EE 179 April 21, 2014 Digital and Analog Communication Systems Handout #16 Homework #2 Solutions

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1 EE 79 April, 04 Digital and Analog Communication Systems Handout #6 Homework # Solutions. Operations on signals (Lathi& Ding.3-3). For the signal g(t) shown below, sketch: a. g(t 4); b. g(t/.5); c. g(t 4); d. g( t). Hint: Recall that replacing t with t T delays the signal by T. Thus g(t 4) is g(t) with t replaced by t. Similarly, g( t) is g( t) with t replaced by t. Solution (0 points). Orthogonal signals (Lathi&Ding.5-5). The energies of two energy signals x(t) and y(t) are E x and E y, respectively. a. If x(t) and y(t) are orthogonal, then show that the energy of the signal x(t)+y(t) is identical to the energy of the signal x(t) y(t) and is given by E x +E y. b. If x(t) and y(t) are orthogonal, find the energies of the signals c x(t)+c y(t) and c x(t) c y(t). c. We define E xy, the correlation of the two energy signals x(t) and y(t), as If z(t) x(t)±y(t), then show that Solution (0 points) E xy x(t)y (t)dt. E z E x +E y ±(E xy +E yx ). a. If x(t) and y(t) are orthogonal, then by definition ( x(t)y (t)dt 0 and x (t)y(t)dt x(t)y (t)dt) 0..

2 Therefore x(t)±y(t) dt E x +E y. (x(t)±y(t))(x(t)±y(t)) dt x(t) dt± x(t) dt+ b. Since c x(t) and c y(t) are orthogonal, c. By part (a), if z(t) x(t)±y(t) then x(t)y (t)dt± y(t) dt E c x±c y E c x +E c y c E x + c E y. E z E x +E y ±(E xy +E yx ). x (t)y(t)dt+ y(t) dt Note that E xy + E yx E xy + E xy ReE xy. If x(t) and y(t) are orthogonal, then E xy 0 and therefore E z E x + E y. Consider z(t) x(t) + y(t). If ReE xy < 0 then E z < E x +E y since x(t) and y(t) partially cancel each other. If ReE xy > 0 then E z < E x +E y since x(t) and y(t) reinforce. 3. Use Fourier transform properties to derive Fourier transforms (Lathi& Ding 3.3-). The Fourier transfer of the triangular pulse g(t) in Fig. P3.3-a is given as G(f) e jπf jπfe jπf ). Use this information, and time-shifting and time-scaling properties, to find the Fourier transforms of the signals shown is Fig. P3.3-b, c, d, e, and f. Solution (0 points) a. By the problem statement, G(f) e jπf jπfe jπf ). Page of 8 EE 79, Spring 04

3 b. Since g (t) g( t), G (f) G( f): G (f) e jπf +jπfe jπf ). c. g (t) g(t )+g (t ). Then by the shift theorem, G (f) e jπf ( e jπf jπfe jπf ) + e jπf ( e jπf +jπfe jπf ) πf) ( jπf e jπf +e j4πf +jπfe j4pif). (πf) d. g 3 (t) g(t )+g (t+). Then G 3 (f) e jπf e jπf jπfe jπf ) + ejπf e jπf +jπfe jπf ) e jπf e jπf) Another formula for G 3 (f) is sinc f. e. g 4 (t) g(t )+g (t+ ). Then cosπf). G 4 (f) e jπf ( e jπf jπfe jπf ) + ejπf ( e jπf +jπfe jπf ) πf) ( e jπf jπfe jπf e jπf +e jπf +jπfe jπf e jπf) (πf) ( jπfe jπf +jπfe jπf) j ( e jπf e jπf) sinπf (πf) πf πf f. g 5 (t) 3 g( (t )). Then G 5 (f) 3 e j4πf G(f) 3e j4πf (4πf) ( e j4πf j4πe j4πf ) 4. Modulation and demodulation. a. Let m(t) be a message signal, f c a constant (carrier frequency), and define x(t) m(t)cos(πf c t+θ 0 ). 3 (4πf) ( j4πf e j4πf ). Let M(f) be the Fourier transform of m(t). Find X(f), the Fourier transform of x(t), in terms of M(f). b. Find the signal (in the time domain), whose Fourier transform is pictured below. Homework # Solutions Page 3 of 8

4 c. A similar relationship can be found for x(t) m(t)sin(πf c t). Find it, and use it to find the Fourier transform of { sin(πt) t < / x(t) 0 t > / without performing any integration. Does the Fourier transform have the properties you would expect (even/odd/neither, real/imaginary/complex)? d. Show that m(t) can be recovered from x(t) m(t)cos(πf c t) by multiplying by cos(πf c t) andpassing theproduct throughalow-pass filter ofbandwidthb Hz, whereb isthebandwidth of m(t). Assume that B f c. Solution (0 points) a. We appeal directly to the definition of the Fourier transform: X(f) x(t)e jπft dt m(t)cos(πf c t+θ 0 )e jπft dt m(t) (ejπfct+θ 0 +e jπfct θ 0 )e jπft dt m(t)e jπ(f fc)t+θ 0 dt+ M(f f c)e jθ 0 + M(f +f c)e jθ 0 m(t)e jπ(f+fc)t θ 0 dt b. With an eye toward using part (a) with θ 0 0, the Fourier transform in this part can be written as ( ) ( ) f 4 f where (f) is the triangle function with height and width. So it looks like we want to find a function m(t) whose Fourier transform is (f/4), for if we then multiply it by cos(π 4 t) the modulation property gives us what we want for the Fourier transform: ( ) ( ) f 4 f 4 F{m(t)cos8πt} Since F{ (t/τ)} τ sinc (τf), we obtain by duality If we thus set we have F { τ sinc (τt) } (f/τ), m(t) 4sinc (t) M(f) (f/4) as desired. So, the signal x(t) whose Fourier transform X(f) in the picture is x(t) 4sinc (t)cos(8πt). Page 4 of 8 EE 79, Spring 04

