EE 314 Signal and Linear System Analysis
Transcript of EE 314 Signal and Linear System Analysis
EE 314 Signal and Linear System Analysis
Graphical Convolution
Lecture 8 EE 314 Signal and Linear System Analysis Slide 1 of 14
Summary of Last Lecture
Lecture 8 EE 314 Signal and Linear System Analysis
• Applying a causal input (𝑥𝑥(𝑡𝑡)) to a causal LTI systems with impulse response ℎ(𝑡𝑡) gives rise to a causal output 𝑦𝑦 𝑡𝑡 :
0
( ) ( ) ( )t
y t x h t dτ τ τ= −∫
( )h t
0
( ) ( )t
h x t dτ τ τ= −∫
Slide 2 of 14
Graphical Convolution
Lecture 8 EE 314 Signal and Linear System Analysis
• Revisiting the prior RC example Let 𝑅𝑅𝑅𝑅 = 1/2 = 𝜏𝜏𝑐𝑐, hence,
Input: 𝑣𝑣𝑖𝑖𝑖𝑖 𝑡𝑡 = 𝑢𝑢 𝑡𝑡 − 𝑢𝑢(𝑡𝑡 − 1)• Analytical soln is: For, 0 < t < 1
2( ) 2 ( )th t e u t−=
0
( ) ( ) ( )t
out inv t v h t dτ τ τ= −∫ ( ) ( )2
0
1 2t
te dτ τ− −= ∫ 2 2
0
2t
te e dτ τ−= ∫
2 2
0
tte e τ− = ( )2 2 1t te e−= − 21 te−= −( )( )21 ( ) ( 1)te u t u t−= − − −
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Graphical Convolution
Lecture 8 EE 314 Signal and Linear System Analysis
For, t > 1
Final result
1
0 1
( ) ( ) ( ) ( ) ( )t
out in inv t v h t d v h t dτ τ τ τ τ τ= − + −∫ ∫1
0
( ) 0h t dτ τ= − +∫12 2
0
te e τ− =
( )2 2 1te e−= − 2( 1) 2t te e− − −= −( )2( 1) 2 ( 1)t te e u t− − −= − −
( )( ) ( )2 2( 1) 2( ) 1 ( ) ( 1) ( 1)t t toutv t e u t u t e e u t− − − −= − − − + − −
( ) ( )2 2( 1)1 ( ) 1 ( 1)t te u t e u t− − −= − + − −
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Graphical Convolution
Lecture 8 EE 314 Signal and Linear System Analysis
( ) ( )2 2( 1)( ) 1 ( ) 1 ( 1)t toutv t e u t e u t− − −= − + − −
Slide 5 of 14
ℎ −𝜏𝜏 + 𝑡𝑡 => Shift then flipℎ(−(𝜏𝜏 − 𝑡𝑡)) => Flip then shift
Graphical Convolution
Lecture 8 EE 314 Signal and Linear System Analysis
• Let’s evaluate the convolution graphically
Overlay a plot of 𝑣𝑣𝑖𝑖𝑖𝑖(𝜏𝜏) with a plot of ℎ(𝑡𝑡 − 𝜏𝜏), compute the area under the product (for 0 ≤ 𝜏𝜏 ≤ 𝑡𝑡), then vary 𝑡𝑡.
0
( ) ( ) ( )t
out inv t v h t dτ τ τ= −∫
( )inv t( )h t
We need 𝑣𝑣𝑖𝑖𝑖𝑖(𝜏𝜏)
We need ℎ(𝑡𝑡 − 𝜏𝜏)
Flip then shift by 𝑡𝑡
The area under the product!!
ORℎ(𝑡𝑡 − 𝜏𝜏) = ℎ(−(𝜏𝜏 − 𝑡𝑡))
Shift by 𝑡𝑡 then flip
Slide 6 of 14
Graphical Convolution
Lecture 8 EE 314 Signal and Linear System Analysis
• Start with 𝑡𝑡 < 0
0
( ) ( ) ( )t
y t x h t dτ τ τ= −∫
( ( ))h tτ− − ( )x τ𝑡𝑡 = −0.5 𝑠𝑠𝑠𝑠𝑠𝑠
0=
( ( 0.5))h τ− −
The area under the product?
0.5
0
( 0.5) ( ) ( )y x h t dτ τ τ−
− = −∫
Slide 7 of 14
Graphical Convolution
Lecture 8 EE 314 Signal and Linear System Analysis
• Now, consider 0 ≤ 𝑡𝑡 < 1
0
( ) ( ) ( )t
y t x h t dτ τ τ= −∫
( ( ))h tτ− − ( )x τ𝑡𝑡 = +0.6 𝑠𝑠𝑠𝑠𝑠𝑠
2( )
0
2t
te dτ τ− −= ∫
2( ) 2 ( )th t e u t−=
( ) ( ( ))x h tτ τ− −
Slide 8 of 14
Graphical Convolution
Lecture 8 EE 314 Signal and Linear System Analysis
• Now, consider 𝑡𝑡 > 1
0
( ) ( ) ( )t
y t x h t dτ τ τ= −∫
( ( ))h tτ− −
( )x τ
𝑡𝑡 = +1.5 𝑠𝑠𝑠𝑠𝑠𝑠
12( )
0
2 te dτ τ− −= ∫
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Graphical Convolution
Lecture 8 EE 314 Signal and Linear System Analysis
MATLAB code
Slide 10 of 14
Graphical Convolution
Lecture 8 EE 314 Signal and Linear System Analysis
• More Convolution Examples
MATLAB code
Slide 11 of 14
Graphical Convolution
Lecture 8 EE 314 Signal and Linear System Analysis
• Systems connected in series
• Systems connected in parallel
( )z t1( ) ( )* ( )z t x t h t=
2( ) ( )* ( )y t z t h t=
( )1 2( )* ( ) * ( )x t h t h t=
( )1 2( )** )( ()x t h t h t=
1( ) ( )* ( ) ( )* ( )Ny t x t h t x t h t= + +
[ ]1( )* ( ) ( )Nh t tx t h+ +=
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Causal LTI System
Lecture 8 EE 314 Signal and Linear System Analysis
• Causality A LTI system is causal if it does NOT rely on future inputs in
order to determine the current output.o All real/physical systems are causal – They can not anticipate
future inputs!!
o i.e., A causal system has an impulse response that is a causal function.
( ) 0 for all 0h t t⇒ = <
( ) ( ) ( )y t x h t dτ τ τ∞
−∞
= −∫ Consider 𝜏𝜏 > 𝑡𝑡 ‼
Would use future values of 𝑥𝑥 𝑡𝑡to determine 𝑦𝑦(𝑡𝑡)!!
Does this system have memory?
Give an example of a system that does NOT have memory?
( )h t
0
( ) ( )t
x h t dτ τ τ= −∫
Slide 13 of 14
Next Lecture
Lecture 8 EE 314 Signal and Linear System Analysis
• LTI Sinusoidal Response
• Reading Assignment: Chap. 2.7
Slide 14 of 14