EE120 Fall 2016 HOMEWORK 3 Solutions. Problem 1 Solution. 1.a. GSI: Phillip Sandborn

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1 EE Fall 6 HOMEWORK 3 Solutions GSI: Phillip Sandborn Problem Solution For each solution, draw xx(ττ) and flip around ττ =, then slide the result across h(ττ)..a.

2 .b Use the same x as part a.

3 .c

4 .d Analytical convolution gives: yy(tt) = xx(tt ττ)h(ττ)dddd = exp (tt ττ) uu(tt ττ) exp( ττ) uu(ττ )dddd The integrand is zero unless tt ττ > ; and ττ > ; which require the integration variable, ττ, to be ττ < tt and ττ >. tt yy(tt) = exp (tt ττ) exp( ττ) dddd uu(tt ) = exp( tt) exp( ττ) dddd uu(tt ) yy(tt) = [exp( tt) exp( tt)] uu(tt ) This gives the sketch above. We include the u(t-) to account for the zero overlap of x and h before t =. tt

5 Problem Solution Π(tt) = uu tt + uu tt cccccccc(tt) = δδ(tt nn) nn= Part a) Π tt = uu tt + uu tt : This is a box of height with width, centered at zero cccccccc tt = 4 nn= δδ( (tt 4nn)) = 4 δδ(tt 4nn) 4 nn= : This is a delta train with spacing 4, and delta height = 4. xx (t) = Π tt cccccccc tt 4 = Π ττ cccccccc tt ττ dddd 4 Using flip and drag method for convolution, xx (t) will be a periodic function with a repeating box of width, height 4, fundamental period TT = 4, and fundamental frequency ωω = : To find aa kk, we use the analysis equation. aa kk = 4 aa kk = 4 aa kk = aa kk = 4 sin kkkk xx (tt) exp jjjj tt dddd 4 exp jjjj tt dddd exp jjjj tt dddd = 4 exp jjjj exp jjjj kkkk jj For kk =, we solve the integral: aa = 4 aa = dddd = 4dddd

6 Part b) Π tt = uu 6 tt + uu 6 tt : This is a box of height with width 6, centered at zero 6 cccccccc tt = 4 nn= δδ( (tt 4nn)) = 4 δδ(tt 4nn) 4 nn= : This is a delta train with spacing 4, and delta height = 4. xx (t) = Π tt 6 cccccccc tt 4 = Π ττ 6 cccccccc tt ττ dddd 4 Using flip and drag method for convolution, xx (t) will be a square wave with a low level of 4 and high level of 8, fundamental period TT = 4, and fundamental frequency ωω = : From this flip-and-slide result, we can see that the function xx (tt) can be written: xx (tt) = 8 4Π tt To find aa kk, we use the analysis equation. aa kk = 4 xx (tt) exp jjjj tt dddd aa kk = 4 aa kk = aa kk = 4 sin(kkkk) kk 8 4Π tt exp jjjj tt dddd exp jjjj tt dddd 4 sin kkkk kkkk aa kk = 4 sin kkkk ; for kk. kkkk For kk =, we can solve the integral: aa = 4 aa = 6 8 4Π tt dddd Π tt exp jjjj tt dddd = Π tt dddd = 8 Π tt dddd = 8 dddd = 6 Using L Hopital s rule on the final expression for aa kk = 4 sin kkkk have to use the previous expression to use L Hopital properly: aa kk = 4 sin(kkkk) kkkk does not give the correct answer. We kk 4 sin kkkk kkkk. This result should also give aa = 6. Even though the boxed answer is correct for non-zero k, it is evident that evaluating at k = will give: / /. This should be a cue that we will need to apply L Hopital s rule on both terms!!