5 c. The computation is very similar to what was done in part (a): X(f) x(t)e jπft dt m(t)sin(πf c t)e jπft dt m(t) j (ejπfct e jπfct )e jπft dt j m(t)e jπ(f fc)t dt j j M(f f c) j M(f +f c) m(t)e jπ(f+fc)t dt (We can also find X(f) using the convolution theorem: X(f) M(f) j( δ(f fc ) δ(f+f c ) ), where j( δ(f fc ) δ(f + f c ) ) is the Fourier transform of sin(πf c t).) To find the Fourier transform of { sin(πt) t / x(t) 0 t / we note that x(t) Π(t)sinπt and we can apply the sine-modulation formula with f c. Since F{Π(f)} sinc(f), X(f) j( ) sinc(f ) sinc(f +). d. The modulated signal is x(t) m(t)cos(πf c t). Multiplying by cos(πf c t) yields m(t)cos (πf c t) m(t) ( +cos(4πf c t) ) m(t)+m(t)cos(4πf c t). Observe that the resulting signal contains the original signal m(t) and a modulated copy of the signal moved toafrequency center of f c. If the bandwidth of theoriginal signal isb < f c, then the modulated copy will not extend further than f c from its center frequency and a low-pass filter from f c to +f c will pass only m(t) and filter out the modulated copy. 5. Essential bandwidth (Lathi& Ding 3.7-4). For the signal g(t) a t +a determine in hertz the essential bandwidth B of g(t) such that the energy contained in the spectral components of g(t) of frequencies below B Hz is 99% of the signal energy E g. Hint: determine G(f) by applying the duality property [Eq. (3.6)] to pair 3 of Table 3.. Solution (0 points) Applying the duality property [Eq. (3.6)] to pair 3 of Table 3., we find { } a F e a f. a +(πt) Homework # Solutions Page 5 of 8

6 Applying the time-scaling property with the constant, we obtain π { } a F πe a πf. a +t The signal energy is easily calculated in the frequency domain: E g G (f) df πe a πf df 4π e 4aπf df 8π 4aπ π a. The energy of the signal contained in the high-pass band beyond B Hz is E HF 4π e 4aπf df π a e 4aπB. Setting E HF 0.0E g and solving for B, we obtain B π a e 4aπB 0.0 π a e 4aπB 0.0 B ln(0.0) 4aπ ln(00) 4aπ. 6. Autocorrelation and PSD. Show that the autocorrelation function of g(t) Acos(πf 0 t+θ 0 ) is R g (τ) A cos(πf 0 τ) and that the corresponding PSD is S g (f) 4 A( δ(f f 0 )+δ(f +f 0 ) ). Solution (0 points) Using the definition of the autocorrelation function for power signals as given in Eq. (3.83a), T/ R g (τ) lim T T lim T lim T T/ A T/ T ( A T T/ T/ g(t)g(t τ)dτ T/ cos(πf 0 t+θ 0 )cos(πf 0 t πf 0 τ +θ 0 )dt cos(πf 0 τ)dt+ A T T/ T/ 0 cos(4πf 0 t πf 0 τ +θ 0 )dt (using trigonometric identity cosacosb (cos(a b)+cos(a+b))) ( ) A lim T T T cos(πf 0τ)+ A sin(πf 0 (T +τ) θ 0 )+sin(πf 0 (T τ)+θ 0 ) T 4πf 0 A cos(πf A f(t) 0τ)+ lim T T 4πf 0 A cos(πf 0τ). Thelimit ofthesecond term inthefifthlineis0because thenumerator f(t)is boundedinabsolute value by. The power spectral density of the signal is the Fourier transform of R g (τ): S g (f) A ( δ(f +f0 )+δ(f f 0 ) ). ) Page 6 of 8 EE 79, Spring 04

7 7. Output SNR of linear system (Lathi& Ding 3.8-5). Consider a linear system with impulse response e t u(t). The linear system input is ( g(t) w(t) cos 6πt+ π ), 3 in which w(t) is a noise signal with power spectral density S w (f) Π(f/4). a. Find the total output power of the linear system. b. Find the output power of the signal component due to the sinusoidal input. c. Find the output power of the noise component. d. Determine the output signal-to-noise ratio (SNR) in decibels. Solution (0 points) a. The power spectral density of the sinusoidal signal is S x (f) (δ(f +3)+δ(f 3)), hence the 4 input PSD is S g (f) S w (f)+s x (f) Π(f/4)+ (δ(f +3)+δ(f 3)). 4 The transfer function of the system is the Fourier transform of the impulse response: The output PSD is F { e t u(t) } S y (f) H(f) S g (f) and so the output power is P y S y (f)df +jπf H(f) 4+4π f. 4+4π f ( Π(f/4)+ 4 (δ(f +3)+δ(f 3))). 4+4π f df + /4 4+4π ( 3) + /4 4+4π (+3) tan (π) π + 8+7π b. The PSD of the output due to the sinusoidal signal is S yx (f) H(f) S x (f) and the total power due to the signal is P yx c. The power due to the noise is P yw S yw (f)df 4+4π f ( 4 (δ(f +3)+δ(f 3))), S yx (f)df 8+7π π f df tan (π) π Homework # Solutions Page 7 of 8

8 d. The signal-to-noise ratio in decibels is P yx SNR 0log 0 0log P 0 7. db yw Note that the negative SNR means that the power of the signal is less than the power of the noise. Page 8 of 8 EE 79, Spring 04