Signal Processing and Linear Systems-B.P.Lathi copy

# The loop equation for this circuit see example 111 or

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Unformatted text preview: t (Fig. 2.7c). Therefore, f (t) a nd f ( r) will have t he s ame graphical representations. S imilar r emarks a pply t o get) a nd g (r) (Fig. 2.7d). T he f unction g (t - r) is n ot as easy t o c omprehend. To u nderstand w hat t his function looks like, let us s tart w ith t he f unction g (r) (Fig. 2.7d). Time-inversion of t his f unction (refiection a bout t he v ertical axis r = 0) yields g( - r) (Fig. 2.7e). L et u s denote this function by <p(r) V' <p(r) = g (-r) E xercise E 2.B Now <p( r ) s hifted by t s econds is <p( r - t), given by Using the convolution table, determine <p(r - t) Answer: (~- e-' + ~e-2') u (t) V' 6 E xercise E 2.9 For an LTIC system with the unit impulse response h(t) = e - 2 'u(t), determine the zero-state response y et) if the input f (t) = sin 3 tu(t). Hint: Use the convolution table Pair 12 with suitable values for a Answer: {3 e, and >.. -h [ 3e-2'+v'i3cos(3t-146.32°)]u(t) 127 Multiple Inputs M ultiple i nputs t o L TI systems c an b e t reated b y a pplying t he s uperposition principle. E ach i nput is considered separately, w ith all o ther i nputs a ssumed t o b e zero. T he s um o f all t hese i ndividual system responses c onstitutes t he t otal s ystem o utput w hen all t he i nputs a re applied simultaneously. 2 .4-2 T he input is f (t) = l Oe- 3'u(t), and the response yet) is = 1 0e- 'u(t) S ystem R esponse t o E xternal I nput: T he Z ero-State Response D f(t) T he impulse response h (t) for this system, as obtained in Example 2.3, is yet) 2.4 Or -h [ 3e-2t-v'i3cos(3t+33.68°)]u(t) V' = g [-(r - t)l = get - r) T herefore, we first time-invert g (r) t o o btain g (-r) a nd t hen t ime-shift g (-r) b y t t o o btain get - r). F or positive t, t he s hift is t o t he r ight (Fig. 2.7f); for negative t, t he s hift is t o t he left (Fig. 2.7g). T he p receding discussion gives us a graphical i nterpretation o f t he f unctions f ( r) a nd 9 (t - r). T he convolution c( t) is t he a rea u nder t he p roduct o f t hese two functions. Thus, t o c ompute c(t) a t some positive i nstant t = t1, we first o btain g( - r) by inverting g (r) a bout t he v ertical axis. Next, we right-shift or delay g( - r) by t1 t o o btain 9(t1 - r) (Fig. 2.7f), a nd t hen we multiply t his f unction by f er), giving us t he p roduct f (r)g(t1 - r) (Fig. 2.7f). T he a rea A l u nder t his p roduct 128 2 Time-Domain Analysis of Continuous-Time Systems -1 1- 1- f"'I' -2 0 2.4 System Response t o E xternal Input: T he Zero-State Response 129 is C (tl), t he value of crt) a t t = t l' We c an therefore plot C (tl) = A l on a curve describing crt), as shown in Fig. 2.7i. Observe t hat t he a rea under the product f (r)g( - r) in Fig. 2.7e is c(O), t he value of t he convolution for t = 0 ( at t he origin). A similar procedure is followed in computing t he value of crt) a t t = t2, where t2 is negative (Fig. 2.7g). In this case, t he function g( - r) is shifted by a negative amount ( that is, left-shifted) t o o btain 9(t2 - r). Multiplication of this function with f (r) yields the product f (r)g(t2 -...
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## This note was uploaded on 04/14/2013 for the course ENG 350 taught by Professor Bayliss during the Spring '13 term at Northwestern.

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