7 Part c) xx 3 (t) = Π tt Π(tt ) cccccccc tt 4 = Π(tt ) xx (tt) xx (tt) is the same as in part a. Π(tt ) is a box function of width =, height =, centered at t =. We know from Part a) that the Fourier coefficients of xx (tt) are:aa kk = 4 sin kkkk If we treat Π(tt ) as an impulse response of an LTI system, h(tt) = Π(tt ), the output of this system for an arbitrary input xx(t) can be written as: yy(t) = xx(tt ττ)h(ττ)dddd For xx(tt) = exp(jjjjjj), yy(t) = yy(t) = exp(jjjjjj) yy(t) = sin(ωω/) ωω/ 3/ / eeeeee( jjjjjj) dddd exp(jjjj) exp(jjjjjj) So, {exp(jjjjtt)} = sin(ωω/) ωω/ xx (tt) = kk= aa kk exp jjjj tt Write output y using xx (tt) eeeeee jjjj(tt ττ) Π(ττ )dddd = exp(jjjjjj) eeeeee jjjj 3 eeeeee jjjj jjjj exp(jjjj). Now, we use our synthesis equation for xx (tt): yy(t) = HH{xx (tt)} = HH aa kk exp jjjj kk= tt = aa kk HH exp jjjj kk= tt Plug in kk for ωω in the transfer function: HH exp jjjj tt = sin kk / kk / exp jjkk exp jjkk tt yy(t) = exp jjkk kk= tt = aa exp jjkk kk= kk exp jjkk tt sin kk aa / kk kk / So, the new aa kk = aa kk sin kk / kk / exp jjkk aa = aa eeeeeeee = 6 sin kk = sin kk 4 kk exp jjkk

8 Problem 3 Solution Part A: Fourier coefficients xx (tt) = cos(ωω tt) xx (tt) = cos(ωω tt) Find Fourier coefficients of x : xx (tt) is periodic with fundamental period, TT = ωω = ωω aa kk = ωω aa kk = ωω aa kk = ωω ωω cos(ωω tt) exp( jjjjωω tt) dddd ωω exp(jjωω tt)+exp( jjωω tt) ωω Solve for k = + and -, aa = ωω ωω exp( jjjjωω tt) dddd [exp(jjωω ( kk)tt) + exp( jjωω ( + kk)tt)]dddd [ + exp( jjωω 4tt)]dddd aa = ωω tt + exp( jjωω 4tt) ωω jjωω 4 aa = ωω ωω = Substituting k = - leads to the same result, aa = Solve for other k, aa kk = ωω aa kk = exp(jjωω ( kk)tt) jjωω ( kk) 4 exp(jj( kk)) ( kk) aa kk = for kk ±. + exp( jjωω (+kk)tt) ωω jjωω (+kk) + exp( jj(+kk)) (+kk) Since the phase of the complex exponentials is always a positive or negative integer multiple of, the sums always evaluate to for all k except k = + or -.

9 Next, find Fourier coefficients for x : xx (tt) is periodic with fundamental period, TT = ωω aa kk = ωω aa kk = ωω aa kk = ωω aa kk = ωω aa kk = ωω ωω cos(ωω tt) exp( jjjjωω tt) dddd ωω exp(jjωω tt)+exp( jjωω tt) ωω ωω ωω exp( jjjjωω tt) dddd [exp(jjωω ( kk)tt) + exp( jjωω ( + kk)tt)]dddd exp(jjωω ( kk)tt) jjωω ( kk) + exp( jjωω (+kk)tt) jjωω (+kk) jj exp jj( kk) exp jj( kk) ( kk) aa kk = sin ( kk) ( kk) aa kk = ( ) kk 4kk + sin (+kk) (+kk) ωω ωω + exp jj(+kk) exp jj(+kk) (+kk) aa =

10 Part B: Time averaged power We will use PP aaaaaa,xx = TT <TT> xx(tt) dddd Find time averaged power for x (t) xx (tt) = cos(ωω tt) ; TT = ωω PP aaaaaa,xx = TT PP aaaaaa,xx = ωω PP aaaaaa,xx = ωω <TT> ωω xx (tt) dddd cos(ωω tt) dddd ωω +cos( ωω tt) PP aaaaaa,xx = ωω tt + sin(4ωω tt) ωω 8ωω PP aaaaaa,xx = ωω = ωω dddd Find time averaged power for x (t) xx (tt) = cos(ωω tt) ; TT = ωω PP aaaaaa,xx = TT PP aaaaaa,xx = ωω PP aaaaaa,xx = ωω <TT> ωω xx (tt) dddd cos(ωω tt) dddd ωω +cos( ωω tt) dddd PP aaaaaa,xx = ωω tt + sin(ωω tt) ωω 4ωω PP aaaaaa,xx = ωω = ωω = ωω ωω = ωω ωω cos (ωω tt) dddd cos (ωω tt) dddd

