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Lecture #10 EGR 261 – Signals and Systems

Lecture #10 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. Representation and Analysis of Systems

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Lecture #10 EGR 261 – Signals and Systems

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  1. Lecture #10 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 • Representation and Analysis of Systems • The analysis of systems this course is limited to linear, time-invariant, • continuous-time (LTIC) systems. • LTIC systems can be characterized (described) in three ways: • 1) Differential equations – a differential equation (DE) for y(t) in terms of x(t) – found using standard D.E. methods (review of Ch. 7-8 in Nilsson text). The DE can be used to find the response to any input, x(t). Special cases include the unit step response (USR) and the impulse response, h(t). • 2) Impulse response – the system output, y(t), can be determined using the impulse response, h(t), through evaluation of the convolution integral • Transfer function – The transfer function, H(s) can be determined through Laplace transform analysis. H(s) can be used to find the USR, h(t), or the output y(t) for any input x(t).

  2. + 4/3 H + x(t) y(t) ¼ F 1  _ _ Lecture #10 EGR 261 – Signals and Systems Differential Equations System Representation As we discussed the classification of systems, it was stated that a linear system can be represented by a linear differential equation of the form: Example: Determine the differential equation for the output y(t) for any general input forcing function x(t). Note that this differential equation completely characterizes the system.

  3. Lecture #10 EGR 261 – Signals and Systems Unit Step Response (USR) and Impulse Response, h(t) Recall that we earlier found the unit step response, USR, and the impulse response, h(t), using s-domain techniques involving the transfer function, H(s). To summarize, we found that: We can also find h(t) and USR from the differential equation that characterizes a system. Determining the USR and h(t) from a Differential Equation We have just reviewed writing and solving differential equations. Recall that if the input, x(t), to the differential equation is a unit step function, then y(t) is called the unit step response (USR). So, USR = y(t) when x(t) = u(t) The impulse response can then be found using the following relationship: Impulse response = h(t) = d/dt[USR(t)] (t) = d/dt[u(t)] Also recall that

  4. Lecture #10 EGR 261 – Signals and Systems Example: The USR for a system is y(t) = USR = [2 + 3e-4t]u(t). Find the impulse response, h(t). Hint: Use the product rule. Answer: h(t) = 5(t) – 12e-4tu(t)

  5. Lecture #10 EGR 261 – Signals and Systems • Example: • Write the differential equation for y(t) that characterizes the system (circuit). • Find the unit step response, USR, by solving the differential equation for x(t) = u(t). • Find the impulse response, h(t), using the relationship h(t) = d/dt[USR(t)] + + 1  y(t) x(t) 1 F _ _

  6. Lecture #10 EGR 261 – Signals and Systems Example: (continued)

  7. Lecture #10 EGR 261 – Signals and Systems • Determining a transfer function from a differential equation • Taking the Laplace Transform of a differential equation will yield the transfer function if: • The differential equation uses a general input, x(t) • The initial conditions are zero Example: Find H(s) for the D.E. shown below. Invalid. Initial conditions are not zero. Invalid. Input is not general. Example: Find H(s) for the D.E. shown below.

  8. H(s) Input = X(s) Y(s) = Output s-domain representation of a circuit or system using a transfer function H(s) Lecture #10 EGR 261 – Signals and Systems The Impulse Response and the Convolution Integral Recall from our earlier study of systems using Laplace transforms that the transfer function, H(s), completely characterizes a system and that the output y(t) can be determined for any input x(t) using H(s).

  9. System Input = x(t) = (t) y(t) = h(t) = impulse response y(t) can be found for any x(t) using h(t) and the convolution integral h(t) y(t) x(t) Lecture #10 EGR 261 – Signals and Systems Just as H(s) completely characterizes a system and can be used to determine the output of the system for any input, the impulse response h(t) also completely characterizes the system. • s-domain approach: • The output of a system can be determined with H(s) and Laplace transform techniques. • time-domain approaches: • Differential equations (just covered) • Convolution. The output of a system can be determined with h(t) and the convolution integral.

  10. Lecture #10 EGR 261 – Signals and Systems • Why use the convolution integral? • It allows us to work directly in the time-domain. This is particularly useful with experimental data where using Laplace transforms might be difficult or impossible. • Insight can be gained as to how closely the output waveform replicates the input waveform. Development of the convolution integral The following development is presented in section 13.6 of Electric Circuits, by Nilsson. A LTIC system can be described by its impulse response, h(t), as shown below.

