MATLAB Signals processing problems with solutions

1. Generate a 1000 Hz Sinusoid time Sequence for 4 cycles with sampling frequency of 16KHz. Note: A sine wave with frequency f_c is given by sin⁡(2πf_c t)

• Use plot Command to display the result.
• How many samples are present in one period of the sinewave?
• What is the total length of the time and amplitude vectors here?

Solution:

Code:

clear all

fc=1000;            %provided for 1000KHz Sinusoid 

fs=16000;           %sampling rate 16KHz

T=1/1000;           %Time period /singel cycle time duration of Sinusoid 

ts=(1/16000);       % sampling time interval for every sample

t1=0:ts:(4*T)-ts;   %time vector for desired 4 cycles.(4*T)

x=sin(2*pi*fc*t1);  %sine wave function with fc frequency

subplot(2,1,1)

plot(t1,x);         % command to plot the 4 cyceles signal only

subplot(2,1,2)      %for subplot window

plot(t1,x);         %to plot again with sampled view

hold on;            %to hold the previous plot

stem(t1,x);         %to see the sampled signal to count number of sample per period

Ln=length(t1)       %to count length of amplitude and time vector

a)

Figure:

Fig(a): plot of 4 cycle sinusoid with 1000 Hz frequency and sampled curve with sampling rate of 16000 Hz.

Explanation:

b) from the second curve its clear to count number of sample is 16 .

again we can also count by theoryticallly time period of a single cycle T=1/1000 , sampling rate is 16000 so time interval of every sample is ts=1/16000. And number of sampler in one period =T/ts=16 .

c) ) length of time vector =64 (formula provide in the code.)

1. Generate a sinusoid at 15 KHz frequency with the same sampling frequency of 16KHz. Use the same number of sample points as obtained in 1(c) above. Use plot to show your result. From the graph, identify the period and frequency of the sine wave.
2. 
   
3. Solution: Code:

clear all

fc=15000;       %frequency of sainusoid given 15KHz

fs=16000;       %sampling rate is 16KHz     

ts=1/16000;     %sampling rate or interval

t=0:ts:(64*ts)-ts;    % stop time is multiple of sampling interval and

                       %number sample we want.

                      %in our preivious solution we found number of sample

                      %is 64. so in this case time duration is (ts*number

                      %of sample)                     

x=sin(2*pi*fc*t);       % sinusoidal signal of frequency fc

plot(t,x);

Explanation:

Given sampling frequency if 16KHz and signal max frequency is 15KHz .

According to Nyquist criteria signal need to sample with sampling rate of at

least ( > 2Fs) two times of max frequency od the signal . But in this problem 16 KHz is not equal or greater than (2*15KHz =30KHz) . So sampling rate does not follow the Nyquist criteria. Although signal is 15 KHz but we found an alias of the signal which is 1000Hz from the curve.

From the we found T=0.001s and frequency F=1/T =1000 Hz

3.

An LTI system is defined by the impulse response h(t)=exp(-1000t). Sample this signal at the same rate as above (16 KHz) and truncate it to 20 samples. Convolve the signal in (1) above with this filter using the MATLAB command conv. Plot the Output .Note: The Output should also be a sine wave but with a starting and ending transients.

Solution:

Code:

%task 3:

clear all

fc=1000; % frequency of sinusoid signal

T=1/fc; %period of sinusoidal

fs=16000;

ts=1/16000;

t=0:ts:(20*ts)-ts;%to truncate into 20 samples

x=sin(2*pi*fc*t);% sinusoid with fc frequency

t1=0:ts:20*ts-ts;% time vector of impulse response with 20 samplse truncate

x1=exp(-1000*t1);%impulse function

subplot(2,1,1)

plot(t,x)

hold on

%plot(t1,x1);  %impulse plot

%x(21:end)=[]

%x1(21:end)=[]

y=conv(x,x1);

subplot(2,1,2)

plot(y)

Explanation:

According to question we need to take all the sampling criteria and after tha to truncate for 20 samples we need to formulate for 2o time interval shown in the code.

Figure:

Figure: Convoluted output of given signals