed t 0 seconds If then System T Delay t 0 System T Delay t 0 x(t) x(t-t 0 ) y(t) y(t-t 0 ) T[x(t-t 0 )] System 1 st Delay 1 st =

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Time invariance example Leo Lam © 2010-2012 T{x(t)}=2x(t) x(t) y(t)= 2x(t) y(t-t 0 ) T Delay x(t-t 0 ) 2x(t-t 0 ) Delay T Identical time invariant!

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Time invariance test Leo Lam © 2010-2012 Test steps: 1.Find y(t) 2.Find y(t-t 0 ) 3.Find T{x(t-t 0 )} 4.Compare! IIf y(t-t 0 ) = T{x(t-t 0 )} Time invariant!

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Time invariance example Leo Lam © 2010-2012 T(x(t)) = x 2 (t) 1.y(t) = x 2 (t) 2.y(t-t 0 ) =x 2 (t-t 0 ) 3.T(x(t-t 0 )) = x 2 (t-t 0 ) 4.y(t-t 0 ) = T(x(t-t 0 )) Time invariant! KEY: In step 2 you replace t by t-t 0. In step 3 you replace x(t) by x(t-t 0 ).

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Time invariance example Leo Lam © 2010-2012 Your turn! T{(x(t)} = t x(t) 1.y(t) = t*x(t) 2.y(t-t 0 ) =(t-t 0 ) x(t-t 0 ) 3.T(x(t-t 0 )) = t x(t-t 0 ) 4.y(t-t 0 )) != T(x(t-t 0 )) Not time invariant! KEY: In step 2 you replace t by t-t 0. In step 3 you replace x(t) by x(t-t 0 ).

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Time invariance example Leo Lam © 2010-2012 Still you… T(x(t)) = 3x(t - 5) 1.y(t) = 3x(t-5) 2.y(t – t 0 ) = 3x(t-t 0 -5) 3.T(x(t – t 0 )) = 3x(t-t 0 -5) 4.y(t-t 0 )) = T(x(t-t 0 )) Time invariant! KEY: In step 2 you replace t by t-t 0. In step 3 you replace x(t) by x(t-t 0 ).

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Time invariance example Leo Lam © 2010-2012 Still you… T(x(t)) = x(5t) 1.y(t) = x(5t) 2.y(t – 3) = x(5(t-3)) = x(5t – 15) 3.T(x(t-3)) = x(5t- 3) 4.Oops… Not time invariant! Does it make sense? KEY: In step 2 you replace t by t-t 0. In step 3 you replace x(t) by x(t-t 0 ). Shift then scale

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Time invariance example Leo Lam © 2010-2012 Graphically: T(x(t)) = x(5t) 1.y(t) = x(5t) 2.y(t – 3) = x(5(t-3)) = x(5t – 15) 3.T(x(t-3)) = x(5t- 3) t 0 system input x(t) 5 t 0 system output y(t) = x(5t) 1 t 0 3 4 shifted system output y(t-3) = x(5(t-3)) t 0 3 8 shifted system input x(t-3) 0.6 1.6 t system output for shifted system input T(x(t-3)) = x(5t-3)

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Time invariance example Leo Lam © 2010-2012 Integral 1.First: 2.Second: 3.Third: 4.Lastly: Time invariant! KEY: In step 2 you replace t by t-t 0. In step 3 you replace x(t) by x(t-t 0 ).

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System properties Leo Lam © 2010-2011 Linearity: A System is Linear if it meets the following two criteria: Together…superposition Ifand Then If Then System Response to a linear combination of inputs is the linear combination of the outputs. Additivity Scaling

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Linearity Leo Lam © 2010-2011 Order of addition and multiplication doesnt matter. = System T System T Linear combination System 1 st Combo 1 st Linear combination

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Linearity Leo Lam © 2010-2011 Positive proof –Prove both scaling & additivity separately –Prove them together with combined formula Negative proof –Show either scaling OR additivity fail (mathematically, or with a counter example) –Show combined formula doesnt hold

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Linearity Proof Leo Lam © 2010-2011 Combo Proof Step 1: find y i (t) Step 2: find y_combo Step 3: find T{x_combo} Step 4: If y_combo = T{x_combo} Linear System T System T Linear combination System 1 st Combo 1 st Linear combination

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Linearity Example Leo Lam © 2010-2011 Is T linear? T x(t)y(t)=cx(t) Equal Linear

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Linearity Example Leo Lam © 2010-2011 Is T linear? Not equal non-linear T x(t)y(t)=(x(t)) 2

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Linearity Example Leo Lam © 2010-2011 Is T linear? Not equal non-linear T x(t)y(t)=x(t)+5

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Linearity Example Leo Lam © 2010-2011 Is T linear? =

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Linearity unique case Leo Lam © 2010-2011 How about scaling with 0? If T{x(t)} is a linear system, then zero input must give a zero output A great negative test

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Spotting non-linearity Leo Lam © 2010-2011 multiplying x(t) by another x() y(t)=g[x(t)] where g() is nonlinear piecewise definition of y(t) in terms of values of x, e.g. (although sometimes ok) NOT Formal Proofs!

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Leo Lam © 2010-2011 Signals and Systems EE235 October 14 th Friday Online version.

Leo Lam © 2010-2011 Signals and Systems EE235 October 14 th Friday Online version.

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