  11. Some general input x(t) is shown. x(t) can be expressed as a series of rectangular pulses. So, x(t) = x0(t) + x0(t) + … + xi(t) + … Each pulse could be expressed in terms of unit step functions (i.e., window functions) as: xi(t) = x(i)[u(t - i) - u(t – (i +  )] As the width of each pulse approaches zero, x(t) can be expressed as a series of impulse functions as follows: x(t) = x(0)  (t - 0) + x(1)  (t - 1) + … + x(i)  (t - i) Area under pulse Lecture #10 EGR 261 – Signals and Systems

  12. The output y(t) consists of a sum of uniformly delayed impulse responses where the strength of each responses depends on the strength of the impulse driving the circuit. y(t) = x(0)  h(t - 0) + x(1)  h(t - 1) + … + x(i)  h(t - i) or y(t) = x(i) h(t - i)   As   approaches zero, this summation approaches a continuous integration, or Recall that if x(t) = Kd(t), then y(t) = Kh(t) Lecture #10 EGR 261 – Signals and Systems This integral is referred to as the convolution integral and is generally expressed as:

  13. Lecture #10 EGR 261 – Signals and Systems Properties of the convolution integral Commutative property: Distributive property: Associative property: Shift property:

  14. Lecture #10 EGR 261 – Signals and Systems Why use graphical evaluation of the convolution integral? The convolution integral can be evaluated directly, but this can be very difficult (especially for piecewise-continuous functions), so a graphical approach is more commonly used. However, even the graphical approach is quite challenging. A detailed example will follow shortly to illustrate the procedure. • Convolution can be understood by examining the graphical interpretation of the convolution integral. The graphical approach is helpful for the following reasons: • It is helpful in evaluating the convolution of complicated signals • It allows us to grasp visually the convolution integral’s result • It allows us to perform convolution with signals that can only be described graphically. Convolution Integral:

  15. Lecture #10 EGR 261 – Signals and Systems • Procedure for evaluating the convolution integral graphically • Graph x() (with  on the horizontal-axis). (see note below) • 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. • Compile the results of y(t) for each range and graph y(t). Note: x(t) and h(t) can be reversed since x(t)*h(t) = h(t)*x(t), i.e., commutative property. In general, it is easiest to invert and delay the simplest function.

  16. h(t) x(t) 2 1 t t 1 1 3 0 0 x() 1  1 0 Lecture #10 EGR 261 – Signals and Systems Example: Find y(t) = x(t)*h(t) for x(t) and h(t) shown below using the graphical method. Follow the procedure listed on the previous slide. Solution: 1. Form x():

  17. Lecture #10 EGR 261 – Signals and Systems 2. Form h(t - ): h(-) h() 2 2 Note: Remember that  is the independent variable, not t. t is simply a constant. h(t - ) = h(-  + t), so t is the amount of time shift (to the left).   -3 -1 1 3 0 0 h(t-) t 2  t-3 t-1 -3 -1 0 3. Consider different ranges of t (time-shift): Each range of t selected should result in different portions of the waveforms overlapping. The convolution y(t) = x(t)*h(t) is the area under the product of the overlapping portions.

  18. h(t-) x() h(t-) sliding as t varies over the range - < t < 1 2 1  t-3 t-1 1 0 No overlap Lecture #10 EGR 261 – Signals and Systems 3A. First range selected: (- <t < 1): This range results in no overlap for all values of  (i.e., - <  < ).

  19. h(t-) x() h(t-) sliding as t varies over the range 1 < t < 2 2 1  t-3 t-1 1 0 Area of overlap between  = 0 and  = t-1 Lecture #10 EGR 261 – Signals and Systems 3B. Second range selected: (1 <t < 2): This range results in an overlap from  = 0 to  = t-1.

  20. h(t-) x() h(t-) sliding as t varies over the range 2 < t < 3 2 1  t-1 t-3 1 0 Area of overlap between  = 0 and  = 1 Lecture #10 EGR 261 – Signals and Systems 3C. Third range selected: (2 <t < 3): This range results in an overlap from  = 0 to  = 1.

  21. h(t-) x() h(t-) sliding as t varies over the range 3 < t < 4 2 1  t-3 t-1 1 0 Area of overlap between  = t-3 and  = 1 Lecture #10 EGR 261 – Signals and Systems 3D. Fourth range selected: (3 <t < 4): This range results in an overlap from  = t-3 to  = 1.

  22. h(t-) sliding as t varies over the range t > 4 x() h(t-) 2 1  t-3 t-1 1 0 No overlap Lecture #10 EGR 261 – Signals and Systems 3E. Fifth range selected: (4 <t): This range results in no overlap for all values of  (i.e., - <  < ).

  23. y(t) 2 t 1 2 3 4 0 Lecture #10 EGR 261 – Signals and Systems 5. Compile the results of y(t) over the five ranges. Also graph y(t).

  24. Lecture #10 EGR 261 – Signals and Systems Example: Find y(t) = x(t)*h(t) for x(t) and h(t) shown below using the graphical method. h(t) x(t) 20 10 t t -1 3 6 0 0 4

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