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    Lab-Assignment 4

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    Lab-Assignment 4

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    Lab report

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    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    4 views14 pages

    Lab-Assignment 4

    Uploaded by

    aajif
    Lab report

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    of 14

    BMED-M3142:- Signals and Systems-Lab

    Lab: Frequency Content of Continuous-Time


    Signal

    Addis Ababa University


    Addis Ababa Institute of Technology (AAiT)
    Center of Biomedical Engineering

    Yidnekachew S. 2024
    Frequency Content
    Fourier Series (periodic signals)

    Time Frequency
    Domain Domain

    Fourier Transform

    *Fourier Transform in MATLab can be obtained by fft().


    **There is a relationship between Fourier Series and Fourier Transform.
    Notations and abbreviations
    Fourier Transform of sin(5t)

    dt = .05;
    t = -10:dt:10;
    x = sin(5*t);
    X = fft(x)*dt;
    X = fftshift(X);
    Nw = length(X);
    k = -(Nw-1)/2:1:(Nw-1)/2;
    w = k*2*pi/Nw/dt; %rad./sample
    plot(w,abs(X)); grid on;
    Frequency response of the continuous-time LTI
    system described by the differential equation:

    First, we define coefficient matrices:

    A=[a3 a2 a1 a0] B=[b3 b2 b1 b0]

    Second, we use the following function:


    [h,w] = freqs(B,A);
    plot(w, abs(h))
    Question 3 – assignment 7:

    a0 = 121868727358.1180, a1 = 48890434.5196
    a2 = 6209.9310, a3 = 1, b0 = 121868727358.1180;
    a0 = 121868727358.1180;
    a1 = 48890434.5196;
    a2 = 6209.9310;
    a3 = 1;
    a = [a3 a2 a1 a0];
    b0 = 121868727358.1180;
    b = [0 0 0 b0];
    [h,w] = freqs(b,a);
    plot(w,abs(h)); grid on;
    Example
    Let x1(1) = sin(5t) and x2(t) = sin(7t), −10 ≤ t ≤ 10, Use matlab to find the
    Fourier transform,

    Compare your results with what you expect.


    clear all; x(t) = x1(t/2) .
    close all;
    %*******************************
    dt = .05;
    t = -10:dt:10;
    Nt = length(t);
    x1 = sin(5*t);
    x = upsample(x1,2);
    X = fft(x,max(1001,Nt))*dt;
    X = fftshift(X);
    Nw = length(X);
    k = -(Nw-1)/2:1:(Nw-1)/2;
    w = k*2*pi/Nw/dt;
    figure(2)
    subplot(211);plot(x);grid
    xlabel('x(t)=x(t/2)')
    subplot(212);plot(w,abs(X));grid
    xlabel('frequency')
    clear all; x(t) = x1(t) + x2(t).
    close all;
    %*******************************
    dt = .05;
    t = -10:dt:10;
    Nt = length(t);
    x1 = sin(5*t);
    x2 = sin(7*t);
    x = x1+x2;
    X = fft(x,max(1001,Nt))*dt;
    X = fftshift(X);
    Nw = length(X);
    k = -(Nw-1)/2:1:(Nw-1)/2;
    w = k*2*pi/Nw/dt;
    figure(2)
    subplot(211);plot(t,x);grid
    xlabel('x1+x2')
    subplot(212);plot(w,abs(X));grid
    xlabel('frequency')
    clear all;
    x(t) = x1(t) . x2(t).
    close all;
    %*******************************
    dt = .05;
    t = -10:dt:10;
    Nt = length(t);
    x1 = sin(5*t);
    x2 = sin(7*t);
    x = x1.*x2;
    X = fft(x,max(1001,Nt))*dt;
    X = fftshift(X);
    Nw = length(X);
    k = -(Nw-1)/2:1:(Nw-1)/2;
    w = k*2*pi/Nw/dt;
    figure(2)
    subplot(211);plot(t,x);grid
    xlabel('x1+x2')
    subplot(212);plot(w,abs(X));grid
    xlabel('frequency')
    Exercise 1
    Find the amplitude spectrum of the two-frequency signal:

    x(t) = cos(2 π 100t) + cos(2 π 500t)

    Begin by creating a vector, x, with sampled values of the continuous time


    function. If we want to sample the signal every 0.0002 seconds and create
    a sequence of length 250, this will cover a time interval of length
    250*0.0002 = 0.05 seconds.
    Exercise 2
    The END

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