ECE 45 Homework 3 Solutions

Transcription

1 UC San Diego J. Connelly ECE 45 Homework 3 Solutions Problem 3. Calculate the Fourier transform of the function (t) { t t / otherwise. Use the statement of Problem 3. to verify your answer. Note: the function (t) is sometimes called the unit triangle function, as it a triangular pulse with height, width, and is centered at. Hint: Recall the trig identity cos(x) sin (x). Let z(t) { t t / otherwise Then (t) z(t) + z( t), so by the linearity and time reversal properties of the Fourier Transform, F(ω) Z(ω)+Z( ω). Therefore D(ω) Z(ω) / z(t)e jωt dt ( t)e jωt dt jω(t )+ (jω) jω e jω/ ω jω e jω/ ω + 4 ( e jω/ +e jω/) ω 4 4 cos(ω/) ω 8sin (ω/4) ω e jωt sinc (ω/4) / t +jω ejω/ ω wheresinc(x) sinx x. This corresponds to the function in Problem 3.3 witha W and t. Please report any typos/errors to jconnelly@ucsd.edu

2 Problem 3. Let A,W, and t be real numbers such that A,W >, and suppose thatg(t) is given by g(t) A t t W t + W Show the Fourier transform ofg(t) is equal to AW sinc (Wω/4) e jωt using the results of Problem 3. and the properties of the Fourier transform. g(t) is a triangular pulse of height A, width W, and is centered at t. (t), from Problem 3., is a triangular pulse of height, width, and is centered at. Thus g(t) is an amplitude-scaled, timescaled, time-shifted version of (t). In particular,g(t) A ( ) t t W. By the amplitude scaling, time scaling, and time shift properties: G(ω) A ) ω D( e jωt W W AW D(Wω)e jωt AW sinc (Wω/4) e jωt

3 Problem 3.3 Find the Fourier transform of the function if t 3 x(t) if t < otherwise. Recall the unit rectangle function rect(t) { t / otherwise. We denote the Fourier transform ofrect(t) by R(ω). Then R(ω) / / rect(t)e jωt dt e jωt dt jω (e jω/ e jω/ ) ω sin(ω/) sinc(ω/). We havex(t) rect ( t 6) rect ( t ), so by the time-scaling property, we have X(ω) 6R(6ω) 4R(ω) 6sinc(3ω) 4sinc(ω). Alternatively,x(t) rect ( ) ( t+ rect t ) ( +rect t ), so by the time-scaling and time-shift properties, we have X(ω) e jω R(ω) 4R(ω)+e jω R(ω) R(ω)(e jω +e jω ) 4R(ω)(cos(ω) ) 4sinc(ω/)(cos(ω) ). It can be shown with trig manipulations that these two functions are equal. Problem 3.4 Find the inverse Fourier transform of the function F(ω) +7jω ω (ω jω )( ω +jω 6) By factoring each of the quadratic polynomials, we have F(ω) (3+jω)(4+jω) (+jω) ( jω)(3+jω) (4+jω) (+jω) ( jω) A (+jω) + B +jω + C jω

4 where(4+jω) A( jω)+b(+jω)( jω)+c(+jω). Therefore which yields,a and B C 3. 4 A+B +C A+B +C B C From Discussion Notes 6, for anya >, we have whereu(t) is the unit step function. Thus by the linearity of the Fourier transform, F(e at u(t)) a+jω F(e at u( t)) a jω F(te at u(t)) (a+jω) f(t) te t u(t)+ 3 e t u(t)+ 3 et u( t) { (t+ 3 )e t t 3 et t <.

5 Problem 3.5 Suppose a functionf(t) has Fourier transform F(ω) πjωe ω. Is f(t) purely real? Is f(t) purely imaginary? Is f(t) even? Is f(t) odd? What is f()? Calculate f(t) and verify these properties. Recall the Fourier transform is a one-to-one mapping, soy(t) z(t) if and only ify(ω) Z(ω). For any functionx(t), x (t) π X (W)e jwt dw π Therefore, the Fourier transform ofx (t) isx ( ω). Ifx(t) is real, then x(t) x (t), so X(ω) X ( ω). Ifx(t) is imaginary, then x(t) x (t), sox(ω) X ( ω). X ( ω)e jωt ( dω) X ( ω)e jωt dω F (X ( ω)) π x( t) X(W)e jwt dw π π Therefore, the Fourier transform ofx( t) isx( ω). If a functionx(t) is even, then x(t) x( t), so X(ω) X( ω). If a functionx(t) is odd, then x(t) x(t), so X(ω) X( ω). For the functionf(t), we have X( ω)e jωt ( dω) X( ω)e jωt dω F (X( ω)) π F ( ω) ( πjωe ω ) πjωe ω F(ω) F( ω) πjωe ω πjωe ω F(ω) Therefore, f(t) is real and odd. For any odd function, f() f( ) f(). Let G(ω) πe ω. Then F(ω) jωg(ω), so by the time-derivative propertyf(t) d dt g(t). g(t) π ] G(ω)e jωt dω e ω e jωt dω + e ω(+jt) dω + e ω e jωt dω e ω(jt ) dω e ω e jωt dω +jt + jt ( jt)+(+jt) (+jt)( jt) Hencef(t) d dt+t t (+t ) +t We also havef(),f(t) f( t), and f(t) is real, thus verifying our claims.