11 Part C: Power at Desired Frequency The desired frequency is ωω. The k th Fourier coefficient for xx corresponds to a frequency of TT. T is the fundamental period of xx, TT = ωω. TT = ωω kk = TT ωω = Sum of square of Fourier coefficients k = + and k = -: aa = 3 = 3 aa = 3 + = 3 aa + aa = By Parseval s Theorem for continuous time periodic signals (OW section 3.5.7, p.5), we know the sum of squares of the fourier coefficients must be equal to the average power in the periodic signal. The average power is shown in Part b: PP aaaaaa,xx = Neglecting aa = means subtracting aa from PP aaaaaa,xx PP aaaaaa,xx aa = Therefore, the fraction of time averaged power at ωω (neglecting the DC term) is approximately aa + aa.947 =.9 =.95 = 95%.947

12 Part D: LTI Filter with x(t) If we have a low pass filter, HH(jjjj) = +jj ωω ωω And we know that xx (tt) is expressed by its Fourier coefficients in the form, xx (tt) = kk= aa kk exp(jjjjωω tt) Then the output of the system, yy(tt), is given by yy(tt) = kk= aa kk HH(jjjjωω ) exp(jjjjωω tt) The Fourier series coefficients of yy(tt) are bb kk : bb kk = aa kk HH(jjjjωω ) = aa kk = ( ) kk +jjjj 4kk +jjjj bb kk = 4 (+kk ) ( 4kk ) The DC term is bb =. The Fourier coefficients of yy(tt) for k = + and - are: bb = 3 +jj bb = 3 jj bb = 9 bb = 9 The fraction of power in the k = + and k = - components is given (where the total power minus DC is given by summing all b k s by computer calculation, no simple closed form solution): bb + bb = 98% Truncating the series at k = +6 and -6 gives a good estimate, since the magnitude of b k is proportional to /k 4. By using a filter, we get a signal where more of the total power is concentrated at the desired frequency.

13 Problem 4 Solution Part A: xx[nn] = δδ[nn + ] + δδ[nn] + δδ[nn ], N = 6 Using the analysis equation for discrete time signals: aa kk NN <NN> xx[nn] exp jjjj NN nn Select a period from n = -7 to n = +8 aa kk = 6 8 nn= 7 δδ[nn + ] + δδ[nn] + δδ[nn ] exp jjjj nn 6 Since x is non-zero only when n = -,, or +, the sum can be written: aa kk = 6 aa kk = 6 + exp jjjj ( ) + exp jjjj () + exp jjjj () aa kk = 6 + cos kk 6 exp jjjj + exp jjjj 6 6

14 Part B: xx[nn] = [,,,,,,,], N = 8 Using the analysis equation for discrete time signals: aa kk NN <NN> xx[nn] exp jjjj NN nn Select a period from n = to n = +7 aa kk = 8 7 nn= 3 (δδ[nn] + δδ[nn ] + δδ[nn ] + δδ[nn 3]) exp jjjj 8 nn aa kk = exp jjjj 8 nn= nn = 4 8 WWnn aa kk = 8 exp( jjjjjj) exp jjjjjj 4 = 8 aa kk = 8 sin kkkk sin kkkk 8 exp jjjj nn= = 8 WW4 WW kkkk sin sin kkkk 8 Since x is non-zero only when n =,,, or 3, the sum could also be written: aa kk = 8 exp jjjj 8 () + exp jjjj 8 () + exp jjjj 8 () + exp jjjj 8 (3) For k =, the sum simplifies: aa = 8 [ ] =

EE 341 A Few Review Problems Fall 2018

EE 341 A Few Review Problems Fall 2018 1. A linear, time-invariant system with input, xx(tt), and output, yy(tt), has the unit-step response, cc(tt), shown below. a. Find, and accurately sketch, the system