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### EECE 301 Note Set 11 CT Convolution

Course: EECE 301, Fall 2011
School: Binghamton
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Word Count: 1588

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301 Signals EECE &amp; Systems Prof. Mark Fowler Note Set #11 C-T Systems: Computing Convolution Reading Assignment: Section 2.6 of Kamen and Heck 1/20 Course Flow Diagram The arrows here show conceptual flow between ideas. Note the parallel structure between the pink blocks (C-T Freq. Analysis) and the blue blocks (D-T Freq. Analysis). New Signal Models Ch. 1 Intro C-T Signal Model Functions on Real Line...

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301 Signals EECE & Systems Prof. Mark Fowler Note Set #11 C-T Systems: Computing Convolution Reading Assignment: Section 2.6 of Kamen and Heck 1/20 Course Flow Diagram The arrows here show conceptual flow between ideas. Note the parallel structure between the pink blocks (C-T Freq. Analysis) and the blue blocks (D-T Freq. Analysis). New Signal Models Ch. 1 Intro C-T Signal Model Functions on Real Line System Properties LTI Causal Etc D-T Signal Model Functions on Integers New Signal Model Powerful Analysis Tool Ch. 3: CT Fourier Signal Models Ch. 5: CT Fourier System Models Ch. 6 & 8: Laplace Models for CT Signals & Systems Fourier Series Periodic Signals Fourier Transform (CTFT) Non-Periodic Signals Frequency Response Based on Fourier Transform Transfer Function New System Model New System Model Ch. 2 Diff Eqs C-T System Model Differential Equations D-T Signal Model Difference Equations Ch. 2 Convolution Zero-State Response C-T System Model Convolution Integral Zero-Input Response Characteristic Eq. D-T System Model Convolution Sum Ch. 4: DT Fourier Signal Models DTFT (for Hand Analysis) DFT & FFT (for Computer Analysis) Ch. 5: DT Fourier System Models Freq. Response for DT Based on DTFT New System Model New System Model Ch. 7: Z Trans. Models for DT Signals & Systems Transfer Function New System Model2/20 C-T convolution properties Many of these are the same as for DT convolution. We only discuss the new ones here. 1. Derivative Property: See the next slide for the others d [ x (t ) v (t )] = x (t ) v (t ) dt = x (t ) v (t ) derivative 2. Integration Property Let y(t) = x(t)*h(t), then t x( )d h(t ) = x(t ) t h( )d y( )d = t The properties of convolution help perform analysis and design tasks that involve convolution. For example, the associative property says that (in theory) we can interchange to order of two linear systems in practice, before we can switch the order we need to check what impact that might have on the physical interface conditions. 3/20 Convolution Properties These are things you can exploit to make it easier to solve convolution problems 1.Commutativity x (t ) h (t ) = h (t ) x (t ) You can choose which signal to flip 2. Associativity x (t ) (v (t ) w(t )) = ( x (t ) v (t )) w(t ) Can change order sometimes one order is easier than another 3. Distributivity OR x (t ) ( h1 (t ) + h2 (t )) = x (t ) h1 (t ) + x (t ) h2 (t ) may be easier to split complicated system h[n] into sum of simple ones we can split complicated input into sum of simple ones (nothing more than linearity) 4. Convolution with impulses x (t ) (t ) = x (t ) 4/20 Computing CT Convolution -For D-T systems, convolution is something we do for analysis and for implementation (either via H/W or S/W). -For C-T systems, we do convolution for analysis nature does convolution for implementation. If we are analyzing a given system (e.g., a circuit) we may need to compute a convolution to determine how it behaves in response to various different input signals If we are designing a system (e.g., a circuit) we may need to be able to visualize how convolution works in order to choose the correct type of system impulse response to make the system work the way we want it to. Well learn how to perform Graphical Convolution, which is nothing more than steps that help you use graphical insight to evaluate the convolution integral. 