Question

Can you please help me answer Task 2.b?

Please show all work.

fs=44100; no_pts=8192;
t=([0:no_pts-1]')/fs;
y1=sin(2*pi*1000*t);
figure;
plot(t,y1);
xlabel('t (second)')
ylabel('y(t)')
axis([0,.004,-1.2,1.2]) % constrain axis so you can actually see the wave
sound(y1,fs); % play sound using windows driver.
%%
% Check the frequency domain signal. fr is the frequency vector and f1 is the magnitude of F{y1}.
fr=([0:no_pts-1]')/no_pts*fs; %in Hz
fr=fr(1:no_pts/2); % single-sided spectrum
f1=abs(fft(y1)); % compute fft
f1=f1(1:no_pts/2)/fs;
%%
% F is the continuous time Fourier. (See derivation notes.) Notice the small
% amount of aliasing due to the fact that the truncated sinusoid is not bandlimited.
frp=fr*2*pi;tmax=max(t);
F1=1/1i*sin((frp-1000*2*pi)*tmax/2).*exp(-1i*(frp-1000*2*pi)*tmax/2)./(frp-1000*2*pi); % li = i, the imaginary partsymbol
F2=1/1i*sin((frp+1000*2*pi)*tmax/2).*exp(-1i*(frp+1000*2*pi)*tmax/2)./(frp+1000*2*pi);
F=abs(F1-F2); % magnitude
figure;
plot(fr, F, fr, f1) % compare the continuous time Fourier with FFT, linear scale
axis([0,1500,0,0.09]) % constrain axis so you can actually see the pulse
xlabel('f (Hz)')
ylabel('|Y(f)|')
legend('Continuous-time FT', 'FFT')
figure;
loglog(fr,F,fr,f1); % compare the continuous time Fourier with FFT, log-log scale
xlabel('f (Hz)')
ylabel('|Y(f)|')
legend('Continuous-time FT', 'FFT')

5. Derivation of Fourier Transform of a Windowed Sinusoid In this lab exercise, we define the signal y(t) sin(wot), for t in [0,Tmaxl. Consequently, the CFT of y(t) is S(w-w)-6(w+ w )] 2sin(wT/2)T F{sin(wot)puls40.(t)} maz max W 1sin((w-Wo )Tm/2),-j(u-wo)T2 sin((wWoTm /2)-j(w+w) max max max: e e w-W This expression is used as a basis to compare the FFT-computed transform. The latter will include aliasing effects, attaining its theoretical value as the sampling interval approaches zero. The following MATLAB code compares the CFT of a windowed sinusoid derived by Equation and the FFT method: Task 2 In the MATLAB code provided in Section 5 of the lab manual, we created a sinusoid with frequency 1 kHz. We used a high sampling frequency of fs 44.1 kHz that represents a fairly good approximation of the continuous time signal. Copy the MATLAB code and run it in a script. Include the plots generated a) X Figure File Edit View Insert Tools Desktop Window Help 0.5 -0.5 -1 0 0.5 3 1.5 2 2,5 3.5 4 t(second) 3 x 10 7 (1)6 X Figure 2 Window File Edit View Insert Tools Desktop Help 0.09 Continuous-time FT FFT 0.08 0.07 0.0 0.05 0.04 0.03 0.02 0.01 0 0 500 1000 1500 f(Hz) 7 Figure 3 File Edit View Insert Tools Desktop Window Help 10-1 Continuous-time FT FFT 10-2 10-3 104 10-5 10-6 100 101 102 103 f (Hz) 4 10 105 Compare the analytical continuous-time Fourier transform with its FFT computation. Discuss your b) observations

Answer 1

Part-b

In continuous time fourier transform i.e. CTFT, the signal time period is extended to infinity in order to make the frequency resolution very very small. This brings the harmonics very close to each other and the spectrum become continuous. However, the CTFT of a periodic signal having a defined period, the CTFT is discrete as like that to the FFT.

Therefore, in present case, the signal is of 1KHz i.e. have a well defined time period of 1 msec, therefore, we see an overlaping FT and CTFT plot.