6 Problem 3.6 Suppose x(t) is the input to an LTI system with transfer function H(ω), and y(t) is the output of this system, where x(t) e t cos(at) and H(ω) +e jω +e 3jω. Find a real numbera > such thaty(). Is your answer unique? We have Y(ω) X(ω)H(ω) X(ω)+e jω X(ω)+e j3ω X(ω). Therefore by the linearity and time-shift properties of the Fourier transform, y(t) x(t)+x(t )+x(t 3). Then y() x()+x( )+x( 3). x() e cos() x( ) e cos( A) x( 3) e 3 cos( 3A). If A π(n + /), for any integer n, we have x( ) x( 3), so y() x().

7 Problem 3.7 Supposeg(t) is the input to an LTI system with transfer functionh(ω), andg(ω) is the Fourier transform ofg(t). Find the output of the systemy(t). 8 G(ω) H(ω) 5 5 H(ω) Let X(ω) { ω ω otherwise In the interval[ 5, 5], H(ω) and H(ω) ω, and H(ω) otherwise. So, Y(ω) H(ω)G(ω) X(ω)e jω, which, by the time-shifting property of the Fourier transform, impliesy(t) x(t ). Thus we have x(t) X(ω)e jωt dω π π jt ejt + πt cos(t) πt 4sin (t/) πt sinc (t/) π (+ω)e jωt dω + π y(t) sinc( ) t π jt e jt πt Alternatively, we can use the Duality Property and our results from Problem 3.3. Duality Property: ( ω)e jωt dω If the Fourier transform off(t) is F(ω), then the Fourier transform off(t) isπf( ω). Let f(t) be a triangular pulse of height, width, centered at. Then F(ω) π π sinc (ω/). We havex(ω) πf(ω), and sincef is an even function,x(ω) πf( ω). Therefore, by the Duality Property, the Fourier transform off(t) isx(ω), sox(t) F(t) π sinc (t/).

8 { ift< Problem 3.8 Recall the unit step functionu(t) is given by u(t) ift. Suppose we have a system for which the outputy(t) is y(t) t x(τ)dτ when the input isx(t). Findy(t) and its Fourier transformy(ω) when the input is y(t) t x(t) u(t+) u(t )+u(t 3). t dτ t < x(τ)dτ dτ t dτ t < 3 otherwise +t t < 3 t t < 3 otherwise ( ) t 4 By Problem 3.3 witha,w 4, andt, we have Y(ω) 4sinc (ω) e jω Problem 3.9 An LTI system has impulse response h(t) e 3t u(t). What was the input x(t), when the output is e 3t u(t) e 4t u(t)? For alla >, we have Therefore H(ω) We also have ThusX(ω) F(e at u(t)) e at u(t)e jωt dt e t(a+jω) dt 3+jω andy(ω) 3+jω 4+jω (3+jω) 4+jω, so x(t) u(t)e 4t. Y(ω) X(ω)H(ω) X(ω) 3+jω a+jω (4+jω).

9 Problem 3. Let x(t) u(t)e 3t. Find y(t) when d y(t) dt + dy(t) dt y(t) x(t). For alla >, we havef(e at u(t)) a+jω and F(eat u( t)) a jω By taking the Fourier transform of both sides of the differential equation, we have Therefore Y(ω) (jω) Y(ω)+jωY(ω) Y(ω) X(ω) 3+jω X(ω) (jω) +jω (3+jω)(jω )(jω +) A 3+jω + B jω + C +jω Solving fora, B, and C gives us which implies A(jω )(+jω)+b(3+jω)(+jω)+c(3+jω)(jω ) A( ω +jω )+B( ω +5jω +6)+C( ω +jω 3) A+B +C A+5B +C A+6B 3C and so Thus A 4, B, C 3 y(t) 4 u(t)e 3t u( t)et 3 u(t)e t.

