Lesson+17 - In your own words explain Kirchoffs two laws 1...

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1 Lesson 17 In your own words explain Kirchoff’s two laws.
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Challenge 16 Steady State Sinusoidal Response Chapter 13.7 Challenge 17 Lesson 17 2 Lesson 17
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Challenge 16 Using time-domain methods, convolve x(t) = e -t u(t) and h(t) = e -2t u(t). That is, determine y(t) = x(t) h(t). 0 ; ) ( ) ( ) ( ); ( ) ( 2 0 2 2 0 2 t e e d e e d e e t y t u e t x t u e t h t t t t t t t t 3 Lesson 17
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Using the convolution theorem, convolve x(t) = e -t u(t) and h(t) = e -2t u(t). That is, determine y(t) = x(t) h(t). 0 ; ) ( ) ) 2 /( 1 ) 1 /( 1 ( )) ( ( ) ( 2 1 1 t e e t y s s L s Y L t y t t (same as before) 4 Lesson 17 )) 1 /( 1 ) 2 /( 1 ( )) 2 /( 1 ))( 1 /( 1 ( ) ( ) ( ) ( s s s s s H s X s Y
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0 1 2 3 4 5 6 7 8 9 10 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 ; ) ( ); ( ) ( ); ( ) ( 2 2 t e e t y t u e t x t u e t h t t t t » t=0:0.01:10; x=exp(-t);h=exp(-2*t); » y=conv(x,h); yy=y(1:1001); » plot(t,x,t,h,t,yy/100) x(t) h(t ) y(t ) Machine production using discrete-time simulation. 5 Lesson 17
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Revisit the importance of convolution We make a bold claim that the physical world behaves as a natural lowpass filter. Can this explain why normal noise (a.k.a., Gaussian) called normal? 1 0 -1 PDF 1 0 -1 h(t) (channel) Random signal x(t) (binary valued heads/tail) y 1 (t)= x(t)*h(t) h(t) x(t) y 1 (t) PDF 6 Lesson 17
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h(t) (channel) y 1 (t) (previous output) y 2 (t)= y 1 (t)*h(t) 1 0 -1 PDF 1 0 -1 h(t)] y 1 (t) y 2 (t) PDF 7 Lesson 17
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Steady State Frequency Response Author’s tale. Assume that the input to an LTI having h(t) H(s) is sinusoidal (consult trig, table): x(t) = A cos( t+ ) = A cos( ) cos( t) - A sin( ) sin( t) It therefore follows that (consult Laplace table): 2 2 2 2 2 2 ) sin( ) cos( ) sin( ) cos( ) ( s s A s A s s A s X 8 Lesson 17
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Author’s tale. According to past studies, the solution should consist of a steady-state component define by the input and Eigenvalues, plus a natural response defined by the Eigenvalue of H(s). If all the Eigenvalues (poles) of H(s) have negative real parts, the natural solution will eventually converge to zero leaving only the only the steady- state solution. At steady-state solution would have the (Heaviside) form: 2 2 ) sin( ) cos( ) ( ) ( ) ( ) ( s s A s H s X s H s Y j s K j s K s Y 2 1 ) ( ; t for 9 Lesson 17
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Author’s tale . 1 2 2 1 K K j s K j s K s Y ; ) ( ω ω j e j H j H Since K 2 =K 1 *, the filter function is: j j s j s e A j H j s s A j H s Y j s K 2 1 sin cos ) ( ) ( 1 10 Lesson 17
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Author’s tale. Examine: where: * 1 2 1 2 K K e j H A K j Interpreting
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