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

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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




• 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−= − − −

Slide 3 of 14




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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( ) ( )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



• 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




• Start with 𝑡𝑡 < 0

0

( ) ( ) ( )t


( ( ))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




• Now, consider 0 ≤ 𝑡𝑡 < 1

0

( ) ( ) ( )t


( ( ))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




• Now, consider 𝑡𝑡 > 1

0

( ) ( ) ( )t


( ( ))h tτ− −

( )x τ

𝑡𝑡 = +1.5 𝑠𝑠𝑠𝑠𝑠𝑠

12( )

0

2 te dτ τ− −= ∫

Slide 9 of 14




MATLAB code

Slide 10 of 14


http://mercury.pr.erau.edu/%7Ebruders/teaching/EE314/MATLAB_code%5Clecture8



• More Convolution Examples

MATLAB code

Slide 11 of 14


http://mercury.pr.erau.edu/%7Ebruders/teaching/EE314/MATLAB_code%5Clecture8



• 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+ +=

Slide 12 of 14


Causal LTI System


• 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


• LTI Sinusoidal Response

• Reading Assignment: Chap. 2.7

Slide 14 of 14


ee 314 signal and linear system analysis

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