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PS-8-2007-Solutions

# PS-8-2007-Solutions - EE 261 The Fourier Transform and its...

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EE 261 The Fourier Transform and its Applications Fall 2007 Problem Set Eight Solutions 1. (20 points) A True Story : Professor Osgood and a graduate student were working on a discrete form of the sampling theorem. This included looking at the DFT of the discrete rect function f [ n ] = braceleftBigg 1 , | n | ≤ N 4 0 , N 2 + 1 n < N 4 , N 4 < n N 2 The grad student, ever eager, said ‘Let me work this out.’ A short time later the student came back saying ‘I took a particular value of N and I plotted the DFT using MATLAB (their FFT routine). Here are plots of the real part and the imaginary part.’ -10 -8 -6 -4 -2 0 2 4 6 8 10 -2 0 2 4 6 8 k real part of F(k) -10 -8 -6 -4 -2 0 2 4 6 8 10 -2 -1 0 1 2 k imaginary part of F(k) (a) Produce these figures. 1

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Professor Osgood said, ‘That can’t be correct.’ (b) Is Professor Osgood right to object? If so, what is the basis of his objection, and produce the correct plot. If not, explain why the student is correct. Solution The student executed the following MATLAB command. f = [0 0 0 0 0 0 0 1 1 1 1 1 1 1 0 0 0 0 0 0 0]; F = fftshift(fft(fftshift(f))); DFT in MATLAB is done assuming DC at the first component of the array. Therefore the command fftshift is used to alter the center of the input array. There is a fine detail here that is worth noting. When the array size is N, fftshift shifts the ceil(N/2)+1 component to be the first component of the array. Therefore, to obtain the desired result, the rect function should be centered around the ceil(N/2)+1 component. The following command will give F without any imaginary compo- nent. f = [0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 0 0 0 0 0 0 0]; F = fftshift(fft(fftshift(f))); Here’s the correct plot: 2
-10 -8 -6 -4 -2 0 2 4 6 8 10 -2 0 2 4 6 8 k real part of F(k) -10 -8 -6 -4 -2 0 2 4 6 8 10 -2 -1 0 1 2 k imaginary part of F(k) 2. (20 points) Linearity and time-invariance. State whether the following systems are linear or non-linear, time-invariant or time-variant, and why. Assume that v ( t ) is the input and w ( t ) is the output for all systems. No credit will be given for answers without explanations or proofs! (a) w ( t ) = v ( t ) cos( ωt ) (b) w ( t ) = sin( v ( t )) (c) w ( t ) = integraltext -∞ v ( τ ) e - 2 πitτ (d) w ( t ) = d dt v ( t ) (e) w ( t ) = cos( ωt + v ( t )) Solution: In each case, we need to use the definition of linearity and time-invariance to prove or disprove the claim. A system T is linear if T { α 1 v 1 ( t ) + α 2 v 2 ( t ) } = α 1 T { v 1 ( t ) } + α 2 T { v 2 ( t ) } . Given that w ( t ) = T { v ( t ) } , a system T is time-invariant if w ( t τ ) = T { v ( t τ ) } . (a) Linearity : ( α 1 v 1 ( t ) + α 2 v 2 ( t )) cos( ωt ) = α 1 v 1 ( t ) cos( ωt ) + α 2 v 2 ( t ) cos( ωt ), so the system is clearly linear. Time-Invariance : v ( t τ ) cos( ωt ) negationslash = w ( t τ ), so the system is time-variant. (b) Linearity : sin( α 1 v 1 ( t )+ α 2 v 2 ( t )) negationslash = α 1 sin( v 1 ( t ))+ α 2 sin( v 2 ( t )), so the system is non-linear. Time-Invariance : sin( v ( t τ )) = w ( t τ ), so the system is time-invariant.

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PS-8-2007-Solutions - EE 261 The Fourier Transform and its...

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