Consider the following MATLAB code: f_s = 1e4; tt = 0.2: 1/f_s:3; xx = cos(1760*

Question

Consider the following MATLAB code: f_s = 1e4; tt = 0.2: 1/f_s:3; xx = cos(1760*pi*tt + 330*cos(4*pi*tt).*cos(2*pi*tt).*exp(-0.01*tt)); sounds(xx, 1e4); If you run this code you will hear some unusual sounds. (a) The variable xx represents a continuous-time signal x(t). Write a mathematical expression for this signal. (b) What is the instantaneous frequency (in Hz) for x(t) at time 1, 5s? (c) Suppose the signal continues indefinitely. Sketch the instantaneous frequency (in Hz) as a function of time t, for 10000

Explanation / Answer

a) x (t) = cos [ 1760.pi.t + 330.cos (4.pi.t ).cos (2.pi.t ) e-0.01t ]

b) now let f(w) = 1760.pi.t + 330.cos (4.pi.t ).cos (2.pi.t ) e-0.01t

instantaneous frequency is given by f = 1/2pi . d/dt (f(w)

= 1/2pi ( 1760.pi -330.4.pi sin(4pi t)cos (2.pi.t ) e-0.01t -330 2pi sin (2pit)330.cos (4.pi.t )e-0.01t - 330*0.01*cos (4.pi.t ).cos (2.pi.t ) e-0.01t )

at t= 1.5

f= 880+330*0.01*e0.01*1.5=883.25

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