Question

down here is my answer is it correct or not?

t=0:0.001:0.1;
xt=sin((2*pi*50).*t);
dt=randn(1,length(t));
yt=xt+dt;
THD=thd(yt)

%%RUN


fs=10000;
t=0:1/fs:10;
yt=1+sin((pi/2)*t)-(1/7)*sin((9*pi/2)*t+120*(pi/180))+2*exp(-(1i*pi/4)*t)+2*exp((1i*pi/4)*t);
L = length(yt);
fft_yt = fftshift(fft(yt));
df = fs/L;
f = -fs/2+df:df:fs/2;
subplot(211)
plot(t,yt,'b','LineWidth',1)
ylabel('y(t)')
xlabel('t')
subplot(212)
plot(f,abs(fft_yt),'b','LineWidth',1)
ylabel('|Y(f)|')
xlabel('f')

Answer 1

Understanding your digital footprint helps you choose and control what you leave online for others to find. Every time you are online you create a trail of any information you have viewed or created. ... Leaving a positive digital footprint can be very beneficial to your reputation and future opportunities.

Answer 2

Answer:

In order to bring it to rest, its velocity has to be decreased to zero, and in order to achieve that, the kinetic energy has to be drained off and sent somewhere else. An external force has to absorb energy from the object, i.e. do negative work on it, equal to its kinetic energy, or -1/2 mv2.