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HW13s09_soln

# HW13s09_soln - SOLUTIONS TO ECE2025 PROBLEM SET#13 PROBLEM...

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1 S OLUTIONS TO ECE2025 P ROBLEM S ET #13, A PRIL 24, 2009 PROBLEM 13.1. (a) Since | j ω | = | ω | , we get the following: (b) Since v ( t ) = x ( t )cos(66 π t ) , the modulation property of Table 11-3 tells us that V ( j ω ) can be found by shifting X ( j ω ) up and down by 66 π , then multiplying by 1/2 , so that: V ( j ω ) = X ( j ( ω 66 π )) + X ( j ( ω + 66 π )): Similarly, since w ( t ) = v ( t )cos(15 π t ) we get: W ( j ω ) = V ( j ( ω 15 π )) + V ( j ( ω + 15 π )): Finally, since z ( t ) = w ( t )cos(51 π t ) we get: Z ( j ω ) = W ( j ( ω 51 π )) + W ( j ( ω + 51 π )): (c) The picture above shows that z ( t ) = x ( t )/4 + high-frequency terms, so we can recover x ( t ) by applying z ( t ) to a low-pass filter which passes – 7 π < ω < 7 π with a gain of 4 , and its stop band must start at 23 π , so that the frequency response is zero for | ω | > 23 π : 7 π 7 π 7 π | X ( j ω ) | ω -- -- 73 π 7 π /2 | V ( j ω ) | ω 7 π /2 59 π 73 π 59 π 66 π 66 π 0 -- -- 58 π | W ( j ω ) | ω 44 π 51 π 81 π 88 π 7 π /4 7 π /4 74 π 81 π 51 π 0 -- -- | Z ( j ω ) | ω 102 π 30 π 7 π /8 7 π /4 132 π 0 7 π 132 π 7 π /8 7 π /8 30 π 102 π 7 π 23 π H ( j ω ) ω 23 π 0 7 π 7 π 23 π

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2 PROBLEM 13.2. From Table 11-2 we see that the Fourier transform of the pulse train p ( t ) = n δ ( t nT s ) with period T s is another pulse train P ( j ω ) = ω s k δ ( ω k ω s ), where ω s = 2 π / T s . Since x s ( t ) = x ( t ) p ( t ) , the multiplication property of Table 11-3 tells us that the Fourier transform X s ( j ω ) can be found by convolving the triangle X ( j ω ) with the pulse train P ( j ω ) . This places a copy of the triangle at every integer multiple of the sampling frequency, yielding: X s ( j ω ) = k X ( ω k ω s ). (a) We must avoid overlapping triangles if we want x r ( t ) = x ( t ) . Therefore, the smallest sampling rate is twice the maximum frequency in x ( t ) , namely ω s = 2 ω max = 160 π . The choice ω s = 2 ω max is known as the Nyquist rate. With this choice, the Fourier transform of x s ( t ) looks like this: (b) Putting a copy of the triangle at every integer multiple of 150 π will result in overlapping triangles, i.e. aliasing, as shown below:
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