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Signals and Systems EE235 Leo Lam Leo Lam ©
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Today’s menu Homework 1 Posted System properties Causality Stability
Invertibility Time invariance Leo Lam ©
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System properties Causality: A System is Causal if it meets this criterion If T{x(t)}=y(t) then y(t+a) depends only on x(t+b) where b<=a The output depends only on current or past values of the input. “The system does not anticipate the input.” (It does not laugh before it’s tickled!) Leo Lam ©
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Test for Causality y(t) = 4x(t) y(t) = x(t –3) y(t) = x(t + 5)
System is causal if output depends only on past and present input signal y(t) = 4x(t) y(t) = x(t –3) y(t) = x(t + 5) y(t) = x(3t) y(t) = (t + 5)x(t) y(t) = x(-t) causal (amplification) causal (delay) non-causal (time-shift forward, y(0)=x(5)) non-causal (speed-up, y(1)=x(3)) causal (ramp times x(t)) non-causal (time reverse, negative time needs future, y(-1)=x(1)) Leo Lam ©
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causal if Causality Example What values of t0 would make T causal?
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Causality Example Is T causal? YES
Depends only on past and present signals Leo Lam ©
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Causality Example What values of a would make T causal?
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Causality Example Causal NOT causal: x(t)’s include t =t+1
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Invertibility test Positive test: find the inverse
For some systems, you need tools that we’ll 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 ©
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Invertibility Example
Is T invertible? NOT Invertible Leo Lam ©
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Invertibility Example
Is T invertible? YES Leo Lam ©
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Invertibility Example
y(t) = 4x(t) y(t) = x(t –3) y(t) = x(t)2 y(t) = x(3t) y(t) = (t + 5)x(t) y(t) = cos(x(t)) invertible: Ti{y(t)}=y(t)/4 invertible: Ti{y(t)}=y(t+3) NOT invertible: don’t know sign of x(t) invertible: Ti{y(t)}=y(t/3) NOT invertible: can’t find x(-5) NOT invertible: x=0,2 π,4 π,… all give cos(x)=1 Leo Lam ©
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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 ©
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Stability test Is it stable?
Bounded input results in a bounded output STABLE! Leo Lam ©
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Stability test How about this, your turn? 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. Not stable Leo Lam ©
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Stability test How about this, your turn? Let for all t Stable
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Stability test How about this, your turn? Stable Stable NOT Stable
Tx(t) is a ramp 1/x(t) blows up at x(t) = 0 Stable NOT Stable Leo Lam ©
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Time invariance The system behaves the same no matter when you use it
Input is delayed by t0 seconds, output is the same but delayed t0 seconds If then System T Delay t0 x(t) x(t-t0) y(t) y(t-t0) T[x(t-t0)] System 1st Delay 1st = If the input is delayed by t0 seconds, then the output is the same but delayed t0 seconds Leo Lam ©
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Time invariance example
T{x(t)}=2x(t) x(t) y(t)= 2x(t) y(t-t0) T Delay x(t-t0) 2x(t-t0) Delay T Amplification of 2 is time invariant Identical time invariant! Leo Lam ©
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Time invariance test Test steps: Find y(t) Find y(t-t0)
Find T{x(t-t0)} Compare! IIf y(t-t0) = T{x(t-t0)} Time invariant! Basically taking BOTH routes should come to the same answer. Leo Lam ©
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Time invariance example
T(x(t)) = x2(t) y(t) = x2(t) y(t-t0) =x2(t-t0) T(x(t-t0)) = x2(t-t0) y(t-t0) = T(x(t-t0)) Time invariant! KEY: In step 2 you replace the t by t-t0. In step 3 you replace the x(t) by x(t-t0). Basically taking BOTH routes should come to the same answer. Leo Lam ©
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Time invariance example
Your turn! T(T(x(t)) = t x(t) y(t) = t*x(t) y(t-t0) =(t-t0) x(t-t0) T(x(t-t0)) = t x(t-t0) y(t-t0)) = T(x(t-t0)) Not time invariant! KEY: In step 2 you replace the t by t-t0. In step 3 you replace the x(t) by x(t-t0). Basically taking BOTH routes should come to the same answer. Leo Lam ©
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Time invariance example
Still you… T(x(t)) = 3x(t - 5) y(t) = 3x(t-5) y(t – t0) = 3x(t-t0-5) T(x(t – t0)) = 3x(t-t0-5) y(t-t0)) = T(x(t-t0)) Time invariant! KEY: In step 2 you replace the t by t-t0. In step 3 you replace the x(t) by x(t-t0). Basically taking BOTH routes should come to the same answer. Leo Lam ©
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Time invariance example
Still you… T(x(t)) = x(5t) y(t) = x(5t) y(t – 3) = x(5(t-3)) = x(5t – 15) T(x(t-3)) = x(5t- 3) Oops… Not time invariant! Does it make sense? KEY: In step 2 you replace the t by t-t0. In step 3 you replace the x(t) by x(t-t0). Shift then scale Basically taking BOTH routes should come to the same answer. Leo Lam ©
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Time invariance example
system output y(t) = x(5t) 1 Graphically: T(x(t)) = x(5t) y(t) = x(5t) y(t – 3) = x(5(t-3)) = x(5t – 15) T(x(t-3)) = x(5t- 3) t shifted system output y(t-3) = x(5(t-3)) t system input x(t) 5 t system output for shifted system input T(x(t-3)) = x(5t-3) t shifted system input x(t-3) Basically taking BOTH routes should come to the same answer. Leo Lam ©
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Time invariance example
Integral First: Second: Third: Lastly: Time invariant! KEY: In step 2 you replace the t by t-t0. In step 3 you replace the x(t) by x(t-t0). Basically taking BOTH routes should come to the same answer. Leo Lam ©
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Summary: System properties Leo Lam ©
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