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For the following code:fs=12e3;Ts=1/fs;tlen=1.5;t=0:Ts:tlen − Ts;fmin=0.1e3;fmax=3e3;fstep=(fmax − fmin)/length(t);f=fmin:fstep:fmax − fstep;x=sin(2*pi*f.*t);sound(x,fs)plot(t,x)xlabel('time, s')ylabel('amplitude, arbitary units')axis([0 1/fmin*5 − 1 1])

Question

For the following code:fs=12e3;Ts=1/fs;tlen=1.5;t=0:Ts:tlen − Ts;fmin=0.1e3;fmax=3e3;fstep=(fmax − fmin)/length(t);f=fmin:fstep:fmax − fstep;x=sin(2pif.t);sound(x,fs)plot(t,x)xlabel('time, s')ylabel('amplitude, arbitary units')axis([0 1/fmin5 − 1 1])

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Solution

This code is written in MATLAB, a high-level language and interactive environment for numerical computation, visualization, and programming. Here's a step-by-step explanation of what the code does:

  1. fs=12e3;: This line sets the sampling frequency to 12 kHz. The sampling frequency is the number of samples taken per second.

  2. Ts=1/fs;: This line calculates the sampling period, which is the reciprocal of the sampling frequency.

  3. tlen=1.5;: This line sets the total time length of the signal to 1.5 seconds.

  4. t=0:Ts:tlen − Ts;: This line creates a time vector from 0 to tlen with a step size of Ts.

  5. fmin=0.1e3;fmax=3e3;: These lines set the minimum and maximum frequencies for the signal to 0.1 kHz and 3 kHz, respectively.

  6. fstep=(fmax − fmin)/length(t);: This line calculates the frequency step size.

  7. f=fmin:fstep:fmax − fstep;: This line creates a frequency vector from fmin to fmax with a step size of fstep.

  8. x=sin(2*pi*f.*t);: This line generates a sinusoidal signal with a frequency that varies over time.

  9. sound(x,fs): This line plays the sound of the signal.

  10. plot(t,x): This line plots the signal in the time domain.

  11. xlabel('time, s')ylabel('amplitude, arbitary units'): These lines label the x-axis as 'time, s' and the y-axis as 'amplitude, arbitrary units'.

  12. axis([0 1/fmin*5 − 1 1]): This line sets the limits of the x-axis and y-axis for the plot. The x-axis ranges from 0 to 5 periods of the minimum frequency, and the y-axis ranges from -1 to 1.

This problem has been solved

Similar Questions

or the following code:fs=18.28e3;Ts=1/fs;tlen=4.2;t=0:Ts:tlen − Ts;fmin=0.1e3;fmax=3e3;fstep=(fmax − fmin)/length(t);f=fmin:fstep:fmax − fstep;x=sin(2*pi*f.*t);sound(x,fs)plot(t,x)xlabel('time, s')ylabel('amplitude, arbitary units')axis([0 1/fmin*5 − 1 1])

For the following:f=.5e3;fs=44.1e3;Ts=1/fs;tlen=.5;t=0:Ts:tlen − Ts;x=sin(2*pi*f*t);sound(x,fs)plot(t,x)xlabel('time, s')ylabel('amplitude, arbitary units')axis([0 1/f*10 − 1 1])What is fs?

For the following, drag and drop the statements to the correct lines: Blank 1 Question 1 Blank 2 Question 1f=.5e3;tlen=.5; Blank 3 Question 1 Blank 4 Question 1sound(x,fs) Blank 5 Question 1xlabel('time, s')ylabel('amplitude, arbitary units')axis([0 1/f*10 − 1 1])t=0:Ts:tlen − Ts; Ts=1/fs; x=sin(2*pi*f*t); fs=44.1e3; plot(t,x)

For the following: f1=1.7e3; fs=38e3; Ts=1/fs; tlen=.4; t=0:Ts:tlen-Ts; N=length(t); x=sin(2*pi*f1*t); standev=1; noise=randn(1,N)*standev; xn=x+noise; fcf=720; Omegacf=2*pi*fcf/fs; fo=f1; Omegao=2*pi*fo/fs; M=1001; n=(0:M-1)-floor(M/2); h=Omegacf/pi*sinc(n*Omegacf/pi); w=hamming(M)'; A=1; B=h.*w.*cos(n*Omegao); y=filter(B,A,xn); subplot(2,1,1) plot(t,x,'LineWidth',3); hold on plot(t,xn,'LineWidth',1); plot(t,y,'g','LineWidth',2); axis([1/f1*100 1/f1*110 -4 4]); xlabel('time,s'); ylabel('amplitude, arbitrary units'); legend('original signal','with noise','filtered'); hold off X=fft(x); Xn=fft(xn); Y=fft(y); subplot(2,1,2) fbin=fs/N; f=0:fbin:fs-fbin; plot(f,20*log10(abs(X)),'LineWidth',3); hold on plot(f,20*log10(abs(Xn)),'LineWidth',1); plot(f,20*log10(abs(Y)),'g','LineWidth',2); xlabel('frequency, Hz'); ylabel('magnitude response, dB'); legend('noisy signal spectrum','filtered spectrum') axis([1e2 fs/2 -40 100]) hold off What band of frequencies will this filter keep? Minimum frequency = Answer field 1 kHz. Maximum frequency = Answer field 2 kHz.

What is the frequency of the function y = sec5x?

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