东南大学2015短学期MATLAB作业3

.
Answer:
N=1024;
dt=1/N;
t=0:dt:1-dt;
x=sin(2*pi*32.*t) + sin(2*pi*137.*t) + sin(2*pi*467.*t);
X=fft(x);
subplot(211);plot(t,x);
xlabel('t/sec');ylabel('x');
.
Answer:
omega=1:pi/100:pi;
z=exp(j .* omega);
H = z ./ (z.^2 + 1.6 * z + 0.9);
subplot(211);
plot(omega, abs(H));
xlabel('\Omega'),ylabel('|H|');
subplot(212);
vi)Connect two oscilloscopes to the circuit, one to the transfer function output and the other to the output of the integrator.
iv)Connect the output of this summer to a squarefunction. This is obtained by usingf(u)in the nonlinear blocks.
v)Connect the output of this squarer to an integrator.
Repeat the experiment but with an initial condition of –2 on the integrator. Again draw the results.
1.2 FIRST ORDER SYSTEM
A single time constant may be simulated using the transfer function block in which you enter the coefficients of the numerator and denominator polynomials. Set up the configuration in a new SIMULINK window to realise the transfer function 1/(s+2) with the input unit step and an oscilloscope connected to the output of the transfer function block.Plotthe block diagram in the space below. Simulate the system for 5 seconds andplotthe response.
i)A unit step input.
ii)Connect the unit step to a transfer function of1/(s2+2ζs+1)withζ= 1.0.
iii)Take a summing block and connect the input step to a +ve input and the transfer function output to a –ve input.
nx=[-2:6];
nh=[0:7];
Plot the functions and convolution results.
Answer:
x=[3, 11, 7, 0, -1, 7, -5, 0, 2];
h=[11, 9, 0, -7, -3, 2, 0 -1];
nx=[-2:6];
nh=[0:7];
MatlabWorksheet3
Part A
ing function conv_m.m to make convolution between the following to functions (x and h):
x=[3, 11, 7, 0, -1, 7, -5, 0, 2];
h=[11, 9, 0, -7, -3, 2, 0 -1];
ylabel('x[n] Hamming win');
X=fft(x1);
df=1;
f=0:df:N-1;
subplot(3,2,4);
plot(f,20*log10(abs(X)));
xlabel('k (Hz)');
ylabel('|X[k]| (db)');
Part B
Simulation Using SIMULINK
.
The hamming window can be codedin Matlabas
for n=1:N
hamming(n)=0.54+0.46*cos((2*n-N+1)*pi/N);
end;
where N is the data length in the FFT.
Answer:
N=1024;
dt=1/N;
1.1 UNDERSTANDING SIMPLE WAVEFORMS AND INTEGRATION
Create a pulse of height 2 units from time 0 to 4 seconds by subtracting two unit steps and adding a gain. Connect this pulse to an integrator with a gain of 0.5 and a zero initial condition. Connect oscilloscopes to show the pulse and the output of the integrator. You may wish to name your simulation (block) diagram; to do so use thesaveasfeature under edit. Your block diagram should be similar as below.
axis([ 0 N/2 0 1.5*max(20*log10(abs(X)))]);
%hamming window
for n=1:N
w=0.54+0.46*cos((2*n-N+1)*pi/N);
end;
x1=x .*w;
subplot(3,2,2);
plot(t,x1);
xlabel('t');
Before simulating you need to pull down the simulation header and double click on parameters. Unless told otherwise always assume that you can use the ode 45 integration algorithm shown in this window and the other default parameters. Typically you will only alter the start and stop times but for this first simulation you can use the default values of 0 and 10 seconds. Double click on the oscilloscopes to get the windows in which the traces will appear, pull down the simulate menu and click onrun.Plotbelow the integrator input and output waveforms.
plot(omega, atan2(imag(H),real(H)));
xlabel('\Omega');ylabel('Phase of H')
efftto analyse following signal by plotting the original signal and its spectrum.
f=0:N-1;
subplot(323);stem(f,abs(X));
xlabel('f/Hz');ylabel('|X|');
axis([ 0 N/2 0 1.5*max(abs(X))]);
subplot(325);stem(f,20*log10(abs(X)));
xlabel('f/Hz');ylabel('20lg|X|');
axis([-6 10 -8 12]);
subplot(3,1,3);
stem(ny,y);
xlabel('n'); ylabel('y[n]');
axis([-6 10 -150 250]);
2.Plot the frequency response over for the following transfer function by letting , where is the frequency (rad/sample)., with appropriate labels and title.
t=0:dt:1-dt;
x=sin(2*pi*32.5.*t) + sin(2*pi*137.4.*t) + sin(2*pi*467.7.*t);
X=fft(x);
subplot(321);plot(t,x);