Understanding Fourier Transforms and Sampling in ECE 340
School
University of Alberta**We aren't endorsed by this school
Course
ECE 340
Subject
Electrical Engineering
Date
Dec 10, 2024
Pages
3
Uploaded by MinisterStar29964
ECE 340 Section A1 Assignment 4 - SolutionsQuestion 1 Ans.Lett=kT, then(1)f[k] =f(kT) = cos(2π·k/3) = cos(23πk)(2)f[k] =f(kT) = sin(43πk)(3)f[k] =f(kT) = sinc(2k/5)or=sin(2πk/5)2πk/5Question 2 Ans.(1) The analog spectrumF(ω) is sketched in Figure 1a, based on which we knowf(t) is bandlimited.(a)F(ω)(b)¯F(ω)Figure 1: Question 2(2) From the table of Fourier transform pairs, the signalf(t), which has a rectangle windowspectrum, is a sinc function of time, i.e.f(t) =Wπsinc(Wtπ), whereWis the bandwidth of thespectrum. Therefore, for this question,f(t) = 8sinc(8t).f(t) has infinite length.(3) The Fourier transform of the sampled signal (i.e.¯F(ω)) is sketched in Figure 1b. It is obtainedby duplicating the analog spectrumF(ω) (scaled by 1/T) with the center at integer multiplesofωs= 2π/T= 20πrad/sec.Sincef(t) is band-limited and the bandwidth is 8π, the Nyquist frequency is 2ωB= 16πrad/sec, or 8 Hz.Question 3 Ans.For the continuous-time signalf(t) = sin(4πt)+2 sin(1.5πt), its continuous-timeFourier transform isF(ω) =πj[δ(ω−4π)−δ(ω+ 4π)] +2πj[δ(ω−1.5π)−δ(ω+ 1.5π)],(1)and it is shown in Figure 2 (the sketch ofF(ω)is optional).(1) After sampling withFs= 3 Hz, the sampled spectrum is as follows.(Note: for simplicity of sketching, I choose to include the factors ‘1/j’ and ‘-1/j’ as part of themagnitude to show the difference instead of drawing the stems as both positive and negativeterms, as these factors only reflect the phase differences. But it is also okay if you choose tosketch them as both positive and negative terms. )
Figure 2: Question 3 -F(ω): the factors ‘1/j’ and ‘-1/j’ are included as part of the magnitude ofthe stemsFigure 3:Question 3 - Sampled¯F(ω) withFs= 3 Hz (red line - the original spectrum beforesampling)¯F(ω) =1T∞Xn=−∞F(ω−nωs) = 3∞Xn=−∞F(ω−6πn),n−integerwhereF(ω) is the analog spectrum shown above.¯F(ω) is shown in Figure 3.Therefore, the impulses±πjδ(ω±4π) will be replicated atω=· · ·±2π,±4π,±8π,±10π· · ·withscaled amplitude 3π(by assumingn=· · ·−2,−1,0,1,2. . .). Similarly, for the two impulsescorresponding to 2 sin(1.5πt), after sampling, will appear atω=±1.5π,±4.5π,±7.5π,· · ·,and their amplitudes equal to±6πj.(2) When the sampled signal passes through a low pass filter with cutoff frequencyωs/2 = 3πrad/sec, the impulses at±2πand±1.5πremain but the rest are removed. The Fourier trans-form of the reconstructed signal isFr(ω) =−πj[δ(ω−2π)−δ(ω+ 2π)] +2πj[δ(ω−1.5π)−δ(ω+ 1.5π)]Figure 4: Question 3 - The reconstructedFr(ω) by the ILPF2
(Optional:the reconstructed spectrum is shown Figure 4.)Therefore the reconstructed signal isfr(t) =−sin(2πt) + 2 sin(1.5πt)̸=f(t)Note that the Nyquist frequency is 2×4π= 8πrad/sec, while the sampling frequency isωs= 6πrad/sec. The reason for the mismatch is because we sample below the Nyquist rate,and thus we observe aliasing. To be specific, the ‘new’ frequency component at 2πrad/sec isdue to aliasing.Question 4 Ans.(1) Givenf1(t) = 5sinc(200t), we knowW= 200π, from the table, its Fourier transform isF1(Ω) = 5·πWrectω2W=140rectω400πwhererectω400π=(1,|ω| ≤200π,0,otherwise.It is a bandlimited signal with bandwidth 200πrad/sec, or 100Hz.Therefore, the Nyquistsampling rate is 2×100 = 200 Hz, which is the minimum sampling rate to choose. Thereforethe maximum sampling period isT= 1/Fs= 5ms.(2) Iff2(t) = 5sinc(200t) sin(100πt), based on convolution property, its Fourier transform isF2(ω) =12πF{sinc(200t)} ∗ F{sin(100πt)}=12π·140rectω400π∗πj[δ(ω−100π)−δ(ω+ 100π)]=1j80rectω−100π400π−1j80rectω+ 100π400πin the above operation, the shifting property is used. Because of the shifting, the bandwidthbecomes 300πrad/sec, or 150 Hz. The Nyquist rate should be 300Hz, which corresponds to amaximum sampling period of 1/300 sec.3