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Lecture #11 EGR 261 – Signals and Systems

Lecture #11 EGR 261 – Signals and Systems. Read : Ch. 2, Sect. 1-8 in Linear Signals & Systems, 2 nd Ed. by Lathi Ch. 13, Sect. 6 in Electric Circuits, 9 th Ed . by Nilsson. Convolution In this class we will cover the following topics: Convolution – graphical approach

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Lecture #11 EGR 261 – Signals and Systems

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  1. Lecture #11 EGR 261 – Signals and Systems Read: Ch. 2, Sect. 1-8 in Linear Signals & Systems, 2nd Ed. by Lathi Ch. 13, Sect. 6 in Electric Circuits, 9th Ed. by Nilsson • Convolution • In this class we will cover the following topics: • Convolution – graphical approach • Convolution – direct integration approach • Convolution – using Laplace transforms to bypass the convolution integral

  2. Lecture #11 EGR 261 – Signals and Systems Procedure for evaluating y(t) = x(t)*h(t) graphically (Note that x and h can be reversed in the steps below (commutative property), depending on which form of the convolution integral is used.) • Graph x() (with  on the horizontal-axis). • Invert h() to form h(-). Then shift h(-) along the  axis t seconds for form h(t - ). Note that as the time shift t varies, the waveform h(t - ) will “slide” across x() and in some cases the waveforms will overlap. • Determine the different ranges of t which result in unique overlapping portions of x() and h(t - ). For each range determine the area under the product of x() and h(to - ). This area is y(t) for the range. • 4. Compile the results of y(t) for each range and graph y(t). Which signal to invert and shift? Since x(t)*h(t) = h(t)*x(t), we can shift and delay either signal (commutative property). Hint: Shift and delay the simplest waveform

  3. y(t) = x(t)*h(t) 2 t 1 2 3 4 0 Lecture #11 EGR 261 – Signals and Systems Length of the response y(t) = x(t)*h(t) In general if the length of the input x(t) is T1 and the length of the impulse response h(t) is T2, then the length of y(t) = x(t)*h(t) will be T1 + T2. h(t) x(t) y(t) = x(t)*h(t) t t t T1 T2 T1 +T2 Example Recall the example worked in the last presentation where we determined y(t) = x(t)*h(t) as shown below. h(t) x(t) 2 1 t t 1 1 3 0 0 Length = 1 Length = 2 Length = 1+2 = 3

  4. Lecture #11 EGR 261 – Signals and Systems Example 1: Find y(t) = x(t)*h(t) using the graphical method. What should be the length of y(t)? h(t) x(t) 6 4 3 t t 2 6 5 4 1 0 0

  5. Lecture #11 EGR 261 – Signals and Systems Example 1: (continued)

  6. h(t) x(t) 2 2 t t 2 1 2 0 0 Lecture #11 EGR 261 – Signals and Systems Example 2: Find y(t) = x(t)*h(t) using the graphical method. Which waveform is easiest to shift and delay? What should be the length of y(t)?

  7. Lecture #11 EGR 261 – Signals and Systems Example 2: (continued)

  8. x(t) h(t) 1 t t -3 0 0 Lecture #11 EGR 261 – Signals and Systems Example 3 (problem 2.4-18d in Lathi text): Find y(t) = x(t)*h(t) using the graphical method. Which waveform is easiest to shift and delay?

  9. Lecture #11 EGR 261 – Signals and Systems Example 3: (continued)

  10. + 1  + x(t) = u(t+3)-u(t) y(t) 1 F _ _ Lecture #11 EGR 261 – Signals and Systems • Example 4: • Sketch x(t). • Find H(s) for the circuit shown below. Also find h(t) from H(s). • Find y(t) using Laplace transform techniques. • Look familiar? This is the same problem used in Example 3. Show that y(t) found using convolution is exactly the same as y(t) found using Laplace transforms.

  11. Lecture #11 EGR 261 – Signals and Systems Example 4: (continued)

  12. x(t) = (t)+2(t-1)+3(t-2) h(t) 2 t t 0 0 3 1 2 Lecture #11 EGR 261 – Signals and Systems Example 5: Find y(t) = x(t)*h(t) using the graphical method.

  13. Lecture #11 EGR 261 – Signals and Systems Example 5: (continued)

  14. u(t - ) = 1 only for  < t so change upper limit to t u() = 1 only for  > 0 so change lower limit to 0 A graphical approach would show that the functions overlap for 0 < t <  or else note that x(t) and y(t) both are non-zero for t > 0. so u(t)*u(t) = tu(t) Lecture #11 EGR 261 – Signals and Systems Convolution by direct integration Convolution can be performed by direct evaluation of the convolution integral (without sketching graphs) for fairly straightforward functions. This becomes difficult for piecewise-continuous functions, such as were used in many of our graphical examples. Example 6: Evaluate y(t) = x(t)*h(t) for: x(t) = u(t), h(t) = u(t) by directly evaluating the convolution integral rather than through a graphical approach.

  15. Lecture #11 EGR 261 – Signals and Systems Example 7: Evaluate y(t) = x(t)*h(t) for: x(t) = tu(t), h(t) = u(t) by directly evaluating the convolution integral rather than through a graphical approach. Example 8: Evaluate y(t) = x(t)*h(t) for: x(t) = e-tu(t), h(t) = u(t) by directly evaluating the convolution integral rather than through a graphical approach.

  16. Lecture #11 EGR 261 Signals and Systems Table 2-1 from Signals & Systems, 2nd Edition, by Lathi

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