For the following code:fs=29.2e3;Ts=1/fs;tlen=9.4;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])How many samples will be in the entire signal?
Question
For the following code:fs=29.2e3;Ts=1/fs;tlen=9.4;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])How many samples will be in the entire signal?
Solution 1
The number of samples in the entire signal can be calculated by multiplying the total length of the signal in seconds (tlen) by the sampling frequency (fs).
In the given code, tlen = 9.4 seconds and fs = 29.2e3 Hz (or 29.2 kHz).
So, the total number of samples = tlen * fs = 9.4 * 29.2e3 = 274480 samples.
Therefore, there will be 274480 samples in the entire signal.
Solution 2
The number of samples in the entire signal can be calculated by the total time of the signal divided by the sampling period. In this case, the total time (tlen) is 9.4 seconds and the sampling period (Ts) is the inverse of the sampling frequency (fs), which is 1/29.2e3.
So, the number of samples in the entire signal is tlen/Ts = 9.4 * 29.2e3 = 274480 samples.
Similar Questions
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: 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.
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)
The time period of a sound wave is 0.020 s. What is its frequency?
Two detectors of sound (D1 and D2) are located on y-axis with co-ordinates (0,3.4√3 km) and (0,5.1√3 km) respectively. A source of sound(S) of natural frequency 100Hz is moving along x− axis with a constant speed of 170m/s. The frequencies observed by D1 and D2 when source is nearest to them is f1 Hz and f2 Hz respectively. Find f2–f1. Take speed of sound = 340 m/s.
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