5/20 Steps for Graphical Convolution x(t)*h(t) y (t ) = x ( )h(t )d 1. Re-Write the signals as functions of : x() and h() 2. Flip just one of the signals around t = 0 to get either x(-) or h(-) a. It is usually best to flip the signal with shorter duration b. For notational purposes here: well flip h() to get h(-) 3. Find Edges of the flipped signal a. Find the left-hand-edge -value of h(-): call it L,0 b. Find the right-hand-edge -value of h(-): call it R,0 4. Shift h(-) by an arbitrary value of t to get h(t - ) and get its edges a. Find the left-hand-edge -value of h(t - ) as a function of t: call it L,t Important: It will always be L,t = t + L,0 b. Note: I use for what the book uses ... It is not a big deal as they are just dummy variables!!! Find the right-hand-edge -value of h(t - ) as a function of t: call it R,t Important: It will always be R,t = t + R,0 Note: If the signal you flipped is NOT finite duration, one or both of L,t and R,t will be infinite (L,t = and/or R,t= ) 6/20 Steps Continued 5. Find Regions of -Overlap a. What you are trying to do here is find intervals of t over which the product x() h(t - ) has a single mathematical form in terms b. of In each region find: Interval of t that makes the identified overlap happen c. Working examples is the best way to learn how this is done Tips: 6. Regions should be contiguous with no gaps!!! Dont worry about < vs. etc. For Each Region: Form the Product x() h(t - ) and Integrate a. Form product x() h(t - ) b. Find the Limits of Integration by finding the interval of over which the product is nonzero i. Found by seeing where the edges of x() and h(t - ) lie ii. Recall that the edges of h(t - ) are L,t and R,t , which often depend on the value of t So the limits of integration may depend on t c. Integrate the product x() h(t - ) over the limits found in 6b i. The result is generally a function of t, but is only valid for the interval of t found for the current region ii. Think of the result as a time-section of the output y(t) 7/20 Steps Continued 7. Assemble the output from the output time-sections for all the regions a. Note: you do NOT add the sections together b. You define the output piecewise c. Finally, if possible, look for a way to write the output in a simpler form 8/20 Example: Graphically Convolve Two Signals y (t ) = h() x(t )d = x()h(t )d Convolve these two signals: By Properties of By Properties of Convolution Convolution these two forms are these two forms are Equal Equal This is why we can This is why we can flip either signal flip either signal x(t) 2 2 t h(t) 3 1 t 9/20 Step #1: Write as Function of x() 2 2 h() 3 1 Step #2: Flip h() to get h(-) x() 2 0 2 h(-) 3 1 0 Usually Easier to Flip the Shorter Signal 10/20 Step #3: Find Edges of Flipped Signal x() 2 0 2 h(-) 3 1 L,0 = 1 0 R,0 = 0 11/20 Motivating Step #4: Shift by t to get h(t-) & Its Edges Just looking at 2 arbitrary t values In Each Case We Get L,t = t + L,0 R,t = t + R,0 For tt= 2 For = 2 For tt= -2 For = -2 h(t-) =h(2-) 3 h(t-) =h(-2-) 3 3 2 1 2 L,t = t + L,0 R,t = t + R,0 L,t = t + L,0 R,t = t + R,0 L,t = t 1 R,t = t + 0 L,t = t 1 R,t = t + 0 L,-2 = -2 1 R ,-2 = 2+0 L,2 = 2 1 R,2 = 2+0 12/20 Doing Step #4: Shift by t to get h(t-) & Its Edges For Arbitrary Shift by tt For Arbitrary Shift by h(t ) 3 t1 t L,t = t + L,0 R,t = t + R,0 L,t = t 1 R,t = t + 0 13/20 Step #5: Find Regions of -Overlap 2 x() 2 Region II Region h(t-) No -Overlap No -Overlap tt< 0 <0 3 L,t R,t= t = t -1 Want R,t< 0 2 t<0 x() 2 h(t-) Partial -Overlap Partial -Overlap 3 t -1 Want L,t 0 t-1 0 t1 t Region II Region II 0 tt 1 0 1 Want R,t 0 t0 14/20 Step #5 (Continued): Find Regions of -Overlap x() 2 Region III Region III 2 h(t-) Total -Overlap Total -Overlap 3 1 < tt 2 1< 2 t -1 Want L,t > 0 t-1 > 0 t Want R,t 2 t>1 t2 x() 2 Region IV Region IV 2 h(t-) Partial -Overlap Partial -Overlap 3 2 < tt 3 2< 3 t -1 Want L,t 2 t-1 2 t3 t Want R,t > 2 t>2 15/20 Step #5 (Continued): Find Regions of -Overlap 2 x() 2 h(t-) No -Overlap No -Overlap tt> 3 >3 3 t -1 Want L,t > 2 Region V Region V t t-1 > 2 t>3 16/20 Step #6: Form Product & Integrate For Each Region 2 h(t-) 3 t -1 2 Region I: tt< 0 Region I: < 0 y (t ) = t x()h(t )d = h(t-) x() = 0 x() 0 d = 0 y (t ) = 0 for all t < 0 Region II: 0 tt 1 Region II: 0 1 2 h(t-) 3 t -1 With 0 integrand the limits dont matter!!! 2 0 y (t ) = t x( )h(t )d t t = 6d = [6 ]0 = 6t 6 0 = 6t h(t-) x() 6 0 y (t ) = 6t for 0 t 1 0t 17/20 Step #6 (Continued): Form Product & Integrate For Each Region x() h(t-) Region III: 1 < tt 2 Region III: 1 < 2 2 2 3 y (t ) = t -1 t x( )h(t )d h(t-) x() t = 6 6d = [6 ]t 1 = 6t 6(t 1) = 6 t t 1 t -1 x() h(t-) t y (t ) = 6 for all t such that : 1 < t 2 Region IV: 2 < tt 3 Region IV: 2 < 3 2 2 3 y (t ) = t -1 t x( )h(t )d 2 h(t-) x() = 6 6d = [6 ]t 1 = 6 2 6(t 1) = 6t + 18 2 t 1 y (t ) = 6t + 18 for 2 < t 3 t -1 2 18/20 Step #6 (Continued): Form Product & Integrate For Each Region x() 2 3 Region V: tt> 3 Region V: > 3 h(t-) 2 t -1 t y (t ) = x()h(t )d = h(t-) x() = 0 0 d = 0 y (t ) = 0 for all t > 3 19/20 Step #7: Assemble Output Signal Region II Region tt< 0 <0 Region II Region II 0 tt 1 0 1 Region III Region III 1 < tt 2 1< 2 Region IV Region IV 2 < tt 3 2< 3 Region V Region V tt> 3 >3 y (t ) = 0 y (t ) = 6t y (t ) = 6 y (t ) = 6t + 18 y (t ) = 0 y (t ) 6 0 1 2 3 t 20/20
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