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Signals and Systems EE235 Leo Lam ©
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Today’s menu Good weekend? System properties Time Invariance Linearity
Superposition! 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
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 t by t-t0. In step 3 you replace 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 t by t-t0. In step 3 you replace 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 t by t-t0. In step 3 you replace x(t) by x(t-t0). Basically taking BOTH routes should come to the same answer. Leo Lam ©
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System properties Linearity: A System is Linear if it meets the following two criteria: Together…superposition If and Additivity Then If Then Scaling “System Response to a linear combination of inputs is the linear combination of the outputs.” Leo Lam ©
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Linearity Order of addition and multiplication doesn’t matter.
System T Linear combination System 1st Combo 1st = If the input is delayed by t0 seconds, then the output is the same but delayed t0 seconds Linear combination Leo Lam ©
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Linearity Positive proof Negative 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 doesn’t hold If the input is delayed by t0 seconds, then the output is the same but delayed t0 seconds Leo Lam ©
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Linearity Proof Combo Proof System 1st Combo 1st Step 1: find yi(t)
Step 2: find y_combo Step 3: find T{x_combo} Step 4: If y_combo = T{x_combo} Linear System T Linear combination System 1st Combo 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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Linearity Example Is T linear? T x(t) y(t)=cx(t) Equal Linear
If the input is delayed by t0 seconds, then the output is the same but delayed t0 seconds Equal Linear Leo Lam ©
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Linearity Example Is T linear? T x(t) y(t)=(x(t))2
If the input is delayed by t0 seconds, then the output is the same but delayed t0 seconds Not equal non-linear Leo Lam ©
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Linearity Example Is T linear? T x(t) y(t)=x(t)+5
If the input is delayed by t0 seconds, then the output is the same but delayed t0 seconds Not equal non-linear Leo Lam ©
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Linearity Example Is T linear? =
If the input is delayed by t0 seconds, then the output is the same but delayed t0 seconds = Leo Lam ©
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Linearity unique case 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” If the input is delayed by t0 seconds, then the output is the same but delayed t0 seconds Leo Lam ©
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Spotting non-linearity
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. NOT Formal Proofs! (although sometimes ok) Leo Lam ©
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Superposition Superposition is…
Weighted sum of inputs weighted sum of outputs “Divide & conquer” Leo Lam ©
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Superposition example
Graphically x1(t) T 1 y1(t) 2 x2(t) y2(t) 3 -1 ? -y2(t) 19 Leo Lam ©
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Superposition example
Slightly aside (same system) Is it time-invariant? No idea: not enough information Single input-output pair cannot test positively x1(t) T 1 y1(t) 2 x2(t) y2(t) 3 Specific input-output pair (or pairs), you CANNOT test positively for system properties. 20 Leo Lam ©
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Superposition example
Unique case can be used negatively x1(t) T 1 y1(t) 2 x2(t) -1 y2(t) -2 NOT Time Invariant: Shift by 1 shift by 2 x1(t)=u(t) S y1(t)=tu(t) NOT Stable: Bounded input gives unbounded output 21 Leo Lam ©
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Summary: System properties
Causal: output does not depend on future input times Invertible: can uniquely find system input for any output Stable: bounded input gives bounded output Time-invariant: Time-shifted input gives a time-shifted output Linear: response to linear combo of inputs is the linear combo of corresponding outputs Leo Lam ©
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