10 MATLAB Problem 4 Output : Joe (jconnel@ucsd.edu) the script (as a.m file) that you used for this problem. In this problem, I am giving you a data file consisting of several amplitude modulated audio signals (similar to AM radio). Let x (t),...,x 9 (t) denote the 9 audio signals. I have provided the signal 9 r(t) cos(πf k t)x k (t) where f k 5(k ). k The signalsx (t),...,x 9 (t) can be viewed as the audio content of a radio station, thecos(πf k t)x k (t) signals can be viewed as what the radio stations are transmitting, and r(t) can be viewed as the signal your antenna is receiving (since by Maxwell s equations, EM waves are additive). Your goal will be to tune into each of the 9 stations and decipher the modulated audio messages. Place the files mod.mat and noisy.mat into your MATLAB directory and load them into your workspace using load mod.mat; and load noisy.mat; (warning, these files are quite large at roughly 7MB) noisy.mat is the signal 9 x k (t). k In other words, noisy.mat is what happens when radio stations DO NOT perform modulation. This is akin to multiple people talking to you at once. It is very difficult to pick out any one of the messages, since they are all communicating in the same frequency band. Output : try running mod play(noisy); and describe what you hear. Note that you will need to use this custom function to play the audio files in this problem (as opposed to the usual sound function). We will make use of the fact that x(t)cos(ω t) X(ω +ω )+X(ω ω ) Each of the signals x (t),...,x 9 (t) was multiplied by a different carrier frequency, and the bandwidth of each signal is smaller than the differences in the carrier frequencies, so 9 ( R(ω) Xk (ω πf k )+X k (ω πf k ) ) k This allows us to spread out the signals in the frequency domain. Plot the modulated signal in the frequency domain using: Len length(mod); Fs 85; f Fs ( Len/ : Len/ )/Len; Mod Freq fft(mod); plot(f,abs(fftshift(mod Freq))); Output : include well-labeled plots of both the modulated (mod) and unmodulated (noisy) signals in the frequency domain, and explain the differences you notice. You may need to adjust the limits of the axes to better analyze these figures.

11 You will be required to submit Outputs 3 & 4 for any three of the nine audio signals, but I encourage you to try tuning into all nine stations. The following steps describe how to recover thekth audio message. Filter out all of the modulated signals, except for the kth. You can do this using the provided filter file, e.g. Filtered Signal Mod Freq. HW3 Filter(f, A, B); You will need to experiment with the values ofaand B to correctly filter out thekth signal. You will now need to convert this filtered signal back to the time domain, which yields the signal cos(πf k t)x k (t). You can do this using filtered signal real(ifft(filtered Signal)); You now need to undo the modulation to recoverx k (t). We will be using the fact x(t)cos (ω t) x(t) +cos(ω t) X(ω) + X(ω +ω )+X(ω ω ) 4 So multiplying cos(πf k t)x k (t) by cos(πf k t) will yield copies of X(ω) in the frequency domain that are centered at and ±πf k. Do this by taking t (:Len-) / F s ; demod signal filtered signal. ( cos( pi f k t)); Our final step is to filter out the high frequency components so that we are left with our desired messagex k (t). We can do this by sending x(t)cos (ω t) through a low-pass filter that zeros out thex k (ω ±ω k ) terms but allows thex k (ω) term to pass through. Demod Signal fft(demod); Message Demod Signal. HW3 Filter(f, A, B); You will need to experiment with the values ofaand B to filter out the high frequency terms. Output 3: include a well-labeled plot of the message in the frequency domain. Now we convert back to the time domain and play using the provided function: message real(ifft(message)); mod play(message); Output 4: What is thekth audio message?

12 Station x (t) cos(πf t) y (t) Station x (t) y (t) Channel r(t) Receiver cos(πf t) cos(πf k t) Band Pass Filter tuned to kth station y k (t) Low Pass Filter x k (t) Station N... cos(πf N t) x N (t) y N (t) A block diagram of the basic amplitude modulation scheme used in this problem. Modulation allows for concurrent communication by dividing up portions of the frequency spectrum. See ECE45 MATLAB3.m for implementation details. When running mod play(noisy); it sounds like 9 people talking at once. This is because the 9 messagesx (t),...,x 9 (t) are simply summed in the noisy file. This results in a jumbled audio file in which it is difficult to pick out any one message. Fourier Transform of Summed Unmodulated Signals Fourier Transform of Summed Modulated Signals x 5 In the modulated figure, the signals are spread out in the frequency domain. Thekth peak corresponds to the (shifted) Fourier transform of the kth message. Since the signals are spread out in frequency, we can filter out undesired signals. Once the undesired signals are filtered out, we can demodulate by multiplying by the carrier signal and filtering.

13 FT of Message # FT of Message # FT of Message # FT of Message #4 4 6 FT of Message #5 4 6 FT of Message # FT of Message #7 4 6 FT of Message # FT of Message # The Fourier transforms of the messages.. I ll be back. - The Terminator in The Terminator, The ring must be destroyed! - Gandalf the Grey in The Lord of the RIngs: The Fellowship of the Ring,. 3. Bueller? Bueller? - Ben Stein in Ferris Bueller s Day Off, You have failed me for the last time. - Darth Vader in Star Wars: The Empire Strikes Back, Mmm. This is a tasty burger. - Jules Winnfield in Pulp Fiction, Game over, man. It s game over! - Private Hudson in Aliens, I m sorry, Dave. I m afraid I can t do that. - HAL 9 in : A Space Odyssey, How am I gonna generated that kind of power? It can t be done. - Doc Brown in Back to the Future, Who died and made you Einstein? - Valentine McKee in Tremors, 99.