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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])

Question

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(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 1

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

  1. fs=18.28e3;: This line sets the sampling frequency fs to 18.28 kHz. This is the rate at which the signal will be sampled.

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

  3. tlen=4.2;: This line sets the length of the time vector tlen to 4.2 seconds.

  4. t=0:Ts:tlen − Ts;: This line creates a time vector t that starts at 0, increments by Ts, and ends at tlen - Ts.

  5. fmin=0.1e3;fmax=3e3;: These lines set the minimum frequency fmin to 0.1 kHz and the maximum frequency fmax to 3 kHz.

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

  7. f=fmin:fstep:fmax − fstep;: This line creates a frequency vector f that starts at fmin, increments by fstep, and ends at fmax - fstep.

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

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

  10. plot(t,x): This line plots the signal x against the time vector t.

  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.

Please note that the code you provided seems to have some syntax errors. The correct code should look like this:

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, arbitrary units')
axis([0 1/fmin*5 -1 1])

This problem has been solved

Solution 2

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

  1. fs=18.28e3;: This line sets the sampling frequency fs to 18.28 kHz. This is the rate at which the signal will be sampled.

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

  3. tlen=4.2;: This line sets the length of the time vector tlen to 4.2 seconds.

  4. t=0:Ts:tlen − Ts;: This line creates a time vector t that starts at 0, increments by Ts, and ends at tlen - Ts.

  5. fmin=0.1e3;fmax=3e3;: These lines set the minimum frequency fmin to 0.1 kHz and the maximum frequency fmax to 3 kHz.

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

  7. f=fmin:fstep:fmax − fstep;: This line creates a frequency vector f that starts at fmin, increments by fstep, and ends at fmax - fstep.

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

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

  10. plot(t,x): This line plots the signal x against the time vector t.

  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.

Please note that the code you provided seems to have some syntax errors. The correct code should look like this:

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, arbitrary units')
axis([0 1/fmin*5 -1 1])

This problem has been solved

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The time period of a sound wave is 0.020 s. What is its frequency?

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