1) Time Differentiation Operation with Discrete Signals. Write a script that shows that the backwards difference of a triangular function (Figure 1) has an FFT of the form given in (3). The final Spectrum should look like Figure 2. Generate Signal Fs = 10000; Tp = 1/Fs; time = 0:T_P:2; % Sampling Frequency % Sampling Period % Sampling Time vector x = zeros (size (time)); x(1:1000) = triang (1000); Sample script for triangle function used in figure 1. Differentiation of the signal in time the main corresponds to multiplying its Fourier transform by in the frequency domain DFT{x(n) x(n-1)} = DFT{x(n)} DFT{x(n-1)} = X(jw) - e-wx(jw) DFT{x(n) x(n 1)} = (1 - e-j)x(jw) (1) (2) (3) Triangle Function Amplitude 0 0 0.05 0.1 0.15 0.2 0.25 Triangle Function 500 10 20 30 40 60 70 80 90 100 50 (1 - e-w)FGW) 0.5 10 20 30 40 60 70 80 90 100 50 Backward Difference 1 Magnitude 0.5 1 80 10 20 30 40 60 70 90 100 50 Time (sec) Figure 2 - Resulting Spectrum for Problem 1 2) Multiplication in Time Domain is Convolution in the Frequency Domain Write a script that shows that the backwards difference of a triangular function (Figure 3) has an FFT of the form given in (4). The final Spectrum should look like Figure 5. Figure 4 shows the spectral data of the functions and the result of the convolution. % Generate Signal Fs = 10000; T_P = 1/Fs; time = -10:T_p:10; % Sampling Frequency & Sampling Period % Sampling Time vector x = sinc(100*time); % Generate signal x sin (100*time); X.*y; % Generate signal y % Product of x and y z Sample script for triangle function used in figure 3. DFT{x(n)y(n)} = DFT{x(n)} * DFT{y(n)} (4) sinc(100t) Function 0.5 Amplitude 0 -0.25 -0.2 -0.15 -0.1 0.1 0.15 0.2 0.25 -0.050 0.05 sin(100t) Function 1 Amplitude A X 0.0367 Y-0.50416 -1 -0.25 -0.2 -0.15 0.05 0.15 0.2 0.25 -0.1 -0.05 0 0.1 sinc(100t)sin(100t) Function % 0.2 Amplitude 0 -0.2 -0.25 -0.2 -0.15 -0.1 -0.05 0.1 0.15 0.2 0.25 0 0.05 Time (sec) Figure 3 - Input Signal for Problem 2 DFT sinc(1004) 100 Magnitude 50 DFT (sinc(1000) 100 Magnitude 50 0 -100 -80 -60 -40 -20 20 40 60 80 100 1 0 DFT sin (1000) 1 10,104 Magnitude 5 -80 -60 -40 0 0 -100 x 106 40 60 80 100 -20 20 DFT sinc(100+)] *DET sin(1001)] T 10 Magnitude 1 0 -100 -80 -60 -40 -20 0 20 40 60 80 100 Figure 4 - Spectrums of signals and the result of convolution (conv () ) using the 'same' option Problem 2 Page 3 of 4 Time Domain 0.4 x(t)*y(t) IDFT conv(X(jw),Y(jw)) 0.2 Amplitude -0.2 -0.4 -0.25 -0.2 -0.15 -0.1 -0.05 0 0.05 0.1 0.15 0.2 0.25 Time (sec) Figure 5 - Time Domain Comparison of Outputs using ifft() 'symmetric property for Problem 2