20115ee102_1_hw5_soln

# 20115ee102_1_hw5_soln - 1 EE102 Systems and Signals Fall...

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Unformatted text preview: 1 EE102 Systems and Signals Fall Quarter 2011 Jin Hyung Lee Homework #5 Solution 1. Each of these signals can be written as a sum of scaled and shifted unit rectangles and triangles, 2-1 1-2 1 2-2-1 x(t) a) 2-1 1-2 1 2-2-1 t x(t) b) Find a simple expression for each signal, and then compute the Fourier transform. Solution: (a) One representation for the first signal is x ( t ) = Δ( t + 1 / 2)- Δ( t + 1 / 2) The Fourier transform is then X ( jω ) = e jω/ 2 sinc 2 ( ω/ 2 π )- e- jω/ 2 sinc 2 ( ω/ 2 π ) = ( e jω/ 2- e- jω/ 2 ) sinc 2 ( ω/ 2 π ) = 2 j sin( ω/ 2) sinc 2 ( ω/ 2 π ) (b) One representation of the second signal is x ( t ) = 2Δ( t/ 2)- Δ( t ) + rect ( t/ 2) The Fourier transform is then X ( jω ) = 4 sinc 2 ( ω/π )- sinc 2 ( ω/ 2 π ) + 2 rect ( ω/π ) 2 2. Determine whether the assertions are true or false, and provide a supporting argument. (a) If ( f * g )( t ) = f ( t ) , then g ( t ) must be an impulse, δ ( t ) . Solution: Time-domain convolution ( f * g )( t ) = f ( t ) corresponds to frequency domain multiplication F ( jω ) G ( jω ) = F ( jω ) . If g ( t ) = δ ( t ) , then G ( jω ) = 1 , so this is sufficient. However, it isn’t necessary. All we require is that G ( jω ) = 1 when F ( jω ) 6 = 0 . For example, if F ( jω ) is bandlimited to ± ω c , then G ( jω ) = rect ( ω/ (2 ω c )) would also work. Hence the assertion is False . F ( j ) G ( j ) 1 c- c This is fortunate. That means that if f ( t ) is a signal, and g ( t ) is the impulse response of a system, we can exactly reproduce f ( t ) provided f ( t ) is bandlimited, and the fre- quency response of g ( t ) is unity across this bandwidth. (b) If the convolution of two functions f 1 ( t ) and f 2 ( t ) is identically zero, ( f 1 * f 2 )( t ) = 0 then either f 1 ( t ) or f 2 ( t ) is identically zero, or both are identically zero. Solution Again, the convolution corresponds to frequency domain multiplication, so F 1 ( jω ) F 2 ( jω ) = 0 . A zero signal corresponds to a zero transform, so the question is whether F 1 ( jω ) or F 2 ( jω ) or both must be identically zero. Certainly, if one or both are zero, the convolution is zero. This is sufficient. However, it is not necessary. All we need is that each be zero where the other is non-zero. For example, if f 1 ( t ) = sinc ( t ) and f 2 ( t ) = sinc ( t )cos(4 πt ) , then Then F 1 ( jω...
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## This note was uploaded on 12/13/2011 for the course EE 102 taught by Professor Levan during the Fall '08 term at UCLA.

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20115ee102_1_hw5_soln - 1 EE102 Systems and Signals Fall...

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