( 1) EE 313 Linear Signals & Systems (Fall 2018) Solution Set for Homework #10 on Laplace Transforms

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1 EE 33 Linear Signal & Sytem (Fall 08) Solution Set for Homework #0 on Laplace Tranform By: Mr. Houhang Salimian & Prof. Brian L. Evan Problem. a) xt () = ut () ut ( ) From lecture Lut { ()} = and { } t L x( t t ) d d = e X( ) e e = =, for all Thi i a finite-amplitude finite-duration ignal, and hence the region of convergence i the entire plane. At firt glance, it would eem that the region of convergence hould have been Re{} > 0 becaue the ignal i caual and the denominator goe to zero when goe to zero. However, when goe to zero, the numerator alo goe to zero. We can ue L Hôpital rule by letting go to zero. The derivative of the numerator with repect to i exp(-) and the derivative of the denominator with repect to i. The limit of a goe to zero i. b) c) xt e ut 3t () = 3 ( ) = and region of convergence i Re{} > -Re{a}. + a 3t 3( t + ) 6 3( t ) xt () = 3 e ut ( ) = 3 e ut ( ) = 3 e e ut ( ) at From L{ e u() t} X () = 3e 6 e + 3 = 3e e And the region of convergence i Re{} > -3. Writing the exp(-) eparately from the ret of the expreion can help highlight the term, which correpond in the time domain to a delay by econd. xt e ut 3( t ) () = 3 ( ) Thi part i imilar to part b. X () = 3e + 3 = e The region of convergence i Re{} > -3. d) ( π ) xt () = 5in ( t ) ut ( ) 0 { in t u( t) } = From L ( ω ) ω 0 + ω0 X () = 5e π + π = 5π + π e And the region of convergence i Re{} > 0. and region of convergence i Re{} > 0.

2 Fall 08 EE 33 Homework 0 olution The Univerity of Texa at Autin Problem. a) ( π ) xt () = co 0 t ut () { co t u( t) } From L ( ω ) = = + = 0 + ω0 ( 0π ) + 400π The region of convergence i Re{} > 0. for Re{} >0. MATLAB code t = -:/0000:; unittep = zero(ize(t)); unittep (t>= 0) = ; x = co(0*pi*t).*unittep; plot(t,x) xlabel('time()') ylabel('x') Zero are root of nominator and pole are root of denominator. zero : = 0, pole : + ( 0π) = 0 =± j0π, In the figure legend, Re mean Re{}.

3 Fall 08 EE 33 Homework 0 olution The Univerity of Texa at Autin b) xt e ut 8t () = () at From L{ e u() t} = for Re{} > -Re{a}. + a = + 8 The region of convergence i Re{} > -8. t = -:/0000:; unittep = zero(ize(t)); unittep (t>= 0) = ; x = exp(-8*t).*unittep; plot(t,x) xlabel('time()') ylabel('x') pole : + 8= 0 = 8 In the figure legend, Re mean Re{}.

4 Fall 08 EE 33 Homework 0 olution The Univerity of Texa at Autin c) xt e ut ut e ut 8 8 () ( t t = ) () = () () = for Re{} > -Re{a}. + a 8 = = at From L{ e u() t} ( ) The region of convergence i Re{} > 0. t = -:/0000:; unittep = zero(ize(t)); unittep (t>= 0) = ; x = (-exp(-8*t)).*unittep; plot(t,x) xlabel('time()') ylabel('x') pole : ( 8) 0 0, 8 + = = = In the figure legend, Re mean Re{}. Problem 3. Uing the property: L " x t = L x t for zero initial condition, we get: a) Y + Y = X => Y + = X => H = = Becaue the ytem i caual, the region of convergence i Re{} > -.

5 Fall 08 EE 33 Homework 0 olution The Univerity of Texa at Autin b) Uing the Laplace tranform pair L e " u t = all, we obtain + H () = = = t ht () = δ () t e ut () for Re{} > -Re{a}, L δ(t) = for c) H jω = " by ubtituting = jω into H() above. Thi ubtitution i valid becaue " the imaginary axi lie within the region of convergence of Re{} > -. d) w = -0:/0000:0; H= j*w./(j*w+); Hmag=ab(H) ; Hphae=angle(H); plot(w,hmag) title('magnitude Repone'); figure plot(w,hphae) title('phae Repone'); According to magnitude repone, the filter notche out zero frequency. Hence, it i a notch filter. It could alo be called a highpa filter, but a DC notch filter would be more decriptive and a better anwer. Here the plot of the magnitude and phae uing the freq command in Matlab, which will plot the frequency repone on a log cale in frequency. The magnitude will alo be on a log cale.

6 Fall 08 EE 33 Homework 0 olution The Univerity of Texa at Autin freq( [ 0], [ ] ); We ee a highpa repone over the frequencie plotted. Pleae note that freq( [], [ ] ) would mean intead of. e) x t = u t => X = u t e " dt = e " dt = have alo obtained the tranform by uing L e " u t = f) Y () = HX () () = + Uing the Laplace tranform pair L e " u t = y t = e u t, for we get for the tranfer function for Re{} > 0. We could and ubtituting a = 0. Problem 4. a) Uing the Laplace tranform pair L e " u t =, we get h t = e u t b) t { } X () = L 7 e u() t =, for Re{} > + 7 Y () = H()X () = From { } r+ 7,for Re{}> ( + ) r at r Lte ut () =,forre{} > Re{} a ( a+ ) y(t) = 7 t e t u(t) c) Uing convolution:

7 Fall 08 EE 33 Homework 0 olution The Univerity of Texa at Autin d) τ ( t τ) yt () = ht ()* ht () = h( ) ht ( ) d = e u( ) e ut ( ) d t τ τ τ τ τ τ 0, t < 0 t τ ( t τ) t t () τ τ () τ τ t t τ 0 = e u e d = e u d = e d = te, t 0 h(t)*h(t) = t e t u(t) For a dicrete-time verion of thi problem, pleae ee Handout F Convolution of Exponential Sequence at ion%0exp%0sequence.pdf r at r L t e u() t =, for Re{} > Re{} a ( a+ ) From { } r+ t yt () = ht ()* ht () Y() = L{ te ut ()} =, forre{} > ( + ) Pole are the root of the denominator: pole :( + ) = 0, = Thu, Y() ha a double pole at = -.